[{"Name":"Intermolecular Forces","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intermolecular Forces","Duration":"6m 38s","ChapterTopicVideoID":23609,"CourseChapterTopicPlaylistID":110286,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.370 ","End":"00:07.755","Text":"In this video, we\u0027ll discuss intermolecular forces between molecules."},{"Start":"00:07.755 ","End":"00:13.650","Text":"We\u0027re going to compare intermolecular interaction versus ionic,"},{"Start":"00:13.650 ","End":"00:16.720","Text":"and covalent bonding interaction."},{"Start":"00:17.210 ","End":"00:21.540","Text":"The first thing to note is that intermolecular forces are much"},{"Start":"00:21.540 ","End":"00:26.790","Text":"weaker than interactions that lead to ionic, or covalent bonding."},{"Start":"00:26.790 ","End":"00:31.980","Text":"Now these forces are attractive at long distances between the molecules,"},{"Start":"00:31.980 ","End":"00:35.485","Text":"and repulsive at short distances."},{"Start":"00:35.485 ","End":"00:39.080","Text":"Now this is just the same as in covalent bonding."},{"Start":"00:39.080 ","End":"00:42.110","Text":"However, the minimum in energy for"},{"Start":"00:42.110 ","End":"00:44.780","Text":"intermolecular interaction occurs at"},{"Start":"00:44.780 ","End":"00:49.070","Text":"longer distances than in the case of ionic and covalent bonding,"},{"Start":"00:49.070 ","End":"00:52.170","Text":"and is less negative."},{"Start":"00:53.740 ","End":"00:58.955","Text":"Here\u0027s the diagram we drew when we talked about covalent interaction."},{"Start":"00:58.955 ","End":"01:01.945","Text":"In this case, between 2 hydrogens."},{"Start":"01:01.945 ","End":"01:06.035","Text":"We see the minimum between the attractive forces,"},{"Start":"01:06.035 ","End":"01:08.645","Text":"and the repulsive forces."},{"Start":"01:08.645 ","End":"01:12.380","Text":"Now in the case of intermolecular interactions,"},{"Start":"01:12.380 ","End":"01:15.080","Text":"we get similar diagram,"},{"Start":"01:17.240 ","End":"01:25.490","Text":"but the minimum occurs at longer distances and is much less negative."},{"Start":"01:25.490 ","End":"01:26.840","Text":"That means it\u0027s weaker,"},{"Start":"01:26.840 ","End":"01:29.465","Text":"we have a much weaker interaction."},{"Start":"01:29.465 ","End":"01:31.421","Text":"This isn\u0027t really to scale,"},{"Start":"01:31.421 ","End":"01:35.760","Text":"it should even be weaker than that."},{"Start":"01:36.160 ","End":"01:40.835","Text":"Now let us recall what we learned about non-ideal gases."},{"Start":"01:40.835 ","End":"01:43.100","Text":"You recall that in an ideal gas,"},{"Start":"01:43.100 ","End":"01:49.010","Text":"you ignore interactions between atoms or molecules."},{"Start":"01:49.010 ","End":"01:53.720","Text":"Deviations from ideal gas behavior occur when we can no"},{"Start":"01:53.720 ","End":"01:57.800","Text":"longer ignore the attractive forces between molecules."},{"Start":"01:57.800 ","End":"01:59.345","Text":"When does this happen?"},{"Start":"01:59.345 ","End":"02:02.630","Text":"At low temperature or high pressure."},{"Start":"02:02.630 ","End":"02:05.885","Text":"Then instead of using the ideal gas equation,"},{"Start":"02:05.885 ","End":"02:09.070","Text":"we can use the van der Waals equation."},{"Start":"02:09.070 ","End":"02:17.630","Text":"Here it is. P plus an^2 over V^2 times V minus nb is equal to nRT."},{"Start":"02:17.630 ","End":"02:20.395","Text":"Just to remind you P is the pressure,"},{"Start":"02:20.395 ","End":"02:22.440","Text":"n is the number of moles,"},{"Start":"02:22.440 ","End":"02:24.645","Text":"V is volume,"},{"Start":"02:24.645 ","End":"02:32.585","Text":"and a and b are coefficients which are different for each individual atom or molecule."},{"Start":"02:32.585 ","End":"02:38.320","Text":"R is the gas constant and T is the temperature in Kelvin."},{"Start":"02:38.320 ","End":"02:45.995","Text":"Now this term, an^2 over V^2 that arises from these attractive intermolecular forces."},{"Start":"02:45.995 ","End":"02:49.010","Text":"They\u0027re often called van der Waals forces"},{"Start":"02:49.010 ","End":"02:54.089","Text":"because of the person who invented this equation."},{"Start":"02:54.430 ","End":"02:57.830","Text":"Now these forces are also important,"},{"Start":"02:57.830 ","End":"03:04.610","Text":"as we\u0027ll see in forming condensed matters a general name for liquids and solids."},{"Start":"03:04.610 ","End":"03:10.655","Text":"Now I want to remind you of 2 concepts that are going to be very important in"},{"Start":"03:10.655 ","End":"03:17.710","Text":"our study of intermolecular forces that are dipole moments and polarizability."},{"Start":"03:17.710 ","End":"03:22.270","Text":"First, let\u0027s talk about dipole moments."},{"Start":"03:22.270 ","End":"03:25.025","Text":"Now in a polar covalent bond,"},{"Start":"03:25.025 ","End":"03:29.000","Text":"the electrons are not shared equally between the atoms."},{"Start":"03:29.000 ","End":"03:33.835","Text":"The electron density is greater for the more electronegative atom."},{"Start":"03:33.835 ","End":"03:40.205","Text":"A dipole moment Mu is formed pointing towards the more electronegative atom."},{"Start":"03:40.205 ","End":"03:43.205","Text":"Some people use the opposite convention here."},{"Start":"03:43.205 ","End":"03:45.110","Text":"We\u0027ll use this convention."},{"Start":"03:45.110 ","End":"03:50.435","Text":"The dipole moment will point towards the more electronegative atom."},{"Start":"03:50.435 ","End":"03:54.139","Text":"Now how do you get the dipole moment of a molecule?"},{"Start":"03:54.139 ","End":"03:56.750","Text":"We find it by summing the dipole moments of"},{"Start":"03:56.750 ","End":"04:02.280","Text":"the individual bonds while taking into account the geometry of the molecule."},{"Start":"04:02.600 ","End":"04:06.270","Text":"Here are 2 examples, CCl_4,"},{"Start":"04:06.270 ","End":"04:10.275","Text":"which is nonpolar, and CH_3Cl, which is polar."},{"Start":"04:10.275 ","End":"04:14.140","Text":"If CCl_4 is tetrahedral,"},{"Start":"04:18.890 ","End":"04:25.195","Text":"if we sum the individual dipole moments point towards Cl,"},{"Start":"04:25.195 ","End":"04:27.535","Text":"then we get 0,"},{"Start":"04:27.535 ","End":"04:29.990","Text":"Mu here is 0."},{"Start":"04:30.330 ","End":"04:38.510","Text":"The other one is CH_3Cl, again tetrahedral."},{"Start":"04:38.540 ","End":"04:46.560","Text":"But we have each atoms and Cl atoms and the dipoles of these bonds are not equal."},{"Start":"04:46.560 ","End":"04:47.750","Text":"When we sum them up,"},{"Start":"04:47.750 ","End":"04:52.175","Text":"we get a net dipole moment pointing in this direction."},{"Start":"04:52.175 ","End":"04:56.030","Text":"Here\u0027s non-polar and polar."},{"Start":"04:56.030 ","End":"05:02.515","Text":"Now the second concept is polarizability. What\u0027s that?"},{"Start":"05:02.515 ","End":"05:05.260","Text":"In the presence of an electrical field or near an ion,"},{"Start":"05:05.260 ","End":"05:06.695","Text":"atom, or molecule,"},{"Start":"05:06.695 ","End":"05:09.650","Text":"the electron charge distribution in a molecule becomes"},{"Start":"05:09.650 ","End":"05:13.580","Text":"distorted and the molecule is polarized."},{"Start":"05:13.580 ","End":"05:16.610","Text":"Now, the polarizability,"},{"Start":"05:16.610 ","End":"05:18.190","Text":"which we write as Alpha,"},{"Start":"05:18.190 ","End":"05:22.980","Text":"is a measure of how easy it is to polarize the molecule."},{"Start":"05:23.350 ","End":"05:28.385","Text":"Polarizability increases with the molar mass."},{"Start":"05:28.385 ","End":"05:33.455","Text":"More electrons, then higher polarizability."},{"Start":"05:33.455 ","End":"05:38.435","Text":"Polarizability is also proportional to the volume of the atom."},{"Start":"05:38.435 ","End":"05:43.865","Text":"The Alpha is proportional to the volume of the atom or molecule."},{"Start":"05:43.865 ","End":"05:46.700","Text":"The loosely bound valence electrons,"},{"Start":"05:46.700 ","End":"05:48.740","Text":"which are usually far from the nucleus,"},{"Start":"05:48.740 ","End":"05:52.040","Text":"contribute most to the polarizability."},{"Start":"05:52.040 ","End":"05:56.720","Text":"That\u0027s why large molecules are more polarizable."},{"Start":"05:56.720 ","End":"05:59.015","Text":"Here\u0027s an example."},{"Start":"05:59.015 ","End":"06:05.135","Text":"The polarizability of CCl_4 is much greater than the polarizability of CH_4."},{"Start":"06:05.135 ","End":"06:11.930","Text":"Because CCl_4 has a higher molar mass than CH_4 and"},{"Start":"06:11.930 ","End":"06:20.390","Text":"the electrons are further from the Cl nucleus than they are from the hydrogen nucleus."},{"Start":"06:20.390 ","End":"06:25.730","Text":"So CH_4 is a much smaller molecule than CCl_4,"},{"Start":"06:25.730 ","End":"06:30.010","Text":"so CCl_4 has a much higher polarizability."},{"Start":"06:30.010 ","End":"06:33.080","Text":"In this video, we began our study of"},{"Start":"06:33.080 ","End":"06:38.940","Text":"intermolecular forces and we\u0027ll continue it in the following videos."}],"ID":24520},{"Watched":false,"Name":"Dipole-Dipole Interactions","Duration":"6m 39s","ChapterTopicVideoID":23668,"CourseChapterTopicPlaylistID":110286,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.470 ","End":"00:04.700","Text":"The first type of attractive inter-molecular force that"},{"Start":"00:04.700 ","End":"00:08.935","Text":"we\u0027ll talk about is dipole-dipole interaction."},{"Start":"00:08.935 ","End":"00:17.445","Text":"Now, dipole-dipole interactions take place between polar molecules such as CH_3Cl."},{"Start":"00:17.445 ","End":"00:22.185","Text":"The dipoles line up so that the positive end of 1 dipole"},{"Start":"00:22.185 ","End":"00:26.670","Text":"is next to the negative end or pole of another dipole."},{"Start":"00:26.670 ","End":"00:29.410","Text":"Here, they are lining up."},{"Start":"00:29.870 ","End":"00:33.450","Text":"Positive pole, negative pole,"},{"Start":"00:33.450 ","End":"00:36.135","Text":"negative pole, positive pole."},{"Start":"00:36.135 ","End":"00:39.975","Text":"This can continue in 3-dimensions."},{"Start":"00:39.975 ","End":"00:42.530","Text":"Now, if this is in a solid,"},{"Start":"00:42.530 ","End":"00:45.920","Text":"for example, where the molecules can\u0027t rotate,"},{"Start":"00:45.920 ","End":"00:55.385","Text":"the potential energy V is proportional to minus mu 1 times mu 2 divided by r^3."},{"Start":"00:55.385 ","End":"00:59.690","Text":"Now, mu 1 and mu 2 are the dipoles of neighboring molecules."},{"Start":"00:59.690 ","End":"01:03.410","Text":"Sometimes, they\u0027ll be different molecules or they have different dipoles,"},{"Start":"01:03.410 ","End":"01:06.030","Text":"mu 1 and mu 2."},{"Start":"01:06.100 ","End":"01:13.850","Text":"Now, the meaning of the negative sign here is that it\u0027s attractive potential,"},{"Start":"01:13.850 ","End":"01:15.920","Text":"that means it has lower energy,"},{"Start":"01:15.920 ","End":"01:20.790","Text":"the energy is lowered and we get greater stability."},{"Start":"01:20.960 ","End":"01:24.725","Text":"Now, if we have a gas or a liquid,"},{"Start":"01:24.725 ","End":"01:27.920","Text":"the situation is slightly different."},{"Start":"01:27.920 ","End":"01:33.590","Text":"Let\u0027s consider dipole-dipole interactions in the gas or liquid."},{"Start":"01:33.590 ","End":"01:35.810","Text":"Now, in a gas and liquid,"},{"Start":"01:35.810 ","End":"01:39.035","Text":"the molecules can rotate freely."},{"Start":"01:39.035 ","End":"01:42.230","Text":"For freely rotating polar molecules,"},{"Start":"01:42.230 ","End":"01:44.930","Text":"the interaction between positive and negative"},{"Start":"01:44.930 ","End":"01:48.950","Text":"nearly cancel the interactions between positive and positive,"},{"Start":"01:48.950 ","End":"01:50.785","Text":"or negative and negative."},{"Start":"01:50.785 ","End":"01:55.505","Text":"The interaction between positive and negative is attractive, of course,"},{"Start":"01:55.505 ","End":"01:57.860","Text":"and between the same signs,"},{"Start":"01:57.860 ","End":"02:02.450","Text":"positive and positive or negative and negative, it\u0027s repulsive."},{"Start":"02:02.450 ","End":"02:08.320","Text":"There\u0027s near cancellation, but is not complete cancellation."},{"Start":"02:08.320 ","End":"02:15.290","Text":"Because the molecules spend slightly more time in the attractive configuration."},{"Start":"02:15.290 ","End":"02:21.475","Text":"That means the overall potential is going to be slightly negative."},{"Start":"02:21.475 ","End":"02:31.205","Text":"The potential energy is proportional to minus mu 1 times mu 2 divided by r^6."},{"Start":"02:31.205 ","End":"02:34.440","Text":"Remember before we had r^3,"},{"Start":"02:34.440 ","End":"02:39.120","Text":"and now we have r^6."},{"Start":"02:39.120 ","End":"02:47.100","Text":"Here, it\u0027s drawn, this is minus 1 divided by r^6."},{"Start":"02:47.100 ","End":"02:53.640","Text":"The next one,"},{"Start":"02:53.640 ","End":"02:55.835","Text":"minus 1 divided by r^3, that\u0027s the green one."},{"Start":"02:55.835 ","End":"03:03.860","Text":"We can add some more potentials to the r^6 and r^3 that we\u0027ve done already."},{"Start":"03:03.860 ","End":"03:09.640","Text":"Here\u0027s V proportional to minus 1 divided by r^2."},{"Start":"03:09.640 ","End":"03:14.000","Text":"This is appropriate for the interaction between an I and a polar molecule."},{"Start":"03:14.000 ","End":"03:16.790","Text":"For example, we might have that in hydration."},{"Start":"03:16.790 ","End":"03:19.280","Text":"We might have sodium plus,"},{"Start":"03:19.280 ","End":"03:21.890","Text":"surrounded by water molecules."},{"Start":"03:21.890 ","End":"03:24.600","Text":"Water is polar, of course."},{"Start":"03:25.880 ","End":"03:28.860","Text":"Or we might have V proportional"},{"Start":"03:28.860 ","End":"03:34.205","Text":"to minus 1 over r. That\u0027s the interaction between positive and negative ions,"},{"Start":"03:34.205 ","End":"03:37.170","Text":"such as in NACL."},{"Start":"03:39.220 ","End":"03:42.140","Text":"Here, they\u0027re drawn as well."},{"Start":"03:42.140 ","End":"03:49.610","Text":"Here\u0027s minus 1 over r^2,"},{"Start":"03:49.610 ","End":"03:54.290","Text":"and here\u0027s minus 1 over r. What can we see here?"},{"Start":"03:54.290 ","End":"03:58.610","Text":"We see that as the power increases,"},{"Start":"03:58.610 ","End":"04:02.780","Text":"the potential takes place or significant at"},{"Start":"04:02.780 ","End":"04:12.540","Text":"lower distances between the molecules and dies out faster."},{"Start":"04:12.540 ","End":"04:20.610","Text":"That means that minus 1 over r^6 is very weak compared to minus 1 over r^3,"},{"Start":"04:20.610 ","End":"04:25.005","Text":"and that\u0027s weaker than minus 1 over r^2,"},{"Start":"04:25.005 ","End":"04:31.250","Text":"and that\u0027s even weaker than minus 1 over r. From all this,"},{"Start":"04:31.250 ","End":"04:34.130","Text":"we can conclude that the interaction between rotating"},{"Start":"04:34.130 ","End":"04:39.125","Text":"polar molecules is weaker than between non-rotating polar molecules."},{"Start":"04:39.125 ","End":"04:45.970","Text":"The molecules that rotate only interact when they are very close."},{"Start":"04:45.970 ","End":"04:50.120","Text":"Now, in liquids, the molecules are closer than in a gas,"},{"Start":"04:50.120 ","End":"04:53.000","Text":"so the interactions there are stronger,"},{"Start":"04:53.000 ","End":"04:56.300","Text":"and that now we can understand why in a gas,"},{"Start":"04:56.300 ","End":"05:00.655","Text":"we can sometimes ignore these interactions altogether."},{"Start":"05:00.655 ","End":"05:03.800","Text":"Now, what\u0027s the influence of this?"},{"Start":"05:03.800 ","End":"05:06.500","Text":"For example, the boiling points of"},{"Start":"05:06.500 ","End":"05:10.024","Text":"polar molecules are higher than those of non-polar molecules,"},{"Start":"05:10.024 ","End":"05:14.245","Text":"even though the molecules have the same mass."},{"Start":"05:14.245 ","End":"05:16.455","Text":"For example, if we have"},{"Start":"05:16.455 ","End":"05:23.770","Text":"cis-dichloroethene compared to trans-dichloroethene, let\u0027s draw them."},{"Start":"05:25.550 ","End":"05:31.620","Text":"Here\u0027s the cis where the 2 hydrogens are the same size,the 2 chlorines are the same-side."},{"Start":"05:31.620 ","End":"05:38.885","Text":"This is trans where the chlorines are on the opposite side,"},{"Start":"05:38.885 ","End":"05:40.970","Text":"and here the hydrogen is opposite sides."},{"Start":"05:40.970 ","End":"05:43.530","Text":"This one is polar,"},{"Start":"05:44.480 ","End":"05:47.920","Text":"and this is nonpolar."},{"Start":"05:49.180 ","End":"05:53.960","Text":"Now what we see is that for the polar molecule,"},{"Start":"05:53.960 ","End":"06:00.525","Text":"the boiling point is higher than for the nonpolar molecule."},{"Start":"06:00.525 ","End":"06:05.120","Text":"The reason for this is that the polar molecules,"},{"Start":"06:05.120 ","End":"06:06.845","Text":"the forces between the molecules,"},{"Start":"06:06.845 ","End":"06:10.085","Text":"are greater than for non-polar molecules,"},{"Start":"06:10.085 ","End":"06:12.935","Text":"so you have to supply more energy,"},{"Start":"06:12.935 ","End":"06:20.285","Text":"more heat to boil the polar molecule than to boil the non-polar molecule."},{"Start":"06:20.285 ","End":"06:24.890","Text":"For cis-dichloroethene, it\u0027s 60.2 degrees Celsius,"},{"Start":"06:24.890 ","End":"06:27.265","Text":"whereas for the trans molecule,"},{"Start":"06:27.265 ","End":"06:31.960","Text":"it\u0027s 48.5 degrees Celsius."},{"Start":"06:31.960 ","End":"06:38.160","Text":"In this video, we\u0027ll discuss dipole-dipole interactions."}],"ID":24573},{"Watched":false,"Name":"Dispersion (London) forces","Duration":"8m 8s","ChapterTopicVideoID":23614,"CourseChapterTopicPlaylistID":110286,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"In the previous video,"},{"Start":"00:02.115 ","End":"00:05.145","Text":"we discussed dipole-dipole interactions."},{"Start":"00:05.145 ","End":"00:06.855","Text":"In this video, we\u0027ll talk about"},{"Start":"00:06.855 ","End":"00:13.500","Text":"attractive forces that exist even between non-polar atoms and molecules."},{"Start":"00:13.500 ","End":"00:17.813","Text":"Now how do we know that such forces exist?"},{"Start":"00:17.813 ","End":"00:20.805","Text":"An example is the noble gases."},{"Start":"00:20.805 ","End":"00:26.585","Text":"For example, noble gases such as helium become liquid at low temperatures,"},{"Start":"00:26.585 ","End":"00:28.100","Text":"at 4 Kelvin,"},{"Start":"00:28.100 ","End":"00:29.555","Text":"in the case of helium,"},{"Start":"00:29.555 ","End":"00:35.160","Text":"so there must be forces between helium atoms."},{"Start":"00:35.230 ","End":"00:39.590","Text":"These forces are called dispersion or sometimes"},{"Start":"00:39.590 ","End":"00:44.485","Text":"London forces after the person who discovered them."},{"Start":"00:44.485 ","End":"00:47.554","Text":"Now, what\u0027s the origin of these forces?"},{"Start":"00:47.554 ","End":"00:52.549","Text":"Atoms and non-polar molecules have instantaneous dipole moments."},{"Start":"00:52.549 ","End":"00:57.260","Text":"That\u0027s because the electron density doesn\u0027t stay in one place but moves"},{"Start":"00:57.260 ","End":"01:02.860","Text":"around so that one can form an instantaneous dipole moment."},{"Start":"01:02.860 ","End":"01:06.380","Text":"Now that instantaneous dipole moment induces"},{"Start":"01:06.380 ","End":"01:10.550","Text":"an instantaneous dipole moment in a neighboring molecule,"},{"Start":"01:10.550 ","End":"01:14.885","Text":"and then the 2 molecules attract each other."},{"Start":"01:14.885 ","End":"01:17.455","Text":"We can draw it here."},{"Start":"01:17.455 ","End":"01:21.585","Text":"Here\u0027s 1 instantaneous dipole,"},{"Start":"01:21.585 ","End":"01:24.000","Text":"another instantaneous dipole,"},{"Start":"01:24.000 ","End":"01:27.255","Text":"and then the attraction between the positive and negative,"},{"Start":"01:27.255 ","End":"01:30.630","Text":"and the other negative and positive."},{"Start":"01:31.060 ","End":"01:33.170","Text":"As we said before,"},{"Start":"01:33.170 ","End":"01:38.790","Text":"the force between these molecules is called the dispersion or London force."},{"Start":"01:38.860 ","End":"01:45.110","Text":"Now, the strength of the interaction depends on the polarizability of the molecules."},{"Start":"01:45.110 ","End":"01:47.075","Text":"The higher the polarizability,"},{"Start":"01:47.075 ","End":"01:50.510","Text":"the more the electron density moves around and"},{"Start":"01:50.510 ","End":"01:54.840","Text":"forms stronger instantaneous dipole moments."},{"Start":"01:55.450 ","End":"02:00.930","Text":"In this case, the potential energy V is proportional to"},{"Start":"02:00.930 ","End":"02:06.285","Text":"minus Alpha_1 up times Alpha_2 divided by r^6."},{"Start":"02:06.285 ","End":"02:12.690","Text":"This is yet another case where we have behavior depending on r^6."},{"Start":"02:12.690 ","End":"02:19.230","Text":"Alpha_1 and Alpha_2 are the polarizabilities of the 2 molecules."},{"Start":"02:19.480 ","End":"02:25.130","Text":"We learned before that the polarizability increases with molar mass,"},{"Start":"02:25.130 ","End":"02:27.920","Text":"so do the London forces."},{"Start":"02:27.920 ","End":"02:30.815","Text":"The more electrons, the higher the molar mass,"},{"Start":"02:30.815 ","End":"02:33.980","Text":"the higher the London forces."},{"Start":"02:33.980 ","End":"02:37.085","Text":"This has important consequences."},{"Start":"02:37.085 ","End":"02:41.675","Text":"For example, the boiling point increases with increasing molar mass."},{"Start":"02:41.675 ","End":"02:44.495","Text":"Because the boiling point is higher,"},{"Start":"02:44.495 ","End":"02:47.320","Text":"the higher the intermolecular forces,"},{"Start":"02:47.320 ","End":"02:49.795","Text":"and the intermolecular forces here,"},{"Start":"02:49.795 ","End":"02:52.145","Text":"increase with increasing molar mass,"},{"Start":"02:52.145 ","End":"02:56.970","Text":"so the boiling point increases with increasing molar mass."},{"Start":"02:57.130 ","End":"02:59.690","Text":"Let\u0027s take an example."},{"Start":"02:59.690 ","End":"03:03.650","Text":"If we compare methane with carbon tetrachloride,"},{"Start":"03:03.650 ","End":"03:07.595","Text":"we see a tremendous difference in the boiling points."},{"Start":"03:07.595 ","End":"03:11.270","Text":"For methane, which is much lighter than the carbon tetrachloride,"},{"Start":"03:11.270 ","End":"03:15.249","Text":"the boiling point is minus 161 degrees Celsius."},{"Start":"03:15.249 ","End":"03:17.445","Text":"So it\u0027s a gas at room temperature."},{"Start":"03:17.445 ","End":"03:23.225","Text":"Whereas the boiling point of carbon tetrachloride is 77 degrees Celsius,"},{"Start":"03:23.225 ","End":"03:26.610","Text":"and that\u0027s a liquid at room temperature."},{"Start":"03:26.770 ","End":"03:30.950","Text":"Another important thing is the shape of the molecules."},{"Start":"03:30.950 ","End":"03:32.525","Text":"For the same molar mass,"},{"Start":"03:32.525 ","End":"03:37.745","Text":"long molecules are more polarizable than round molecules."},{"Start":"03:37.745 ","End":"03:42.995","Text":"Here\u0027s an example: the isomers of C_5H_12,"},{"Start":"03:42.995 ","End":"03:49.045","Text":"pentane and neopentane. Let\u0027s draw them."},{"Start":"03:49.045 ","End":"03:53.214","Text":"Pentane is a long molecule,"},{"Start":"03:53.214 ","End":"03:59.545","Text":"like a zigzag, whereas neopentane is like a ball."},{"Start":"03:59.545 ","End":"04:06.860","Text":"We see that pentane boils at 36.1 degrees Celsius,"},{"Start":"04:06.860 ","End":"04:12.110","Text":"whereas neopentane boils at 9.5 degrees Celsius."},{"Start":"04:12.110 ","End":"04:16.175","Text":"The reason for this is when the molecules are long,"},{"Start":"04:16.175 ","End":"04:21.110","Text":"they have more points of contact with the neighboring molecules and"},{"Start":"04:21.110 ","End":"04:27.000","Text":"can induce higher instantaneous dipole moments."},{"Start":"04:27.590 ","End":"04:32.775","Text":"So far we\u0027ve had induced dipole-induced dipole,"},{"Start":"04:32.775 ","End":"04:38.860","Text":"and now a hybrid, dipole-induced-dipole interaction."},{"Start":"04:38.890 ","End":"04:45.870","Text":"This takes place when a polar molecule interacts with a non-polar molecule."},{"Start":"04:46.190 ","End":"04:51.325","Text":"The polar molecule induces a dipole in a non-polar molecule."},{"Start":"04:51.325 ","End":"04:53.605","Text":"We have a polar molecule,"},{"Start":"04:53.605 ","End":"04:56.665","Text":"and then we have another molecule that doesn\u0027t have a dipole,"},{"Start":"04:56.665 ","End":"04:59.860","Text":"and then 1 is induced in it,"},{"Start":"04:59.860 ","End":"05:03.050","Text":"and then of course there\u0027s attraction."},{"Start":"05:04.280 ","End":"05:10.880","Text":"The potential energy is again proportional to 1 over r^6,"},{"Start":"05:10.880 ","End":"05:13.810","Text":"except the numerator is different."},{"Start":"05:13.810 ","End":"05:15.550","Text":"Now it\u0027s Mu 1 squared,"},{"Start":"05:15.550 ","End":"05:19.165","Text":"the dipole moment of the polar molecule squared,"},{"Start":"05:19.165 ","End":"05:23.510","Text":"times the polarizability of the other molecule."},{"Start":"05:24.410 ","End":"05:26.610","Text":"Now the 2 very,"},{"Start":"05:26.610 ","End":"05:29.010","Text":"very important points you have to remember."},{"Start":"05:29.010 ","End":"05:32.180","Text":"The London forces exist for all molecules,"},{"Start":"05:32.180 ","End":"05:34.160","Text":"whether polar or non-polar."},{"Start":"05:34.160 ","End":"05:37.520","Text":"This is the basic force that exists."},{"Start":"05:37.520 ","End":"05:42.180","Text":"It doesn\u0027t matter where the molecule is polar or non-polar."},{"Start":"05:42.670 ","End":"05:46.040","Text":"The dipole-dipole interactions or"},{"Start":"05:46.040 ","End":"05:50.170","Text":"dipole-induced-dipole interactions are additional forces."},{"Start":"05:50.170 ","End":"05:52.295","Text":"First, you have the London forces,"},{"Start":"05:52.295 ","End":"05:54.005","Text":"and then in addition,"},{"Start":"05:54.005 ","End":"05:57.300","Text":"you have dipole-dipole forces."},{"Start":"05:57.970 ","End":"06:01.530","Text":"Let\u0027s take an example."},{"Start":"06:01.760 ","End":"06:08.625","Text":"Why is the boiling point of HCl lower than that of HI?"},{"Start":"06:08.625 ","End":"06:12.585","Text":"Here all the relevant data."},{"Start":"06:12.585 ","End":"06:16.650","Text":"The mass of HCl is lower than that of HI."},{"Start":"06:16.650 ","End":"06:22.310","Text":"The polarizability consequently is lower for HCl than HI."},{"Start":"06:22.310 ","End":"06:26.315","Text":"However, the dipole moments are in the opposite behavior."},{"Start":"06:26.315 ","End":"06:31.325","Text":"HCl has a higher dipole moment than HI."},{"Start":"06:31.325 ","End":"06:35.329","Text":"Let\u0027s first consider the dispersion forces."},{"Start":"06:35.329 ","End":"06:40.630","Text":"We see that the polarizability of HCl is lower than HI,"},{"Start":"06:40.630 ","End":"06:46.595","Text":"so we expect the dispersion to be lower for HCl than HI,"},{"Start":"06:46.595 ","End":"06:52.525","Text":"and so the boiling point of HCl should be lower than that of HI."},{"Start":"06:52.525 ","End":"07:03.580","Text":"This HCl has a lower boiling point than HI when we consider the dispersion forces."},{"Start":"07:03.580 ","End":"07:08.270","Text":"However, HCl is more polar than HI."},{"Start":"07:08.270 ","End":"07:11.960","Text":"The dipole moment of HCl is 1.08 debye,"},{"Start":"07:11.960 ","End":"07:15.535","Text":"and HI, 0.44 debye."},{"Start":"07:15.535 ","End":"07:21.390","Text":"This would suggest that the dipole-dipole forces are greater for"},{"Start":"07:21.390 ","End":"07:27.590","Text":"HCl and that would suggest a higher boiling point for HCl,"},{"Start":"07:27.590 ","End":"07:33.040","Text":"so HI should have a lower boiling point than HCl."},{"Start":"07:33.040 ","End":"07:40.750","Text":"In practice, the boiling point of HCl is lower than that of HI."},{"Start":"07:40.760 ","End":"07:45.060","Text":"So if the boiling point of HCl is lower than HI,"},{"Start":"07:45.060 ","End":"07:49.820","Text":"that means the dispersion forces are the dominant forces."},{"Start":"07:49.820 ","End":"07:52.400","Text":"It doesn\u0027t mean to say the other forces don\u0027t exist,"},{"Start":"07:52.400 ","End":"07:56.800","Text":"but the dominant forces are the dispersion forces."},{"Start":"07:56.800 ","End":"08:01.445","Text":"In this video, we discussed dispersion intermolecular forces,"},{"Start":"08:01.445 ","End":"08:03.635","Text":"also called London forces,"},{"Start":"08:03.635 ","End":"08:08.040","Text":"and also dipole-induced-dipole interactions."}],"ID":24525},{"Watched":false,"Name":"Hydrogen Bonding","Duration":"6m 48s","ChapterTopicVideoID":23612,"CourseChapterTopicPlaylistID":110286,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:02.280","Text":"In the previous video,"},{"Start":"00:02.280 ","End":"00:05.999","Text":"we talked about dispersion on London intermolecular forces."},{"Start":"00:05.999 ","End":"00:09.730","Text":"In this video, we\u0027ll talk about hydrogen bonding."},{"Start":"00:09.740 ","End":"00:13.840","Text":"What\u0027s the evidence for hydrogen bonding?"},{"Start":"00:13.910 ","End":"00:20.760","Text":"Let\u0027s consider the boiling points in degrees Celsius of the hydrides of Group 16 and 17."},{"Start":"00:20.760 ","End":"00:24.360","Text":"Here\u0027s Group 16 starts with water with a boiling point"},{"Start":"00:24.360 ","End":"00:27.810","Text":"of 100 degrees Celsius because water is, of course,"},{"Start":"00:27.810 ","End":"00:29.745","Text":"a liquid at room temperature,"},{"Start":"00:29.745 ","End":"00:33.840","Text":"then H_2S has a much lower boiling point,"},{"Start":"00:33.840 ","End":"00:35.970","Text":"H_2Se a little higher,"},{"Start":"00:35.970 ","End":"00:39.135","Text":"and H_2Te a little higher than that."},{"Start":"00:39.135 ","End":"00:43.145","Text":"In Group 17, we have the same behavior."},{"Start":"00:43.145 ","End":"00:46.460","Text":"HF has a high boiling point,"},{"Start":"00:46.460 ","End":"00:51.205","Text":"relatively of 19.5 degrees Celsius."},{"Start":"00:51.205 ","End":"00:54.645","Text":"HCl has a much lower boiling point."},{"Start":"00:54.645 ","End":"00:59.270","Text":"HBr has a little higher and HI higher than that."},{"Start":"00:59.270 ","End":"01:03.390","Text":"We can draw this behavior in a graph."},{"Start":"01:12.610 ","End":"01:21.880","Text":"We\u0027re plotting here the boiling point versus the mass."},{"Start":"01:22.190 ","End":"01:25.160","Text":"The first hydride of the series has"},{"Start":"01:25.160 ","End":"01:29.420","Text":"a high boiling point then the second one has a much lower one,"},{"Start":"01:29.420 ","End":"01:33.570","Text":"and gradually the boiling points increase."},{"Start":"01:33.710 ","End":"01:37.085","Text":"Apart from H_2O and HF,"},{"Start":"01:37.085 ","End":"01:40.370","Text":"the boiling points increase with molar mass,"},{"Start":"01:40.370 ","End":"01:43.550","Text":"molecular mass, that\u0027s called dispersion,"},{"Start":"01:43.550 ","End":"01:45.440","Text":"that\u0027s the reason for it."},{"Start":"01:45.440 ","End":"01:51.424","Text":"However, the boiling points of H_2O on HF are much higher than the rest of their series,"},{"Start":"01:51.424 ","End":"01:56.360","Text":"and this suggests strong intermolecular forces."},{"Start":"01:56.360 ","End":"02:00.905","Text":"Now, Group 15, which begins with ammonia,"},{"Start":"02:00.905 ","End":"02:03.755","Text":"NH_3, behaves the same way."},{"Start":"02:03.755 ","End":"02:08.350","Text":"But Group 14, which begins with methane, does not."},{"Start":"02:08.350 ","End":"02:15.600","Text":"Now, the intermolecular force that causes this behavior is called hydrogen bonding."},{"Start":"02:15.670 ","End":"02:19.805","Text":"What happens is that in a hydrogen bond,"},{"Start":"02:19.805 ","End":"02:24.440","Text":"a hydrogen atom bonded to a small electronegative atom,"},{"Start":"02:24.440 ","End":"02:27.485","Text":"such as N, O, or F,"},{"Start":"02:27.485 ","End":"02:31.820","Text":"is attracted to the lone pair on another N,"},{"Start":"02:31.820 ","End":"02:33.590","Text":"F, or O atom."},{"Start":"02:33.590 ","End":"02:39.360","Text":"It can be in the same molecule or usually in a different molecule."},{"Start":"02:39.730 ","End":"02:45.810","Text":"Let\u0027s take the examples of water and HF."},{"Start":"02:46.670 ","End":"02:49.060","Text":"Here is water."},{"Start":"02:49.060 ","End":"02:56.015","Text":"1 molecule, we have an HO bond, another HO bond."},{"Start":"02:56.015 ","End":"02:58.815","Text":"On the second molecule an HO bond,"},{"Start":"02:58.815 ","End":"03:01.775","Text":"and another HO bond."},{"Start":"03:01.775 ","End":"03:07.000","Text":"What happens is that we have attraction between H which is Delta"},{"Start":"03:07.000 ","End":"03:13.555","Text":"positive and O in the other molecule which is Delta minus."},{"Start":"03:13.555 ","End":"03:17.440","Text":"There\u0027s an attraction between this H,"},{"Start":"03:17.440 ","End":"03:23.050","Text":"and the lone pair on the oxygen atom of the neighboring molecule."},{"Start":"03:23.050 ","End":"03:25.540","Text":"We indicate this by 3 dots."},{"Start":"03:25.540 ","End":"03:27.460","Text":"This is a hydrogen bond."},{"Start":"03:27.460 ","End":"03:30.680","Text":"Let\u0027s look at HF."},{"Start":"03:31.910 ","End":"03:37.420","Text":"Here are the actual bonds in the HF molecules."},{"Start":"03:40.220 ","End":"03:46.924","Text":"In addition, we have attraction between this hydrogen, for example,"},{"Start":"03:46.924 ","End":"03:52.760","Text":"Delta plus, and the neighboring fluorine on the next molecule,"},{"Start":"03:52.760 ","End":"03:54.500","Text":"which is Delta minus."},{"Start":"03:54.500 ","End":"03:57.625","Text":"We can indicate this by 3 dots."},{"Start":"03:57.625 ","End":"04:02.510","Text":"Then between this H of HF molecule,"},{"Start":"04:02.510 ","End":"04:06.179","Text":"and the next FH molecule."},{"Start":"04:06.820 ","End":"04:10.260","Text":"Then the same thing here."},{"Start":"04:12.680 ","End":"04:20.135","Text":"O and F atoms are very electronegative and pull the electrons in the bond strongly."},{"Start":"04:20.135 ","End":"04:25.250","Text":"This leaves the proton of the H atom almost unshielded."},{"Start":"04:25.250 ","End":"04:28.650","Text":"It\u0027s almost H plus."},{"Start":"04:28.650 ","End":"04:37.040","Text":"The positive charge on the hydrogen attracts the lone pair electrons on O or F. Now,"},{"Start":"04:37.040 ","End":"04:42.283","Text":"these hydrogen bonds are stronger than other intermolecular forces,"},{"Start":"04:42.283 ","End":"04:44.825","Text":"but weaker than a covalent bond."},{"Start":"04:44.825 ","End":"04:50.875","Text":"The hydrogen bond tends to be between 15 and 40 kilojoules per mole,"},{"Start":"04:50.875 ","End":"04:56.540","Text":"whereas the covalent bond is greater than 150 kilojoules per mole."},{"Start":"04:56.540 ","End":"04:58.310","Text":"These are intermediate."},{"Start":"04:58.310 ","End":"05:02.540","Text":"The hydrogen bond is intermediate in strength between"},{"Start":"05:02.540 ","End":"05:08.340","Text":"the other intermolecular forces and a proper covalent bond."},{"Start":"05:08.770 ","End":"05:11.840","Text":"Now, it\u0027s called a hydrogen bond,"},{"Start":"05:11.840 ","End":"05:14.270","Text":"but it\u0027s not a real chemical bond,"},{"Start":"05:14.270 ","End":"05:17.435","Text":"but an electrostatic attraction."},{"Start":"05:17.435 ","End":"05:21.513","Text":"Some people call it dipole-dipole bond,"},{"Start":"05:21.513 ","End":"05:28.400","Text":"dipole-dipole attraction because HO"},{"Start":"05:28.400 ","End":"05:33.030","Text":"has a dipole and this other HO also has a dipole."},{"Start":"05:33.650 ","End":"05:38.990","Text":"Now it\u0027s strongest when H lies on a straight line between 2 O,"},{"Start":"05:38.990 ","End":"05:41.550","Text":"or F or N atoms."},{"Start":"05:41.550 ","End":"05:43.580","Text":"We can see that here."},{"Start":"05:43.580 ","End":"05:52.180","Text":"This hydrogen is in-between the 2 O\u0027s in a straight line. The same is here."},{"Start":"05:52.180 ","End":"05:56.810","Text":"This hydrogen is in a straight line between 2 F\u0027s."},{"Start":"05:59.430 ","End":"06:05.420","Text":"Now the hydrogen bond is denoted by 3 dots."},{"Start":"06:07.350 ","End":"06:13.165","Text":"Another point to note is that the hydrogen bond is longer than a covalent bond."},{"Start":"06:13.165 ","End":"06:17.540","Text":"For example, it\u0027s 175 picometers in ice as"},{"Start":"06:17.540 ","End":"06:22.390","Text":"opposed to 101 picometers for an O-H single bond."},{"Start":"06:22.390 ","End":"06:25.884","Text":"This is in tune with what we\u0027ve learned before;"},{"Start":"06:25.884 ","End":"06:27.995","Text":"the longer the bond,"},{"Start":"06:27.995 ","End":"06:30.400","Text":"the weaker it is."},{"Start":"06:30.400 ","End":"06:40.220","Text":"That means that the H bond in ice is weaker than an ordinary O-H single bond."},{"Start":"06:40.220 ","End":"06:43.760","Text":"In this video, we\u0027ve talked about hydrogen bonds."},{"Start":"06:43.760 ","End":"06:48.750","Text":"In the next video will show examples of hydrogen bonding."}],"ID":24523},{"Watched":false,"Name":"Examples of Hydrogen Bonding","Duration":"5m 37s","ChapterTopicVideoID":23611,"CourseChapterTopicPlaylistID":110286,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.740 ","End":"00:02.760","Text":"In the previous video,"},{"Start":"00:02.760 ","End":"00:04.785","Text":"we learned about hydrogen bonding."},{"Start":"00:04.785 ","End":"00:09.490","Text":"In this video, we\u0027ll discuss a few examples of this bonding."},{"Start":"00:10.130 ","End":"00:13.245","Text":"We\u0027re going to begin by talking about"},{"Start":"00:13.245 ","End":"00:18.375","Text":"intramolecular hydrogen bonding that\u0027s between 2 molecules."},{"Start":"00:18.375 ","End":"00:26.230","Text":"The case we\u0027ll discuss is acetic acid, CH_3CO_2H."},{"Start":"00:26.300 ","End":"00:30.555","Text":"Here are 2 acetic acid molecules."},{"Start":"00:30.555 ","End":"00:33.585","Text":"There\u0027s 1 on the left, CH_3,"},{"Start":"00:33.585 ","End":"00:35.490","Text":"a double bond to O,"},{"Start":"00:35.490 ","End":"00:40.205","Text":"single bond to O H and the same thing on the right-hand side."},{"Start":"00:40.205 ","End":"00:42.980","Text":"Now here we have the hydrogen bonds."},{"Start":"00:42.980 ","End":"00:46.025","Text":"Oxygen on the left-hand molecule,"},{"Start":"00:46.025 ","End":"00:48.560","Text":"hydrogen from the right-hand molecule,"},{"Start":"00:48.560 ","End":"00:50.900","Text":"and oxygen from the right-hand molecule."},{"Start":"00:50.900 ","End":"00:54.350","Text":"Here is our hydrogen bond."},{"Start":"00:54.350 ","End":"00:56.660","Text":"Recall that it\u0027s a straight line."},{"Start":"00:56.660 ","End":"00:59.089","Text":"The hydrogen\u0027s lying in-between"},{"Start":"00:59.089 ","End":"01:03.470","Text":"the oxygen of 1 molecule and the oxygen of the other molecule"},{"Start":"01:03.470 ","End":"01:11.105","Text":"and the hydrogen bond is longer than the normal single bond between oxygen and hydrogen."},{"Start":"01:11.105 ","End":"01:13.205","Text":"The same thing here,"},{"Start":"01:13.205 ","End":"01:21.350","Text":"so 2 acetic acid molecules are held together by 2 hydrogen bonds."},{"Start":"01:21.520 ","End":"01:25.130","Text":"Now, as you\u0027ve probably met in your biology courses,"},{"Start":"01:25.130 ","End":"01:26.525","Text":"we also have H bonding,"},{"Start":"01:26.525 ","End":"01:31.290","Text":"hydrogen bonding between the base pairs in DNA."},{"Start":"01:31.340 ","End":"01:35.300","Text":"The 2 strands of the double helix in DNA are kept"},{"Start":"01:35.300 ","End":"01:39.035","Text":"together by hydrogen bonding between the base pairs."},{"Start":"01:39.035 ","End":"01:41.420","Text":"We have a pair, adenine, thymine,"},{"Start":"01:41.420 ","End":"01:45.470","Text":"which we\u0027ll call A-T, and cytosine, guanine,"},{"Start":"01:45.470 ","End":"01:51.380","Text":"C-T. On the left-hand side,"},{"Start":"01:51.380 ","End":"01:56.435","Text":"we have adenine here and thiamine on the right-hand side."},{"Start":"01:56.435 ","End":"01:59.915","Text":"We can see the hydrogen bonds N, H,"},{"Start":"01:59.915 ","End":"02:04.070","Text":"O, the N on the left,"},{"Start":"02:04.070 ","End":"02:13.425","Text":"on the adenine and the O on the thymine and here\u0027s another N on A,"},{"Start":"02:13.425 ","End":"02:17.715","Text":"and this nitrogen on T. We have"},{"Start":"02:17.715 ","End":"02:23.705","Text":"2 hydrogen bonds holding these very important base pairs together."},{"Start":"02:23.705 ","End":"02:27.125","Text":"Now if we look at guanine and cytosine,"},{"Start":"02:27.125 ","End":"02:29.945","Text":"we see there are 3 hydrogen bonds,"},{"Start":"02:29.945 ","End":"02:33.745","Text":"1 oxygen and nitrogen."},{"Start":"02:33.745 ","End":"02:40.275","Text":"Another 1 between nitrogen and guanine and nitrogen on cytosine."},{"Start":"02:40.275 ","End":"02:48.170","Text":"Another 1 between nitrogen on guanine and oxygen on cytosine."},{"Start":"02:48.170 ","End":"02:54.305","Text":"Here we have 2 hydrogen bonds and here we have 3 hydrogen bonds."},{"Start":"02:54.305 ","End":"02:58.235","Text":"Now let\u0027s look at the case of ice."},{"Start":"02:58.235 ","End":"03:03.235","Text":"As we all know ice is a solid phase of water."},{"Start":"03:03.235 ","End":"03:09.230","Text":"Sometimes, it\u0027s amorphous that means without any crystalline structure."},{"Start":"03:09.230 ","End":"03:11.845","Text":"But usually it\u0027s crystalline."},{"Start":"03:11.845 ","End":"03:15.460","Text":"Now, ice is extremely complicated and can have"},{"Start":"03:15.460 ","End":"03:18.385","Text":"at least 18 different crystal structures"},{"Start":"03:18.385 ","End":"03:22.240","Text":"depending on the pressure and temperature which it is formed."},{"Start":"03:22.240 ","End":"03:26.765","Text":"There are many hydrogen bonds in crystalline ice."},{"Start":"03:26.765 ","End":"03:30.640","Text":"Now we\u0027re going to take the example of ice VI."},{"Start":"03:30.640 ","End":"03:32.860","Text":"We\u0027re going to take the unit cell."},{"Start":"03:32.860 ","End":"03:38.259","Text":"That\u0027s the unit that if we repeat it in all 3 directions,"},{"Start":"03:38.259 ","End":"03:42.410","Text":"you will get the whole structure of the crystal."},{"Start":"03:43.020 ","End":"03:50.890","Text":"Here we see that oxygen is in red and hydrogen is in gray."},{"Start":"03:50.890 ","End":"03:58.460","Text":"We can see an example of a hydrogen bond perhaps between this hydrogen and this oxygen."},{"Start":"03:58.460 ","End":"04:01.085","Text":"There are lots and lots of examples here."},{"Start":"04:01.085 ","End":"04:05.660","Text":"Now, the point about it is that we see that there\u0027s lots of"},{"Start":"04:05.660 ","End":"04:11.030","Text":"hydrogen bonding and there\u0027s lots of open space."},{"Start":"04:11.030 ","End":"04:15.980","Text":"Let\u0027s talk about what happens when this ice melts."},{"Start":"04:15.980 ","End":"04:18.875","Text":"Now because of the empty space in ice,"},{"Start":"04:18.875 ","End":"04:21.080","Text":"it\u0027s less dense than water."},{"Start":"04:21.080 ","End":"04:22.865","Text":"This is very rare."},{"Start":"04:22.865 ","End":"04:27.080","Text":"Usually solids are more dense than water."},{"Start":"04:27.080 ","End":"04:31.790","Text":"We\u0027re going to talk about the anomalous behavior of water."},{"Start":"04:31.790 ","End":"04:35.480","Text":"At 0 degrees Celsius, ice melts."},{"Start":"04:35.480 ","End":"04:37.969","Text":"We all know that. What happens?"},{"Start":"04:37.969 ","End":"04:40.430","Text":"Some of the hydrogen bonds break."},{"Start":"04:40.430 ","End":"04:45.335","Text":"Now there\u0027s less free space and the density is greater than that of ice."},{"Start":"04:45.335 ","End":"04:48.380","Text":"The molecules of water can move closer to each"},{"Start":"04:48.380 ","End":"04:51.710","Text":"other because some of the hydrogen bonds are broken."},{"Start":"04:51.710 ","End":"04:54.380","Text":"That\u0027s the reason that ice floats on water."},{"Start":"04:54.380 ","End":"04:58.580","Text":"Ice is less dense than water, so it floats."},{"Start":"04:58.580 ","End":"05:02.540","Text":"Now as we heat from 0-4 degrees Celsius,"},{"Start":"05:02.540 ","End":"05:05.195","Text":"the density of the water increases."},{"Start":"05:05.195 ","End":"05:09.545","Text":"That\u0027s because more and more hydrogen bonds are being broken,"},{"Start":"05:09.545 ","End":"05:12.485","Text":"allowing the water molecules to get closer."},{"Start":"05:12.485 ","End":"05:14.780","Text":"This is anomalous behavior."},{"Start":"05:14.780 ","End":"05:16.250","Text":"Generally when you heat,"},{"Start":"05:16.250 ","End":"05:20.125","Text":"the density decreases, here it\u0027s increasing."},{"Start":"05:20.125 ","End":"05:25.130","Text":"But the story changes above 4 degrees Celsius,"},{"Start":"05:25.130 ","End":"05:29.255","Text":"then the density starts decreasing with increasing temperature,"},{"Start":"05:29.255 ","End":"05:31.940","Text":"as is the usual behavior."},{"Start":"05:31.940 ","End":"05:38.190","Text":"In this video, we saw some examples of hydrogen bonding."}],"ID":24522},{"Watched":false,"Name":"Definition of Van Der Waals Forces","Duration":"1m 24s","ChapterTopicVideoID":23669,"CourseChapterTopicPlaylistID":110286,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:05.680","Text":"In this video, we will define Van der Waals forces."},{"Start":"00:05.840 ","End":"00:12.060","Text":"Now there are several different definitions of Van der Waals interactions of forces,"},{"Start":"00:12.060 ","End":"00:16.440","Text":"and you must check with your lecturers which they prefer."},{"Start":"00:16.440 ","End":"00:23.400","Text":"1 definition is that Van der Waals forces are all attractive forces that depend"},{"Start":"00:23.400 ","End":"00:31.410","Text":"on r^-6, 1/r^6."},{"Start":"00:31.410 ","End":"00:33.390","Text":"These include London forces,"},{"Start":"00:33.390 ","End":"00:40.720","Text":"dipole-dipole interactions in a gas or liquid, and dipole-induced-dipole interactions."},{"Start":"00:41.450 ","End":"00:44.530","Text":"Another definition is that,"},{"Start":"00:44.530 ","End":"00:51.725","Text":"Van der Waals forces are all weak attractive forces involving dipole-dipole interactions."},{"Start":"00:51.725 ","End":"00:56.000","Text":"These include London forces, dipole-dipole interactions,"},{"Start":"00:56.000 ","End":"01:01.115","Text":"a gas liquid or solid, and dipole-induced-dipole interactions."},{"Start":"01:01.115 ","End":"01:05.930","Text":"Hydrogen bonding involves dipole-dipole interactions as well,"},{"Start":"01:05.930 ","End":"01:09.350","Text":"and it\u0027s often included in this definition."},{"Start":"01:09.350 ","End":"01:14.990","Text":"In this video, we gave 2 alternative definitions of Van der Waals forces."},{"Start":"01:14.990 ","End":"01:20.990","Text":"You\u0027d be well advised to talk to your lecturers and ask them which they prefer,"},{"Start":"01:20.990 ","End":"01:24.630","Text":"or perhaps they even have another definition."}],"ID":24574}],"Thumbnail":null,"ID":110286},{"Name":"Properties of Liquids","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Surface tension","Duration":"8m 5s","ChapterTopicVideoID":23618,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.380","Text":"In previous videos, we learned about inter-molecular forces."},{"Start":"00:04.380 ","End":"00:06.075","Text":"In the next few videos,"},{"Start":"00:06.075 ","End":"00:12.370","Text":"we\u0027ll discuss some properties of liquids that result from intermolecular forces."},{"Start":"00:12.620 ","End":"00:16.900","Text":"We\u0027re going, to begin with, surface tension."},{"Start":"00:17.690 ","End":"00:23.660","Text":"Molecules in the bulk of a liquid are attracted by surrounding molecules."},{"Start":"00:23.660 ","End":"00:27.995","Text":"Here\u0027s a picture. Here\u0027s a molecule in the center of the red one,"},{"Start":"00:27.995 ","End":"00:32.190","Text":"and it\u0027s attracted by surrounding molecules."},{"Start":"00:36.010 ","End":"00:40.970","Text":"However, if we have a molecule at the surface of a liquid,"},{"Start":"00:40.970 ","End":"00:45.840","Text":"it\u0027s only attracted by molecules inside the liquid."},{"Start":"00:46.390 ","End":"00:48.995","Text":"Here\u0027s a little illustration."},{"Start":"00:48.995 ","End":"00:57.060","Text":"Here\u0027s a molecule at the surface and it\u0027s attracted by molecules inside the liquid."},{"Start":"00:57.790 ","End":"01:02.300","Text":"This means that molecules at the surface are less stable."},{"Start":"01:02.300 ","End":"01:06.570","Text":"They have higher energy than those in the bulk."},{"Start":"01:07.690 ","End":"01:12.530","Text":"Because of that, because the energy at the surface is higher,"},{"Start":"01:12.530 ","End":"01:17.495","Text":"liquids tend to minimize their surface area."},{"Start":"01:17.495 ","End":"01:23.164","Text":"It is as if an invisible skin forms on the surface of the liquid."},{"Start":"01:23.164 ","End":"01:29.449","Text":"You can\u0027t see it but it\u0027s like a little skin, thin skin."},{"Start":"01:29.449 ","End":"01:33.800","Text":"Surface tension or sometimes called surface energy,"},{"Start":"01:33.800 ","End":"01:36.440","Text":"indicated by the letter Gamma,"},{"Start":"01:36.440 ","End":"01:42.550","Text":"is the amount of energy or work required to increase the surface area of a liquid."},{"Start":"01:42.550 ","End":"01:48.480","Text":"To increase the surface area takes work or energy."},{"Start":"01:48.620 ","End":"01:52.520","Text":"The units are joules per meter squared."},{"Start":"01:52.520 ","End":"01:55.520","Text":"If we analyze joules per meter squared,"},{"Start":"01:55.520 ","End":"01:59.999","Text":"we\u0027ll see it\u0027s the same as Newtons per meter."},{"Start":"02:02.980 ","End":"02:10.040","Text":"Let\u0027s do that. A joule is kilograms times meter per second, all squared."},{"Start":"02:10.040 ","End":"02:12.530","Text":"Here we have the meter to the power minus 2,"},{"Start":"02:12.530 ","End":"02:14.210","Text":"meter to the power minus 2."},{"Start":"02:14.210 ","End":"02:16.310","Text":"If we multiply this out,"},{"Start":"02:16.310 ","End":"02:21.010","Text":"Here\u0027s meters squared times meter to the power minus 2 is just one."},{"Start":"02:21.010 ","End":"02:24.245","Text":"We\u0027re left with kilograms per second square."},{"Start":"02:24.245 ","End":"02:28.625","Text":"Now if we put in meter times meter to the power minus one,"},{"Start":"02:28.625 ","End":"02:30.480","Text":"which is just of course 1."},{"Start":"02:30.480 ","End":"02:33.590","Text":"Then we can divide this into 2 parts."},{"Start":"02:33.590 ","End":"02:40.385","Text":"Kilograms times meters per second squared is a newton force,"},{"Start":"02:40.385 ","End":"02:43.970","Text":"and we\u0027re left with the meter to the power minus 1."},{"Start":"02:43.970 ","End":"02:46.655","Text":"This is Newtons per meter."},{"Start":"02:46.655 ","End":"02:51.514","Text":"Here are units of energy joules per meter squared,"},{"Start":"02:51.514 ","End":"02:56.490","Text":"and here are units of force, Newtons per meter."},{"Start":"02:56.780 ","End":"02:59.390","Text":"Here are some examples."},{"Start":"02:59.390 ","End":"03:06.215","Text":"At 25 degrees Celsius and the units are millinewtons per meter."},{"Start":"03:06.215 ","End":"03:12.350","Text":"For ethanol, C_2H_5 OH, is 22.8."},{"Start":"03:12.350 ","End":"03:20.570","Text":"For water, 72.75 water has more hydrogen bonds than ethanol."},{"Start":"03:20.570 ","End":"03:23.625","Text":"Water is 4 ethanol has 3,"},{"Start":"03:23.625 ","End":"03:28.820","Text":"and it\u0027s very high for Mercury, 472."},{"Start":"03:28.820 ","End":"03:35.320","Text":"Mercury has very strong metallic bonds between the atoms."},{"Start":"03:35.320 ","End":"03:40.640","Text":"Now the surface tension decreases with increasing temperature."},{"Start":"03:40.640 ","End":"03:42.125","Text":"Why is this so,"},{"Start":"03:42.125 ","End":"03:44.990","Text":"it because the molecules have greater kinetic energy"},{"Start":"03:44.990 ","End":"03:49.080","Text":"and can overcome the intermolecular forces."},{"Start":"03:49.450 ","End":"03:56.150","Text":"Let\u0027s talk about some phenomena resulting from surface tension."},{"Start":"03:56.150 ","End":"04:01.655","Text":"The first is it a needle or insect floats on water,"},{"Start":"04:01.655 ","End":"04:03.560","Text":"doesn\u0027t really float on the water."},{"Start":"04:03.560 ","End":"04:05.150","Text":"It\u0027s on top of the water,"},{"Start":"04:05.150 ","End":"04:11.705","Text":"and the reason for this is the surface tension overcomes the gravitational forces."},{"Start":"04:11.705 ","End":"04:13.550","Text":"Here\u0027s a little picture."},{"Start":"04:13.550 ","End":"04:21.510","Text":"Here\u0027s our very clean needle on water and we see that it\u0027s on the surface of the water."},{"Start":"04:21.860 ","End":"04:25.150","Text":"The surface tension is overcoming"},{"Start":"04:25.150 ","End":"04:32.150","Text":"the gravitational forces that would force the needle to sink into the water."},{"Start":"04:33.020 ","End":"04:37.565","Text":"Now the phenomenon is the wetting of a surface."},{"Start":"04:37.565 ","End":"04:41.170","Text":"Here is our glass slide."},{"Start":"04:41.170 ","End":"04:44.630","Text":"On one side, it\u0027s clean."},{"Start":"04:44.640 ","End":"04:49.855","Text":"The other side is dirty or has oil on it."},{"Start":"04:49.855 ","End":"04:52.660","Text":"Now in the left-hand side,"},{"Start":"04:52.660 ","End":"04:57.025","Text":"where there is where it\u0027s clean."},{"Start":"04:57.025 ","End":"05:00.399","Text":"We can wet this slide,"},{"Start":"05:00.399 ","End":"05:02.545","Text":"we can wet this part of the slide."},{"Start":"05:02.545 ","End":"05:04.600","Text":"We get wetting."},{"Start":"05:04.600 ","End":"05:07.310","Text":"It\u0027s in the side where it\u0027s dirty."},{"Start":"05:07.310 ","End":"05:16.425","Text":"The water forms globules which lie on the surface of the slide."},{"Start":"05:16.425 ","End":"05:18.940","Text":"There\u0027s 2 different behaviors."},{"Start":"05:18.940 ","End":"05:22.430","Text":"This is like we have in an oil skin,"},{"Start":"05:22.430 ","End":"05:28.470","Text":"why we have oil skin coats that repel the water."},{"Start":"05:28.990 ","End":"05:36.749","Text":"Now the forces between liquid molecules in a drop are called cohesive sticking together."},{"Start":"05:38.560 ","End":"05:42.950","Text":"Whereas the forces between liquid molecules and surfaces"},{"Start":"05:42.950 ","End":"05:47.340","Text":"are called adhesive sticking to each other."},{"Start":"05:47.600 ","End":"05:53.950","Text":"Now if the cohesive forces are greater than the adhesive, there\u0027s no wetting."},{"Start":"05:53.950 ","End":"05:58.490","Text":"Here\u0027s the cohesive is greater than adhesives."},{"Start":"05:58.530 ","End":"06:05.890","Text":"There\u0027s no wetting. But if the adhesive force are greater than cohesive,"},{"Start":"06:05.890 ","End":"06:09.890","Text":"then we get wetting as in this slide."},{"Start":"06:10.320 ","End":"06:15.235","Text":"Another phenomenon where we can relate to"},{"Start":"06:15.235 ","End":"06:22.910","Text":"cohesive and adhesive forces is we\u0027ve compared water and mercury meniscus in a test tube."},{"Start":"06:23.940 ","End":"06:26.830","Text":"Here are 2 test tubes."},{"Start":"06:26.830 ","End":"06:30.230","Text":"The left-hand side we have water."},{"Start":"06:30.630 ","End":"06:34.470","Text":"The right-hand side we have mercury."},{"Start":"06:35.650 ","End":"06:43.475","Text":"Now the water meniscus goes down and up at the sides, it\u0027s convex."},{"Start":"06:43.475 ","End":"06:47.165","Text":"The mercury meniscus is concave."},{"Start":"06:47.165 ","End":"06:50.660","Text":"It\u0027s higher at the middle than the sides."},{"Start":"06:50.660 ","End":"06:52.835","Text":"What\u0027s happening here?"},{"Start":"06:52.835 ","End":"06:58.970","Text":"In the water, we have very strong adhesive forces to the sides of the test tube,"},{"Start":"06:58.970 ","End":"07:01.430","Text":"and lower cohesive forces."},{"Start":"07:01.430 ","End":"07:05.210","Text":"Whereas in the mercury the way for high cohesive forces in"},{"Start":"07:05.210 ","End":"07:10.220","Text":"the center and lower adhesive forces at the sides."},{"Start":"07:10.220 ","End":"07:14.340","Text":"In water, adhesive is greater than cohesive,"},{"Start":"07:14.340 ","End":"07:18.995","Text":"and in the mercury, cohesive is greater than adhesive."},{"Start":"07:18.995 ","End":"07:23.420","Text":"Related phenomenon is capillary action."},{"Start":"07:23.420 ","End":"07:27.155","Text":"Here\u0027s our capillary tube."},{"Start":"07:27.155 ","End":"07:30.570","Text":"Very narrow tube."},{"Start":"07:30.830 ","End":"07:36.350","Text":"Water rises in this thin capillary tube."},{"Start":"07:36.350 ","End":"07:39.320","Text":"That\u0027s because it\u0027s very thin."},{"Start":"07:39.320 ","End":"07:45.090","Text":"Very few cohesive forces are very strong adhesive forces."},{"Start":"07:45.500 ","End":"07:48.295","Text":"As you probably know from Biology,"},{"Start":"07:48.295 ","End":"07:52.010","Text":"there are many phenomena in nature which result"},{"Start":"07:52.010 ","End":"07:57.290","Text":"from the fact that water rises in very thin tubes,"},{"Start":"07:57.290 ","End":"08:00.110","Text":"it\u0027s called capillary action."},{"Start":"08:00.110 ","End":"08:05.489","Text":"In this video, we talked about surface tension."}],"ID":30036},{"Watched":false,"Name":"Viscosity","Duration":"4m 53s","ChapterTopicVideoID":23620,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"In the previous video,"},{"Start":"00:01.890 ","End":"00:03.945","Text":"we talked about surface tension."},{"Start":"00:03.945 ","End":"00:07.720","Text":"In this video we\u0027ll discuss viscosity."},{"Start":"00:08.360 ","End":"00:12.015","Text":"Viscosity, which is written as mu or eta,"},{"Start":"00:12.015 ","End":"00:15.494","Text":"is the liquid\u0027s resistance to flow."},{"Start":"00:15.494 ","End":"00:19.860","Text":"We can estimate the viscosity by measuring the time taken for"},{"Start":"00:19.860 ","End":"00:25.050","Text":"a steel ball or a marble to drop through a certain distance."},{"Start":"00:25.050 ","End":"00:28.725","Text":"We have a bowl inside the liquid,"},{"Start":"00:28.725 ","End":"00:32.410","Text":"and we measure how long it takes to drop."},{"Start":"00:32.410 ","End":"00:36.300","Text":"The slower the ball moves, the greater viscosity."},{"Start":"00:36.300 ","End":"00:39.799","Text":"Obviously will move more slowly in"},{"Start":"00:39.799 ","End":"00:44.520","Text":"molasses or treacle something like that than it will in water."},{"Start":"00:44.620 ","End":"00:49.490","Text":"Now a liquid with a high viscosity is said to be viscous."},{"Start":"00:49.490 ","End":"00:55.355","Text":"Honey and treacle or molasses are viscous liquids."},{"Start":"00:55.355 ","End":"01:01.525","Text":"Units used to measure them is this rather strange unit called a centipoise."},{"Start":"01:01.525 ","End":"01:05.910","Text":"A centipoise, the cent is 10^minus 2."},{"Start":"01:05.910 ","End":"01:12.985","Text":"The poise is 10^minus 1 times kilograms per meter per second."},{"Start":"01:12.985 ","End":"01:16.650","Text":"We can multiply 10^minus 2 by 10^minus 1,"},{"Start":"01:16.650 ","End":"01:18.425","Text":"you get 10^minus 3."},{"Start":"01:18.425 ","End":"01:21.490","Text":"We still have the kilogram times the meter to"},{"Start":"01:21.490 ","End":"01:24.970","Text":"the power minus 1 times the second power minus 1."},{"Start":"01:24.970 ","End":"01:32.470","Text":"Now, we can write the kilogram per meter per second in a slightly different way."},{"Start":"01:32.470 ","End":"01:36.160","Text":"Instead of writing meter to the power minus 1,"},{"Start":"01:36.160 ","End":"01:38.710","Text":"we could write meter/meter squared."},{"Start":"01:38.710 ","End":"01:41.620","Text":"Instead of writing second to the power minus 1,"},{"Start":"01:41.620 ","End":"01:44.770","Text":"we can write second to the power minus 2 times"},{"Start":"01:44.770 ","End":"01:49.550","Text":"second together when they\u0027re multiplied that second minus 1."},{"Start":"01:49.550 ","End":"01:51.110","Text":"If we do that,"},{"Start":"01:51.110 ","End":"01:59.250","Text":"we notice that the numerator is a Newton and the denominator is area."},{"Start":"01:59.250 ","End":"02:04.095","Text":"Divide 1 by the other, we get Pascal."},{"Start":"02:04.095 ","End":"02:06.750","Text":"Here we have 10^minus 3, that\u0027s milli,"},{"Start":"02:06.750 ","End":"02:10.200","Text":"so we have millipascal times second,"},{"Start":"02:10.200 ","End":"02:12.540","Text":"and this is the SI units."},{"Start":"02:12.540 ","End":"02:14.925","Text":"We can have 2 alternative ways,"},{"Start":"02:14.925 ","End":"02:21.885","Text":"either centipoise or millipascals times second."},{"Start":"02:21.885 ","End":"02:25.235","Text":"Now, why is a liquid viscous?"},{"Start":"02:25.235 ","End":"02:30.845","Text":"The viscosity of liquid is indicative of the strength of forces between the molecules."},{"Start":"02:30.845 ","End":"02:37.810","Text":"A molecule needs to overcome these forces to slip past other molecules."},{"Start":"02:37.810 ","End":"02:43.325","Text":"Now viscosity is a complicated phenomenon and what we\u0027re saying here is very simple."},{"Start":"02:43.325 ","End":"02:47.810","Text":"It\u0027s difficult to predict because other factors are also important,"},{"Start":"02:47.810 ","End":"02:52.890","Text":"such as how easily a molecule can take up a new position after moving."},{"Start":"02:53.860 ","End":"02:55.925","Text":"One thing is for sure,"},{"Start":"02:55.925 ","End":"02:59.510","Text":"viscosity decreases with increasing temperature."},{"Start":"02:59.510 ","End":"03:06.490","Text":"Because the molecules have more energy and can move more easily, overcoming the forces."},{"Start":"03:06.490 ","End":"03:09.700","Text":"Let\u0027s take 3 examples, water,"},{"Start":"03:09.700 ","End":"03:14.430","Text":"glycerol and long hydrocarbon chains."},{"Start":"03:15.340 ","End":"03:18.965","Text":"Now water has a low viscosity,"},{"Start":"03:18.965 ","End":"03:25.265","Text":"0.89 centipoise at 25 degrees Celsius."},{"Start":"03:25.265 ","End":"03:28.415","Text":"Even although it forms hydrogen bonds."},{"Start":"03:28.415 ","End":"03:33.755","Text":"Remember that each water molecule forms 4 hydrogen bonds."},{"Start":"03:33.755 ","End":"03:36.350","Text":"Now water molecules easily form"},{"Start":"03:36.350 ","End":"03:41.495","Text":"the new hydrogen bonds after breaking them to allow them to slip past other molecules."},{"Start":"03:41.495 ","End":"03:44.420","Text":"That\u0027s the reason it has low viscosity."},{"Start":"03:44.420 ","End":"03:50.825","Text":"Once it\u0027s escaped, the hydrogen bonds very easily forms new ones."},{"Start":"03:50.825 ","End":"03:56.915","Text":"Now glycerol is a long molecule with lots of OHs in it."},{"Start":"03:56.915 ","End":"03:59.720","Text":"It has high viscosity,"},{"Start":"03:59.720 ","End":"04:04.460","Text":"1,200 centipoise at 20 degrees Celsius."},{"Start":"04:04.460 ","End":"04:10.790","Text":"The reason for the high viscosity is because it can form many hydrogen bonds."},{"Start":"04:10.790 ","End":"04:16.040","Text":"It also has difficulty in reforming hydrogen bonds once broken unlike water,"},{"Start":"04:16.040 ","End":"04:19.835","Text":"so this has high viscosity."},{"Start":"04:19.835 ","End":"04:23.830","Text":"Now if we have long hydrocarbon chains,"},{"Start":"04:23.830 ","End":"04:27.375","Text":"such as in petroleum or fuel oil,"},{"Start":"04:27.375 ","End":"04:33.300","Text":"these are viscous even although they do not form any H bonds at all this just CHs here,"},{"Start":"04:33.300 ","End":"04:35.600","Text":"there\u0027s no Os, no oxygens."},{"Start":"04:35.600 ","End":"04:40.745","Text":"The reason is the long change become entangled like cooked spaghetti."},{"Start":"04:40.745 ","End":"04:44.615","Text":"We have long chains becoming entangled."},{"Start":"04:44.615 ","End":"04:49.759","Text":"They have to become disentangled in order for them to move."},{"Start":"04:49.759 ","End":"04:53.970","Text":"In this video, we talked about viscosity."}],"ID":30037},{"Watched":false,"Name":"Enthalpy of vaporization","Duration":"5m 3s","ChapterTopicVideoID":23736,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:02.490","Text":"In previous videos,"},{"Start":"00:02.490 ","End":"00:05.310","Text":"we discussed surface tension of viscosity."},{"Start":"00:05.310 ","End":"00:07.965","Text":"In this video, we\u0027ll talk about the influence of"},{"Start":"00:07.965 ","End":"00:12.730","Text":"intermolecular forces on the enthalpy of vaporization."},{"Start":"00:14.810 ","End":"00:19.660","Text":"We\u0027re going to talk about the enthalpy of vaporization."},{"Start":"00:20.630 ","End":"00:23.160","Text":"The enthalpy of vaporization,"},{"Start":"00:23.160 ","End":"00:26.370","Text":"Delta H_vap is the heat required"},{"Start":"00:26.370 ","End":"00:30.639","Text":"to vaporize the liquid at a certain pressure and temperature."},{"Start":"00:31.100 ","End":"00:39.289","Text":"Its units are kilojoules per mole and it\u0027s an endothermic process."},{"Start":"00:39.289 ","End":"00:42.635","Text":"Delta H of vaporization is greater than 0,"},{"Start":"00:42.635 ","End":"00:46.950","Text":"it takes energy to vaporize the liquid."},{"Start":"00:47.990 ","End":"00:51.800","Text":"Now the speeds in kinetic energies of molecules are"},{"Start":"00:51.800 ","End":"00:55.564","Text":"distributed over a wide range at every temperature."},{"Start":"00:55.564 ","End":"00:59.550","Text":"We talked about this when we talked about gases."},{"Start":"01:00.670 ","End":"01:04.920","Text":"Here\u0027s the Maxwell-Boltzmann distribution."},{"Start":"01:04.930 ","End":"01:11.495","Text":"Replot the percentage of molecules with a particular speed versus the speed."},{"Start":"01:11.495 ","End":"01:18.320","Text":"It\u0027s this red curve we see it rises steeply and then decreases more slowly."},{"Start":"01:18.320 ","End":"01:22.810","Text":"So the average is not quite at the top but slightly displaced."},{"Start":"01:22.810 ","End":"01:26.520","Text":"That\u0027s u average this average speed."},{"Start":"01:30.340 ","End":"01:32.455","Text":"Now at every temperature,"},{"Start":"01:32.455 ","End":"01:36.140","Text":"there are some molecules with sufficient kinetic energy to overcome"},{"Start":"01:36.140 ","End":"01:41.525","Text":"the intermolecular forces and escape from the liquid into the gas phase."},{"Start":"01:41.525 ","End":"01:47.790","Text":"See here these molecules beyond this black line."},{"Start":"01:52.130 ","End":"01:58.840","Text":"That\u0027s the reason that we get evaporation at any temperature."},{"Start":"02:02.840 ","End":"02:05.405","Text":"As we increase the temperature,"},{"Start":"02:05.405 ","End":"02:09.540","Text":"more molecules have enough energy to escape."},{"Start":"02:09.610 ","End":"02:12.080","Text":"At a higher temperature,"},{"Start":"02:12.080 ","End":"02:16.565","Text":"the curve rises has an average later on,"},{"Start":"02:16.565 ","End":"02:23.040","Text":"and then decreases with a larger tail, longer tail."},{"Start":"02:23.920 ","End":"02:30.750","Text":"The number of molecules with particular speed is now greater."},{"Start":"02:31.460 ","End":"02:34.564","Text":"Whereas the surface area increases,"},{"Start":"02:34.564 ","End":"02:36.965","Text":"more molecules can escape."},{"Start":"02:36.965 ","End":"02:40.670","Text":"If we want to evaporate something rapidly,"},{"Start":"02:40.670 ","End":"02:43.770","Text":"we spread it over a larger area."},{"Start":"02:44.990 ","End":"02:52.280","Text":"Now, Delta H vaporization is higher for liquids with stronger intermolecular forces."},{"Start":"02:52.280 ","End":"02:54.740","Text":"If liquids have strong intermolecular forces,"},{"Start":"02:54.740 ","End":"03:02.470","Text":"we have to address more energy to release them from the liquid into the gas phase,"},{"Start":"03:03.710 ","End":"03:11.060","Text":"and of course, hydrogen bonding has a significant influence on intermolecular forces."},{"Start":"03:11.060 ","End":"03:14.660","Text":"It\u0027s very important as an intermolecular force so it has"},{"Start":"03:14.660 ","End":"03:19.560","Text":"a strong influence on Delta H of vaporization."},{"Start":"03:24.290 ","End":"03:26.870","Text":"Let\u0027s take some examples."},{"Start":"03:26.870 ","End":"03:31.140","Text":"Dimethyl ether, water, and ethanol."},{"Start":"03:32.590 ","End":"03:35.585","Text":"Let\u0027s start with dimethyl ether,"},{"Start":"03:35.585 ","End":"03:44.150","Text":"that (CH_3)_2O oxygen in the center and 2 CH_3 groups,"},{"Start":"03:44.150 ","End":"03:47.100","Text":"2 methyl groups."},{"Start":"03:47.570 ","End":"03:54.230","Text":"Here, Delta H of vaporization is 18.5 kilojoules per mole."},{"Start":"03:54.230 ","End":"03:58.590","Text":"There\u0027s no possibility here of hydrogen bonding."},{"Start":"04:02.090 ","End":"04:07.395","Text":"Now water, that\u0027s H_2O of course."},{"Start":"04:07.395 ","End":"04:11.720","Text":"The water molecule can form 4H bonds so we have"},{"Start":"04:11.720 ","End":"04:15.500","Text":"a significantly larger Delta H of vaporization."},{"Start":"04:15.500 ","End":"04:21.275","Text":"Delta H of vaporization is 44.0 kilojoules per mole,"},{"Start":"04:21.275 ","End":"04:25.500","Text":"and all this is at 25 degrees Celsius."},{"Start":"04:28.240 ","End":"04:32.405","Text":"Now, ethanol has less hydrogen bonds."},{"Start":"04:32.405 ","End":"04:36.390","Text":"It can form 3 hydrogen bonds."},{"Start":"04:36.430 ","End":"04:40.820","Text":"Delta H of vaporization is pretty high,"},{"Start":"04:40.820 ","End":"04:47.580","Text":"but less than that of water is 42.3 kilojoules per mole."},{"Start":"04:55.130 ","End":"05:01.050","Text":"In this video, we talked about the enthalpy of vaporization."}],"ID":30038},{"Watched":false,"Name":"Vapor pressure","Duration":"5m 19s","ChapterTopicVideoID":23619,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.100","Text":"In the previous video,"},{"Start":"00:02.100 ","End":"00:04.920","Text":"we learned about the enthalpy of vaporization."},{"Start":"00:04.920 ","End":"00:07.740","Text":"In this video we\u0027ll discuss the related topic of"},{"Start":"00:07.740 ","End":"00:12.550","Text":"vapor pressure and its temperature dependence."},{"Start":"00:12.710 ","End":"00:16.904","Text":"We\u0027re going to talk about vapor pressure."},{"Start":"00:16.904 ","End":"00:22.740","Text":"We all know that when liquids open to the atmosphere, it gradually evaporates."},{"Start":"00:22.740 ","End":"00:25.890","Text":"If at least for glass of water overnight,"},{"Start":"00:25.890 ","End":"00:29.920","Text":"by the morning there will be less water in it."},{"Start":"00:30.080 ","End":"00:34.860","Text":"However, liquid in a closed vessel evaporates,"},{"Start":"00:34.860 ","End":"00:36.990","Text":"then condenses and so on,"},{"Start":"00:36.990 ","End":"00:42.330","Text":"until we get a dynamic equilibrium between the liquid and its vapor."},{"Start":"00:42.330 ","End":"00:45.015","Text":"Here\u0027s a closed vessel,"},{"Start":"00:45.015 ","End":"00:47.685","Text":"put some water in the bottom,"},{"Start":"00:47.685 ","End":"00:50.900","Text":"the water gradually evaporates,"},{"Start":"00:50.900 ","End":"00:56.264","Text":"then condenses again until after a certain amount of time,"},{"Start":"00:56.264 ","End":"00:59.060","Text":"we have a fixed amount of"},{"Start":"00:59.060 ","End":"01:03.680","Text":"molecules in the gas phase and a fixed amount in the liquid phase."},{"Start":"01:03.680 ","End":"01:07.380","Text":"This is called a dynamic equilibrium."},{"Start":"01:09.980 ","End":"01:13.610","Text":"Now the vapor pressure is the pressure exerted by"},{"Start":"01:13.610 ","End":"01:17.885","Text":"the vapor that\u0027s in dynamic equilibrium with its liquid."},{"Start":"01:17.885 ","End":"01:23.510","Text":"It\u0027s the pressure of this gas above the liquid."},{"Start":"01:24.700 ","End":"01:27.665","Text":"Here\u0027s our equilibrium liquid."},{"Start":"01:27.665 ","End":"01:30.365","Text":"This is equilibrium sign and vapor."},{"Start":"01:30.365 ","End":"01:39.630","Text":"Liquid to vapor is evaporation or vaporization and going the other way is condensation."},{"Start":"01:51.020 ","End":"01:55.340","Text":"We say that liquids with high vapor pressure at room temperature are"},{"Start":"01:55.340 ","End":"02:02.040","Text":"volatile and those with low vapor pressure are nonvolatile."},{"Start":"02:02.620 ","End":"02:08.100","Text":"Now the vapor pressure increases with temperature."},{"Start":"02:09.010 ","End":"02:12.110","Text":"The normal boiling point of a liquid is"},{"Start":"02:12.110 ","End":"02:15.170","Text":"the temperature at which the vapor pressure is equal to"},{"Start":"02:15.170 ","End":"02:21.390","Text":"atmospheric pressure of 760 millimeters of mercury."},{"Start":"02:21.940 ","End":"02:25.250","Text":"Let\u0027s take some examples."},{"Start":"02:25.250 ","End":"02:29.000","Text":"Acetone is volatile, we can smell it."},{"Start":"02:29.000 ","End":"02:33.245","Text":"The molecules in the liquid escape to the vapor,"},{"Start":"02:33.245 ","End":"02:36.710","Text":"get to our noses and we smell it rather strongly."},{"Start":"02:36.710 ","End":"02:41.840","Text":"The vapor pressure is 231 millimeters of mercury."},{"Start":"02:41.840 ","End":"02:44.690","Text":"Water is moderately volatile."},{"Start":"02:44.690 ","End":"02:49.655","Text":"The vapor pressure is 23.8 millimeters of mercury."},{"Start":"02:49.655 ","End":"02:53.885","Text":"Example of a non-volatile substance is mercury,"},{"Start":"02:53.885 ","End":"03:01.460","Text":"which has an extremely low vapor pressure of 0.0018 millimeters of mercury."},{"Start":"03:01.460 ","End":"03:03.560","Text":"Is very little gas,"},{"Start":"03:03.560 ","End":"03:06.305","Text":"but this gas is very poisonous."},{"Start":"03:06.305 ","End":"03:11.670","Text":"Now let\u0027s look at the temperature dependence of the vapor pressure of water."},{"Start":"03:11.920 ","End":"03:19.910","Text":"Here\u0027s the vapor pressure plotted against the temperature in degrees Celsius."},{"Start":"03:19.910 ","End":"03:25.670","Text":"It gradually increases, then more steeply."},{"Start":"03:25.670 ","End":"03:28.469","Text":"Here at this point,"},{"Start":"03:29.290 ","End":"03:33.919","Text":"where it\u0027s equal to 760 millimeters of mercury,"},{"Start":"03:33.919 ","End":"03:36.095","Text":"that\u0027s the boiling point of water."},{"Start":"03:36.095 ","End":"03:39.510","Text":"We see it\u0027s 100 degrees Celsius, that\u0027s boiling point."},{"Start":"03:41.830 ","End":"03:46.535","Text":"Now, chemists and physicists love straight lines."},{"Start":"03:46.535 ","End":"03:49.235","Text":"They discovered that if you plot the line,"},{"Start":"03:49.235 ","End":"03:56.720","Text":"the natural logarithm of the pressure versus 1 over the temperature in Kelvin,"},{"Start":"03:56.720 ","End":"04:00.030","Text":"you get a straight line."},{"Start":"04:00.880 ","End":"04:03.515","Text":"Here\u0027s our straight line."},{"Start":"04:03.515 ","End":"04:11.255","Text":"When we plot line of P versus 1 over T. We could write it mathematically."},{"Start":"04:11.255 ","End":"04:16.850","Text":"Line P is equal to minus Delta H_vaporization divided by R,"},{"Start":"04:16.850 ","End":"04:19.530","Text":"R is the gas constant,"},{"Start":"04:23.170 ","End":"04:26.135","Text":"multiplied by 1 over T,"},{"Start":"04:26.135 ","End":"04:27.680","Text":"is the temperature in Kelvin,"},{"Start":"04:27.680 ","End":"04:30.685","Text":"plus B, which is the intercept."},{"Start":"04:30.685 ","End":"04:37.830","Text":"This is like Y is equal to MX plus C or B,"},{"Start":"04:37.830 ","End":"04:39.930","Text":"whatever you used to writing."},{"Start":"04:39.930 ","End":"04:41.970","Text":"This is Y,"},{"Start":"04:41.970 ","End":"04:43.740","Text":"this is X,"},{"Start":"04:43.740 ","End":"04:45.210","Text":"this is a slope,"},{"Start":"04:45.210 ","End":"04:49.835","Text":"M and B is the intercept."},{"Start":"04:49.835 ","End":"04:53.390","Text":"The slope here is minus a negative slope,"},{"Start":"04:53.390 ","End":"04:58.835","Text":"Delta H_vaporization divided by R. So the slope is minus"},{"Start":"04:58.835 ","End":"05:07.190","Text":"Delta H_vaporization divided by the gas constant R. In this video,"},{"Start":"05:07.190 ","End":"05:10.490","Text":"we learned about vapor pressure and its temperature dependence."},{"Start":"05:10.490 ","End":"05:11.810","Text":"In the next video,"},{"Start":"05:11.810 ","End":"05:19.110","Text":"we\u0027ll derive the Clausius-Clapeyron equation for the temperature dependence."}],"ID":30039},{"Watched":false,"Name":"Exercise 1","Duration":"1m ","ChapterTopicVideoID":28543,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.465","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.465 ","End":"00:10.200","Text":"How much heat is required to vaporize 2.45 moles of methanol at 298 kelvin?"},{"Start":"00:10.200 ","End":"00:16.485","Text":"The enthalpy of vaporization of methanol at 298 K equals 38 kilojoules per mole."},{"Start":"00:16.485 ","End":"00:20.640","Text":"We want to calculate the heat required to vaporize the methanol."},{"Start":"00:20.640 ","End":"00:30.105","Text":"Q the heat equals the enthalpy of vaporization times the number of moles."},{"Start":"00:30.105 ","End":"00:32.835","Text":"In our case, we\u0027re talking about the methanol."},{"Start":"00:32.835 ","End":"00:37.335","Text":"The enthalpy of vaporization equals 38 kilojoules per mole."},{"Start":"00:37.335 ","End":"00:41.745","Text":"We also know that we have 2.45 moles of methanol."},{"Start":"00:41.745 ","End":"00:44.165","Text":"The moles cancel out,"},{"Start":"00:44.165 ","End":"00:50.480","Text":"and this equals 93.1 and we\u0027re left with kilojoules."},{"Start":"00:50.480 ","End":"00:57.065","Text":"The heat that is required to vaporize the methanol equals 93.1 kilojoules."},{"Start":"00:57.065 ","End":"00:58.550","Text":"That is our final answer."},{"Start":"00:58.550 ","End":"01:01.050","Text":"Thank you very much for watching."}],"ID":30040},{"Watched":false,"Name":"Exercise 2","Duration":"4m 55s","ChapterTopicVideoID":28544,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:03.615","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.615 ","End":"00:07.530","Text":"The enthalpy of vaporization of liquid acetonitrile equals"},{"Start":"00:07.530 ","End":"00:12.255","Text":"29.75 kilojoules per mole at 81.6 degrees Celsius."},{"Start":"00:12.255 ","End":"00:18.764","Text":"What is the volume of acetonitrile gas formed at 81.6 degrees Celsius in 1 atmosphere"},{"Start":"00:18.764 ","End":"00:22.790","Text":"when 2.5 kilojoules of heat are absorbed at"},{"Start":"00:22.790 ","End":"00:27.080","Text":"a constant temperature of 81.6 degrees Celsius."},{"Start":"00:27.080 ","End":"00:29.935","Text":"Here we begin with liquid acetonitrile."},{"Start":"00:29.935 ","End":"00:33.395","Text":"Then we have heat which is absorbed,"},{"Start":"00:33.395 ","End":"00:35.390","Text":"and the liquid turns into gas."},{"Start":"00:35.390 ","End":"00:40.460","Text":"Now we\u0027re asked what is the volume of the acetonitrile gas in this case."},{"Start":"00:40.460 ","End":"00:43.190","Text":"To calculate the volume of the acetonitrile gas,"},{"Start":"00:43.190 ","End":"00:49.960","Text":"we\u0027re going to use the ideal gas law, PV equals nRT."},{"Start":"00:50.720 ","End":"00:53.090","Text":"Remember, P is the pressure,"},{"Start":"00:53.090 ","End":"00:54.304","Text":"V is the volume,"},{"Start":"00:54.304 ","End":"00:56.140","Text":"n is the number of moles,"},{"Start":"00:56.140 ","End":"00:57.690","Text":"R is the gas constant,"},{"Start":"00:57.690 ","End":"00:58.890","Text":"it has different values,"},{"Start":"00:58.890 ","End":"01:02.085","Text":"and T is the temperature in Kelvin."},{"Start":"01:02.085 ","End":"01:03.660","Text":"You want to calculate the volume."},{"Start":"01:03.660 ","End":"01:06.530","Text":"We\u0027re just going to divide both sides by the pressure."},{"Start":"01:06.530 ","End":"01:14.300","Text":"The volume equals nRT divided by P. As you can see,"},{"Start":"01:14.300 ","End":"01:17.180","Text":"we have the temperature in degrees Celsius."},{"Start":"01:17.180 ","End":"01:22.070","Text":"We\u0027re going to turn it into Kelvin and we also have the pressure we have 1 atmosphere."},{"Start":"01:22.070 ","End":"01:24.560","Text":"What we\u0027re missing is the number of moles."},{"Start":"01:24.560 ","End":"01:27.140","Text":"To find the number of moles of the gas,"},{"Start":"01:27.140 ","End":"01:28.700","Text":"we\u0027re going to use the enthalpy of vaporization,"},{"Start":"01:28.700 ","End":"01:31.760","Text":"which is given it equals 29.75 kilojoules per"},{"Start":"01:31.760 ","End":"01:35.870","Text":"mole and we\u0027re going to use the amount of heat absorbed,"},{"Start":"01:35.870 ","End":"01:38.790","Text":"which is 2.5 kilojoules."},{"Start":"01:38.920 ","End":"01:47.575","Text":"The heat equals the enthalpy of vaporization times the number of moles."},{"Start":"01:47.575 ","End":"01:49.670","Text":"Since we\u0027re looking before the number of moles,"},{"Start":"01:49.670 ","End":"01:53.350","Text":"we\u0027re going to divide both sides by the enthalpy of vaporization."},{"Start":"01:53.350 ","End":"02:01.200","Text":"The number of moles equals the heat divided by the enthalpy of vaporization."},{"Start":"02:01.200 ","End":"02:08.870","Text":"This equals in our case 2.5 kilojoules divided by 29.75 kilojoules per mole."},{"Start":"02:08.870 ","End":"02:10.730","Text":"This equals the heat absorbed,"},{"Start":"02:10.730 ","End":"02:13.955","Text":"which is 2.5 kilojoules in our case,"},{"Start":"02:13.955 ","End":"02:17.300","Text":"divided by the enthalpy of vaporization,"},{"Start":"02:17.300 ","End":"02:21.260","Text":"which equals 29.75 kilojoules per mole."},{"Start":"02:21.260 ","End":"02:28.865","Text":"This equals 8.4 times 10^minus 2 mole."},{"Start":"02:28.865 ","End":"02:31.010","Text":"Because again, remember if we look at our units,"},{"Start":"02:31.010 ","End":"02:35.854","Text":"we have kilojoules divided by kilojoules per mole."},{"Start":"02:35.854 ","End":"02:42.830","Text":"Remember that when you divide by a fraction,"},{"Start":"02:42.830 ","End":"02:46.715","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"02:46.715 ","End":"02:53.270","Text":"This equals kilojoules times mole"},{"Start":"02:53.270 ","End":"02:58.430","Text":"divided by kilojoules and the kilojoules cancel out and we\u0027re left with mole."},{"Start":"02:58.430 ","End":"03:05.510","Text":"The number of moles of acetonitrile gas equals 8.4 times 10^negative 2 mole."},{"Start":"03:05.510 ","End":"03:09.115","Text":"Now we can take our ideal gas law."},{"Start":"03:09.115 ","End":"03:13.640","Text":"We have V equals nRT divided by P. We\u0027re just going to write that again,"},{"Start":"03:13.640 ","End":"03:20.195","Text":"V equals nRT divided by P. N is the number of moles,"},{"Start":"03:20.195 ","End":"03:26.720","Text":"8.4 times 10 to the negative 2 mole times the gas constant R,"},{"Start":"03:26.720 ","End":"03:33.845","Text":"which is 0.082 liters times atmosphere divided by mole times Kelvin."},{"Start":"03:33.845 ","End":"03:37.715","Text":"Now we need to multiply this by the temperature."},{"Start":"03:37.715 ","End":"03:40.850","Text":"The temperature is 81.6 degrees Celsius,"},{"Start":"03:40.850 ","End":"03:43.250","Text":"but we need the temperature in Kelvin."},{"Start":"03:43.250 ","End":"03:47.510","Text":"We\u0027re just going to do that right here on the side."},{"Start":"03:47.510 ","End":"03:51.950","Text":"We start with 81.6 degrees Celsius,"},{"Start":"03:51.950 ","End":"03:55.410","Text":"so T equals 81.6 degrees Celsius and we add"},{"Start":"03:55.410 ","End":"04:01.250","Text":"273 in order to get our temperature in Kelvin."},{"Start":"04:01.250 ","End":"04:06.520","Text":"This equals 354.6 Kelvin."},{"Start":"04:06.520 ","End":"04:09.765","Text":"Again, the temperature equals 354.6 Kelvin."},{"Start":"04:09.765 ","End":"04:14.765","Text":"We\u0027re going to multiply it here and we can also see that the moles already cancel out."},{"Start":"04:14.765 ","End":"04:20.060","Text":"This is divided by the pressure."},{"Start":"04:20.060 ","End":"04:24.470","Text":"The pressure that was given in the question is 1 atmosphere,"},{"Start":"04:24.470 ","End":"04:26.690","Text":"so it\u0027s divided by 1 atmosphere."},{"Start":"04:26.690 ","End":"04:31.585","Text":"After multiplying this equals 2.44,"},{"Start":"04:31.585 ","End":"04:33.815","Text":"and if we look at our units,"},{"Start":"04:33.815 ","End":"04:38.360","Text":"we\u0027re left with liters times atmosphere divided by atmosphere."},{"Start":"04:38.360 ","End":"04:45.425","Text":"The atmosphere canceled out and we\u0027re left with 2.44 liters."},{"Start":"04:45.425 ","End":"04:51.395","Text":"The volume of the gas that we found equals 2.44 liters."},{"Start":"04:51.395 ","End":"04:53.210","Text":"That is our final answer."},{"Start":"04:53.210 ","End":"04:55.800","Text":"Thank you very much for watching."}],"ID":30041},{"Watched":false,"Name":"Exercise 3","Duration":"4m 22s","ChapterTopicVideoID":28545,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.030 ","End":"00:08.280","Text":"3.5 liters of benzene gas form when a liquid sample of benzene absorb"},{"Start":"00:08.280 ","End":"00:16.230","Text":"0.61 kilojoules of heat at 298 kelvin and 95.1 millimeters mercury."},{"Start":"00:16.230 ","End":"00:22.005","Text":"Calculate the enthalpy of vaporization of benzene at 298 kelvin."},{"Start":"00:22.005 ","End":"00:26.910","Text":"First of all, we\u0027re going to calculate the number of moles using the ideal gas law."},{"Start":"00:26.910 ","End":"00:29.430","Text":"Then when we have the number of moles,"},{"Start":"00:29.430 ","End":"00:32.520","Text":"we\u0027re going to calculate the enthalpy of vaporization using the number of"},{"Start":"00:32.520 ","End":"00:36.450","Text":"moles and the amount of heat absorbed."},{"Start":"00:36.450 ","End":"00:41.070","Text":"First of all, the ideal gas law is PV equals"},{"Start":"00:41.070 ","End":"00:45.230","Text":"nRT and we need to calculate the number of moles."},{"Start":"00:45.230 ","End":"00:47.480","Text":"Remember, P is the pressure, V is the volume,"},{"Start":"00:47.480 ","End":"00:48.530","Text":"n is the number of moles,"},{"Start":"00:48.530 ","End":"00:51.680","Text":"R is the gas constant and T is the temperature in Kelvin."},{"Start":"00:51.680 ","End":"00:53.600","Text":"Now if we look at the question,"},{"Start":"00:53.600 ","End":"00:55.325","Text":"we know that we have the volume,"},{"Start":"00:55.325 ","End":"00:56.974","Text":"we have our temperature,"},{"Start":"00:56.974 ","End":"01:00.140","Text":"and we have the pressure,"},{"Start":"01:00.140 ","End":"01:02.000","Text":"so we can calculate the number of moles."},{"Start":"01:02.000 ","End":"01:03.515","Text":"The number of moles, n,"},{"Start":"01:03.515 ","End":"01:05.510","Text":"we\u0027re going to divide both sides by RT."},{"Start":"01:05.510 ","End":"01:11.400","Text":"N equals PV divided by RT."},{"Start":"01:11.400 ","End":"01:13.910","Text":"This equals P, the pressure,"},{"Start":"01:13.910 ","End":"01:21.095","Text":"we have 95.1 millimeters of mercury times the volume,"},{"Start":"01:21.095 ","End":"01:25.900","Text":"which is 3.5 liters,"},{"Start":"01:25.900 ","End":"01:29.450","Text":"divided by the gas constant R,"},{"Start":"01:29.450 ","End":"01:34.970","Text":"which is 0.082 liters times atmosphere,"},{"Start":"01:34.970 ","End":"01:39.260","Text":"divided by mole times kelvin."},{"Start":"01:39.260 ","End":"01:43.310","Text":"Now you can use whichever the values of the gas constant you like."},{"Start":"01:43.310 ","End":"01:45.395","Text":"I\u0027m using the 0.082,"},{"Start":"01:45.395 ","End":"01:50.430","Text":"and this is times the temperature which we know is 298 kelvin."},{"Start":"01:54.500 ","End":"01:57.740","Text":"Now we can see the kelvins cancel out."},{"Start":"01:57.740 ","End":"02:02.524","Text":"Now since I used the gas constant that I have atmosphere in the units,"},{"Start":"02:02.524 ","End":"02:05.120","Text":"I\u0027m going to multiply my pressure,"},{"Start":"02:05.120 ","End":"02:08.045","Text":"which is in millimeters of mercury,"},{"Start":"02:08.045 ","End":"02:18.260","Text":"times 1 atmosphere, divided by 760 millimeters of mercury because in 1 atmosphere,"},{"Start":"02:18.260 ","End":"02:20.600","Text":"we have 760 millimeters of mercury."},{"Start":"02:20.600 ","End":"02:25.265","Text":"Now the millimeters of mercury unit is going to cancel out."},{"Start":"02:25.265 ","End":"02:29.485","Text":"Now, after multiplying and dividing,"},{"Start":"02:29.485 ","End":"02:33.225","Text":"this comes to 0.018."},{"Start":"02:33.225 ","End":"02:36.350","Text":"Let\u0027s see, let\u0027s look at our units, we\u0027ll see what we have left."},{"Start":"02:36.350 ","End":"02:43.190","Text":"We have in our numerator atmosphere times liters and in the denominator,"},{"Start":"02:43.190 ","End":"02:47.120","Text":"we also have liters times atmospheres divided by mole."},{"Start":"02:47.120 ","End":"02:49.460","Text":"Now remember that when you\u0027re dividing by a fraction,"},{"Start":"02:49.460 ","End":"02:51.290","Text":"it\u0027s the same as multiplying by the reciprocal of"},{"Start":"02:51.290 ","End":"02:53.795","Text":"the fraction so we\u0027re just going to move this to here."},{"Start":"02:53.795 ","End":"02:57.080","Text":"We have atmosphere times liters"},{"Start":"02:57.080 ","End":"03:00.270","Text":"and this is multiplied by the reciprocal of this fraction."},{"Start":"03:00.270 ","End":"03:04.250","Text":"It\u0027s going to be mole divided by liters times atmosphere."},{"Start":"03:04.250 ","End":"03:08.750","Text":"The liters cancel out and the atmosphere also cancel out and we\u0027re left with mole."},{"Start":"03:08.750 ","End":"03:15.090","Text":"N equals 0.018 mole."},{"Start":"03:15.090 ","End":"03:17.330","Text":"We have the number of moles of the benzene and we want to"},{"Start":"03:17.330 ","End":"03:20.885","Text":"calculate the enthalpy of vaporization of the benzene."},{"Start":"03:20.885 ","End":"03:23.620","Text":"We\u0027re going to use the number of moles that we calculated,"},{"Start":"03:23.620 ","End":"03:26.465","Text":"and we\u0027re going to also use the amount of heat absorbed."},{"Start":"03:26.465 ","End":"03:29.225","Text":"Remember q, which is the amount of heat,"},{"Start":"03:29.225 ","End":"03:35.595","Text":"equals the enthalpy of vaporization times the number of moles."},{"Start":"03:35.595 ","End":"03:39.425","Text":"Since we\u0027re interested in the enthalpy of vaporization,"},{"Start":"03:39.425 ","End":"03:42.310","Text":"so that\u0027s Delta H vaporization,"},{"Start":"03:42.310 ","End":"03:45.860","Text":"we\u0027re going to divide both sides by the number of moles, so it\u0027s going to be q,"},{"Start":"03:45.860 ","End":"03:47.690","Text":"which is the heat divide by n,"},{"Start":"03:47.690 ","End":"03:49.700","Text":"which is the number of moles."},{"Start":"03:49.700 ","End":"03:51.470","Text":"This was from the question,"},{"Start":"03:51.470 ","End":"03:56.740","Text":"the heat equals 0.61 kilojoules given in the question."},{"Start":"03:56.740 ","End":"04:00.170","Text":"This is divided by the number of moles that we calculated,"},{"Start":"04:00.170 ","End":"04:04.190","Text":"so that\u0027s 0.018 moles."},{"Start":"04:04.190 ","End":"04:12.870","Text":"This equals 33.89 kilojoules per mole."},{"Start":"04:12.870 ","End":"04:19.610","Text":"The enthalpy of vaporization that we calculated equals 33.89 kilojoules per mole."},{"Start":"04:19.610 ","End":"04:20.930","Text":"That is our final answer."},{"Start":"04:20.930 ","End":"04:23.520","Text":"Thank you very much for watching."}],"ID":30042},{"Watched":false,"Name":"Exercise 4","Duration":"7m 57s","ChapterTopicVideoID":28546,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.555","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.555 ","End":"00:06.480","Text":"Calculate the volume of methane gas measured at"},{"Start":"00:06.480 ","End":"00:10.380","Text":"35.7 degrees Celsius and 740 millimeters of"},{"Start":"00:10.380 ","End":"00:13.529","Text":"mercury that must be burned in order to vaporize"},{"Start":"00:13.529 ","End":"00:17.115","Text":"5.24 liters of water at 100 degrees Celsius."},{"Start":"00:17.115 ","End":"00:22.020","Text":"We have the enthalpy of combustion of methane and the enthalpy of vaporization of water,"},{"Start":"00:22.020 ","End":"00:28.780","Text":"and we also have the density of water at 100 degrees Celsius."},{"Start":"00:29.240 ","End":"00:32.520","Text":"We need to calculate the volume of the methane gas."},{"Start":"00:32.520 ","End":"00:35.850","Text":"Now, what\u0027s happening here is we have the combustion of methane gas."},{"Start":"00:35.850 ","End":"00:39.630","Text":"Since the combustion of methane gas is exothermic,"},{"Start":"00:39.630 ","End":"00:40.920","Text":"we have heat being released,"},{"Start":"00:40.920 ","End":"00:45.345","Text":"and this heat is used to vaporize 5.24 liters of water."},{"Start":"00:45.345 ","End":"00:48.360","Text":"First we\u0027re going to start by calculating the amount of heat we"},{"Start":"00:48.360 ","End":"00:51.375","Text":"need in order to vaporize 5.24 liters of water."},{"Start":"00:51.375 ","End":"00:55.890","Text":"The amount of heat q equals the enthalpy of vaporization"},{"Start":"00:55.890 ","End":"01:00.570","Text":"of water times the number of moles of water."},{"Start":"01:00.570 ","End":"01:02.610","Text":"The enthalpy of vaporization is given,"},{"Start":"01:02.610 ","End":"01:06.690","Text":"it equals 40.7 kilojoules per mole."},{"Start":"01:06.690 ","End":"01:10.215","Text":"We need to calculate the number of moles of the water."},{"Start":"01:10.215 ","End":"01:13.140","Text":"For this purpose, we\u0027re going to use the density of water which is"},{"Start":"01:13.140 ","End":"01:16.395","Text":"given and also the volume of the water."},{"Start":"01:16.395 ","End":"01:17.940","Text":"Remember the density d,"},{"Start":"01:17.940 ","End":"01:19.703","Text":"equals m, the mass,"},{"Start":"01:19.703 ","End":"01:21.870","Text":"divided by v, the volume."},{"Start":"01:21.870 ","End":"01:26.160","Text":"Now, we want to calculate the mass because we know the density and the volume."},{"Start":"01:26.160 ","End":"01:28.215","Text":"We\u0027re going to multiply both sides by the volume."},{"Start":"01:28.215 ","End":"01:32.496","Text":"M equals d times v. D is given,"},{"Start":"01:32.496 ","End":"01:35.565","Text":"0.96 grams per milliliter,"},{"Start":"01:35.565 ","End":"01:36.870","Text":"times v, the volume,"},{"Start":"01:36.870 ","End":"01:39.060","Text":"which is 5.24 liters."},{"Start":"01:39.060 ","End":"01:41.805","Text":"We\u0027re going to convert these liters into milliliters."},{"Start":"01:41.805 ","End":"01:46.020","Text":"We\u0027re going to multiply by remembering 1 liter we have 1,000 milliliters,"},{"Start":"01:46.020 ","End":"01:49.635","Text":"so that\u0027s times 1,000 milliliters divided by 1 liter."},{"Start":"01:49.635 ","End":"01:51.885","Text":"The liters cancel out,"},{"Start":"01:51.885 ","End":"01:53.805","Text":"and so do the milliliters."},{"Start":"01:53.805 ","End":"01:56.865","Text":"We\u0027re going to be left with grams in our units."},{"Start":"01:56.865 ","End":"02:02.295","Text":"This equals 5,030.4 grams."},{"Start":"02:02.295 ","End":"02:04.065","Text":"That\u0027s the mass of our water."},{"Start":"02:04.065 ","End":"02:07.080","Text":"Now, we want to calculate the number of moles."},{"Start":"02:07.080 ","End":"02:11.235","Text":"Remember the number of moles equals the mass divided by the molar mass."},{"Start":"02:11.235 ","End":"02:13.080","Text":"The molar mass of water,"},{"Start":"02:13.080 ","End":"02:14.835","Text":"which is H_2O,"},{"Start":"02:14.835 ","End":"02:18.330","Text":"equals the molar mass of hydrogen times 2,"},{"Start":"02:18.330 ","End":"02:19.680","Text":"since we have 2 hydrogens,"},{"Start":"02:19.680 ","End":"02:22.770","Text":"plus the molar mass of oxygen."},{"Start":"02:22.770 ","End":"02:25.455","Text":"That\u0027s 2 times 1 gram per mole,"},{"Start":"02:25.455 ","End":"02:28.020","Text":"plus 16 grams per mole,"},{"Start":"02:28.020 ","End":"02:30.420","Text":"which is the molar mass of oxygen,"},{"Start":"02:30.420 ","End":"02:34.290","Text":"and this equals 18 grams per mole."},{"Start":"02:34.290 ","End":"02:37.095","Text":"Again, the number of moles of the water"},{"Start":"02:37.095 ","End":"02:40.230","Text":"equals the mass of the water divided by the molar mass."},{"Start":"02:40.230 ","End":"02:48.060","Text":"The mass we calculated equals 5,030.4 grams divided by the molar mass,"},{"Start":"02:48.060 ","End":"02:50.550","Text":"which equals 18 grams per mole."},{"Start":"02:50.550 ","End":"02:57.030","Text":"This equals 279.47 moles."},{"Start":"02:57.030 ","End":"02:58.470","Text":"Now, just taking a look at the units,"},{"Start":"02:58.470 ","End":"03:00.780","Text":"we have grams divided by grams per mole,"},{"Start":"03:00.780 ","End":"03:05.220","Text":"dividing by a fraction is the same as multiplying by the reciprocal of the fraction,"},{"Start":"03:05.220 ","End":"03:06.990","Text":"so we can just look at it here on the side."},{"Start":"03:06.990 ","End":"03:10.650","Text":"It\u0027s going to be grams times mole per grams,"},{"Start":"03:10.650 ","End":"03:13.890","Text":"the grams cancel out and we\u0027re left with mole."},{"Start":"03:13.890 ","End":"03:16.140","Text":"Now that we have the number of moles of water,"},{"Start":"03:16.140 ","End":"03:19.140","Text":"we can go back to our heat."},{"Start":"03:19.140 ","End":"03:25.440","Text":"Again, the heat equals the enthalpy of vaporization times the number of moles of water."},{"Start":"03:25.440 ","End":"03:27.615","Text":"We\u0027re going to copy that down here again."},{"Start":"03:27.615 ","End":"03:29.070","Text":"Q, the amount of heat,"},{"Start":"03:29.070 ","End":"03:35.310","Text":"equals the enthalpy of vaporization times the number of moles."},{"Start":"03:35.310 ","End":"03:38.025","Text":"Everything of water, of course."},{"Start":"03:38.025 ","End":"03:42.765","Text":"We have 40.7 kilojoules per mole,"},{"Start":"03:42.765 ","End":"03:46.110","Text":"which is the enthalpy of vaporization, times the number of moles,"},{"Start":"03:46.110 ","End":"03:50.550","Text":"which is 279.47 moles."},{"Start":"03:50.550 ","End":"03:53.160","Text":"The moles cancel out."},{"Start":"03:53.160 ","End":"04:03.765","Text":"This equals 11,374.29 kilojoules."},{"Start":"04:03.765 ","End":"04:10.790","Text":"Remember, this is the amount of heat that is needed to vaporize 5.24 liters of water,"},{"Start":"04:10.790 ","End":"04:13.470","Text":"so we have the amount of heat now."},{"Start":"04:13.670 ","End":"04:18.350","Text":"Now we have the amount of heat which is used to vaporize the water."},{"Start":"04:18.350 ","End":"04:21.440","Text":"Remember, this amount of heat is the same amount of heat which is"},{"Start":"04:21.440 ","End":"04:24.830","Text":"released by the combustion of methane gas."},{"Start":"04:24.830 ","End":"04:27.995","Text":"Now I want to calculate the volume of the methane gas."},{"Start":"04:27.995 ","End":"04:29.900","Text":"We\u0027re going to use the ideal gas law for this,"},{"Start":"04:29.900 ","End":"04:33.800","Text":"which is PV equals nRT."},{"Start":"04:33.800 ","End":"04:36.365","Text":"Now remember P is the pressure, V is the volume,"},{"Start":"04:36.365 ","End":"04:38.090","Text":"n is the number of moles,"},{"Start":"04:38.090 ","End":"04:41.254","Text":"R is the gas constant and T is the temperature."},{"Start":"04:41.254 ","End":"04:43.190","Text":"Now, we want to calculate the volume,"},{"Start":"04:43.190 ","End":"04:45.834","Text":"meaning we\u0027re going to divide both sides by the pressure."},{"Start":"04:45.834 ","End":"04:51.290","Text":"Volume equals nRT divided by P. Now,"},{"Start":"04:51.290 ","End":"04:57.620","Text":"we know the temperature and we know the pressure because these are given in the question."},{"Start":"04:57.620 ","End":"04:59.390","Text":"We\u0027re missing the number of moles."},{"Start":"04:59.390 ","End":"05:02.800","Text":"The number of moles we\u0027re going to calculate from the heat."},{"Start":"05:02.800 ","End":"05:10.755","Text":"Remember that the heat equals the enthalpy of combustion times the number of moles."},{"Start":"05:10.755 ","End":"05:12.520","Text":"Since we\u0027re looking for the number of moles,"},{"Start":"05:12.520 ","End":"05:15.310","Text":"we\u0027re going to divide both sides by the enthalpy of combustion."},{"Start":"05:15.310 ","End":"05:16.643","Text":"N, the number of moles,"},{"Start":"05:16.643 ","End":"05:22.269","Text":"equals heat divided by the enthalpy of combustion."},{"Start":"05:22.269 ","End":"05:25.914","Text":"Now remember, the enthalpy of combustion of methane is negative."},{"Start":"05:25.914 ","End":"05:29.590","Text":"It\u0027s a negative value since heat is being released."},{"Start":"05:29.590 ","End":"05:32.530","Text":"Therefore, we\u0027re also going to take the negative value of the heat,"},{"Start":"05:32.530 ","End":"05:35.470","Text":"which we calculated from the water since heat is being released,"},{"Start":"05:35.470 ","End":"05:42.895","Text":"so that\u0027s going to be minus 11,374.29 kilojoules"},{"Start":"05:42.895 ","End":"05:51.980","Text":"divided by minus 890 kilojoules per mole."},{"Start":"05:51.990 ","End":"05:58.270","Text":"This equals 12.78 moles."},{"Start":"05:58.270 ","End":"06:02.335","Text":"The number of moles of the methane gas equals 12.78 moles."},{"Start":"06:02.335 ","End":"06:09.190","Text":"Now, we\u0027re going back to the V equals nRT divided by P equation. We\u0027ll write that again."},{"Start":"06:09.190 ","End":"06:15.510","Text":"The volume equals nRT divided by P. Remember,"},{"Start":"06:15.510 ","End":"06:21.280","Text":"we calculated n and"},{"Start":"06:21.280 ","End":"06:26.870","Text":"it equals 12.78 moles."},{"Start":"06:27.060 ","End":"06:31.390","Text":"We\u0027re going to use, for the gas constant, 0.082,"},{"Start":"06:31.390 ","End":"06:34.570","Text":"which is liters times atmosphere,"},{"Start":"06:34.570 ","End":"06:38.780","Text":"divided by mole times Kelvin."},{"Start":"06:39.460 ","End":"06:42.665","Text":"Now, the temperature that was given in the question,"},{"Start":"06:42.665 ","End":"06:48.620","Text":"because we need the temperature now is 35.7 degrees Celsius."},{"Start":"06:48.620 ","End":"06:50.960","Text":"We\u0027re going to add 273."},{"Start":"06:50.960 ","End":"06:54.590","Text":"Therefore, we will have the temperature in Kelvin."},{"Start":"06:54.590 ","End":"06:57.665","Text":"This is all divided by the pressure."},{"Start":"06:57.665 ","End":"07:04.550","Text":"The pressure that was given in the question is 740 millimeters of mercury."},{"Start":"07:04.550 ","End":"07:06.760","Text":"We\u0027re going to convert these into atmosphere."},{"Start":"07:06.760 ","End":"07:09.530","Text":"We have atmosphere in our numerator and in our denominator."},{"Start":"07:09.530 ","End":"07:18.670","Text":"That\u0027s times 1 atmosphere divided by 760 millimeters of mercury."},{"Start":"07:18.670 ","End":"07:22.045","Text":"The millimeters of mercury cancel out."},{"Start":"07:22.045 ","End":"07:25.385","Text":"We can see that the kelvin cancels out too,"},{"Start":"07:25.385 ","End":"07:28.295","Text":"and so does the moles."},{"Start":"07:28.295 ","End":"07:31.880","Text":"We\u0027re left with liters times atmospheres divided by atmosphere."},{"Start":"07:31.880 ","End":"07:33.560","Text":"Atmosphere is also going to cancel out."},{"Start":"07:33.560 ","End":"07:38.610","Text":"Therefore, our answer will be in units of liters."},{"Start":"07:38.610 ","End":"07:47.660","Text":"After multiplying and dividing this equals 333.51 liters."},{"Start":"07:47.660 ","End":"07:53.690","Text":"The volume of the methane gas that we found equals 333.51 liters."},{"Start":"07:53.690 ","End":"07:55.190","Text":"That is our final answer."},{"Start":"07:55.190 ","End":"07:57.720","Text":"Thank you very much for watching."}],"ID":30043},{"Watched":false,"Name":"Exercise 5","Duration":"6m 30s","ChapterTopicVideoID":28547,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:03.390","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.390 ","End":"00:08.910","Text":"How much heat is required to raise the temperature of 157 grams of ethanol from"},{"Start":"00:08.910 ","End":"00:11.310","Text":"15 degrees Celsius to 25 degrees"},{"Start":"00:11.310 ","End":"00:16.020","Text":"Celsius and then vaporize the ethanol at 25 degrees Celsius?"},{"Start":"00:16.020 ","End":"00:23.220","Text":"The enthalpy of vaporization of ethanol at 298 = 42.6 kilojoules per mole and we also"},{"Start":"00:23.220 ","End":"00:31.240","Text":"know the specific heat capacity of ethanol equals 2.46 joule per gram degree Celsius."},{"Start":"00:31.400 ","End":"00:37.310","Text":"We need to calculate the amount of heat required for 2 purposes."},{"Start":"00:37.310 ","End":"00:40.700","Text":"The first purpose is to raise the temperature of the ethanol,"},{"Start":"00:40.700 ","End":"00:42.530","Text":"and the second one is to vaporize the ethanol."},{"Start":"00:42.530 ","End":"00:45.110","Text":"So q the heat equals,"},{"Start":"00:45.110 ","End":"00:46.970","Text":"in order to raise the temperature of the ethanol,"},{"Start":"00:46.970 ","End":"00:52.830","Text":"we\u0027re going to use m times s times Delta t. Now,"},{"Start":"00:52.830 ","End":"00:54.180","Text":"remember m is the mass,"},{"Start":"00:54.180 ","End":"00:55.520","Text":"s is the specific heat,"},{"Start":"00:55.520 ","End":"00:59.750","Text":"which is given and Delta t is the difference in the temperature,"},{"Start":"00:59.750 ","End":"01:01.860","Text":"which we also know."},{"Start":"01:01.940 ","End":"01:05.570","Text":"Now s the specific heat in many other places is"},{"Start":"01:05.570 ","End":"01:09.290","Text":"given the letter c, just for you to know."},{"Start":"01:09.290 ","End":"01:11.900","Text":"That\u0027s the part of the raising the temperature."},{"Start":"01:11.900 ","End":"01:14.165","Text":"Now, we need to vaporize the ethanol."},{"Start":"01:14.165 ","End":"01:16.640","Text":"In order to vaporize, we\u0027re going to add,"},{"Start":"01:16.640 ","End":"01:18.410","Text":"and this is going to be the enthalpy of"},{"Start":"01:18.410 ","End":"01:22.205","Text":"vaporization times the number of moles of the ethanol."},{"Start":"01:22.205 ","End":"01:27.480","Text":"That\u0027s the enthalpy of vaporization of the ethanol times the number of moles."},{"Start":"01:27.480 ","End":"01:32.240","Text":"That\u0027s going to give us the part where we need to vaporize the ethanol."},{"Start":"01:32.240 ","End":"01:36.380","Text":"Again, q equals we have the mass of the ethanol,"},{"Start":"01:36.380 ","End":"01:38.905","Text":"it equals 157 grams."},{"Start":"01:38.905 ","End":"01:44.070","Text":"So that\u0027s 157 grams times the specific heat,"},{"Start":"01:44.070 ","End":"01:50.310","Text":"which equals 2.46 joule divided by grams times degrees Celsius."},{"Start":"01:50.310 ","End":"01:53.390","Text":"We can already see that the grams cancel out and"},{"Start":"01:53.390 ","End":"01:56.550","Text":"we have to multiply this by Delta t. Now,"},{"Start":"01:56.550 ","End":"02:01.880","Text":"Delta t is the final temperature minus the initial temperature."},{"Start":"02:01.880 ","End":"02:05.030","Text":"We know that the final temperature is 25 degrees"},{"Start":"02:05.030 ","End":"02:08.150","Text":"Celsius and the initial temperature is 15 degrees Celsius."},{"Start":"02:08.150 ","End":"02:13.180","Text":"That\u0027s going to be 25 minus 15."},{"Start":"02:13.180 ","End":"02:15.590","Text":"We\u0027re going to be left with degrees Celsius,"},{"Start":"02:15.590 ","End":"02:18.950","Text":"which cancels out with the degrees Celsius."},{"Start":"02:18.950 ","End":"02:21.680","Text":"Remember, we\u0027re just looking at our first part."},{"Start":"02:21.680 ","End":"02:23.270","Text":"We\u0027re going to divide these into 2."},{"Start":"02:23.270 ","End":"02:27.185","Text":"We\u0027ll be looking at the raising the temperature of the ethanol from 15-25."},{"Start":"02:27.185 ","End":"02:33.780","Text":"This equals 3,862.2 joules."},{"Start":"02:33.780 ","End":"02:36.050","Text":"That\u0027s the first part of the question,"},{"Start":"02:36.050 ","End":"02:37.925","Text":"the raising the temperature of the ethanol."},{"Start":"02:37.925 ","End":"02:42.065","Text":"The second part we\u0027re going to look at right now is to vaporize the ethanol."},{"Start":"02:42.065 ","End":"02:44.600","Text":"As we said, so let\u0027s call this q1 and"},{"Start":"02:44.600 ","End":"02:47.450","Text":"q2 and we\u0027re just going to sum them up at the end of the question."},{"Start":"02:47.450 ","End":"02:53.015","Text":"q2 equals the enthalpy of vaporization times the number of moles."},{"Start":"02:53.015 ","End":"02:59.643","Text":"The enthalpy of vaporization is 42.6 kilojoules per mole,"},{"Start":"02:59.643 ","End":"03:02.450","Text":"and we need it times the number of moles."},{"Start":"03:02.450 ","End":"03:04.580","Text":"To calculate the number of moles,"},{"Start":"03:04.580 ","End":"03:07.310","Text":"we\u0027re going to use the mass that we were given in the question."},{"Start":"03:07.310 ","End":"03:10.850","Text":"We have 157 grams of ethanol."},{"Start":"03:10.850 ","End":"03:16.325","Text":"Remember, the number of moles equals the mass divided by the molar mass."},{"Start":"03:16.325 ","End":"03:17.820","Text":"We have the mass,"},{"Start":"03:17.820 ","End":"03:19.580","Text":"and we need to calculate the molar mass."},{"Start":"03:19.580 ","End":"03:24.860","Text":"To calculate the molar mass of ethanol,"},{"Start":"03:24.860 ","End":"03:28.285","Text":"so ethanol is C_2H_6O,"},{"Start":"03:28.285 ","End":"03:34.550","Text":"so it equals the molar mass of carbon times 2 because we have 2 carbons plus"},{"Start":"03:34.550 ","End":"03:38.820","Text":"the molar mass of hydrogen times 6 because we have"},{"Start":"03:38.820 ","End":"03:43.790","Text":"6 hydrogens plus the molar mass of oxygen, we only have 1."},{"Start":"03:43.790 ","End":"03:49.550","Text":"This equals 2 times 12.01 grams per"},{"Start":"03:49.550 ","End":"03:57.847","Text":"mole plus 6 times 1 grams per mole plus 16 grams per mole."},{"Start":"03:57.847 ","End":"04:04.840","Text":"This equals 46.02 grams per mole."},{"Start":"04:04.840 ","End":"04:08.835","Text":"That\u0027s the molar mass of ethanol."},{"Start":"04:08.835 ","End":"04:12.470","Text":"The number of moles of ethanol equals the mass divided by"},{"Start":"04:12.470 ","End":"04:15.740","Text":"the molar mass of ethanol and that equals as we said,"},{"Start":"04:15.740 ","End":"04:18.649","Text":"157 grams, which was given in our question,"},{"Start":"04:18.649 ","End":"04:25.960","Text":"the mass, divided by the molar mass which is 46.02 grams per mole,"},{"Start":"04:26.720 ","End":"04:32.240","Text":"and this equals 3.41 moles."},{"Start":"04:32.240 ","End":"04:33.830","Text":"If we look at the units,"},{"Start":"04:33.830 ","End":"04:35.840","Text":"remember that when you\u0027re dividing by a fraction,"},{"Start":"04:35.840 ","End":"04:38.720","Text":"it equals the same as multiplying by the reciprocal of the fraction,"},{"Start":"04:38.720 ","End":"04:43.530","Text":"so grams divided by grams per mole equals grams times mole per grams,"},{"Start":"04:43.530 ","End":"04:44.660","Text":"the grams cancel out,"},{"Start":"04:44.660 ","End":"04:46.750","Text":"and we\u0027re left with mole."},{"Start":"04:46.750 ","End":"04:48.860","Text":"Now that we know the number of moles of the ethanol,"},{"Start":"04:48.860 ","End":"04:52.310","Text":"we can go back to q2 to the heat."},{"Start":"04:52.310 ","End":"04:55.640","Text":"The heat we said equals the enthalpy of vaporization,"},{"Start":"04:55.640 ","End":"04:59.270","Text":"which is 42.6 kilojoules per mole,"},{"Start":"04:59.270 ","End":"05:01.070","Text":"times the number of moles,"},{"Start":"05:01.070 ","End":"05:02.840","Text":"which is 3.41 moles,"},{"Start":"05:02.840 ","End":"05:05.520","Text":"which we calculated right now."},{"Start":"05:05.710 ","End":"05:08.240","Text":"The moles will cancel out,"},{"Start":"05:08.240 ","End":"05:17.060","Text":"and this equals 145.27 kilojoules."},{"Start":"05:17.060 ","End":"05:18.795","Text":"Now, we\u0027re going to sum up the heat."},{"Start":"05:18.795 ","End":"05:23.750","Text":"The general heat that q = q_1 plus q2,"},{"Start":"05:23.750 ","End":"05:32.539","Text":"q_1 = 3,862.2 joules,"},{"Start":"05:32.539 ","End":"05:37.040","Text":"and q_2"},{"Start":"05:37.040 ","End":"05:43.205","Text":"= 145.27 kilojoules."},{"Start":"05:43.205 ","End":"05:45.890","Text":"We\u0027re going to convert the joules into kilojoules."},{"Start":"05:45.890 ","End":"05:50.180","Text":"We\u0027re going to multiply the q_1 by,"},{"Start":"05:50.180 ","End":"05:56.505","Text":"remember that in 1 kilojoule we have 1,000 joules."},{"Start":"05:56.505 ","End":"06:03.655","Text":"The joules cancel out and this equals 3.86"},{"Start":"06:03.655 ","End":"06:10.591","Text":"kilojoules plus 145.27 kilojoules,"},{"Start":"06:10.591 ","End":"06:19.025","Text":"and together this comes to 149.13 kilojoules."},{"Start":"06:19.025 ","End":"06:22.640","Text":"The heat that was required to raise the temperature of the ethanol and to"},{"Start":"06:22.640 ","End":"06:27.125","Text":"vaporize the ethanol equals 149.13 kilojoules."},{"Start":"06:27.125 ","End":"06:28.550","Text":"That is our final answer."},{"Start":"06:28.550 ","End":"06:31.200","Text":"Thank you very much for watching."}],"ID":30044},{"Watched":false,"Name":"Clausius-Clapeyron equation","Duration":"8m 18s","ChapterTopicVideoID":23616,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:02.460","Text":"In the previous video,"},{"Start":"00:02.460 ","End":"00:06.540","Text":"we discussed vapor pressure and its temperature dependence."},{"Start":"00:06.540 ","End":"00:12.070","Text":"In this video, we\u0027ll derive a useful equation for the temperature dependence."},{"Start":"00:12.920 ","End":"00:18.000","Text":"This equation is called the Clausius-Clapeyron equation."},{"Start":"00:18.000 ","End":"00:20.970","Text":"Now, we\u0027re going to write the equation that we had at"},{"Start":"00:20.970 ","End":"00:24.495","Text":"the end of the last video in 2 different ways."},{"Start":"00:24.495 ","End":"00:27.210","Text":"First, the first situation."},{"Start":"00:27.210 ","End":"00:34.470","Text":"ln P_1 is equal to minus Delta H_vaporization divided by R, the gas constant,"},{"Start":"00:34.470 ","End":"00:36.690","Text":"times 1 over T_1,"},{"Start":"00:36.690 ","End":"00:40.180","Text":"so this is the pressure of the first situation and"},{"Start":"00:40.180 ","End":"00:44.485","Text":"the temperature in the first situation, plus B."},{"Start":"00:44.485 ","End":"00:47.285","Text":"Then we\u0027re going to write for another situation;"},{"Start":"00:47.285 ","End":"00:55.735","Text":"ln P_2 is equal to minus Delta H_vaporization divided by R multiplied by 1 over T_2,"},{"Start":"00:55.735 ","End":"00:58.125","Text":"the temperature in the second situation,"},{"Start":"00:58.125 ","End":"01:02.100","Text":"plus B, the same constant as before."},{"Start":"01:02.100 ","End":"01:05.475","Text":"Now, we want to get rid of this B,"},{"Start":"01:05.475 ","End":"01:08.090","Text":"so we\u0027re going to subtract the equations."},{"Start":"01:08.090 ","End":"01:13.595","Text":"We\u0027re going to subtract the first equation from the second equation."},{"Start":"01:13.595 ","End":"01:16.624","Text":"Let\u0027s take the second equation."},{"Start":"01:16.624 ","End":"01:20.990","Text":"We have ln P_2 on the left-hand side,"},{"Start":"01:20.990 ","End":"01:23.600","Text":"in the first equation we have ln P_1,"},{"Start":"01:23.600 ","End":"01:26.800","Text":"so that\u0027s ln P_2 minus ln P_1,"},{"Start":"01:26.800 ","End":"01:32.220","Text":"which is used to ln of the ratio of P_2 and P_1,"},{"Start":"01:32.220 ","End":"01:34.845","Text":"so it\u0027s P_2 divided by P_1."},{"Start":"01:34.845 ","End":"01:41.120","Text":"Now, Delta H_vaporization over R is in both equations,"},{"Start":"01:41.120 ","End":"01:42.770","Text":"so we can take that out."},{"Start":"01:42.770 ","End":"01:48.720","Text":"We have Delta H_vaporization over R times minus 1 over T_2."},{"Start":"01:48.720 ","End":"01:51.105","Text":"Here\u0027s the minus 1 over T_2."},{"Start":"01:51.105 ","End":"01:53.595","Text":"We\u0027re going to subtract the first equation."},{"Start":"01:53.595 ","End":"01:55.320","Text":"That\u0027s minus minus,"},{"Start":"01:55.320 ","End":"01:58.140","Text":"so this will come as plus."},{"Start":"01:58.140 ","End":"02:04.560","Text":"Then we have Delta H_vaporization divide by R times 1 over T_1."},{"Start":"02:04.560 ","End":"02:05.840","Text":"Here\u0027s our equation."},{"Start":"02:05.840 ","End":"02:09.570","Text":"This is the Clausius-Clapeyron equation."},{"Start":"02:11.170 ","End":"02:14.815","Text":"Now, a few points about this equation."},{"Start":"02:14.815 ","End":"02:17.100","Text":"The first thing to note is that"},{"Start":"02:17.100 ","End":"02:21.770","Text":"Delta H_vaporization is usually given in kilojoules per mole,"},{"Start":"02:21.770 ","End":"02:23.900","Text":"whereas the gas constant, R,"},{"Start":"02:23.900 ","End":"02:27.850","Text":"is 8.3145 joules per mole."},{"Start":"02:27.850 ","End":"02:32.300","Text":"To calculate Delta H_vaporization over R,"},{"Start":"02:32.300 ","End":"02:35.090","Text":"we have to convert both to the same units."},{"Start":"02:35.090 ","End":"02:41.480","Text":"Either convert Delta H_vaporization to joules or convert R to kilojoules."},{"Start":"02:41.480 ","End":"02:44.260","Text":"It\u0027s a very, very common mistake."},{"Start":"02:44.260 ","End":"02:49.105","Text":"Now, Delta H_vaporization decreases with increasing temperature,"},{"Start":"02:49.105 ","End":"02:53.005","Text":"but the equation assumes it\u0027s constant."},{"Start":"02:53.005 ","End":"02:59.470","Text":"This leads to some inaccuracy in the predictions of the equation."},{"Start":"02:59.470 ","End":"03:03.310","Text":"We\u0027re going to consider a numerical example."},{"Start":"03:03.310 ","End":"03:10.510","Text":"If the vapor pressure of water at 20-degrees Celsius is 17.5 millimeters of mercury,"},{"Start":"03:10.510 ","End":"03:14.615","Text":"what is the vapor pressure at 80-degrees Celsius?"},{"Start":"03:14.615 ","End":"03:16.530","Text":"We\u0027re going to do it in 2 ways."},{"Start":"03:16.530 ","End":"03:22.810","Text":"First, we use the value of Delta H_vaporization at 25-degrees Celsius,"},{"Start":"03:22.810 ","End":"03:26.060","Text":"that\u0027s 44.0 kilojoules per mole,"},{"Start":"03:26.060 ","End":"03:32.300","Text":"and then we\u0027re going to take the average of its values at 20 and 80-degrees Celsius,"},{"Start":"03:32.300 ","End":"03:36.570","Text":"and that\u0027s 42.89 kilojoules per mole."},{"Start":"03:36.910 ","End":"03:40.655","Text":"The first thing to do is to write a little table."},{"Start":"03:40.655 ","End":"03:45.800","Text":"The pressure in the first situation is 17.5 millimeters of mercury,"},{"Start":"03:45.800 ","End":"03:50.420","Text":"and the temperature is 293.15 kelvin."},{"Start":"03:50.420 ","End":"03:57.240","Text":"Remember that to go from Celsius to kelvin, we add 273.15."},{"Start":"03:57.850 ","End":"04:00.080","Text":"In the second situation,"},{"Start":"04:00.080 ","End":"04:01.370","Text":"we don\u0027t know the pressure."},{"Start":"04:01.370 ","End":"04:02.690","Text":"That\u0027s what we want to find."},{"Start":"04:02.690 ","End":"04:04.720","Text":"We want to find P_2,"},{"Start":"04:04.720 ","End":"04:09.705","Text":"and T_2 is 353.15 kelvin."},{"Start":"04:09.705 ","End":"04:15.600","Text":"This is 20-degrees Celsius and this is 80-degrees Celsius."},{"Start":"04:16.180 ","End":"04:23.775","Text":"Let\u0027s rewrite our Clausius-Clapeyron equation and let\u0027s put in the numbers."},{"Start":"04:23.775 ","End":"04:26.600","Text":"We have the ln of P_2 millimeters of mercury,"},{"Start":"04:26.600 ","End":"04:28.265","Text":"that\u0027s what we want to find,"},{"Start":"04:28.265 ","End":"04:29.810","Text":"divided by P_1,"},{"Start":"04:29.810 ","End":"04:33.370","Text":"which is 17.5 millimeters of mercury."},{"Start":"04:33.370 ","End":"04:38.550","Text":"The millimeters of mercury will go on the top and the bottom because,"},{"Start":"04:38.550 ","End":"04:44.420","Text":"of course, the logarithm is unitless, it\u0027s dimensionless."},{"Start":"04:44.420 ","End":"04:46.745","Text":"On the right-hand side,"},{"Start":"04:46.745 ","End":"04:51.870","Text":"we\u0027re going to write Delta H_vaporization in joules,"},{"Start":"04:51.870 ","End":"04:56.315","Text":"so it\u0027s 44.0 times 1,000 joules per mole."},{"Start":"04:56.315 ","End":"05:03.395","Text":"That\u0027s divide by 8.3145 joules per mole per kelvin."},{"Start":"05:03.395 ","End":"05:07.790","Text":"Joules per mole goes with joules per mole,"},{"Start":"05:07.790 ","End":"05:11.550","Text":"and we\u0027re left with 1 over K^minus 1."},{"Start":"05:12.370 ","End":"05:17.220","Text":"Now, we have the temperatures 1 over T_1,"},{"Start":"05:17.450 ","End":"05:24.495","Text":"which is 1 over 293.15 minus 1 over 353.15,"},{"Start":"05:24.495 ","End":"05:28.310","Text":"and the units of this is kelvin^minus 1 because"},{"Start":"05:28.310 ","End":"05:31.990","Text":"it\u0027s 1 over temperature and this goes with this."},{"Start":"05:31.990 ","End":"05:38.110","Text":"Both sides, we have dimensionless units."},{"Start":"05:38.380 ","End":"05:46.267","Text":"Now, if we work out the right-hand side, we have 5,291.96."},{"Start":"05:46.267 ","End":"05:48.185","Text":"That\u0027s for this ratio."},{"Start":"05:48.185 ","End":"05:54.910","Text":"The difference between these 2 numbers is 5.795 times 10^minus 3."},{"Start":"05:54.910 ","End":"06:01.430","Text":"You must be careful here because 1 over these numbers is very, very small number."},{"Start":"06:01.430 ","End":"06:04.490","Text":"The difference has to be quite accurate,"},{"Start":"06:04.490 ","End":"06:10.299","Text":"so take quite a lot of significant figures."},{"Start":"06:10.910 ","End":"06:16.785","Text":"Here it\u0027s 5.795 times 10^ minus 3."},{"Start":"06:16.785 ","End":"06:22.375","Text":"When we multiply that, we get 3.0667."},{"Start":"06:22.375 ","End":"06:35.420","Text":"Now, we can write P_2 is 17.5 millimeters of mercury times the exponential of 3.0667."},{"Start":"06:35.420 ","End":"06:38.315","Text":"We\u0027re taking the exponential of both sides,"},{"Start":"06:38.315 ","End":"06:49.420","Text":"so we just get P_2 over 17.5 is equal to E^3.0667."},{"Start":"06:49.420 ","End":"06:51.440","Text":"When we work this out,"},{"Start":"06:51.440 ","End":"06:57.160","Text":"we get 375.44 millimeters of mercury."},{"Start":"06:57.160 ","End":"07:02.090","Text":"The pressure at 80-degrees Celsius is"},{"Start":"07:02.090 ","End":"07:07.970","Text":"375.44 millimeters of mercury."},{"Start":"07:07.970 ","End":"07:16.320","Text":"You see that the vapor pressure has increased with temperature."},{"Start":"07:16.490 ","End":"07:22.890","Text":"It was 17.5 and now it\u0027s 375.44."},{"Start":"07:22.890 ","End":"07:28.030","Text":"Now, if instead of using that particular value of Delta H,"},{"Start":"07:28.030 ","End":"07:32.200","Text":"we use the average and do the arithmetic again,"},{"Start":"07:32.200 ","End":"07:38.720","Text":"we\u0027ll get to 347.6 millimeters of mercury, a lower volume."},{"Start":"07:38.850 ","End":"07:45.320","Text":"Now, the experimental answer is 355.1,"},{"Start":"07:46.100 ","End":"07:54.950","Text":"which is much closer to 347.6 than it is to 375.44."},{"Start":"07:54.950 ","End":"07:59.515","Text":"We get a better answer using the average"},{"Start":"07:59.515 ","End":"08:06.050","Text":"of the value at 20-degrees Celsius and 80-degrees Celsius."},{"Start":"08:06.140 ","End":"08:09.980","Text":"In fact, the error is only 2 percent."},{"Start":"08:09.980 ","End":"08:18.330","Text":"We\u0027ve learned here about the Clausius-Clapeyron equation and solved an example."}],"ID":30045},{"Watched":false,"Name":"Exercise 6","Duration":"4m 41s","ChapterTopicVideoID":28541,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.775","Text":"We are going to solve the following exercise."},{"Start":"00:02.775 ","End":"00:07.335","Text":"The enthalpy of vaporization of ethanol equals 38.6 kilojoules per mole."},{"Start":"00:07.335 ","End":"00:10.470","Text":"Calculate the vapor pressure of ethanol at 60 degrees Celsius."},{"Start":"00:10.470 ","End":"00:14.895","Text":"The normal boiling point of ethanol equals 78.4 degrees Celsius."},{"Start":"00:14.895 ","End":"00:16.545","Text":"In this question, first of all,"},{"Start":"00:16.545 ","End":"00:20.730","Text":"we\u0027re given the normal boiling point of ethanol equals 78.4 degrees Celsius,"},{"Start":"00:20.730 ","End":"00:25.860","Text":"so we know that T equals 78.4 degrees Celsius,"},{"Start":"00:25.860 ","End":"00:28.920","Text":"but since we know that it\u0027s the normal boiling point of ethanol,"},{"Start":"00:28.920 ","End":"00:30.240","Text":"we also note that the pressure,"},{"Start":"00:30.240 ","End":"00:33.105","Text":"the vapor pressure, equals 1 atmosphere."},{"Start":"00:33.105 ","End":"00:35.940","Text":"We\u0027re going to call these T_1 and T_2."},{"Start":"00:35.940 ","End":"00:40.130","Text":"We\u0027re going to call these T_1 and P_1 because that\u0027s going to be our first point."},{"Start":"00:40.130 ","End":"00:44.735","Text":"Now we have another point that we know that the temperature equals 60 degrees Celsius,"},{"Start":"00:44.735 ","End":"00:46.685","Text":"but we have to calculate the vapor pressure."},{"Start":"00:46.685 ","End":"00:50.725","Text":"We know that T_2 equals 60 degrees Celsius,"},{"Start":"00:50.725 ","End":"00:53.420","Text":"and we don\u0027t know P_2, that\u0027s what we\u0027re trying to find."},{"Start":"00:53.420 ","End":"00:55.520","Text":"We\u0027re trying to find the vapor pressure."},{"Start":"00:55.520 ","End":"00:58.010","Text":"I just call them P instead of VP."},{"Start":"00:58.010 ","End":"01:01.460","Text":"Now, we also know that the enthalpy of vaporization of ethanol,"},{"Start":"01:01.460 ","End":"01:09.140","Text":"so that\u0027s Delta H vaporization equals 38.6 kilojoules per mole."},{"Start":"01:09.140 ","End":"01:15.589","Text":"Now, whenever you have 2 points and we have 2 temperatures and 2 pressures,"},{"Start":"01:15.589 ","End":"01:17.540","Text":"and we\u0027re also talking about vapor pressures"},{"Start":"01:17.540 ","End":"01:20.610","Text":"and enthalpy of vaporization or boiling point,"},{"Start":"01:20.610 ","End":"01:23.270","Text":"this is a major hint that we need to use"},{"Start":"01:23.270 ","End":"01:28.475","Text":"the Clausius-Clapeyron equation in order to solve this question."},{"Start":"01:28.475 ","End":"01:34.040","Text":"The Clausius-Clapeyron equation states that the natural logarithm of"},{"Start":"01:34.040 ","End":"01:40.400","Text":"P_2 divided by P_1 equals the enthalpy of vaporization divided by R,"},{"Start":"01:40.400 ","End":"01:41.795","Text":"which is the gas constant,"},{"Start":"01:41.795 ","End":"01:47.005","Text":"times 1 divided by T_1 minus 1 divided by T_2."},{"Start":"01:47.005 ","End":"01:50.210","Text":"That\u0027s again the natural logarithm of P_2,"},{"Start":"01:50.210 ","End":"01:53.630","Text":"which we don\u0027t know, so it\u0027s going to be P_2 divided by P_1."},{"Start":"01:53.630 ","End":"01:58.310","Text":"P_1 is 1 atmosphere equals the enthalpy of vaporization,"},{"Start":"01:58.310 ","End":"02:02.510","Text":"which we said equals 38.6 kilojoules per mole,"},{"Start":"02:02.510 ","End":"02:05.029","Text":"divided by the gas constant,"},{"Start":"02:05.029 ","End":"02:13.490","Text":"which we\u0027re going to take 8.3145 joule per mole Kelvin times 1 divided by T_1."},{"Start":"02:13.490 ","End":"02:15.770","Text":"T_1 equals 78.4 degrees Celsius."},{"Start":"02:15.770 ","End":"02:19.165","Text":"We\u0027re just going to convert this into Kelvin,"},{"Start":"02:19.165 ","End":"02:22.035","Text":"so that\u0027s plus 273."},{"Start":"02:22.035 ","End":"02:26.550","Text":"We\u0027ll also convert T_2 into Kelvin, 273."},{"Start":"02:26.550 ","End":"02:30.610","Text":"T_1 comes out to 351.4"},{"Start":"02:31.250 ","End":"02:37.950","Text":"degrees Kelvin minus 1 divided by T_2."},{"Start":"02:37.950 ","End":"02:41.760","Text":"T_2 comes out to 333 Kelvin."},{"Start":"02:41.760 ","End":"02:45.020","Text":"Now we can see that we use the enthalpy of vaporization,"},{"Start":"02:45.020 ","End":"02:46.475","Text":"we have it in kilojoules,"},{"Start":"02:46.475 ","End":"02:49.100","Text":"and in our denominator with a gas constant, we have joules,"},{"Start":"02:49.100 ","End":"02:52.835","Text":"so we\u0027re just going to convert this kilo-joules into joules to make it more convenient."},{"Start":"02:52.835 ","End":"02:57.735","Text":"We\u0027re going to multiply this by 1,000 joules for every 1 kilojoule."},{"Start":"02:57.735 ","End":"03:02.855","Text":"The kilojoules are going to cancel out and we\u0027ll be left with joules divided by joules."},{"Start":"03:02.855 ","End":"03:05.582","Text":"The natural logarithm of P_2,"},{"Start":"03:05.582 ","End":"03:07.595","Text":"this is after dividing by 1,"},{"Start":"03:07.595 ","End":"03:09.170","Text":"and P_2 when we find it,"},{"Start":"03:09.170 ","End":"03:12.865","Text":"it\u0027s going to be an atmosphere since P_1 was an atmosphere."},{"Start":"03:12.865 ","End":"03:15.410","Text":"This equals. After multiplying and dividing,"},{"Start":"03:15.410 ","End":"03:18.815","Text":"this comes out to minus 0.73."},{"Start":"03:18.815 ","End":"03:21.575","Text":"Let\u0027s just take a look at our units."},{"Start":"03:21.575 ","End":"03:27.175","Text":"We have joule per mole divided by joule per mole Kelvin,"},{"Start":"03:27.175 ","End":"03:30.650","Text":"and this is times 1 divided by Kelvin."},{"Start":"03:30.650 ","End":"03:35.180","Text":"Remember that when we\u0027re dividing by a fraction,"},{"Start":"03:35.180 ","End":"03:37.600","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction,"},{"Start":"03:37.600 ","End":"03:39.410","Text":"so let just going to look at this."},{"Start":"03:39.410 ","End":"03:47.420","Text":"This is going to be joule divided by mole times mole times Kelvin divided by joule."},{"Start":"03:47.420 ","End":"03:49.340","Text":"That\u0027s only this part here on the left."},{"Start":"03:49.340 ","End":"03:50.660","Text":"The joule are going to cancel out,"},{"Start":"03:50.660 ","End":"03:51.710","Text":"the moles are going to cancel out,"},{"Start":"03:51.710 ","End":"03:52.805","Text":"and we\u0027re left with Kelvin,"},{"Start":"03:52.805 ","End":"03:55.520","Text":"and then we multiply it by 1 divided by Kelvin."},{"Start":"03:55.520 ","End":"03:58.475","Text":"The Kelvin also cancels out."},{"Start":"03:58.475 ","End":"04:05.750","Text":"Therefore, we are left with the natural logarithm of P_2 equals minus 0.73."},{"Start":"04:05.750 ","End":"04:12.845","Text":"Now, remember that the natural logarithm is the log with base e of P_2, again,"},{"Start":"04:12.845 ","End":"04:17.960","Text":"equals minus 0.73. e to"},{"Start":"04:17.960 ","End":"04:23.685","Text":"the power of minus 0.73 is going to equal P_2,"},{"Start":"04:23.685 ","End":"04:26.120","Text":"and you can check this in your calculator,"},{"Start":"04:26.120 ","End":"04:30.860","Text":"this equals 0.482 atmosphere."},{"Start":"04:30.860 ","End":"04:38.270","Text":"The vapor pressure for our second point that we found equals 0.482 atmosphere."},{"Start":"04:38.270 ","End":"04:39.620","Text":"That is our final answer."},{"Start":"04:39.620 ","End":"04:42.360","Text":"Thank you very much for watching."}],"ID":30046},{"Watched":false,"Name":"Exercise 7","Duration":"5m 51s","ChapterTopicVideoID":28540,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.715","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.715 ","End":"00:08.400","Text":"The vapor pressure of benzene equals 88.72 kilopascal at"},{"Start":"00:08.400 ","End":"00:17.085","Text":"349 Kelvin and 137.46 kilopascal at 363.5 Kelvin."},{"Start":"00:17.085 ","End":"00:21.285","Text":"Calculate the enthalpy of vaporization of benzene."},{"Start":"00:21.285 ","End":"00:28.380","Text":"We know the vapor pressure of benzene equals 88.72 at 349 K,"},{"Start":"00:28.380 ","End":"00:33.555","Text":"meaning the temperature equals 349 K,"},{"Start":"00:33.555 ","End":"00:40.635","Text":"and the vapor pressure equals 88.72 kilopascal."},{"Start":"00:40.635 ","End":"00:46.240","Text":"We also know that when the temperature equals 363.5 Kelvin,"},{"Start":"00:46.240 ","End":"00:54.045","Text":"the vapor pressure equals 137.46 kilopascal."},{"Start":"00:54.045 ","End":"01:00.045","Text":"We\u0027re going to call these T_1 and T_2 and P_1 and P_2."},{"Start":"01:00.045 ","End":"01:03.965","Text":"We want to calculate the enthalpy of vaporization of benzene."},{"Start":"01:03.965 ","End":"01:07.040","Text":"But to calculate the enthalpy of vaporization of benzene,"},{"Start":"01:07.040 ","End":"01:10.025","Text":"we\u0027re going to use the Clausius-Clapeyron equation,"},{"Start":"01:10.025 ","End":"01:15.710","Text":"which states that the natural logarithm of P_2 divided"},{"Start":"01:15.710 ","End":"01:21.170","Text":"by P_1 equals the enthalpy of vaporization divided by R,"},{"Start":"01:21.170 ","End":"01:23.075","Text":"which is the gas constant,"},{"Start":"01:23.075 ","End":"01:27.680","Text":"times 1 divided by T_1 minus 1 divided by T_2."},{"Start":"01:27.680 ","End":"01:30.000","Text":"Now in the question,"},{"Start":"01:30.000 ","End":"01:32.930","Text":"we were given 2 different temperatures and 2 different pressures,"},{"Start":"01:32.930 ","End":"01:37.055","Text":"meaning we have 2 points and we know the temperature and pressure of each one."},{"Start":"01:37.055 ","End":"01:40.010","Text":"Many times when you see 2 different points which have temperatures and"},{"Start":"01:40.010 ","End":"01:43.669","Text":"pressures and we\u0027re talking about the vapor pressures,"},{"Start":"01:43.669 ","End":"01:46.820","Text":"or we\u0027re talking about the enthalpy of vaporization that should be"},{"Start":"01:46.820 ","End":"01:51.100","Text":"a hint that we need to use the Clausius-Clapeyron equation."},{"Start":"01:51.100 ","End":"01:56.840","Text":"If P_2 and P_1 are vapor pressures and T_1 and T_2 are the temperatures,"},{"Start":"01:56.840 ","End":"02:01.550","Text":"R is the gas constant and we\u0027re going to use R equals"},{"Start":"02:01.550 ","End":"02:09.490","Text":"8.3145 joule per mole Kelvin for this equation."},{"Start":"02:09.490 ","End":"02:13.710","Text":"Delta H vaporization is the enthalpy of vaporization."},{"Start":"02:13.710 ","End":"02:18.425","Text":"The natural logarithm of P_2 divided by P_1."},{"Start":"02:18.425 ","End":"02:20.540","Text":"We\u0027re going to take P_2,"},{"Start":"02:20.540 ","End":"02:27.840","Text":"137.46 kilopascal divided by P_1,"},{"Start":"02:27.840 ","End":"02:32.175","Text":"which equals 88.72 kilopascal."},{"Start":"02:32.175 ","End":"02:36.010","Text":"First of all, the kilopascal already cancel out."},{"Start":"02:36.080 ","End":"02:38.929","Text":"This equals the enthalpy of vaporization,"},{"Start":"02:38.929 ","End":"02:43.310","Text":"which we don\u0027t know because we need to calculate divided by the gas constant,"},{"Start":"02:43.310 ","End":"02:49.340","Text":"which we said we\u0027re going to take 8.3145 joule divided by mole Kelvin."},{"Start":"02:49.340 ","End":"02:53.815","Text":"This is times 1 divided by T_1,"},{"Start":"02:53.815 ","End":"03:01.980","Text":"which is 1 divided by 349 K minus 1 divided by T_2,"},{"Start":"03:01.980 ","End":"03:08.990","Text":"which equals 363.5 K. This comes to the natural logarithm of"},{"Start":"03:08.990 ","End":"03:13.970","Text":"1.55 equals the enthalpy of"},{"Start":"03:13.970 ","End":"03:22.300","Text":"vaporization times 1.56 times 10 to the negative 5."},{"Start":"03:22.300 ","End":"03:25.100","Text":"Now we\u0027re going to take a look at our units."},{"Start":"03:25.100 ","End":"03:29.270","Text":"First of all, our units, we have 1 divided by joule,"},{"Start":"03:29.270 ","End":"03:30.746","Text":"divided by mole K,"},{"Start":"03:30.746 ","End":"03:38.485","Text":"and this is times 1 divided by K. Meaning Kelvin on our right side."},{"Start":"03:38.485 ","End":"03:42.080","Text":"Remember as always, when we\u0027re dividing by a fraction,"},{"Start":"03:42.080 ","End":"03:45.065","Text":"we\u0027re going to multiply by the reciprocal of the fraction."},{"Start":"03:45.065 ","End":"03:53.715","Text":"This is going to equal 1 times mole times Kelvin divided by joule."},{"Start":"03:53.715 ","End":"03:55.455","Text":"I\u0027m just going to write that again."},{"Start":"03:55.455 ","End":"04:00.440","Text":"The natural logarithm of 1.55 equals enthalpy of"},{"Start":"04:00.440 ","End":"04:07.125","Text":"vaporization times 1.56 times 10 to the negative 5."},{"Start":"04:07.125 ","End":"04:11.430","Text":"We\u0027re left with moles times"},{"Start":"04:11.430 ","End":"04:18.075","Text":"Kelvin divided by joule times 1 divided by Kelvin."},{"Start":"04:18.075 ","End":"04:24.080","Text":"The Kelvin is going to cancel out and we\u0027re left with mole divided by joule in our units."},{"Start":"04:24.080 ","End":"04:29.250","Text":"Now we\u0027re going to divide both sides by 1.56 times 10 to the negative 5."},{"Start":"04:29.250 ","End":"04:31.610","Text":"This is going to leave us with the enthalpy of"},{"Start":"04:31.610 ","End":"04:37.970","Text":"vaporization equals the natural logarithm of 1.55"},{"Start":"04:37.970 ","End":"04:47.880","Text":"divided by 1.56 times 10 to the negative 5 mole per joule."},{"Start":"04:47.880 ","End":"04:54.800","Text":"After dividing everything, this equals 28,093."},{"Start":"04:56.960 ","End":"05:02.530","Text":"Remember again, we divide by a fraction is the same as multiplying by the reciprocal."},{"Start":"05:02.530 ","End":"05:06.655","Text":"This is going to be joule divided by mole joule per mole."},{"Start":"05:06.655 ","End":"05:08.355","Text":"We\u0027re going to write that again here."},{"Start":"05:08.355 ","End":"05:16.080","Text":"The enthalpy of vaporization equals 28,093 joule per mole."},{"Start":"05:16.080 ","End":"05:18.420","Text":"We\u0027re just going to convert this into kilojoules."},{"Start":"05:18.420 ","End":"05:24.225","Text":"We\u0027re going to multiply this by 1 kilojoule divided by 1,000 joules,"},{"Start":"05:24.225 ","End":"05:26.265","Text":"cancel out the joules."},{"Start":"05:26.265 ","End":"05:30.345","Text":"After dividing by 1,000 this is going to equal 28."},{"Start":"05:30.345 ","End":"05:38.190","Text":"After dividing by 1,000 this is going to equal 28.1 kilojoules per mole."},{"Start":"05:38.190 ","End":"05:42.620","Text":"The enthalpy of vaporization that we found for benzene in"},{"Start":"05:42.620 ","End":"05:47.495","Text":"this question equals 28.1 kilojoules per mole."},{"Start":"05:47.495 ","End":"05:49.040","Text":"That is our final answer."},{"Start":"05:49.040 ","End":"05:51.540","Text":"Thank you very much for watching."}],"ID":30047},{"Watched":false,"Name":"Exercise 8","Duration":"10m 25s","ChapterTopicVideoID":28542,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.740 ","End":"00:04.320","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:04.320 ","End":"00:08.490","Text":"Chloroform has a vapor pressure of 197 millimeters mercury at"},{"Start":"00:08.490 ","End":"00:15.030","Text":"23 degrees Celsius and 448 millimeters mercury at 45 degrees Celsius."},{"Start":"00:15.030 ","End":"00:19.770","Text":"Calculate the vapor pressure of chloroform at 30 degrees Celsius."},{"Start":"00:19.770 ","End":"00:23.655","Text":"If we take a look at the question and we can see that we have 2 points,"},{"Start":"00:23.655 ","End":"00:27.210","Text":"and we know the vapor pressures at these points."},{"Start":"00:27.210 ","End":"00:30.120","Text":"We also know the temperatures at these points."},{"Start":"00:30.120 ","End":"00:32.925","Text":"We\u0027re going to call it point 1 and we\u0027re going to say,"},{"Start":"00:32.925 ","End":"00:38.080","Text":"T1 equals, so we have 23 degrees Celsius here."},{"Start":"00:38.080 ","End":"00:42.255","Text":"We\u0027re going to go ahead and convert this into kelvin,"},{"Start":"00:42.255 ","End":"00:44.085","Text":"because we need this later."},{"Start":"00:44.085 ","End":"00:46.630","Text":"We\u0027re going to add 273,"},{"Start":"00:47.210 ","End":"00:52.040","Text":"and this comes to 296 Kelvin."},{"Start":"00:52.040 ","End":"00:53.750","Text":"We know that the vapor pressure,"},{"Start":"00:53.750 ","End":"00:59.940","Text":"we\u0027re just going to have to call this P1 equals 197 millimeters of mercury."},{"Start":"01:00.200 ","End":"01:02.350","Text":"Now to our second point,"},{"Start":"01:02.350 ","End":"01:06.140","Text":"we know that T2 equals 45 degrees Celsius, and again,"},{"Start":"01:06.140 ","End":"01:12.550","Text":"we\u0027re going to add 273 in order to convert the degrees Celsius into Kelvin."},{"Start":"01:12.550 ","End":"01:17.235","Text":"This equals 318 Kelvin,"},{"Start":"01:17.235 ","End":"01:26.160","Text":"and we know that the vapor pressure P2 equals 448 millimeters mercury."},{"Start":"01:26.160 ","End":"01:31.395","Text":"That\u0027s 448 millimeters mercury,"},{"Start":"01:31.395 ","End":"01:36.005","Text":"and we\u0027re asked to calculate a vapor pressure at a third point,"},{"Start":"01:36.005 ","End":"01:39.900","Text":"where we know that the temperature equals 30 degrees Celsius."},{"Start":"01:40.280 ","End":"01:44.585","Text":"First of all, we have 2 points here and we know their vapor pressures,"},{"Start":"01:44.585 ","End":"01:46.825","Text":"we also know their temperatures."},{"Start":"01:46.825 ","End":"01:53.490","Text":"This gives us a hint that we need to use the Clausius-Clapeyron equation."},{"Start":"01:54.040 ","End":"02:01.400","Text":"This equation states that the natural logarithm of P2 divided by P1,"},{"Start":"02:01.400 ","End":"02:03.565","Text":"that\u0027s the vapor pressures,"},{"Start":"02:03.565 ","End":"02:08.600","Text":"equals the enthalpy of vaporization divided by R,"},{"Start":"02:08.600 ","End":"02:10.460","Text":"which is the gas constant,"},{"Start":"02:10.460 ","End":"02:20.950","Text":"and R equals 8.3145 joules divided by mole times Kelvin."},{"Start":"02:21.050 ","End":"02:24.800","Text":"This is multiplied by 1 divided by T1,"},{"Start":"02:24.800 ","End":"02:28.475","Text":"which is the first temperature minus 1 divided by T2,"},{"Start":"02:28.475 ","End":"02:30.895","Text":"which is the second temperature."},{"Start":"02:30.895 ","End":"02:34.220","Text":"Now, in this question, we\u0027re given a third"},{"Start":"02:34.220 ","End":"02:36.860","Text":"and we need to calculate the vapor pressure of this third point."},{"Start":"02:36.860 ","End":"02:40.040","Text":"So what we\u0027re going to do is, first we\u0027re going to find the enthalpy of"},{"Start":"02:40.040 ","End":"02:44.555","Text":"vaporization of chloroform from our 2 first points,"},{"Start":"02:44.555 ","End":"02:49.900","Text":"and then we\u0027re going to use this data to find the vapor pressure at the third point."},{"Start":"02:49.900 ","End":"02:54.830","Text":"The natural log, we have of P2 divided by P1,"},{"Start":"02:54.830 ","End":"03:00.900","Text":"we said that P2 equals 448 millimeters mercury,"},{"Start":"03:00.900 ","End":"03:08.355","Text":"divided by P1, which is 197 millimeters mercury."},{"Start":"03:08.355 ","End":"03:15.320","Text":"Millimeters mercury is canceled out and this equals the enthalpy of vaporization,"},{"Start":"03:15.320 ","End":"03:16.595","Text":"which we don\u0027t know yet,"},{"Start":"03:16.595 ","End":"03:18.605","Text":"this is what we\u0027re actually calculating,"},{"Start":"03:18.605 ","End":"03:26.340","Text":"divided by 8.3145 joule per mole times Kelvin,"},{"Start":"03:26.340 ","End":"03:29.775","Text":"and this is all times 1 divided by T1."},{"Start":"03:29.775 ","End":"03:35.460","Text":"T1 equals 296 k minus 1 divided by T2,"},{"Start":"03:35.460 ","End":"03:41.975","Text":"which is 318 K. The natural log of 2.27,"},{"Start":"03:41.975 ","End":"03:44.060","Text":"which is after dividing,"},{"Start":"03:44.060 ","End":"03:47.720","Text":"equals the enthalpy of vaporization of"},{"Start":"03:47.720 ","End":"03:56.040","Text":"chloroform times 2.81 times 10^ negative 5,"},{"Start":"03:56.040 ","End":"04:00.214","Text":"and this is after subtracting and dividing."},{"Start":"04:00.214 ","End":"04:02.405","Text":"Now let\u0027s just take a look at our units,"},{"Start":"04:02.405 ","End":"04:07.410","Text":"so we have 1 divided by joules per"},{"Start":"04:07.410 ","End":"04:14.055","Text":"mole Kelvin multiplied by 1 divided by Kelvin."},{"Start":"04:14.055 ","End":"04:17.120","Text":"Remember, when we\u0027re dividing by a fraction,"},{"Start":"04:17.120 ","End":"04:20.285","Text":"it\u0027s the same as multiplying by the reciprocal."},{"Start":"04:20.285 ","End":"04:23.460","Text":"We\u0027re just going to take a look at the units here,"},{"Start":"04:23.460 ","End":"04:30.090","Text":"so that\u0027s 1 times moles times Kelvin divided by"},{"Start":"04:30.090 ","End":"04:39.750","Text":"joule times 1 divided by k. The kelvin cancels out and we\u0027re left with mole per joule."},{"Start":"04:39.750 ","End":"04:45.440","Text":"Again, the natural log of 2.27 equals the enthalpy of"},{"Start":"04:45.440 ","End":"04:54.825","Text":"vaporization times 2.81 times 10 to the negative 5 mole per joule."},{"Start":"04:54.825 ","End":"04:58.622","Text":"We\u0027re going to divide both sides by 2.81 times 10 to the negative 5,"},{"Start":"04:58.622 ","End":"05:02.285","Text":"and that\u0027s going to give us the enthalpy of vaporization"},{"Start":"05:02.285 ","End":"05:11.190","Text":"equal to 29,173.66,"},{"Start":"05:11.190 ","End":"05:13.640","Text":"and if we look at our units,"},{"Start":"05:13.640 ","End":"05:15.335","Text":"when we divided both sides,"},{"Start":"05:15.335 ","End":"05:18.220","Text":"we also divided both sides by mole per joule."},{"Start":"05:18.220 ","End":"05:22.280","Text":"Therefore, we get joule per mole because remember,"},{"Start":"05:22.280 ","End":"05:25.460","Text":"as we said, when you divide by a fraction,"},{"Start":"05:25.460 ","End":"05:27.815","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"05:27.815 ","End":"05:31.840","Text":"We get the reciprocal of the fraction when we divide by mole per joule."},{"Start":"05:31.840 ","End":"05:33.650","Text":"We\u0027re just going to write this again."},{"Start":"05:33.650 ","End":"05:42.975","Text":"The enthalpy of vaporization equals 29,173.66 joule per mole."},{"Start":"05:42.975 ","End":"05:47.085","Text":"Let\u0027s just convert this into kilojoules,"},{"Start":"05:47.085 ","End":"05:53.520","Text":"so times 1 kilojoule for 1,000 joules."},{"Start":"05:53.520 ","End":"06:02.835","Text":"The joules cancel out and we get 29.17 kilojoules per mole."},{"Start":"06:02.835 ","End":"06:07.060","Text":"We found the enthalpy of vaporization of chloroform but remember,"},{"Start":"06:07.060 ","End":"06:10.460","Text":"we have a third point that we need to find the vapor pressure for."},{"Start":"06:10.460 ","End":"06:13.760","Text":"We\u0027re going to use the enthalpy of vaporization that we found and"},{"Start":"06:13.760 ","End":"06:18.635","Text":"one other point in order to find the vapor pressure of our third point."},{"Start":"06:18.635 ","End":"06:20.480","Text":"Let\u0027s use point 1."},{"Start":"06:20.480 ","End":"06:25.520","Text":"So remember that T1 equals 296"},{"Start":"06:25.520 ","End":"06:34.995","Text":"K and P1 equals 197 millimeters mercury."},{"Start":"06:34.995 ","End":"06:38.810","Text":"Now our T2, we can call it T2 or T3, however you want,"},{"Start":"06:38.810 ","End":"06:41.780","Text":"but we\u0027re going to T2 equals,"},{"Start":"06:41.780 ","End":"06:44.300","Text":"we know that it\u0027s 30 degrees Celsius."},{"Start":"06:44.300 ","End":"06:49.260","Text":"We\u0027re going to add 273 to get it in Kelvin,"},{"Start":"06:49.260 ","End":"06:55.200","Text":"and this is going to give us 303 Kelvin."},{"Start":"06:55.200 ","End":"07:00.635","Text":"We have our enthalpy of vaporization, which we calculated."},{"Start":"07:00.635 ","End":"07:04.235","Text":"I\u0027m going to take the ones in joule per mole,"},{"Start":"07:04.235 ","End":"07:07.805","Text":"it\u0027s just going to be more convenient."},{"Start":"07:07.805 ","End":"07:10.010","Text":"You\u0027ll see in a minute why."},{"Start":"07:10.010 ","End":"07:13.415","Text":"That\u0027s 0.66 joule per mole."},{"Start":"07:13.415 ","End":"07:17.450","Text":"So we\u0027re going to use again the Clausius-Clapeyron equation."},{"Start":"07:17.450 ","End":"07:19.990","Text":"Again, the natural log,"},{"Start":"07:19.990 ","End":"07:28.160","Text":"of P2 divided by P1 equals the enthalpy of vaporization divided by R,"},{"Start":"07:28.160 ","End":"07:34.550","Text":"which is the gas constant times 1 divided by T1 minus 1 divided by T2."},{"Start":"07:34.550 ","End":"07:38.585","Text":"In our case, we have the natural log of P2, which we don\u0027t know,"},{"Start":"07:38.585 ","End":"07:43.760","Text":"divided by 197 millimeters of mercury,"},{"Start":"07:43.760 ","End":"07:46.280","Text":"equal to the enthalpy of vaporization,"},{"Start":"07:46.280 ","End":"07:53.985","Text":"which is 29,173.66 joule per mole,"},{"Start":"07:53.985 ","End":"07:57.244","Text":"divided by R, the gas constant,"},{"Start":"07:57.244 ","End":"08:03.395","Text":"which is 8.3145 joule per mole Kelvin."},{"Start":"08:03.395 ","End":"08:07.990","Text":"Now, the reason I used the enthalpy of vaporization in joule per mole,"},{"Start":"08:07.990 ","End":"08:13.070","Text":"it\u0027s just more convenient because that way I have that numerator in joule per mole and"},{"Start":"08:13.070 ","End":"08:19.970","Text":"the denominator in joule per mole K. That\u0027s times 1 divided by T1,"},{"Start":"08:19.970 ","End":"08:27.010","Text":"which we know is 296K minus 1 divided by T2, which equals 303K."},{"Start":"08:27.560 ","End":"08:34.110","Text":"The natural log of P2 divided by"},{"Start":"08:34.110 ","End":"08:43.100","Text":"197 millimeters of mercury equals after subtracting and multiplying and dividing,"},{"Start":"08:43.100 ","End":"08:47.145","Text":"we get 0.27,"},{"Start":"08:47.145 ","End":"08:48.980","Text":"and if we look at our units,"},{"Start":"08:48.980 ","End":"08:56.705","Text":"we have joule per mole divided by joules per mole times K,"},{"Start":"08:56.705 ","End":"09:01.450","Text":"Kelvin, and this is all times 1 divided by Kelvin."},{"Start":"09:01.450 ","End":"09:04.610","Text":"Again, dividing by a fraction equals the same"},{"Start":"09:04.610 ","End":"09:07.670","Text":"as multiplying by the reciprocal of the fraction."},{"Start":"09:07.670 ","End":"09:10.190","Text":"We can take a look here."},{"Start":"09:10.190 ","End":"09:13.370","Text":"Joule per mole times the reciprocal,"},{"Start":"09:13.370 ","End":"09:16.940","Text":"which is mole times Kelvin divided by joules."},{"Start":"09:16.940 ","End":"09:19.205","Text":"The joules cancel out, the moles cancel out,"},{"Start":"09:19.205 ","End":"09:21.920","Text":"and all this is times 1 divided by K,"},{"Start":"09:21.920 ","End":"09:24.680","Text":"and the Kelvin also cancels out."},{"Start":"09:24.680 ","End":"09:26.775","Text":"We\u0027re actually left with no units,"},{"Start":"09:26.775 ","End":"09:29.240","Text":"we\u0027re just equals 0.207."},{"Start":"09:29.240 ","End":"09:36.590","Text":"The natural log also equals log base e of"},{"Start":"09:36.590 ","End":"09:47.705","Text":"P2 divided by 197 millimeters of mercury equals 0.27."},{"Start":"09:47.705 ","End":"09:53.450","Text":"So e to the power 0.27 equals"},{"Start":"09:53.450 ","End":"10:01.980","Text":"P2 divided by 197 millimeters of mercury."},{"Start":"10:02.330 ","End":"10:12.870","Text":"After multiplying, we get P2 equals 258 millimeters of mercury."},{"Start":"10:12.870 ","End":"10:21.350","Text":"The vapor pressure that we found for our third point equals 258 millimeters of mercury."},{"Start":"10:21.350 ","End":"10:22.880","Text":"That is our final answer."},{"Start":"10:22.880 ","End":"10:25.530","Text":"Thank you very much for watching."}],"ID":30048},{"Watched":false,"Name":"Boiling point and critical point","Duration":"5m 8s","ChapterTopicVideoID":23672,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.420 ","End":"00:02.650","Text":"In the previous videos,"},{"Start":"00:02.650 ","End":"00:04.675","Text":"we talked about vapor pressure."},{"Start":"00:04.675 ","End":"00:09.530","Text":"In this video we\u0027ll learn about the boiling point and critical point."},{"Start":"00:09.930 ","End":"00:14.540","Text":"Let\u0027s start with the boiling point."},{"Start":"00:14.960 ","End":"00:18.640","Text":"Now, the normal boiling point is the temperature at which"},{"Start":"00:18.640 ","End":"00:22.990","Text":"the vapor pressure is 760 millimeters of mercury."},{"Start":"00:22.990 ","End":"00:28.570","Text":"That\u0027s a normal or standard atmospheric pressure."},{"Start":"00:28.570 ","End":"00:34.120","Text":"Now here\u0027s the vapor pressure versus temperature curve,"},{"Start":"00:34.120 ","End":"00:36.400","Text":"we showed in a previous video."},{"Start":"00:36.400 ","End":"00:40.540","Text":"We see that 760 millimeters of mercury,"},{"Start":"00:40.540 ","End":"00:46.610","Text":"it crosses this curve at a 100 degrees Celsius,"},{"Start":"00:46.610 ","End":"00:52.610","Text":"so a 100 degrees Celsius is the normal boiling point of water."},{"Start":"00:52.610 ","End":"00:57.590","Text":"Now what we didn\u0027t stress in a previous video was that a liquid boils in"},{"Start":"00:57.590 ","End":"01:03.110","Text":"a container that\u0027s open to the atmosphere must be open."},{"Start":"01:03.110 ","End":"01:10.860","Text":"The temperature stays constant until all the liquid is escaped into the gas phase."},{"Start":"01:10.900 ","End":"01:16.445","Text":"Now the conversion of a liquid to a gas, or vice versa,"},{"Start":"01:16.445 ","End":"01:20.380","Text":"a gas to a liquid that\u0027s called condensation,"},{"Start":"01:20.380 ","End":"01:24.770","Text":"is an example of a phase change."},{"Start":"01:25.370 ","End":"01:28.960","Text":"Now, supposing the atmospheric pressure is low,"},{"Start":"01:28.960 ","End":"01:30.850","Text":"we may be on the top of a mountain,"},{"Start":"01:30.850 ","End":"01:36.770","Text":"then the water will boil at a temperature lower than a 100 degrees Celsius."},{"Start":"01:36.950 ","End":"01:40.305","Text":"We can see it from the graph we drew."},{"Start":"01:40.305 ","End":"01:43.575","Text":"Supposing that the pressure is lower say it\u0027s here,"},{"Start":"01:43.575 ","End":"01:45.040","Text":"this is the pressure."},{"Start":"01:45.040 ","End":"01:47.590","Text":"Then the temperature will also be lower."},{"Start":"01:47.590 ","End":"01:51.050","Text":"It will be lower than a 100 degrees Celsius."},{"Start":"01:51.050 ","End":"01:56.095","Text":"If on the other hand the pressure is high supposing we\u0027re cooking in the pressure cooker,"},{"Start":"01:56.095 ","End":"02:00.985","Text":"the water will boil at a temperature higher than a 100 degrees Celsius."},{"Start":"02:00.985 ","End":"02:05.105","Text":"The pressure suppose is higher than 760,"},{"Start":"02:05.105 ","End":"02:12.780","Text":"suppose here, then the temperature of boiling will be higher than a 100 degrees Celsius."},{"Start":"02:13.000 ","End":"02:18.455","Text":"That\u0027s the reason that a pressure cooker is used because in the pressure cooker,"},{"Start":"02:18.455 ","End":"02:24.210","Text":"we can cook at a higher temperature for a smaller period of time."},{"Start":"02:24.640 ","End":"02:28.880","Text":"Now let\u0027s talk about the critical point."},{"Start":"02:30.110 ","End":"02:34.925","Text":"But we said that boiling takes place in an open container."},{"Start":"02:34.925 ","End":"02:38.120","Text":"If we have not an open container but a sealed container,"},{"Start":"02:38.120 ","End":"02:42.230","Text":"the pressure and temperature continue to rise as the liquid is heated."},{"Start":"02:42.230 ","End":"02:46.320","Text":"The temperature and pressure go higher and higher."},{"Start":"02:48.050 ","End":"02:50.610","Text":"Now on the left hand side,"},{"Start":"02:50.610 ","End":"02:59.795","Text":"we have a closed container and we have here liquid and gas or vapor."},{"Start":"02:59.795 ","End":"03:02.015","Text":"Now as we heat,"},{"Start":"03:02.015 ","End":"03:04.520","Text":"the density of the liquid decreases."},{"Start":"03:04.520 ","End":"03:09.140","Text":"The density of the gas above it increases until the densities become equal."},{"Start":"03:09.140 ","End":"03:10.925","Text":"Here they\u0027ve become equal."},{"Start":"03:10.925 ","End":"03:15.390","Text":"We can no longer distinguish between a gas and a liquid."},{"Start":"03:15.400 ","End":"03:19.250","Text":"The surface tension of the liquid decreases until"},{"Start":"03:19.250 ","End":"03:22.205","Text":"there\u0027s no interface between liquid and gas."},{"Start":"03:22.205 ","End":"03:25.530","Text":"Here there\u0027s an interface, this line."},{"Start":"03:25.530 ","End":"03:28.355","Text":"Here after we\u0027ve heated sufficiently,"},{"Start":"03:28.355 ","End":"03:33.685","Text":"we can see no interface between the liquid and the gas."},{"Start":"03:33.685 ","End":"03:41.130","Text":"Now, this point at which there\u0027s no longer a need to face is called the critical point."},{"Start":"03:41.420 ","End":"03:48.090","Text":"The temperature and pressure at which this happens are called T_c critical temperature,"},{"Start":"03:48.090 ","End":"03:51.700","Text":"and P_c critical pressure."},{"Start":"03:51.700 ","End":"03:56.825","Text":"This is the highest point in the pressure versus temperature curve."},{"Start":"03:56.825 ","End":"04:04.140","Text":"The curve stops at this point, T_c, P_c."},{"Start":"04:04.670 ","End":"04:07.640","Text":"Now, if we have a fluid,"},{"Start":"04:07.640 ","End":"04:11.675","Text":"we could call it a fluid because we don\u0027t know whether it\u0027s a liquid or a gas,"},{"Start":"04:11.675 ","End":"04:14.570","Text":"at temperature and pressure above the critical point,"},{"Start":"04:14.570 ","End":"04:18.209","Text":"it\u0027s called a supercritical fluid."},{"Start":"04:20.600 ","End":"04:26.150","Text":"Here are 2 examples, water and oxygen."},{"Start":"04:26.150 ","End":"04:32.180","Text":"Now for water, the critical temperature is 647.3 Kelvin,"},{"Start":"04:32.180 ","End":"04:36.575","Text":"and the critical pressure is 218.3 bar."},{"Start":"04:36.575 ","End":"04:40.620","Text":"You see these very high temperatures and pressures."},{"Start":"04:41.210 ","End":"04:45.825","Text":"Now, oxygen which has a gas at room temperature,"},{"Start":"04:45.825 ","End":"04:48.080","Text":"the critical temperatures is not so high,"},{"Start":"04:48.080 ","End":"04:51.035","Text":"it\u0027s a 154.8 Kelvin,"},{"Start":"04:51.035 ","End":"04:55.220","Text":"but the critical pressure is very high as well."},{"Start":"04:55.220 ","End":"04:57.290","Text":"It\u0027s 50.1 bar."},{"Start":"04:57.290 ","End":"04:59.645","Text":"Not as high as in the case of water,"},{"Start":"04:59.645 ","End":"05:02.255","Text":"but still rather high."},{"Start":"05:02.255 ","End":"05:08.340","Text":"In this video, we talked about the boiling point and critical point."}],"ID":30049},{"Watched":false,"Name":"Exercise 9","Duration":"4m 56s","ChapterTopicVideoID":28538,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.590 ","End":"00:04.140","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:04.140 ","End":"00:10.905","Text":"Liquid ethanol and ethanol vapor are in dynamic equilibrium at 63.5 degrees Celsius."},{"Start":"00:10.905 ","End":"00:17.700","Text":"A 670-milliliter sample of the ethanol vapor weighs 0.598 grams."},{"Start":"00:17.700 ","End":"00:23.070","Text":"Calculate the vapor pressure of ethanol at 63.5 degrees Celsius."},{"Start":"00:23.070 ","End":"00:27.020","Text":"First, we know that the ethanol is in dynamic equilibrium."},{"Start":"00:27.020 ","End":"00:31.820","Text":"We have dynamic equilibrium between the vapor and the liquid of ethanol."},{"Start":"00:31.820 ","End":"00:35.660","Text":"Now, this means that the rate of"},{"Start":"00:35.660 ","End":"00:39.830","Text":"condensation of the ethanol equals the rate of vaporization of the ethanol."},{"Start":"00:39.830 ","End":"00:42.365","Text":"This happens in a closed container."},{"Start":"00:42.365 ","End":"00:45.980","Text":"Now since the rate of condensation equals the rate of vaporization,"},{"Start":"00:45.980 ","End":"00:50.810","Text":"there\u0027s no net change and the amount of vapor stays constant."},{"Start":"00:50.810 ","End":"00:56.060","Text":"Now in the state of dynamic equilibrium with the pressure which is exerted by the vapor,"},{"Start":"00:56.060 ","End":"00:58.535","Text":"is called the vapor pressure."},{"Start":"00:58.535 ","End":"01:01.835","Text":"Now we want to calculate the vapor pressure of the ethanol."},{"Start":"01:01.835 ","End":"01:09.005","Text":"We know the volume of the vapor and we also know how much it weighs."},{"Start":"01:09.005 ","End":"01:14.525","Text":"In addition, we know that the temperature equals 63.5 degrees Celsius."},{"Start":"01:14.525 ","End":"01:17.345","Text":"We\u0027re going to use the ideal gas law in order to"},{"Start":"01:17.345 ","End":"01:21.290","Text":"calculate the vapor pressure of the ethanol."},{"Start":"01:21.290 ","End":"01:25.100","Text":"Remember, PV equals nRT."},{"Start":"01:25.100 ","End":"01:26.885","Text":"P is the pressure, V is the volume,"},{"Start":"01:26.885 ","End":"01:28.295","Text":"n is the number of moles,"},{"Start":"01:28.295 ","End":"01:29.600","Text":"R is the gas constant,"},{"Start":"01:29.600 ","End":"01:30.965","Text":"and T is the temperature."},{"Start":"01:30.965 ","End":"01:33.350","Text":"Now we want to calculate P, the pressure."},{"Start":"01:33.350 ","End":"01:36.593","Text":"We\u0027re going to divide both sides by the volume."},{"Start":"01:36.593 ","End":"01:45.300","Text":"P, the pressure equals nRT divided by V. We know the volume,"},{"Start":"01:45.300 ","End":"01:50.865","Text":"again, is 670 milliliters and we know the weight of the vapor."},{"Start":"01:50.865 ","End":"01:53.060","Text":"Now we\u0027re going to calculate the number of moles."},{"Start":"01:53.060 ","End":"01:55.284","Text":"The number of moles, n,"},{"Start":"01:55.284 ","End":"01:58.415","Text":"equals the mass divided by the molar mass,"},{"Start":"01:58.415 ","End":"02:00.050","Text":"and we\u0027re talking about ethanol."},{"Start":"02:00.050 ","End":"02:04.315","Text":"The mass is given in the question and it equals 0.598 grams,"},{"Start":"02:04.315 ","End":"02:07.040","Text":"and this is divided by the molar mass."},{"Start":"02:07.040 ","End":"02:09.365","Text":"We\u0027re going to calculate the molar mass of the ethanol."},{"Start":"02:09.365 ","End":"02:11.180","Text":"The molar mass of ethanol,"},{"Start":"02:11.180 ","End":"02:13.439","Text":"ethanol is C_2H_6O,"},{"Start":"02:14.540 ","End":"02:19.760","Text":"and this equals the molar mass of carbon times 2,"},{"Start":"02:19.760 ","End":"02:21.758","Text":"since we have 2 carbons,"},{"Start":"02:21.758 ","End":"02:25.530","Text":"plus the molar mass of hydrogen times 6,"},{"Start":"02:25.530 ","End":"02:27.470","Text":"since we have 6 hydrogens in ethanol,"},{"Start":"02:27.470 ","End":"02:30.707","Text":"plus the molar mass of oxygen,"},{"Start":"02:30.707 ","End":"02:32.580","Text":"and we have 1."},{"Start":"02:32.580 ","End":"02:36.410","Text":"This equals 2 times 12 grams per mole,"},{"Start":"02:36.410 ","End":"02:38.435","Text":"the molar mass of carbon,"},{"Start":"02:38.435 ","End":"02:41.645","Text":"plus 6 times 1 gram per mole,"},{"Start":"02:41.645 ","End":"02:44.072","Text":"which is the molar mass of hydrogen,"},{"Start":"02:44.072 ","End":"02:47.165","Text":"plus 16 grams per mole,"},{"Start":"02:47.165 ","End":"02:49.745","Text":"which is the molar mass of oxygen."},{"Start":"02:49.745 ","End":"02:54.140","Text":"This equals 46 grams per mole."},{"Start":"02:54.140 ","End":"02:58.430","Text":"The molar mass of ethanol that we calculated is 46 grams per mole."},{"Start":"02:58.430 ","End":"03:01.910","Text":"Now we can go back to calculate the number of moles of the ethanol,"},{"Start":"03:01.910 ","End":"03:03.965","Text":"which is the mass divided by the molar mass."},{"Start":"03:03.965 ","End":"03:05.375","Text":"We wrote down the mass,"},{"Start":"03:05.375 ","End":"03:09.765","Text":"so that\u0027s divided by 46 grams per mole."},{"Start":"03:09.765 ","End":"03:15.385","Text":"This equals 0.013 moles."},{"Start":"03:15.385 ","End":"03:17.330","Text":"Just take a look at the units for a second."},{"Start":"03:17.330 ","End":"03:19.129","Text":"We have grams divided by grams per mole."},{"Start":"03:19.129 ","End":"03:21.350","Text":"Remember when you divide by a fraction,"},{"Start":"03:21.350 ","End":"03:24.560","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction means"},{"Start":"03:24.560 ","End":"03:28.250","Text":"this equals grams times mole per grams."},{"Start":"03:28.250 ","End":"03:31.790","Text":"Grams cancel out and we\u0027re left with mole."},{"Start":"03:31.790 ","End":"03:33.950","Text":"Now that we have the number of moles,"},{"Start":"03:33.950 ","End":"03:36.185","Text":"we can go ahead and continue."},{"Start":"03:36.185 ","End":"03:38.150","Text":"P equals nRT divided by V,"},{"Start":"03:38.150 ","End":"03:41.330","Text":"which equals 0.013 moles times R,"},{"Start":"03:41.330 ","End":"03:45.965","Text":"which is the gas constant that one of its values is"},{"Start":"03:45.965 ","End":"03:52.695","Text":"0.082 liters times atmosphere divided by mole times Kelvin."},{"Start":"03:52.695 ","End":"03:55.820","Text":"This is multiplied by the temperature."},{"Start":"03:55.820 ","End":"04:03.219","Text":"We\u0027re going to multiply this times 63.5 plus 273,"},{"Start":"04:03.219 ","End":"04:04.940","Text":"because we need the temperature in Kelvin."},{"Start":"04:04.940 ","End":"04:09.560","Text":"We can already see that the kelvin cancel out and so do the moles,"},{"Start":"04:09.560 ","End":"04:13.950","Text":"and this is all divided by the volume."},{"Start":"04:13.950 ","End":"04:17.655","Text":"We know that the volume is 670 milliliters,"},{"Start":"04:17.655 ","End":"04:20.765","Text":"and since we have liters in our numerator,"},{"Start":"04:20.765 ","End":"04:28.340","Text":"we\u0027re going to multiply the milliliters by 1 liter divided by 1,000 milliliters,"},{"Start":"04:28.340 ","End":"04:30.860","Text":"because we know that in 1 liter we have 1,000 milliliters,"},{"Start":"04:30.860 ","End":"04:34.670","Text":"and that way the milliliters cancel out and we\u0027re left with liters."},{"Start":"04:34.670 ","End":"04:40.685","Text":"After dividing, this equals 0.535,"},{"Start":"04:40.685 ","End":"04:41.960","Text":"and if we look at our units,"},{"Start":"04:41.960 ","End":"04:44.360","Text":"we have liters times atmosphere divided by liters."},{"Start":"04:44.360 ","End":"04:47.920","Text":"The liters cancel out and we\u0027re left with atmosphere."},{"Start":"04:47.920 ","End":"04:53.090","Text":"P, the vapor pressure that we calculated equals 0.535 atmospheres."},{"Start":"04:53.090 ","End":"04:54.110","Text":"That is our final answer."},{"Start":"04:54.110 ","End":"04:56.640","Text":"Thank you very much for watching."}],"ID":30050},{"Watched":false,"Name":"Exercise 10","Duration":"6m 43s","ChapterTopicVideoID":28539,"CourseChapterTopicPlaylistID":110287,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.315","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.315 ","End":"00:06.540","Text":"Calculate the density of methanol vapor in equilibrium"},{"Start":"00:06.540 ","End":"00:09.930","Text":"with liquid methanol at 25 degrees Celsius"},{"Start":"00:09.930 ","End":"00:17.325","Text":"if the vapor pressure of methanol at 25 degrees Celsius equals 16.85 kilopascal."},{"Start":"00:17.325 ","End":"00:20.730","Text":"We know that we have dynamic equilibrium"},{"Start":"00:20.730 ","End":"00:23.640","Text":"between the methanol vapor and the methanol liquid."},{"Start":"00:23.640 ","End":"00:27.450","Text":"Now, remember that means that the rate of vaporization is equal to"},{"Start":"00:27.450 ","End":"00:31.650","Text":"the rate of condensation of the methanol and therefore,"},{"Start":"00:31.650 ","End":"00:37.470","Text":"there\u0027s no net change and the amount of vapor is constant."},{"Start":"00:37.470 ","End":"00:40.415","Text":"Now, we\u0027re only looking at the methanol vapor"},{"Start":"00:40.415 ","End":"00:43.775","Text":"and we know that the temperature is 25 degrees Celsius."},{"Start":"00:43.775 ","End":"00:46.280","Text":"We also know that our vapor pressure,"},{"Start":"00:46.280 ","End":"00:49.850","Text":"which is the pressure that is exerted by the vapor in"},{"Start":"00:49.850 ","End":"00:54.840","Text":"dynamic equilibrium equals 16.85 kilopascal."},{"Start":"00:54.840 ","End":"00:57.620","Text":"You want to calculate the density of the methanol vapor,"},{"Start":"00:57.620 ","End":"00:59.260","Text":"so density remember,"},{"Start":"00:59.260 ","End":"01:01.320","Text":"d equals m,"},{"Start":"01:01.320 ","End":"01:04.200","Text":"which is the mass divided by v, the volume."},{"Start":"01:04.200 ","End":"01:06.845","Text":"Here we\u0027re going to use the ideal gas law,"},{"Start":"01:06.845 ","End":"01:10.370","Text":"which is PV equals nRT,"},{"Start":"01:10.370 ","End":"01:13.310","Text":"P is the pressure, V is the volume,"},{"Start":"01:13.310 ","End":"01:14.390","Text":"n is the number of moles,"},{"Start":"01:14.390 ","End":"01:15.590","Text":"R is the gas constant,"},{"Start":"01:15.590 ","End":"01:16.940","Text":"and T is the temperature."},{"Start":"01:16.940 ","End":"01:20.465","Text":"Now we want to calculate the density,"},{"Start":"01:20.465 ","End":"01:23.780","Text":"d. Again, d equals m,"},{"Start":"01:23.780 ","End":"01:25.610","Text":"the mass divided by the volume."},{"Start":"01:25.610 ","End":"01:32.385","Text":"Remember that n, the number of moles also equals the mass divided by the molar mass,"},{"Start":"01:32.385 ","End":"01:34.335","Text":"so this is times RT."},{"Start":"01:34.335 ","End":"01:36.230","Text":"Let\u0027s write this again,"},{"Start":"01:36.230 ","End":"01:44.270","Text":"PV equals mRT mass times R times T divided by the molar mass."},{"Start":"01:44.270 ","End":"01:47.030","Text":"Now we can take both sides and divided them by"},{"Start":"01:47.030 ","End":"01:50.635","Text":"the volume in order to have our pressure alone."},{"Start":"01:50.635 ","End":"01:54.425","Text":"Now we\u0027re going to divide both sides by the volume."},{"Start":"01:54.425 ","End":"01:59.375","Text":"That way, we\u0027ll have our pressure on one side so P equals,"},{"Start":"01:59.375 ","End":"02:01.055","Text":"and then we have m,"},{"Start":"02:01.055 ","End":"02:03.575","Text":"which is the mass times R times T,"},{"Start":"02:03.575 ","End":"02:08.015","Text":"divided by the volume times the molar mass."},{"Start":"02:08.015 ","End":"02:12.335","Text":"Because remember, we divided both sides by the volume."},{"Start":"02:12.335 ","End":"02:15.980","Text":"If we look here, we have m divided by v,"},{"Start":"02:15.980 ","End":"02:21.230","Text":"which is the density, times RT divided by the molar mass."},{"Start":"02:21.230 ","End":"02:23.825","Text":"Again, n divided by V equals the density."},{"Start":"02:23.825 ","End":"02:32.720","Text":"This equals the density times RT divided by the molar mass,"},{"Start":"02:32.720 ","End":"02:35.015","Text":"which is what is left."},{"Start":"02:35.015 ","End":"02:39.245","Text":"Now, if we take the density to one side, d equals,"},{"Start":"02:39.245 ","End":"02:47.390","Text":"we\u0027re going to multiply both sides by the molar mass so that\u0027s p times the molar mass of"},{"Start":"02:47.390 ","End":"02:52.035","Text":"methanol going to divide both sides by R times T."},{"Start":"02:52.035 ","End":"02:57.680","Text":"D or density equals the pressure times the molar mass of methanol divided by R,"},{"Start":"02:57.680 ","End":"03:00.895","Text":"which is the gas constant times the temperature."},{"Start":"03:00.895 ","End":"03:05.150","Text":"We\u0027re just going to write this again, D equals P,"},{"Start":"03:05.150 ","End":"03:09.290","Text":"the pressure times the molar mass, divided by R,"},{"Start":"03:09.290 ","End":"03:10.955","Text":"the gas constant, times T,"},{"Start":"03:10.955 ","End":"03:13.955","Text":"temperature, and this equals."},{"Start":"03:13.955 ","End":"03:22.485","Text":"Now the pressure that was given in the question is 16.85 kilopascal."},{"Start":"03:22.485 ","End":"03:25.010","Text":"We\u0027re going to leave a little room for a conversion factor,"},{"Start":"03:25.010 ","End":"03:26.570","Text":"which we\u0027re going to use in a minute."},{"Start":"03:26.570 ","End":"03:30.055","Text":"This is times the molar mass of methanol."},{"Start":"03:30.055 ","End":"03:32.440","Text":"We\u0027re going to calculate the molar mass of methanol."},{"Start":"03:32.440 ","End":"03:34.010","Text":"The molar mass of methanol,"},{"Start":"03:34.010 ","End":"03:40.850","Text":"which is CH_3 OH equals the molar mass of carbon times 1,"},{"Start":"03:40.850 ","End":"03:46.799","Text":"because we have 1 carbon plus the molar mass of hydrogen times 4 because we have 1,"},{"Start":"03:46.799 ","End":"03:51.725","Text":"2, 3, 4 hydrogens plus the molar mass of oxygen."},{"Start":"03:51.725 ","End":"03:57.320","Text":"That equals 12 grams per mole plus 4 times"},{"Start":"03:57.320 ","End":"04:03.960","Text":"1 so that\u0027s plus 4 grams per mole plus 16 grams per mole for oxygen."},{"Start":"04:04.310 ","End":"04:09.050","Text":"This equals 32 grams per mole."},{"Start":"04:09.050 ","End":"04:12.500","Text":"The molar mass of methanol is around 32 grams per mole."},{"Start":"04:12.500 ","End":"04:16.730","Text":"I\u0027m saying around because the carbon I took 12 that you could have"},{"Start":"04:16.730 ","End":"04:21.170","Text":"taken 12.01 from the periodic table of elements and so on."},{"Start":"04:21.170 ","End":"04:23.180","Text":"We have 32 grams per mole,"},{"Start":"04:23.180 ","End":"04:30.090","Text":"which is the molar mass of the methanol so that\u0027s times 32 grams per mole."},{"Start":"04:30.090 ","End":"04:35.330","Text":"Again, I\u0027m leaving place here for a conversion factor because I wouldn\u0027t use it soon."},{"Start":"04:35.330 ","End":"04:38.850","Text":"This is all divided by R,"},{"Start":"04:38.850 ","End":"04:40.275","Text":"which is the gas constant,"},{"Start":"04:40.275 ","End":"04:44.930","Text":"0.082 liters times atmosphere,"},{"Start":"04:44.930 ","End":"04:52.440","Text":"divided by mole times Kelvin and this is multiplied by the temperature."},{"Start":"04:53.080 ","End":"04:55.610","Text":"Now the temperature in the question,"},{"Start":"04:55.610 ","End":"04:57.380","Text":"it was 25 degrees Celsius,"},{"Start":"04:57.380 ","End":"05:00.770","Text":"so that\u0027s 25 plus 273 Kelvin."},{"Start":"05:00.770 ","End":"05:03.260","Text":"Remember we need the temperature in Kelvin,"},{"Start":"05:03.260 ","End":"05:06.750","Text":"so the Kelvins cancel out we can see that immediately."},{"Start":"05:07.250 ","End":"05:10.940","Text":"Since I have atmosphere in my denominator,"},{"Start":"05:10.940 ","End":"05:12.560","Text":"I want to have atmosphere in my numerator."},{"Start":"05:12.560 ","End":"05:16.445","Text":"I\u0027m going to multiply this by 1 atmosphere"},{"Start":"05:16.445 ","End":"05:25.780","Text":"divided by 101.325 kilopascal."},{"Start":"05:25.780 ","End":"05:30.300","Text":"Because we have 101.325 kilopascals in 1 atmosphere."},{"Start":"05:30.300 ","End":"05:40.250","Text":"The kilopascals cancel out and after multiplying and dividing, I get 0.218."},{"Start":"05:40.940 ","End":"05:43.540","Text":"If we take a look at our units,"},{"Start":"05:43.540 ","End":"05:50.005","Text":"we have atmosphere times grams divided by mole in the numerator,"},{"Start":"05:50.005 ","End":"05:56.590","Text":"divided by liters times atmosphere divided by mole."},{"Start":"05:56.590 ","End":"06:00.115","Text":"Now, remember dividing by a fraction"},{"Start":"06:00.115 ","End":"06:04.165","Text":"equals the same as multiplying by the reciprocal of the fraction,"},{"Start":"06:04.165 ","End":"06:13.640","Text":"so this equals 0.218 times atmosphere times grams divided by mole times the reciprocal,"},{"Start":"06:13.640 ","End":"06:18.555","Text":"meaning mole divided by liters times atmosphere."},{"Start":"06:18.555 ","End":"06:21.440","Text":"The moles cancel out, and the atmosphere also"},{"Start":"06:21.440 ","End":"06:24.725","Text":"cancels out and we\u0027re left with grams divided by liters."},{"Start":"06:24.725 ","End":"06:33.355","Text":"This equals 0.218 grams per liter."},{"Start":"06:33.355 ","End":"06:39.710","Text":"The density of the methanol vapor that we found equals 0.218 grams per liter."},{"Start":"06:39.710 ","End":"06:41.300","Text":"That is our final answer."},{"Start":"06:41.300 ","End":"06:43.860","Text":"Thank you very much for watching."}],"ID":30051}],"Thumbnail":null,"ID":110287},{"Name":"Properties of Solids","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Melting","Duration":"3m 44s","ChapterTopicVideoID":23621,"CourseChapterTopicPlaylistID":110288,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23621.jpeg","UploadDate":"2020-12-24T09:43:35.5070000","DurationForVideoObject":"PT3M44S","Description":null,"MetaTitle":"Melting: Video + Workbook | Proprep","MetaDescription":"Intermolecular Forces Liquids and Solids - Properties of Solids. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/general-chemistry/intermolecular-forces-liquids-and-solids/properties-of-solids/vid30032","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.905","Text":"In the previous video,"},{"Start":"00:01.905 ","End":"00:04.514","Text":"we talked about the boiling at critical points."},{"Start":"00:04.514 ","End":"00:07.965","Text":"In this video, we\u0027ll talk about melting."},{"Start":"00:07.965 ","End":"00:11.220","Text":"When a crystalline solid is heated,"},{"Start":"00:11.220 ","End":"00:16.095","Text":"the crystal vibrates and eventually melts, forming a liquid."},{"Start":"00:16.095 ","End":"00:21.735","Text":"The temperature at which melting occurs is called the melting point."},{"Start":"00:21.735 ","End":"00:27.675","Text":"Now the temperature stays constant until all the solid melts."},{"Start":"00:27.675 ","End":"00:32.670","Text":"The heat required to melt the solid is the enthalpy of fusion,"},{"Start":"00:32.670 ","End":"00:36.705","Text":"fusion is another word for melting."},{"Start":"00:36.705 ","End":"00:40.480","Text":"Fusion is the same as melting."},{"Start":"00:44.720 ","End":"00:51.289","Text":"Note the Delta H of fusion is positive since it\u0027s an endothermic process,"},{"Start":"00:51.289 ","End":"00:55.350","Text":"it takes energy to melt solid."},{"Start":"00:55.570 ","End":"00:59.690","Text":"The conversion of a solid to liquid or vice versa,"},{"Start":"00:59.690 ","End":"01:01.535","Text":"which is called freezing of course,"},{"Start":"01:01.535 ","End":"01:03.140","Text":"is a phase change."},{"Start":"01:03.140 ","End":"01:07.770","Text":"We\u0027ve met several examples of phase changes before."},{"Start":"01:09.850 ","End":"01:14.405","Text":"Let\u0027s take two examples, water and naphthalene."},{"Start":"01:14.405 ","End":"01:16.060","Text":"Naphthalene is C_10H_8,"},{"Start":"01:16.060 ","End":"01:20.150","Text":"it has 2 rings."},{"Start":"01:27.890 ","End":"01:33.920","Text":"First, water. The melting point of water as everybody knows and the freezing point is"},{"Start":"01:33.920 ","End":"01:41.870","Text":"0.0 degrees Celsius and Delta H of fusion is 6.01 kilojoules per mole,"},{"Start":"01:41.870 ","End":"01:48.650","Text":"so we have to invest 6.01 kilojoules per mole to melt ice."},{"Start":"01:48.650 ","End":"01:50.525","Text":"Now what about naphthalene?"},{"Start":"01:50.525 ","End":"01:52.430","Text":"Its melting point is higher,"},{"Start":"01:52.430 ","End":"01:55.055","Text":"it\u0027s 80.2 degrees Celsius,"},{"Start":"01:55.055 ","End":"01:59.540","Text":"so that means it\u0027s a solid at room temperature and Delta H"},{"Start":"01:59.540 ","End":"02:04.920","Text":"of fusion is 18.98 kilojoules per mole."},{"Start":"02:04.930 ","End":"02:08.420","Text":"Now we can draw heating and cooling curves,"},{"Start":"02:08.420 ","End":"02:11.520","Text":"that\u0027s temperature versus time."},{"Start":"02:12.170 ","End":"02:16.110","Text":"Here are 2 cooling curves for heating and cooling."},{"Start":"02:16.110 ","End":"02:18.630","Text":"First let\u0027s start with heating."},{"Start":"02:18.630 ","End":"02:22.035","Text":"First, we heat the solid."},{"Start":"02:22.035 ","End":"02:24.525","Text":"At a certain temperature,"},{"Start":"02:24.525 ","End":"02:25.920","Text":"the solid melts,"},{"Start":"02:25.920 ","End":"02:28.004","Text":"we have solid here plus liquid,"},{"Start":"02:28.004 ","End":"02:32.660","Text":"and the temperature stays constant and then begins to"},{"Start":"02:32.660 ","End":"02:37.790","Text":"rise when all the solid has melted and here we have liquid."},{"Start":"02:37.790 ","End":"02:41.675","Text":"Now the cooling curve is the inverse of course."},{"Start":"02:41.675 ","End":"02:44.365","Text":"First of all, we start off with liquid,"},{"Start":"02:44.365 ","End":"02:51.705","Text":"then we get a solid plus a liquid when it begins to crystallize,"},{"Start":"02:51.705 ","End":"02:57.550","Text":"and then as the temperature decreases further, we have solid."},{"Start":"02:57.550 ","End":"03:02.840","Text":"However, sometimes the liquid doesn\u0027t crystallize"},{"Start":"03:02.840 ","End":"03:09.609","Text":"at the freezing point and we get this behavior."},{"Start":"03:09.609 ","End":"03:15.749","Text":"In other words, the temperature is decreasing but the liquid isn\u0027t yet solidifying."},{"Start":"03:15.749 ","End":"03:18.490","Text":"It\u0027s called supercooling."},{"Start":"03:19.900 ","End":"03:24.545","Text":"Now the reason for this is we need a nucleus"},{"Start":"03:24.545 ","End":"03:29.170","Text":"for the crystals to form and sometimes this is absent."},{"Start":"03:29.170 ","End":"03:34.655","Text":"The nucleus could be a tiny crystallite or a speck of dust,"},{"Start":"03:34.655 ","End":"03:40.285","Text":"something like that and when this is absent we get super cooling."},{"Start":"03:40.285 ","End":"03:44.380","Text":"In this video, we talked about melting."}],"ID":30032},{"Watched":false,"Name":"Exercise 1","Duration":"3m 37s","ChapterTopicVideoID":28526,"CourseChapterTopicPlaylistID":110288,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.194","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.194 ","End":"00:08.085","Text":"How much heat must be absorbed in order to melt 700 grams of ice?"},{"Start":"00:08.085 ","End":"00:13.785","Text":"The enthalpy of fusion of ice equals 6.01 kilojoules per mole."},{"Start":"00:13.785 ","End":"00:19.715","Text":"We know that the enthalpy of fusion of ice equals 6.01 kilojoules per mole,"},{"Start":"00:19.715 ","End":"00:23.095","Text":"meaning that, in order to melt 1 mole of ice,"},{"Start":"00:23.095 ","End":"00:26.355","Text":"we need 6.01 kilojoules of heat."},{"Start":"00:26.355 ","End":"00:31.825","Text":"6.01 kilojoules of heat need to be absorbed in order to melt 1 mole of ice."},{"Start":"00:31.825 ","End":"00:35.640","Text":"In the question, we have 700 grams of ice."},{"Start":"00:35.640 ","End":"00:41.923","Text":"The first thing that we need to do is to convert this amount of ice into number of moles."},{"Start":"00:41.923 ","End":"00:43.395","Text":"When we have the number of moles,"},{"Start":"00:43.395 ","End":"00:48.160","Text":"we can multiply it by the molar enthalpy of fusion of ice."},{"Start":"00:48.160 ","End":"00:50.600","Text":"That way, we\u0027re going to find the total amount of"},{"Start":"00:50.600 ","End":"00:54.710","Text":"heat which must be absorbed in order to melt 700 grams of ice."},{"Start":"00:54.710 ","End":"00:57.290","Text":"So n, the number of moles,"},{"Start":"00:57.290 ","End":"00:59.870","Text":"equals the mass, which is m,"},{"Start":"00:59.870 ","End":"01:02.200","Text":"divided by the molar mass."},{"Start":"01:02.200 ","End":"01:03.980","Text":"We know the mass of ice."},{"Start":"01:03.980 ","End":"01:06.370","Text":"We have 700 grams of ice."},{"Start":"01:06.370 ","End":"01:09.050","Text":"Now we need to calculate the molar mass."},{"Start":"01:09.050 ","End":"01:13.010","Text":"The molar mass of ice is the molar mass of water, which is H_2O,"},{"Start":"01:13.010 ","End":"01:21.140","Text":"and this equals 2 times the molar mass of hydrogen plus the molar mass of oxygen."},{"Start":"01:21.140 ","End":"01:25.928","Text":"The molar masses are taken from the periodic table of elements."},{"Start":"01:25.928 ","End":"01:35.680","Text":"So this equals 2 times 1 gram per mole plus 16 grams per mole."},{"Start":"01:35.680 ","End":"01:40.665","Text":"So the molar mass of water comes to 18 grams per mole."},{"Start":"01:40.665 ","End":"01:43.070","Text":"Now that we have the molar mass of water,"},{"Start":"01:43.070 ","End":"01:45.674","Text":"we can calculate the number of moles of ice."},{"Start":"01:45.674 ","End":"01:47.705","Text":"So again, n, the number of moles,"},{"Start":"01:47.705 ","End":"01:51.530","Text":"equals the mass of ice divided by the molar mass."},{"Start":"01:51.530 ","End":"01:58.590","Text":"The mass equals 700 grams that was given to us in the question,"},{"Start":"01:58.590 ","End":"02:07.220","Text":"and this is divided by the molar mass which is 18 grams per mole, which we calculated."},{"Start":"02:07.220 ","End":"02:13.555","Text":"So this comes out to 38.89 moles."},{"Start":"02:13.555 ","End":"02:15.530","Text":"Now just to take a look at the unit,"},{"Start":"02:15.530 ","End":"02:18.405","Text":"we have grams divided by grams per mole."},{"Start":"02:18.405 ","End":"02:19.760","Text":"As always, dividing by"},{"Start":"02:19.760 ","End":"02:23.765","Text":"a fraction is the same as multiplying by the reciprocal of the fraction."},{"Start":"02:23.765 ","End":"02:26.389","Text":"So if we have grams divided by grams per mole,"},{"Start":"02:26.389 ","End":"02:32.090","Text":"it\u0027s the same as grams times mole per gram."},{"Start":"02:32.090 ","End":"02:33.860","Text":"The grams cancel out,"},{"Start":"02:33.860 ","End":"02:36.680","Text":"and we\u0027re left with moles."},{"Start":"02:36.680 ","End":"02:41.750","Text":"So the moles of ice that we calculated equals 38.89 moles."},{"Start":"02:41.750 ","End":"02:44.420","Text":"Now that we know the number of moles of ice,"},{"Start":"02:44.420 ","End":"02:48.565","Text":"we\u0027re going to multiply this by the molar enthalpy of fusion of ice,"},{"Start":"02:48.565 ","End":"02:52.520","Text":"and we\u0027re going to find the amount of heat which must be absorbed."},{"Start":"02:52.520 ","End":"02:55.370","Text":"So the amount of heat, q,"},{"Start":"02:55.370 ","End":"03:03.376","Text":"equals the molar enthalpy of fusion of ice times the number of moles of ice."},{"Start":"03:03.376 ","End":"03:13.030","Text":"This equals 6.01 kilojoules per mole times 38.89 moles."},{"Start":"03:13.030 ","End":"03:17.670","Text":"So our moles cancel out, and we\u0027re going to be left with kilojoules in our units."},{"Start":"03:17.670 ","End":"03:25.551","Text":"After multiplying, this comes to 233.7 kilojoules."},{"Start":"03:25.551 ","End":"03:28.400","Text":"So the amount of heat which must be absorbed in order to melt"},{"Start":"03:28.400 ","End":"03:33.935","Text":"ice equals 233.7 kilojoules."},{"Start":"03:33.935 ","End":"03:35.510","Text":"That is our final answer."},{"Start":"03:35.510 ","End":"03:38.130","Text":"Thank you very much for watching."}],"ID":30033},{"Watched":false,"Name":"Exercise 2","Duration":"5m 41s","ChapterTopicVideoID":28527,"CourseChapterTopicPlaylistID":110288,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.060","Text":"Here we\u0027re going to solve the following exercise?"},{"Start":"00:03.060 ","End":"00:09.585","Text":"A, how much heat is released when a 0.67 kilogram sample of molten aluminum freezes?"},{"Start":"00:09.585 ","End":"00:13.620","Text":"B, how much heat is required to heat a 270 gram sample of"},{"Start":"00:13.620 ","End":"00:18.180","Text":"aluminum from 30 degrees Celsius to 660 degrees Celsius,"},{"Start":"00:18.180 ","End":"00:20.475","Text":"the normal melting point of aluminum,"},{"Start":"00:20.475 ","End":"00:22.710","Text":"and convert it all to liquid?"},{"Start":"00:22.710 ","End":"00:27.725","Text":"We know that the enthalpy of fusion of aluminum equals 10.7 kilojoules per mole,"},{"Start":"00:27.725 ","End":"00:29.910","Text":"and we also know the specific heat of aluminum,"},{"Start":"00:29.910 ","End":"00:33.780","Text":"which equals 0.9 joule per gram degree Celsius."},{"Start":"00:33.780 ","End":"00:35.460","Text":"Let\u0027s begin with A."},{"Start":"00:35.460 ","End":"00:42.060","Text":"Again, how much heat is released when a 0.67 kilogram sample of molten aluminum freezes?"},{"Start":"00:42.060 ","End":"00:46.765","Text":"So in A, we have liquid which is turning into solid."},{"Start":"00:46.765 ","End":"00:51.785","Text":"Let\u0027s write this. We have liquid which turns into solid."},{"Start":"00:51.785 ","End":"00:53.525","Text":"Now, if we look,"},{"Start":"00:53.525 ","End":"00:55.970","Text":"we\u0027re given the enthalpy of fusion of aluminum,"},{"Start":"00:55.970 ","End":"00:58.175","Text":"which is 10.7 kilojoules per mole."},{"Start":"00:58.175 ","End":"01:04.400","Text":"Now the enthalpy of fusion gives us the amount of heat per"},{"Start":"01:04.400 ","End":"01:10.190","Text":"mole that must be absorbed by a solid to turn into a liquid,"},{"Start":"01:10.190 ","End":"01:12.415","Text":"meaning to melt the solid."},{"Start":"01:12.415 ","End":"01:16.640","Text":"So solid to liquid is the Delta H fusion,"},{"Start":"01:16.640 ","End":"01:18.650","Text":"meaning the enthalpy of fusion."},{"Start":"01:18.650 ","End":"01:21.690","Text":"In our case, we have a liquid which turns into solid,"},{"Start":"01:21.690 ","End":"01:24.305","Text":"and this is called solidification."},{"Start":"01:24.305 ","End":"01:29.080","Text":"We need the enthalpy of solidification."},{"Start":"01:29.080 ","End":"01:34.715","Text":"Now, since melting a solid into a liquid is an endothermic process,"},{"Start":"01:34.715 ","End":"01:36.739","Text":"we know that heat is absorbed."},{"Start":"01:36.739 ","End":"01:39.710","Text":"When we have a liquid turning into solid,"},{"Start":"01:39.710 ","End":"01:43.340","Text":"heat is released, and this is an exothermic process."},{"Start":"01:43.340 ","End":"01:50.435","Text":"Actually, the enthalpy of fusion is the same in value as the enthalpy of solidification."},{"Start":"01:50.435 ","End":"01:53.405","Text":"However, they have opposite signs."},{"Start":"01:53.405 ","End":"02:01.935","Text":"So the enthalpy of fusion equals minus the enthalpy of solidification."},{"Start":"02:01.935 ","End":"02:06.854","Text":"Again, the enthalpy of fusion and enthalpy of solidification have the same value."},{"Start":"02:06.854 ","End":"02:09.535","Text":"However, they have opposite signs."},{"Start":"02:09.535 ","End":"02:13.625","Text":"In A, we have a sample which is liquid,"},{"Start":"02:13.625 ","End":"02:15.470","Text":"and it turns into a solid."},{"Start":"02:15.470 ","End":"02:19.250","Text":"It freezes, so we need to use the enthalpy of solidification."},{"Start":"02:19.250 ","End":"02:29.375","Text":"So the enthalpy of solidification of aluminum equals minus 10.7 kilojoules per mole."},{"Start":"02:29.375 ","End":"02:34.325","Text":"That\u0027s the negative of the enthalpy of fusion of aluminum."},{"Start":"02:34.325 ","End":"02:39.542","Text":"First of all, we have the enthalpy of solidification of aluminum now,"},{"Start":"02:39.542 ","End":"02:42.515","Text":"and we need to calculate how much heat is released when a"},{"Start":"02:42.515 ","End":"02:48.174","Text":"0.67 kilograms sample of molten aluminum freezes."},{"Start":"02:48.174 ","End":"02:53.300","Text":"The enthalpy of solidification is per mole of aluminum."},{"Start":"02:53.300 ","End":"02:55.970","Text":"So we have to calculate the number of moles of aluminum"},{"Start":"02:55.970 ","End":"02:58.400","Text":"that we have in our sample and multiply"},{"Start":"02:58.400 ","End":"03:05.450","Text":"this by the enthalpy of solidification in order to achieve the amount of heat released."},{"Start":"03:05.450 ","End":"03:09.050","Text":"So q, the heat, equals the enthalpy of"},{"Start":"03:09.050 ","End":"03:15.440","Text":"solidification of aluminum times the number of moles of aluminum."},{"Start":"03:15.440 ","End":"03:19.730","Text":"Again, the enthalpy of solidification, remember n,"},{"Start":"03:19.730 ","End":"03:24.535","Text":"the number of moles equals the mass divided by the molar mass,"},{"Start":"03:24.535 ","End":"03:27.635","Text":"so that\u0027s times the mass divided by the molar mass."},{"Start":"03:27.635 ","End":"03:36.450","Text":"This equals minus 10.7 kilojoules per mole times the mass."},{"Start":"03:36.450 ","End":"03:42.130","Text":"The mass that was given in the question is 0.67 kilograms."},{"Start":"03:42.130 ","End":"03:46.705","Text":"This is divided by the molar mass of aluminum,"},{"Start":"03:46.705 ","End":"03:52.740","Text":"which equals 29.98 grams per mole."},{"Start":"03:52.740 ","End":"03:57.126","Text":"Since we have kilograms and our numerator and grams in our denominator,"},{"Start":"03:57.126 ","End":"03:59.500","Text":"we\u0027re just going to multiply the numerator by"},{"Start":"03:59.500 ","End":"04:02.868","Text":"our converting factor to convert the kilograms into grams."},{"Start":"04:02.868 ","End":"04:11.408","Text":"Remember that we have 1,000 grams in every 1 kilogram."},{"Start":"04:11.408 ","End":"04:13.640","Text":"So the kilograms cancel out,"},{"Start":"04:13.640 ","End":"04:15.340","Text":"and we\u0027re going to write this again."},{"Start":"04:15.340 ","End":"04:23.695","Text":"We have minus 10.7 kilojoules per mole times 0.67"},{"Start":"04:23.695 ","End":"04:28.619","Text":"times 1,000 grams divided"},{"Start":"04:28.619 ","End":"04:34.305","Text":"by 29.98 grams per mole."},{"Start":"04:34.305 ","End":"04:43.210","Text":"This equals minus 239.13 kilojoules."},{"Start":"04:43.210 ","End":"04:46.240","Text":"Now let\u0027s take a look at our unit."},{"Start":"04:46.240 ","End":"04:53.925","Text":"We have here kilojoules per mole times grams per grams per mole."},{"Start":"04:53.925 ","End":"04:59.860","Text":"So that\u0027s the same as being kilojoules per mole times mole per grams because,"},{"Start":"04:59.860 ","End":"05:02.320","Text":"remember, when we divide by a fraction,"},{"Start":"05:02.320 ","End":"05:04.780","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"05:04.780 ","End":"05:06.880","Text":"So the grams are going to cancel out,"},{"Start":"05:06.880 ","End":"05:08.650","Text":"and the moles cancel out,"},{"Start":"05:08.650 ","End":"05:11.700","Text":"and we\u0027re left with kilojoules."},{"Start":"05:11.700 ","End":"05:16.510","Text":"So q, the heat, equals minus 239.13 kilojoules."},{"Start":"05:16.510 ","End":"05:23.530","Text":"Now, the value of the amount of heat is 239.13 kilojoules."},{"Start":"05:23.530 ","End":"05:29.487","Text":"The negative sign just means that the heat is being released and not absorbed."},{"Start":"05:29.487 ","End":"05:36.935","Text":"So again, the amount of heat for A equals minus 239.13 kilojoules."},{"Start":"05:36.935 ","End":"05:38.840","Text":"That is our final answer."},{"Start":"05:38.840 ","End":"05:41.640","Text":"Thank you very much for watching."}],"ID":30034},{"Watched":false,"Name":"Sublimation","Duration":"4m 45s","ChapterTopicVideoID":23622,"CourseChapterTopicPlaylistID":110288,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:05.400","Text":"In previous videos we talked about vaporization and melting."},{"Start":"00:05.400 ","End":"00:10.410","Text":"In this video we\u0027ll learn about the related concept of sublimation."},{"Start":"00:10.410 ","End":"00:15.779","Text":"Sublimation is a direct conversion of solid to vapor."},{"Start":"00:15.779 ","End":"00:21.220","Text":"The reverse process, vapor to solid is called deposition."},{"Start":"00:21.220 ","End":"00:23.744","Text":"Now, if we have a closed vessel,"},{"Start":"00:23.744 ","End":"00:29.139","Text":"there is a dynamic equilibrium between a solid and the vapor."},{"Start":"00:29.510 ","End":"00:32.760","Text":"When we go from the solid to vapor,"},{"Start":"00:32.760 ","End":"00:35.650","Text":"it is called sublimation."},{"Start":"00:38.000 ","End":"00:45.370","Text":"If we go in the reverse direction, it\u0027s called deposition."},{"Start":"00:48.590 ","End":"00:54.775","Text":"Sublimation, like some of the other processes we\u0027ve talked about in previous videos,"},{"Start":"00:54.775 ","End":"00:58.010","Text":"is a phase change."},{"Start":"00:58.400 ","End":"01:03.565","Text":"We are going from the solid phase to the vapor phase."},{"Start":"01:03.565 ","End":"01:07.580","Text":"Now, if we have a closed vessel,"},{"Start":"01:10.980 ","End":"01:16.645","Text":"then the vapor pressure above the solid is called sublimation pressure."},{"Start":"01:16.645 ","End":"01:20.570","Text":"Just like we had the vapor pressure above a liquid."},{"Start":"01:21.940 ","End":"01:28.140","Text":"Delta H_sub is the enthalpy of sublimation."},{"Start":"01:28.810 ","End":"01:32.150","Text":"Delta H is a state function,"},{"Start":"01:32.150 ","End":"01:33.200","Text":"as we learned before,"},{"Start":"01:33.200 ","End":"01:35.140","Text":"enthalpy is a state function."},{"Start":"01:35.140 ","End":"01:40.430","Text":"It\u0027s not depending on the exact path we choose to go from the beginning to the end."},{"Start":"01:40.430 ","End":"01:43.550","Text":"If we went through solid to liquid to vapor,"},{"Start":"01:43.550 ","End":"01:48.680","Text":"we could see that Delta H sublimation was equal to delta H of fusion,"},{"Start":"01:48.680 ","End":"01:50.255","Text":"that\u0027s solid to liquid,"},{"Start":"01:50.255 ","End":"01:52.375","Text":"plus delta H of vaporization,"},{"Start":"01:52.375 ","End":"01:54.300","Text":"that\u0027s liquid to vapor,"},{"Start":"01:54.300 ","End":"01:57.930","Text":"and so delta H of sublimation is"},{"Start":"01:57.930 ","End":"02:05.405","Text":"greater than Delta H of vaporization,"},{"Start":"02:05.405 ","End":"02:07.175","Text":"they\u0027re not the same."},{"Start":"02:07.175 ","End":"02:12.830","Text":"Of course, since both delta H of fusion and delta H of vaporization are positive,"},{"Start":"02:12.830 ","End":"02:15.890","Text":"Delta H of sublimation is also positive,"},{"Start":"02:15.890 ","End":"02:23.100","Text":"takes energy to heat a solid and to turn it into vapor."},{"Start":"02:23.980 ","End":"02:29.195","Text":"Now, if we plot line of the vapor pressure,"},{"Start":"02:29.195 ","End":"02:32.635","Text":"or sublimation pressure, versus 1/T,"},{"Start":"02:32.635 ","End":"02:36.635","Text":"that was Clausius-Clapeyron that we learned about in the previous video,"},{"Start":"02:36.635 ","End":"02:41.120","Text":"which we get a straight line and this time the slope is minus"},{"Start":"02:41.120 ","End":"02:47.250","Text":"Delta H of sublimation divided by R. Let\u0027s draw that."},{"Start":"02:53.420 ","End":"02:58.825","Text":"Here\u0027s length P plotted against 1/T."},{"Start":"02:58.825 ","End":"03:03.920","Text":"The slope is minus Delta H of sublimation divided by"},{"Start":"03:03.920 ","End":"03:10.310","Text":"R. Here are some examples."},{"Start":"03:10.310 ","End":"03:13.400","Text":"For example, dry ice at solid carbon dioxide,"},{"Start":"03:13.400 ","End":"03:16.100","Text":"often used in labs to cool things."},{"Start":"03:16.100 ","End":"03:19.630","Text":"Another example, solid iodine."},{"Start":"03:19.630 ","End":"03:25.200","Text":"Solid carbon dioxide is sublimated minus 78.5"},{"Start":"03:25.200 ","End":"03:31.380","Text":"degrees Celsius under normal atmospheric pressure of 1 atmosphere."},{"Start":"03:31.780 ","End":"03:33.920","Text":"Here\u0027s a little picture."},{"Start":"03:33.920 ","End":"03:40.475","Text":"Here\u0027s our block of solid carbon dioxide and here are the fumes that you\u0027ll see above it,"},{"Start":"03:40.475 ","End":"03:47.375","Text":"which is the carbon dioxide vaporized and the water vapor in the air that has condensed."},{"Start":"03:47.375 ","End":"03:53.400","Text":"Here we have solid carbon dioxide sublimating."},{"Start":"03:53.840 ","End":"04:03.230","Text":"Solid iodine sublimates below 113.5 degrees Celsius but even at lower temperature,"},{"Start":"04:03.230 ","End":"04:05.030","Text":"say 70 degrees Celsius,"},{"Start":"04:05.030 ","End":"04:08.635","Text":"it has significant sublimation pressure."},{"Start":"04:08.635 ","End":"04:15.620","Text":"Here\u0027s a picture of a flask with solid iodine at the bottom."},{"Start":"04:15.620 ","End":"04:19.069","Text":"This will vaporize is supposed to be the vapor,"},{"Start":"04:19.069 ","End":"04:24.440","Text":"but then will condense on the sides of the container."},{"Start":"04:24.440 ","End":"04:30.980","Text":"Here we\u0027re heating just the bottom and here it\u0027s cool so we heat."},{"Start":"04:30.980 ","End":"04:40.890","Text":"We get sublimation and then cooling on the sides of the vessel."},{"Start":"04:41.050 ","End":"04:45.810","Text":"In this video, we learned about sublimation."}],"ID":30035}],"Thumbnail":null,"ID":110288},{"Name":"Phase Diagrams","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Phase diagram iodine","Duration":"4m 53s","ChapterTopicVideoID":23623,"CourseChapterTopicPlaylistID":110289,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:02.490","Text":"In previous videos,"},{"Start":"00:02.490 ","End":"00:04.398","Text":"we talked about phase changes,"},{"Start":"00:04.398 ","End":"00:08.550","Text":"and in this video, we\u0027ll learn about phase diagrams."},{"Start":"00:08.550 ","End":"00:13.620","Text":"Now phase diagram is a map that shows which phase is"},{"Start":"00:13.620 ","End":"00:18.120","Text":"more stable at all the different temperatures and pressures."},{"Start":"00:18.120 ","End":"00:24.890","Text":"It also shows us the temperatures and pressures in which phase changes occur,"},{"Start":"00:24.890 ","End":"00:27.860","Text":"liquid to solid and so on."},{"Start":"00:27.860 ","End":"00:31.715","Text":"Let\u0027s take an example of iodine."},{"Start":"00:31.715 ","End":"00:35.254","Text":"Here\u0027s the phase diagram for iodine."},{"Start":"00:35.254 ","End":"00:38.870","Text":"We\u0027re plotting pressure versus temperature."},{"Start":"00:38.870 ","End":"00:46.799","Text":"The temperature here is given in degrees Celsius and it\u0027s very much not to scale."},{"Start":"00:51.970 ","End":"00:57.080","Text":"The first thing we need to do is to identify where the solid is,"},{"Start":"00:57.080 ","End":"00:59.255","Text":"where the liquid is, and where the gas is."},{"Start":"00:59.255 ","End":"01:01.625","Text":"We need these 3 regions."},{"Start":"01:01.625 ","End":"01:04.250","Text":"Now we know that at low temperature,"},{"Start":"01:04.250 ","End":"01:05.930","Text":"we\u0027re more likely to have a solid than"},{"Start":"01:05.930 ","End":"01:10.085","Text":"a liquid and as we heat will go from solid to liquid to gas."},{"Start":"01:10.085 ","End":"01:15.440","Text":"Here is solid at low temperatures,"},{"Start":"01:15.440 ","End":"01:19.320","Text":"liquid at higher temperatures,"},{"Start":"01:20.110 ","End":"01:25.205","Text":"and gas at even higher temperatures."},{"Start":"01:25.205 ","End":"01:30.520","Text":"The next thing we need to do is to indicate the vapor pressure,"},{"Start":"01:30.520 ","End":"01:32.300","Text":"that\u0027s liquid to gas,"},{"Start":"01:32.300 ","End":"01:33.800","Text":"and sublimation pressure,"},{"Start":"01:33.800 ","End":"01:36.050","Text":"that\u0027s solid to gas curves."},{"Start":"01:36.050 ","End":"01:42.365","Text":"These are the curves that divide between liquid and gas and between solid and gas."},{"Start":"01:42.365 ","End":"01:47.390","Text":"Let\u0027s start off between solid and gas, that\u0027s this one,"},{"Start":"01:47.390 ","End":"01:51.028","Text":"this is sublimation,"},{"Start":"01:51.028 ","End":"01:54.125","Text":"and then between the liquid and gas,"},{"Start":"01:54.125 ","End":"01:56.940","Text":"that\u0027s called vapor pressure."},{"Start":"02:02.090 ","End":"02:08.090","Text":"That\u0027s a curve. Now there\u0027s"},{"Start":"02:08.090 ","End":"02:13.775","Text":"one more red line in this phase diagram that we haven\u0027t talked about and that\u0027s this one."},{"Start":"02:13.775 ","End":"02:19.970","Text":"This is between solid and liquid and it\u0027s called the fusion curve,"},{"Start":"02:19.970 ","End":"02:22.100","Text":"it divides the solid and the liquid."},{"Start":"02:22.100 ","End":"02:24.510","Text":"That\u0027s fusion."},{"Start":"02:28.580 ","End":"02:33.995","Text":"Now we see for iodine, it\u0027s almost vertical."},{"Start":"02:33.995 ","End":"02:39.155","Text":"Now we\u0027re going to indicate the triple point O."},{"Start":"02:39.155 ","End":"02:45.050","Text":"It\u0027s this point here where we have solid,"},{"Start":"02:45.050 ","End":"02:48.410","Text":"gas, and liquid at the same point."},{"Start":"02:48.410 ","End":"02:53.775","Text":"At this point, which we called the triple point,"},{"Start":"02:53.775 ","End":"02:55.820","Text":"the solid, the liquid,"},{"Start":"02:55.820 ","End":"02:59.070","Text":"and the gas co-exist."},{"Start":"02:59.120 ","End":"03:01.635","Text":"Now for iodine,"},{"Start":"03:01.635 ","End":"03:11.004","Text":"the triple point is at a 113.6 degrees Celsius and 91.6 millimeters of mercury,"},{"Start":"03:11.004 ","End":"03:14.760","Text":"so here\u0027s the point, the triple point."},{"Start":"03:15.440 ","End":"03:23.545","Text":"Now the normal boiling point at 1 atmosphere is 184.4 degrees Celsius."},{"Start":"03:23.545 ","End":"03:27.650","Text":"Here\u0027s a line at 1 atmosphere and it crosses"},{"Start":"03:27.650 ","End":"03:33.575","Text":"the fusion curve almost at the same point as the triple point,"},{"Start":"03:33.575 ","End":"03:43.350","Text":"and it crosses the vapor pressure curve at 184.4 degrees Celsius."},{"Start":"03:45.350 ","End":"03:47.835","Text":"The normal boiling point,"},{"Start":"03:47.835 ","End":"03:50.075","Text":"that\u0027s the boiling point at 1 atmosphere,"},{"Start":"03:50.075 ","End":"03:54.635","Text":"is 184.4 degrees Celsius."},{"Start":"03:54.635 ","End":"03:58.820","Text":"Now another point that we should mention is the critical point,"},{"Start":"03:58.820 ","End":"04:05.195","Text":"that\u0027s C. Now that\u0027s the highest point of the vapor pressure curve,"},{"Start":"04:05.195 ","End":"04:09.270","Text":"vapor pressure curve stops at that point."},{"Start":"04:10.480 ","End":"04:17.089","Text":"For iodine, the critical point is at 546 degrees"},{"Start":"04:17.089 ","End":"04:24.470","Text":"Celsius and the pressure is 115.5 atmospheres."},{"Start":"04:24.470 ","End":"04:29.970","Text":"You can certainly see this diagram is not to scale."},{"Start":"04:30.260 ","End":"04:33.630","Text":"Now beyond C,"},{"Start":"04:33.630 ","End":"04:35.570","Text":"we have supercritical fluid,"},{"Start":"04:35.570 ","End":"04:38.625","Text":"we\u0027ll talk about that in a later video."},{"Start":"04:38.625 ","End":"04:42.200","Text":"Above C, we have supercritical fluid."},{"Start":"04:42.200 ","End":"04:46.500","Text":"Here we can write supercritical fluid."},{"Start":"04:46.900 ","End":"04:52.980","Text":"In this video, we talked about phase diagram of iodine."}],"ID":24534},{"Watched":false,"Name":"Phase diagram carbon dioxide","Duration":"3m 46s","ChapterTopicVideoID":23625,"CourseChapterTopicPlaylistID":110289,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:02.160","Text":"In the previous video,"},{"Start":"00:02.160 ","End":"00:05.100","Text":"we talked about the phase diagram of iodine."},{"Start":"00:05.100 ","End":"00:10.990","Text":"In this video, we\u0027ll learn about the phase diagram of carbon dioxide."},{"Start":"00:11.180 ","End":"00:16.810","Text":"So we\u0027re going to talk about the phase diagram of carbon dioxide."},{"Start":"00:18.770 ","End":"00:22.270","Text":"Here\u0027s the phase diagram."},{"Start":"00:22.280 ","End":"00:26.673","Text":"The first thing we need to do is to identify the solid,"},{"Start":"00:26.673 ","End":"00:29.670","Text":"liquid, and gas regions."},{"Start":"00:29.990 ","End":"00:35.750","Text":"Here is the solid at the lowest temperatures,"},{"Start":"00:35.750 ","End":"00:42.570","Text":"then the liquid lies between 2 red lines,"},{"Start":"00:42.570 ","End":"00:45.430","Text":"and then the gas."},{"Start":"00:45.460 ","End":"00:52.160","Text":"Then we need to indicate the vapor pressure and sublimation pressure curves."},{"Start":"00:52.160 ","End":"00:55.370","Text":"The vapor pressure curve is between the liquid and"},{"Start":"00:55.370 ","End":"01:04.110","Text":"the gas and the sublimation curve is between the solid and the gas."},{"Start":"01:12.290 ","End":"01:15.750","Text":"The third curve is the fusion curve."},{"Start":"01:15.750 ","End":"01:18.725","Text":"So we need to indicate the fusion curve."},{"Start":"01:18.725 ","End":"01:23.329","Text":"Here\u0027s the fusion curve that\u0027s between the solid and liquid,"},{"Start":"01:23.329 ","End":"01:28.410","Text":"that\u0027s called fusion curve."},{"Start":"01:28.600 ","End":"01:34.220","Text":"We see that as the pressure increases,"},{"Start":"01:34.220 ","End":"01:38.850","Text":"melting occurs at a higher temperature."},{"Start":"01:39.800 ","End":"01:44.505","Text":"Now there\u0027s one important difference from iodine."},{"Start":"01:44.505 ","End":"01:49.940","Text":"Unlike iodine, the triple point O lies above 1 atmosphere."},{"Start":"01:49.940 ","End":"01:57.590","Text":"In fact, it\u0027s at 5.1 atmospheres at a temperature of minus 56.7 degrees Celsius."},{"Start":"01:57.590 ","End":"02:00.110","Text":"Here\u0027s the triple point."},{"Start":"02:00.110 ","End":"02:03.380","Text":"You see it\u0027s 5.1 atmospheres,"},{"Start":"02:03.380 ","End":"02:06.540","Text":"well above 1 atmosphere."},{"Start":"02:06.790 ","End":"02:09.635","Text":"Now below the triple point,"},{"Start":"02:09.635 ","End":"02:13.295","Text":"sublimation occurs rather than melting."},{"Start":"02:13.295 ","End":"02:15.380","Text":"Below the triple point,"},{"Start":"02:15.380 ","End":"02:19.830","Text":"we go from solid to gas sublimation."},{"Start":"02:20.920 ","End":"02:26.165","Text":"However, above the triple point, melting can occur."},{"Start":"02:26.165 ","End":"02:36.535","Text":"We go from solid to liquid above the triple point, above 5.1 atmospheres."},{"Start":"02:36.535 ","End":"02:39.590","Text":"Now if we have an open container,"},{"Start":"02:39.590 ","End":"02:42.140","Text":"which is of course at 1 atmosphere,"},{"Start":"02:42.140 ","End":"02:48.450","Text":"sublimation will occur at minus 78.5 degrees Celsius."},{"Start":"02:49.030 ","End":"02:52.655","Text":"If we have a pressure of 1 atmosphere,"},{"Start":"02:52.655 ","End":"02:54.935","Text":"and here, we only have sublimation,"},{"Start":"02:54.935 ","End":"03:01.080","Text":"sublimation will occur at minus 78.5 degrees Celsius."},{"Start":"03:01.420 ","End":"03:05.765","Text":"Now we need to indicate also the critical point."},{"Start":"03:05.765 ","End":"03:11.545","Text":"That\u0027s the highest point in the vapor pressure curve, that\u0027s C here."},{"Start":"03:11.545 ","End":"03:20.420","Text":"In this case, it occurs at 72.9 atmospheres and 31 degrees Celsius,"},{"Start":"03:20.420 ","End":"03:26.010","Text":"so relatively low temperature but very high pressure."},{"Start":"03:26.240 ","End":"03:29.520","Text":"Here\u0027s our critical point."},{"Start":"03:29.520 ","End":"03:32.120","Text":"Of course, above the critical point,"},{"Start":"03:32.120 ","End":"03:40.115","Text":"we have a supercritical fluid, SCF."},{"Start":"03:40.115 ","End":"03:46.350","Text":"In this video, we learned about the phase diagram of carbon dioxide."}],"ID":24536},{"Watched":false,"Name":"Supercritical fluids","Duration":"3m 31s","ChapterTopicVideoID":23624,"CourseChapterTopicPlaylistID":110289,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.130","Text":"In previous videos,"},{"Start":"00:02.130 ","End":"00:04.140","Text":"we talked about the critical point."},{"Start":"00:04.140 ","End":"00:08.890","Text":"In this video, we\u0027ll talk about supercritical fluids."},{"Start":"00:09.380 ","End":"00:14.320","Text":"Let\u0027s recall what we learned about the critical point."},{"Start":"00:14.480 ","End":"00:19.135","Text":"It\u0027s the highest point in the vapor pressure curve,"},{"Start":"00:19.135 ","End":"00:21.290","Text":"and at critical point,"},{"Start":"00:21.290 ","End":"00:26.570","Text":"the liquid and gas have the same pressure and there\u0027s no interface between them."},{"Start":"00:26.570 ","End":"00:28.710","Text":"That means they\u0027re indistinguishable."},{"Start":"00:28.710 ","End":"00:33.410","Text":"Liquid and gas are indistinguishable at the critical point."},{"Start":"00:33.410 ","End":"00:36.155","Text":"Here\u0027s a phase diagram."},{"Start":"00:36.155 ","End":"00:43.260","Text":"We have pressure versus temperature."},{"Start":"00:44.140 ","End":"00:47.335","Text":"Here\u0027s the critical point."},{"Start":"00:47.335 ","End":"00:50.580","Text":"In the region between these 2 curves,"},{"Start":"00:50.580 ","End":"00:52.125","Text":"we have liquid,"},{"Start":"00:52.125 ","End":"00:55.075","Text":"and beyond that, we have gas."},{"Start":"00:55.075 ","End":"00:57.080","Text":"Above the critical point,"},{"Start":"00:57.080 ","End":"01:00.960","Text":"we have supercritical fluid."},{"Start":"01:01.850 ","End":"01:07.009","Text":"Let\u0027s consider what happens below the critical temperature."},{"Start":"01:07.009 ","End":"01:08.930","Text":"That\u0027s this arrow."},{"Start":"01:08.930 ","End":"01:13.550","Text":"We go from gas to liquid."},{"Start":"01:13.550 ","End":"01:18.455","Text":"The gas liquefies as the pressure increases."},{"Start":"01:18.455 ","End":"01:21.140","Text":"On this arrow, we\u0027re increasing the pressure,"},{"Start":"01:21.140 ","End":"01:25.170","Text":"so we go from gas to liquid."},{"Start":"01:25.450 ","End":"01:31.175","Text":"However, above the critical temperature, that\u0027s this arrow."},{"Start":"01:31.175 ","End":"01:33.780","Text":"The gas does not liquefy,"},{"Start":"01:33.780 ","End":"01:35.595","Text":"as the pressure increases."},{"Start":"01:35.595 ","End":"01:38.685","Text":"We go up, but this is the highest point."},{"Start":"01:38.685 ","End":"01:43.560","Text":"We\u0027re going beyond that point to the supercritical fluid."},{"Start":"01:43.790 ","End":"01:47.175","Text":"At pressures above P_c,"},{"Start":"01:47.175 ","End":"01:49.335","Text":"temperatures above T_c,"},{"Start":"01:49.335 ","End":"01:51.475","Text":"and pressures above P_c,"},{"Start":"01:51.475 ","End":"01:55.470","Text":"a supercritical fluid is produced."},{"Start":"01:56.620 ","End":"02:05.340","Text":"Now you may have heard about the importance of supercritical carbon dioxide as a solvent."},{"Start":"02:05.420 ","End":"02:09.120","Text":"Now for supercritical carbon dioxide,"},{"Start":"02:09.120 ","End":"02:15.950","Text":"P_c is 72.9 atmospheres and T_c is 31 degrees Celsius."},{"Start":"02:15.950 ","End":"02:19.380","Text":"We learned about that in the previous video."},{"Start":"02:19.730 ","End":"02:23.990","Text":"Now supercritical fluid is dense."},{"Start":"02:23.990 ","End":"02:28.685","Text":"In fact, if you do a calculation at 250 degrees Celsius,"},{"Start":"02:28.685 ","End":"02:33.335","Text":"its density is almost 1 gram per cubic centimeter,"},{"Start":"02:33.335 ","End":"02:36.860","Text":"and that\u0027s the same as water at room temperature."},{"Start":"02:36.860 ","End":"02:41.760","Text":"That\u0027s the same density as water at room temperature."},{"Start":"02:42.580 ","End":"02:46.520","Text":"But unlike water at room temperature,"},{"Start":"02:46.520 ","End":"02:48.440","Text":"here the temperature is high."},{"Start":"02:48.440 ","End":"02:51.799","Text":"The molecules move fast and can dissolve"},{"Start":"02:51.799 ","End":"02:56.400","Text":"organic molecules such as caffeine from coffee beans."},{"Start":"02:57.650 ","End":"03:04.475","Text":"Now, why is carbon dioxide a good solvent or an important solvent?"},{"Start":"03:04.475 ","End":"03:10.325","Text":"It\u0027s preferable to use it than to use harmful organic solvents."},{"Start":"03:10.325 ","End":"03:16.520","Text":"You don\u0027t want some organic solvent left in your coffee."},{"Start":"03:16.520 ","End":"03:19.490","Text":"It\u0027s also very important in green chemistry."},{"Start":"03:19.490 ","End":"03:23.960","Text":"Chemistry carried out with safe components."},{"Start":"03:23.960 ","End":"03:26.945","Text":"A very important field nowadays."},{"Start":"03:26.945 ","End":"03:31.530","Text":"In this video, we learned about supercritical fluids."}],"ID":24535},{"Watched":false,"Name":"Phase Diagram Water","Duration":"5m 30s","ChapterTopicVideoID":23842,"CourseChapterTopicPlaylistID":110289,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:02.340","Text":"In previous videos,"},{"Start":"00:02.340 ","End":"00:06.615","Text":"we talked about the phase diagrams of iodine and carbon dioxide,"},{"Start":"00:06.615 ","End":"00:10.990","Text":"and in this video, we\u0027ll learn about the phase diagram of water."},{"Start":"00:11.450 ","End":"00:14.625","Text":"Here\u0027s the phase diagram of water."},{"Start":"00:14.625 ","End":"00:17.490","Text":"Let\u0027s indicate the various phases."},{"Start":"00:17.490 ","End":"00:20.955","Text":"Here there\u0027s Ice 1,"},{"Start":"00:20.955 ","End":"00:24.640","Text":"one version of ice, water,"},{"Start":"00:24.890 ","End":"00:32.770","Text":"liquid water, water vapor,"},{"Start":"00:33.860 ","End":"00:37.125","Text":"and other versions of ice."},{"Start":"00:37.125 ","End":"00:42.430","Text":"This is 3, 2,"},{"Start":"00:42.430 ","End":"00:52.410","Text":"5, 6, and 7."},{"Start":"00:52.410 ","End":"00:53.970","Text":"The triple point O,"},{"Start":"00:53.970 ","End":"00:55.985","Text":"that\u0027s this point here,"},{"Start":"00:55.985 ","End":"01:04.629","Text":"is 0.00098 degrees Celsius and 4.58 millimeters of mercury."},{"Start":"01:04.629 ","End":"01:08.420","Text":"You see it\u0027s almost 0."},{"Start":"01:08.420 ","End":"01:12.580","Text":"The normal melting point of Ice 1 is at"},{"Start":"01:12.580 ","End":"01:17.484","Text":"exactly 0 degrees Celsius and 760 millimeters mercury,"},{"Start":"01:17.484 ","End":"01:22.790","Text":"because that\u0027s how 0 degrees Celsius is defined."},{"Start":"01:22.790 ","End":"01:25.755","Text":"That\u0027s this point here."},{"Start":"01:25.755 ","End":"01:30.764","Text":"The 1 atmosphere crosses this line,"},{"Start":"01:30.764 ","End":"01:33.790","Text":"ice between ice and water."},{"Start":"01:33.850 ","End":"01:39.935","Text":"Now if we look at the fusion curve between Ice 1 and liquid water,"},{"Start":"01:39.935 ","End":"01:42.135","Text":"we\u0027ll see that it has a negative slope,"},{"Start":"01:42.135 ","End":"01:44.090","Text":"here is a negative slope."},{"Start":"01:44.090 ","End":"01:49.420","Text":"That means that ice melts at lower temperatures when the pressure increases."},{"Start":"01:49.420 ","End":"01:55.325","Text":"In fact, large changes in pressure lead to small changes in melting point."},{"Start":"01:55.325 ","End":"02:01.490","Text":"It\u0027s a very small effect really and explains the movement of glaciers,"},{"Start":"02:01.490 ","End":"02:05.270","Text":"but not as previously thought ice skating."},{"Start":"02:05.270 ","End":"02:09.365","Text":"It was thought that when you skate on the ice,"},{"Start":"02:09.365 ","End":"02:14.915","Text":"the pressure of the skates melts some of the ice allowing you to glide."},{"Start":"02:14.915 ","End":"02:17.870","Text":"That\u0027s not really the reason."},{"Start":"02:17.870 ","End":"02:20.960","Text":"Now the critical point C,"},{"Start":"02:20.960 ","End":"02:23.470","Text":"that\u0027s this point here,"},{"Start":"02:23.470 ","End":"02:31.260","Text":"is at 374.1 degrees Celsius and 218.2 atmospheres,"},{"Start":"02:31.260 ","End":"02:35.470","Text":"so supercritical water is here."},{"Start":"02:37.430 ","End":"02:42.850","Text":"Now you saw that we had all sorts of ice."},{"Start":"02:42.850 ","End":"02:46.430","Text":"Solid water exists in many different forms."},{"Start":"02:46.430 ","End":"02:50.020","Text":"Some are crystalline, like Ice 1,"},{"Start":"02:50.020 ","End":"02:54.220","Text":"and others are morphous without a crystalline shape."},{"Start":"02:54.220 ","End":"03:00.885","Text":"Now there\u0027re 18 known forms of crystalline ice and these are called polymorphs,"},{"Start":"03:00.885 ","End":"03:03.785","Text":"the phenomenon is polymorphism,"},{"Start":"03:03.785 ","End":"03:09.305","Text":"and Ice 1 is the most common form of ice on Earth."},{"Start":"03:09.305 ","End":"03:13.055","Text":"Now there\u0027s 2 forms of amorphous ice,"},{"Start":"03:13.055 ","End":"03:15.065","Text":"low density and high density,"},{"Start":"03:15.065 ","End":"03:18.355","Text":"and that\u0027s called polyamorphism."},{"Start":"03:18.355 ","End":"03:26.650","Text":"There\u0027s polymorphism for crystals and polyamorphism for amorphous materials."},{"Start":"03:26.740 ","End":"03:33.035","Text":"Now the most common form of ice is amorphous ice in outer space."},{"Start":"03:33.035 ","End":"03:35.495","Text":"Let\u0027s take an example."},{"Start":"03:35.495 ","End":"03:39.260","Text":"There are many triple points because there are many forms of ice,"},{"Start":"03:39.260 ","End":"03:41.210","Text":"many phases of ice."},{"Start":"03:41.210 ","End":"03:45.920","Text":"That\u0027s where 3 phases of water co-exist."},{"Start":"03:45.920 ","End":"03:49.980","Text":"An example is D. Let\u0027s look at this."},{"Start":"03:49.980 ","End":"03:54.675","Text":"Here\u0027s D and that\u0027s between Ice 1,"},{"Start":"03:54.675 ","End":"03:57.945","Text":"Ice 3, and liquid water."},{"Start":"03:57.945 ","End":"04:05.090","Text":"That\u0027s at minus 22 degrees Celsius and 2,045 atmospheres,"},{"Start":"04:05.090 ","End":"04:07.415","Text":"a very high pressure."},{"Start":"04:07.415 ","End":"04:12.185","Text":"Now if we look at the other fusion curves in the water phase diagram,"},{"Start":"04:12.185 ","End":"04:15.560","Text":"we\u0027ll see that they all have positive slopes."},{"Start":"04:15.560 ","End":"04:18.815","Text":"All of these have positive slopes."},{"Start":"04:18.815 ","End":"04:22.040","Text":"Whereas the one between Ice 1 and"},{"Start":"04:22.040 ","End":"04:26.630","Text":"liquid water is very unusual in that it has a negative slope."},{"Start":"04:26.630 ","End":"04:31.620","Text":"Now a few words about what a phase transition is."},{"Start":"04:31.620 ","End":"04:36.155","Text":"A phase transition is a transition from one phase to another"},{"Start":"04:36.155 ","End":"04:40.820","Text":"by crossing a line at 2 phase equilibrium line,"},{"Start":"04:40.820 ","End":"04:44.975","Text":"such as those in red in this diagram."},{"Start":"04:44.975 ","End":"04:48.919","Text":"Now if we draw a line at constant pressure in the diagram,"},{"Start":"04:48.919 ","End":"04:50.555","Text":"for example 1 atmosphere,"},{"Start":"04:50.555 ","End":"04:53.045","Text":"it\u0027s called an isobar."},{"Start":"04:53.045 ","End":"04:56.345","Text":"Here this blue line is an isobar."},{"Start":"04:56.345 ","End":"04:59.060","Text":"We have a phase transition from ice to"},{"Start":"04:59.060 ","End":"05:03.860","Text":"liquid water and another one from liquid water to water vapor."},{"Start":"05:03.860 ","End":"05:08.090","Text":"If instead of drawing a line at constant pressure,"},{"Start":"05:08.090 ","End":"05:10.190","Text":"we draw one at constant temperature,"},{"Start":"05:10.190 ","End":"05:12.380","Text":"it\u0027s called an isotherm."},{"Start":"05:12.380 ","End":"05:14.555","Text":"Let\u0027s take an example."},{"Start":"05:14.555 ","End":"05:17.460","Text":"For example, if it\u0027s here,"},{"Start":"05:17.740 ","End":"05:25.204","Text":"we\u0027re going from water vapor to liquid water at this particular temperature."},{"Start":"05:25.204 ","End":"05:30.540","Text":"In this video, we learned about the phase diagram of water."}],"ID":24784}],"Thumbnail":null,"ID":110289},{"Name":"Bonding in Solids","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Network Covalent Solids","Duration":"5m 50s","ChapterTopicVideoID":23843,"CourseChapterTopicPlaylistID":112678,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.680 ","End":"00:02.910","Text":"In the next few videos,"},{"Start":"00:02.910 ","End":"00:05.235","Text":"we\u0027ll talk about bonding in solids."},{"Start":"00:05.235 ","End":"00:10.560","Text":"In this video, we\u0027ll discuss network covalent solids."},{"Start":"00:10.560 ","End":"00:15.405","Text":"First, a few words about bonding and solids."},{"Start":"00:15.405 ","End":"00:20.080","Text":"Now, solids can be held together by covalent, ionic,"},{"Start":"00:20.080 ","End":"00:24.645","Text":"or metallic bonding, or by intermolecular forces."},{"Start":"00:24.645 ","End":"00:27.900","Text":"It\u0027s found that solids held together by covalent,"},{"Start":"00:27.900 ","End":"00:31.550","Text":"ionic, or metallic bonding are usually not always,"},{"Start":"00:31.550 ","End":"00:34.430","Text":"but usually much stronger and have"},{"Start":"00:34.430 ","End":"00:40.290","Text":"higher melting points than those held together by intermolecular forces."},{"Start":"00:40.400 ","End":"00:44.490","Text":"Let\u0027s discuss network covalent solids."},{"Start":"00:44.490 ","End":"00:48.830","Text":"That\u0027s solids held together by covalent bonds."},{"Start":"00:48.830 ","End":"00:51.785","Text":"Now, network covalent solids,"},{"Start":"00:51.785 ","End":"00:58.315","Text":"covalent bonds extend throughout the crystalline solid in all 3 directions."},{"Start":"00:58.315 ","End":"01:05.857","Text":"We\u0027re going to consider first 2 examples of allotropes of carbon,"},{"Start":"01:05.857 ","End":"01:12.150","Text":"different versions of carbon."},{"Start":"01:12.920 ","End":"01:17.830","Text":"Diamond and graphite are allotropes of carbon."},{"Start":"01:17.830 ","End":"01:19.600","Text":"Now, in the previous video,"},{"Start":"01:19.600 ","End":"01:21.580","Text":"we talked about polymorphs."},{"Start":"01:21.580 ","End":"01:23.440","Text":"Polymorphs, a general name,"},{"Start":"01:23.440 ","End":"01:26.750","Text":"it can be molecules or atoms,"},{"Start":"01:26.750 ","End":"01:32.260","Text":"and allotropes are used for substances composed of a single element."},{"Start":"01:32.260 ","End":"01:37.280","Text":"Allotrope is a special case of polymorph."},{"Start":"01:37.320 ","End":"01:40.585","Text":"Let\u0027s start with diamond."},{"Start":"01:40.585 ","End":"01:43.180","Text":"Here\u0027s a picture of diamond."},{"Start":"01:43.180 ","End":"01:51.645","Text":"Each carbon atom is sp3 hybridized and it\u0027s bonded to 4 carbon atoms."},{"Start":"01:51.645 ","End":"01:55.870","Text":"Just as in the ethene that we discussed previously."},{"Start":"01:55.870 ","End":"02:00.980","Text":"Here\u0027s CH_4, C in the center, 4 hydrogens."},{"Start":"02:00.980 ","End":"02:03.170","Text":"Now, in diamond,"},{"Start":"02:03.170 ","End":"02:06.470","Text":"we replace each of these hydrogens by a carbon."},{"Start":"02:06.470 ","End":"02:13.630","Text":"Then each carbon is the center of being bonded to another 4 carbons,"},{"Start":"02:13.630 ","End":"02:15.915","Text":"and so on, so forth."},{"Start":"02:15.915 ","End":"02:20.180","Text":"Here\u0027s an example. Here\u0027s a carbon bonded to 1,"},{"Start":"02:20.180 ","End":"02:24.470","Text":"2, 3, and this one behind here, 4 carbons."},{"Start":"02:24.470 ","End":"02:29.540","Text":"Another thing to note is there\u0027s also non planar hexagons."},{"Start":"02:29.540 ","End":"02:36.630","Text":"Here\u0027s an example here, 1, 2, 3, 4, 5, 6."},{"Start":"02:39.220 ","End":"02:43.440","Text":"This is a non planar hexagon."},{"Start":"02:44.110 ","End":"02:49.190","Text":"Because of this extensive strong bonding,"},{"Start":"02:49.190 ","End":"02:52.250","Text":"diamond is the hardest solid substance and"},{"Start":"02:52.250 ","End":"02:56.270","Text":"its melting point is above 3,500 degrees Celsius."},{"Start":"02:56.270 ","End":"03:00.905","Text":"Now, if we replace half of the carbon atoms with silicon,"},{"Start":"03:00.905 ","End":"03:03.080","Text":"that\u0027s called carborundum,"},{"Start":"03:03.080 ","End":"03:07.310","Text":"it sublimes at 2,700 degrees Celsius."},{"Start":"03:07.310 ","End":"03:09.935","Text":"It\u0027s also extremely hard."},{"Start":"03:09.935 ","End":"03:15.160","Text":"Now, let\u0027s talk about another allotrope of carbon, graphite."},{"Start":"03:15.160 ","End":"03:21.980","Text":"Here, each carbon atom is sp^2 hybridized and it\u0027s bonded to 3 atoms,"},{"Start":"03:21.980 ","End":"03:23.450","Text":"just as in benzene,"},{"Start":"03:23.450 ","End":"03:27.670","Text":"and forming a plane composed of hexagons."},{"Start":"03:27.670 ","End":"03:31.650","Text":"Here\u0027s benzene, here\u0027s a hexagon in benzene,"},{"Start":"03:31.650 ","End":"03:36.420","Text":"and each place, each carbon is sp2 hybridized."},{"Start":"03:36.420 ","End":"03:40.135","Text":"Each angle is 120 degrees."},{"Start":"03:40.135 ","End":"03:42.905","Text":"That\u0027s the same here."},{"Start":"03:42.905 ","End":"03:48.185","Text":"Here\u0027s an example of a hexagon."},{"Start":"03:48.185 ","End":"03:52.570","Text":"We have sheets composed of hexagons."},{"Start":"03:52.570 ","End":"03:57.660","Text":"Now, just as in benzene,"},{"Start":"03:57.660 ","End":"04:04.035","Text":"we have p orbitals perpendicular to the plane of the molecule."},{"Start":"04:04.035 ","End":"04:09.015","Text":"There\u0027s a p orbital perpendicular to the plane of the molecule."},{"Start":"04:09.015 ","End":"04:13.545","Text":"These overlap on all the carbons."},{"Start":"04:13.545 ","End":"04:15.915","Text":"Of course, in graphite,"},{"Start":"04:15.915 ","End":"04:18.850","Text":"these are also carbons,"},{"Start":"04:19.750 ","End":"04:24.425","Text":"to form extensive delocalization."},{"Start":"04:24.425 ","End":"04:28.610","Text":"The p orbital\u0027s perpendicular plane overlap as in benzene,"},{"Start":"04:28.610 ","End":"04:33.060","Text":"leading to delocalization of the p electrons."},{"Start":"04:33.740 ","End":"04:38.195","Text":"These strong bonds are within the layers,"},{"Start":"04:38.195 ","End":"04:45.935","Text":"and the layers themselves interact with each other by weak van der Waals forces."},{"Start":"04:45.935 ","End":"04:54.650","Text":"Here\u0027s the layer, and it\u0027s bound to the next layer by weak van der Waals forces."},{"Start":"04:54.650 ","End":"04:57.030","Text":"The bond length within the layers,"},{"Start":"04:57.030 ","End":"05:02.160","Text":"142 picometers and between the layers 335 picometers."},{"Start":"05:02.160 ","End":"05:09.115","Text":"We see this strong bonding and much weaker interaction between the layers."},{"Start":"05:09.115 ","End":"05:10.953","Text":"Now unlike diamond,"},{"Start":"05:10.953 ","End":"05:16.480","Text":"graphite conducts electricity that\u0027s due to the delocalized electrons."},{"Start":"05:16.480 ","End":"05:19.970","Text":"Now, if we have impure graphite, for example,"},{"Start":"05:19.970 ","End":"05:22.610","Text":"containing nitrogen and oxygen from the air,"},{"Start":"05:22.610 ","End":"05:27.290","Text":"the bonds or the interaction between the layers is even weaker."},{"Start":"05:27.290 ","End":"05:33.095","Text":"Graphite is a good lubricant as the layers slide over each other."},{"Start":"05:33.095 ","End":"05:36.335","Text":"We\u0027re all familiar with the led and pencils,"},{"Start":"05:36.335 ","End":"05:39.785","Text":"and that consists of graphite mixed with clay."},{"Start":"05:39.785 ","End":"05:45.004","Text":"As we write, the graphite layers flake off onto the paper."},{"Start":"05:45.004 ","End":"05:50.790","Text":"In this video, we discuss network covalent solids."}],"ID":24785},{"Watched":false,"Name":"Allotropes of Carbon","Duration":"6m 4s","ChapterTopicVideoID":23870,"CourseChapterTopicPlaylistID":112678,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.980 ","End":"00:03.060","Text":"In the previous video,"},{"Start":"00:03.060 ","End":"00:07.890","Text":"we learned about 2 allotropes of carbon: diamond and graphite."},{"Start":"00:07.890 ","End":"00:12.800","Text":"In this video, we\u0027ll learn about more recently discovered allotropes of carbon."},{"Start":"00:12.800 ","End":"00:16.930","Text":"I\u0027m going to start off with graphene."},{"Start":"00:17.090 ","End":"00:20.265","Text":"Now in 2004,"},{"Start":"00:20.265 ","End":"00:24.495","Text":"Andre Geim and Konstantin Novoselov"},{"Start":"00:24.495 ","End":"00:30.180","Text":"managed to separate the layers of graphite using sellotape or scotch tape."},{"Start":"00:30.180 ","End":"00:33.419","Text":"Now, people often tried to separate"},{"Start":"00:33.419 ","End":"00:38.670","Text":"the layers and it was a great surprise that such a simple method worked."},{"Start":"00:38.670 ","End":"00:41.210","Text":"For this work and of course,"},{"Start":"00:41.210 ","End":"00:46.250","Text":"for all their revelations concerning graphene,"},{"Start":"00:46.250 ","End":"00:51.835","Text":"these 2 people got the Nobel Prize in 2010."},{"Start":"00:51.835 ","End":"00:55.620","Text":"Here\u0027s a single layer of graphite,"},{"Start":"00:55.620 ","End":"00:57.605","Text":"and it\u0027s called graphene."},{"Start":"00:57.605 ","End":"01:02.370","Text":"You can see the little hexagons everywhere."},{"Start":"01:02.680 ","End":"01:07.990","Text":"Now, this is a new material and it consists of single sheets,"},{"Start":"01:07.990 ","End":"01:11.345","Text":"and each single sheet is called a graphene sheet."},{"Start":"01:11.345 ","End":"01:13.460","Text":"Together, it\u0027s called graphene."},{"Start":"01:13.460 ","End":"01:15.380","Text":"If we have several layers,"},{"Start":"01:15.380 ","End":"01:19.895","Text":"they can be stacked together using water to form a very strong,"},{"Start":"01:19.895 ","End":"01:22.390","Text":"thin, and flexible material."},{"Start":"01:22.390 ","End":"01:26.315","Text":"It\u0027s 100 times stronger than steel of the same thickness,"},{"Start":"01:26.315 ","End":"01:31.105","Text":"so we have something that\u0027s very thin and yet, very, very strong."},{"Start":"01:31.105 ","End":"01:37.040","Text":"Now, unlike graphite which is black because it absorbs all visible frequencies,"},{"Start":"01:37.040 ","End":"01:42.485","Text":"a graphene sheet is nearly transparent because it\u0027s very thin."},{"Start":"01:42.485 ","End":"01:47.105","Text":"Now, graphene has extremely unusual properties."},{"Start":"01:47.105 ","End":"01:52.315","Text":"The more common ones are that it conducts heat and electricity along its plane,"},{"Start":"01:52.315 ","End":"01:58.729","Text":"but it has also many extraordinary physical properties and many potential applications."},{"Start":"01:58.729 ","End":"02:04.915","Text":"It\u0027s a very lively field of research nowadays."},{"Start":"02:04.915 ","End":"02:08.570","Text":"Now, here\u0027s another allotrope of carbon,"},{"Start":"02:08.570 ","End":"02:15.410","Text":"it\u0027s called Buckminster fullerenes or fullerenes for short, or even buckyballs."},{"Start":"02:15.410 ","End":"02:19.710","Text":"We\u0027ll soon see the reason for such a name."},{"Start":"02:20.000 ","End":"02:24.950","Text":"Now, 1985, Harold Kroto, Robert Curl,"},{"Start":"02:24.950 ","End":"02:32.495","Text":"Richard Smalley created the spherical C_60 molecules in the lab."},{"Start":"02:32.495 ","End":"02:37.715","Text":"Later, it was found that they also exist in soot and in outer space."},{"Start":"02:37.715 ","End":"02:42.325","Text":"For all their work on these fullerenes,"},{"Start":"02:42.325 ","End":"02:48.905","Text":"the 3 inventors got the Nobel Prize in 1996."},{"Start":"02:48.905 ","End":"02:52.520","Text":"Now, why did they call it Buckminster fullerene?"},{"Start":"02:52.520 ","End":"02:56.225","Text":"This was after the architect, Buckminster Fuller,"},{"Start":"02:56.225 ","End":"03:02.125","Text":"who constructed geodesic domes in the 1940s."},{"Start":"03:02.125 ","End":"03:06.845","Text":"This was reminiscent, although it\u0027s not exactly the same."},{"Start":"03:06.845 ","End":"03:12.260","Text":"This was shortened to fullerenes or buckyballs."},{"Start":"03:12.260 ","End":"03:19.680","Text":"Later, they also discovered C_70 buckyballs and many others."},{"Start":"03:19.680 ","End":"03:23.975","Text":"Now, the C_60 ball resembles a soccer ball,"},{"Start":"03:23.975 ","End":"03:28.310","Text":"and it has 20 hexagons and 20 pentagons."},{"Start":"03:28.310 ","End":"03:32.720","Text":"Each carbon, just like in graphite,"},{"Start":"03:32.720 ","End":"03:39.910","Text":"each carbon is sp^2 hybridized and is bonded to 3 carbon atoms."},{"Start":"03:40.270 ","End":"03:48.780","Text":"Here\u0027s a picture. Here\u0027s a pentagon and here\u0027s a hexagon."},{"Start":"03:48.780 ","End":"03:52.910","Text":"Now, these fullerenes are black, just like graphite."},{"Start":"03:52.910 ","End":"03:55.730","Text":"But unlike graphite, they can dissolve in"},{"Start":"03:55.730 ","End":"04:00.235","Text":"organic solvents that don\u0027t contain oxygen, for example, toluene."},{"Start":"04:00.235 ","End":"04:03.965","Text":"Then they have all sorts of colors like purple, magenta,"},{"Start":"04:03.965 ","End":"04:09.085","Text":"or green, depending on the concentrations."},{"Start":"04:09.085 ","End":"04:18.920","Text":"Now, here is another allotrope of carbon called carbon nanotubes or CNT for short."},{"Start":"04:18.920 ","End":"04:23.675","Text":"Now, if we\u0027re to take a graphene sheet and roll it into a cylinder,"},{"Start":"04:23.675 ","End":"04:25.820","Text":"can be rolled in all different ways,"},{"Start":"04:25.820 ","End":"04:29.730","Text":"and then cap it at each end by half a buckyball,"},{"Start":"04:29.730 ","End":"04:32.029","Text":"we get a carbon nanotube."},{"Start":"04:32.029 ","End":"04:34.505","Text":"Now, they\u0027re not actually made in the lab that way,"},{"Start":"04:34.505 ","End":"04:39.290","Text":"but that\u0027s what they look like. Here\u0027s a picture."},{"Start":"04:39.290 ","End":"04:41.210","Text":"You can see in the center,"},{"Start":"04:41.210 ","End":"04:42.935","Text":"the bit that\u0027s been rolled up,"},{"Start":"04:42.935 ","End":"04:45.470","Text":"you can see hexagons,"},{"Start":"04:45.470 ","End":"04:50.250","Text":"and at the edge, you can see also pentagons."},{"Start":"04:51.440 ","End":"04:55.770","Text":"Now, there are single-walled CNTs,"},{"Start":"04:55.770 ","End":"05:00.960","Text":"that\u0027s SWCNT, with diameter less than 1 nanometer."},{"Start":"05:00.960 ","End":"05:07.322","Text":"There are also multi-walled CNTs that\u0027s several layers of graphene,"},{"Start":"05:07.322 ","End":"05:08.990","Text":"1 inside the other,"},{"Start":"05:08.990 ","End":"05:11.960","Text":"with a diameter of about 200 nanometers,"},{"Start":"05:11.960 ","End":"05:15.930","Text":"and the length could be micrometers or even millimeters."},{"Start":"05:16.060 ","End":"05:22.370","Text":"Now, these were discovered in 1991 by Sumio Iijima."},{"Start":"05:22.370 ","End":"05:26.180","Text":"He reported the discovery of multi-walled nanotubes."},{"Start":"05:26.180 ","End":"05:28.160","Text":"But if you do your research,"},{"Start":"05:28.160 ","End":"05:33.835","Text":"you\u0027ll find that they existed in nature for quite a long time before that."},{"Start":"05:33.835 ","End":"05:41.220","Text":"Now, these carbon nanotubes are ultra-high strength materials."},{"Start":"05:41.220 ","End":"05:45.410","Text":"They can make van der Waals forces between the CNTs."},{"Start":"05:45.410 ","End":"05:50.510","Text":"They\u0027re also very light and they\u0027re excellent conductors of heat and electricity."},{"Start":"05:50.510 ","End":"05:54.860","Text":"They have many applications: electrochemistry,"},{"Start":"05:54.860 ","End":"05:57.830","Text":"catalysis, and many more."},{"Start":"05:57.830 ","End":"06:04.290","Text":"This video, we got a glimpse of the modern allotropes of carbon."}],"ID":24813},{"Watched":false,"Name":"Molecular and Metallic Solids","Duration":"4m 58s","ChapterTopicVideoID":23850,"CourseChapterTopicPlaylistID":112678,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.130","Text":"In the previous video,"},{"Start":"00:02.130 ","End":"00:05.025","Text":"we discussed network covalent solids."},{"Start":"00:05.025 ","End":"00:09.940","Text":"In this video, we\u0027ll learn about molecular and metallic solids."},{"Start":"00:09.980 ","End":"00:14.350","Text":"Let\u0027s begin with molecular solids."},{"Start":"00:14.510 ","End":"00:18.890","Text":"Molecular solids are assemblies of discrete atoms or"},{"Start":"00:18.890 ","End":"00:23.645","Text":"molecules held in place by inter-molecular forces."},{"Start":"00:23.645 ","End":"00:26.605","Text":"Let\u0027s take a few examples."},{"Start":"00:26.605 ","End":"00:30.185","Text":"We start with non-polar molecules,"},{"Start":"00:30.185 ","End":"00:32.900","Text":"such as rare gases and methane."},{"Start":"00:32.900 ","End":"00:35.465","Text":"We\u0027ll talk about methane later."},{"Start":"00:35.465 ","End":"00:39.965","Text":"These are held together by dispersion forces."},{"Start":"00:39.965 ","End":"00:46.670","Text":"Now polar molecules such as dimethyl ether or hydrogen chloride are"},{"Start":"00:46.670 ","End":"00:53.435","Text":"held together by dispersion forces and in addition, dipole-dipole interactions."},{"Start":"00:53.435 ","End":"00:59.710","Text":"These are stronger solids than the non-polar molecules."},{"Start":"00:59.710 ","End":"01:02.255","Text":"Now going a stage further,"},{"Start":"01:02.255 ","End":"01:05.615","Text":"molecules such as water or ammonia,"},{"Start":"01:05.615 ","End":"01:08.585","Text":"are held together by hydrogen bonds."},{"Start":"01:08.585 ","End":"01:11.335","Text":"These are even stronger."},{"Start":"01:11.335 ","End":"01:16.355","Text":"Now many organic molecules form molecular crystals."},{"Start":"01:16.355 ","End":"01:20.459","Text":"You\u0027ll meet many of them in organic labs."},{"Start":"01:20.620 ","End":"01:25.310","Text":"Let\u0027s talk about the crystal structure of methane I."},{"Start":"01:25.310 ","End":"01:29.345","Text":"Now, methane forms several different crystalline solids."},{"Start":"01:29.345 ","End":"01:33.600","Text":"It\u0027s another example of polymorphism."},{"Start":"01:33.890 ","End":"01:37.980","Text":"We\u0027re going to talk about CH_4I."},{"Start":"01:38.960 ","End":"01:41.575","Text":"Well here\u0027s the structure."},{"Start":"01:41.575 ","End":"01:44.125","Text":"In the corners of this cube,"},{"Start":"01:44.125 ","End":"01:46.630","Text":"we have white balls,"},{"Start":"01:46.630 ","End":"01:50.255","Text":"and on each face we have a red ball."},{"Start":"01:50.255 ","End":"01:53.100","Text":"In each of these positions,"},{"Start":"01:53.100 ","End":"01:56.905","Text":"the ones at the corners and the ones in the face,"},{"Start":"01:56.905 ","End":"02:00.185","Text":"we have CH_4 molecules."},{"Start":"02:00.185 ","End":"02:07.390","Text":"You remember that each CH_4 molecule is tetrahedral in shape."},{"Start":"02:07.390 ","End":"02:09.790","Text":"Here we have this structure,"},{"Start":"02:09.790 ","End":"02:12.445","Text":"it\u0027s called a face-centered cube."},{"Start":"02:12.445 ","End":"02:18.655","Text":"Now, each of these CH_4 molecules can rotate freely."},{"Start":"02:18.655 ","End":"02:21.675","Text":"It\u0027s called a plastic crystal."},{"Start":"02:21.675 ","End":"02:30.830","Text":"The center of mass of each of these CH_4 molecules is in the position of the spheres,"},{"Start":"02:30.830 ","End":"02:33.395","Text":"but within that region,"},{"Start":"02:33.395 ","End":"02:36.270","Text":"they can rotate freely."},{"Start":"02:36.610 ","End":"02:41.150","Text":"Now these molecules interact through dispersion forces,"},{"Start":"02:41.150 ","End":"02:45.200","Text":"so we have a very low melting point because the forces are"},{"Start":"02:45.200 ","End":"02:51.695","Text":"weak and the melting point is minus 182.5 degrees Celsius."},{"Start":"02:51.695 ","End":"02:57.605","Text":"That\u0027s the reason that methane is a gas at room temperature."},{"Start":"02:57.605 ","End":"03:03.055","Text":"Now let\u0027s go onto metallic solids or just metals."},{"Start":"03:03.055 ","End":"03:10.655","Text":"Now the very simplest model of a metallic solid or metal is a free electron model."},{"Start":"03:10.655 ","End":"03:15.515","Text":"In this model, the cations form a close packed structure."},{"Start":"03:15.515 ","End":"03:17.400","Text":"What we had before,"},{"Start":"03:17.400 ","End":"03:23.525","Text":"this face-centered cube is an example of a close packed structure,"},{"Start":"03:23.525 ","End":"03:29.785","Text":"and we\u0027ll talk about close packed structures in several of the coming videos."},{"Start":"03:29.785 ","End":"03:33.300","Text":"Now this model predicts"},{"Start":"03:33.300 ","End":"03:36.870","Text":"high electrical and thermal conductivity but"},{"Start":"03:36.870 ","End":"03:41.620","Text":"has to be modified to explain many of the other properties of metals."},{"Start":"03:41.620 ","End":"03:45.150","Text":"The cations are fixed in position,"},{"Start":"03:45.150 ","End":"03:51.180","Text":"and the valence electrons are free to move throughout the crystal."},{"Start":"03:52.400 ","End":"03:55.220","Text":"Now apart from mercury,"},{"Start":"03:55.220 ","End":"03:58.220","Text":"metals are solids at room temperature,"},{"Start":"03:58.220 ","End":"04:02.880","Text":"and so of higher melting points than molecular solids."},{"Start":"04:04.060 ","End":"04:07.670","Text":"Here\u0027s an example, sodium."},{"Start":"04:07.670 ","End":"04:15.080","Text":"The sodium plus cations form a close packed structure called a body-centered cube, bcc."},{"Start":"04:15.080 ","End":"04:19.555","Text":"Before we had a face-centered cube this is a body-centered cube."},{"Start":"04:19.555 ","End":"04:26.165","Text":"The 1 valence electron from each atom forms a sea of electrons."},{"Start":"04:26.165 ","End":"04:29.030","Text":"Here\u0027s our body-centered cube."},{"Start":"04:29.030 ","End":"04:33.940","Text":"We have spheres at each corner of the cube,"},{"Start":"04:33.940 ","End":"04:39.930","Text":"and another sphere in the center of the cube."},{"Start":"04:39.930 ","End":"04:50.280","Text":"The Na+ cations are in the places indicated by these spheres."},{"Start":"04:51.220 ","End":"04:58.050","Text":"In this video, we talked about molecular and metallic solids."}],"ID":24811},{"Watched":false,"Name":"Ionic Crystals and Lattice Energies","Duration":"6m 59s","ChapterTopicVideoID":23869,"CourseChapterTopicPlaylistID":112678,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:02.310","Text":"In the previous videos,"},{"Start":"00:02.310 ","End":"00:06.435","Text":"we talked about covalent molecular and metallic solids."},{"Start":"00:06.435 ","End":"00:10.785","Text":"Going to talk about ionic solids very briefly."},{"Start":"00:10.785 ","End":"00:15.300","Text":"In later videos we\u0027ll talk about them in far more detail."},{"Start":"00:15.300 ","End":"00:18.215","Text":"What do we know about ionic solids?"},{"Start":"00:18.215 ","End":"00:21.515","Text":"Ionic solids are crystalline in nature."},{"Start":"00:21.515 ","End":"00:24.530","Text":"In order to achieve maximum stability,"},{"Start":"00:24.530 ","End":"00:30.050","Text":"oppositely charged ions are next to each other because they attract each other"},{"Start":"00:30.050 ","End":"00:32.240","Text":"and ions of the same charge are far as"},{"Start":"00:32.240 ","End":"00:36.365","Text":"possible from each other because they repel each other."},{"Start":"00:36.365 ","End":"00:40.400","Text":"Now here\u0027s an example that we met in a very early video."},{"Start":"00:40.400 ","End":"00:48.870","Text":"It\u0027s NaCl and it has a cubic crystal structure and it\u0027s often called rock salt structure."},{"Start":"00:50.170 ","End":"00:52.830","Text":"Na+ is in green,"},{"Start":"00:52.830 ","End":"00:56.610","Text":"is much smaller than Cl- which is in"},{"Start":"00:56.610 ","End":"01:03.265","Text":"violet and we can see the cubic structure quite clearly here."},{"Start":"01:03.265 ","End":"01:06.780","Text":"Now we can see that each Na+ is surrounded by"},{"Start":"01:06.780 ","End":"01:14.750","Text":"6 chloride ions and each chloride ion is surrounded by 6 sodium ions."},{"Start":"01:15.030 ","End":"01:20.320","Text":"Once again, we note that the ions with opposite signs attract each"},{"Start":"01:20.320 ","End":"01:25.375","Text":"other whereas the ions with the same sign repel each other."},{"Start":"01:25.375 ","End":"01:31.160","Text":"Now the inter ionic distance d is the sum of the ionic radii."},{"Start":"01:31.160 ","End":"01:35.245","Text":"The radius of Na+ is 102 picometers."},{"Start":"01:35.245 ","End":"01:41.650","Text":"The radius of chloride ion is 181 picometers and if we add these two"},{"Start":"01:41.650 ","End":"01:49.010","Text":"together we\u0027ll get that the inter ionic distance is 283 picometers."},{"Start":"01:49.010 ","End":"01:51.335","Text":"That\u0027s this distance,"},{"Start":"01:51.335 ","End":"01:53.610","Text":"one to the other."},{"Start":"01:53.860 ","End":"02:02.100","Text":"Now an important quantity is the lattice energy or the formation energy of a crystal."},{"Start":"02:03.100 ","End":"02:05.690","Text":"Now the lattice energy,"},{"Start":"02:05.690 ","End":"02:07.910","Text":"Delta H lattice for crystal is"},{"Start":"02:07.910 ","End":"02:14.045","Text":"the enthalpy change when a crystals formed from its ions in the gas phase."},{"Start":"02:14.045 ","End":"02:17.480","Text":"The lattice energy is negative since the formation of"},{"Start":"02:17.480 ","End":"02:20.755","Text":"the crystal is an exothermic process."},{"Start":"02:20.755 ","End":"02:24.920","Text":"A crystal is actually ever formed in this way."},{"Start":"02:24.920 ","End":"02:30.830","Text":"It\u0027s an artificial way of finding the strength of the crystal."},{"Start":"02:30.830 ","End":"02:33.865","Text":"It\u0027s called the lattice energy."},{"Start":"02:33.865 ","End":"02:36.465","Text":"Let\u0027s take an example."},{"Start":"02:36.465 ","End":"02:39.700","Text":"Once again we have NaCl."},{"Start":"02:39.700 ","End":"02:43.040","Text":"Here is the lattice energy."},{"Start":"02:43.040 ","End":"02:50.060","Text":"It\u0027s a process in which Na+ in the gas phase combines with Cl- in the gas phase to"},{"Start":"02:50.060 ","End":"02:59.370","Text":"form NaCl solid and Delta H lattice is minus 787 kilojoules per mole."},{"Start":"03:02.300 ","End":"03:06.335","Text":"How can we estimate the lattice energy?"},{"Start":"03:06.335 ","End":"03:08.555","Text":"Later on in this chapter,"},{"Start":"03:08.555 ","End":"03:16.320","Text":"we\u0027ll see how you can determine this value indirectly using Hess\u0027s law."},{"Start":"03:16.540 ","End":"03:20.840","Text":"We\u0027re going to estimate the lattice energy."},{"Start":"03:20.840 ","End":"03:26.780","Text":"We can use the Coulomb equation to estimate the lattice energy."},{"Start":"03:26.780 ","End":"03:33.990","Text":"Now the electrostatic potential energy that\u0027s the Coulomb equation for two ions,"},{"Start":"03:33.990 ","End":"03:43.260","Text":"is V equal to z_1 times z_2 times e^2 divided by 4 pi epsilon_0 d. Now,"},{"Start":"03:43.260 ","End":"03:45.090","Text":"what are all these quantities?"},{"Start":"03:45.090 ","End":"03:47.940","Text":"z_1 and z_2 are the ionic charges,"},{"Start":"03:47.940 ","End":"03:54.305","Text":"e is electron charge and Epsilon_0 is a constant called the vacuum permittivity."},{"Start":"03:54.305 ","End":"03:57.335","Text":"Its actual values is not really important here."},{"Start":"03:57.335 ","End":"04:02.495","Text":"What we need to see is that V increases as z_1 and z_2 increase,"},{"Start":"04:02.495 ","End":"04:08.760","Text":"and if we increase d then V decreases."},{"Start":"04:08.760 ","End":"04:12.555","Text":"V increases if z_1,2"},{"Start":"04:12.555 ","End":"04:20.125","Text":"increase and V decreases if d increases."},{"Start":"04:20.125 ","End":"04:27.580","Text":"We can write this as V is equal to minus the absolute value of z_1 times z_2."},{"Start":"04:27.580 ","End":"04:31.860","Text":"These are oppositely charged ions so they have opposite signs so that\u0027s"},{"Start":"04:31.860 ","End":"04:36.685","Text":"accounts for the negative sign times e^2 over 4 pi"},{"Start":"04:36.685 ","End":"04:46.790","Text":"Epsilon_0 d. Now with a mole of ionic crystal,"},{"Start":"04:47.520 ","End":"04:52.345","Text":"V is equal to minus A times N_A that\u0027s Avogadro\u0027s number."},{"Start":"04:52.345 ","End":"04:56.325","Text":"The absolute value of z_1 times z_2,"},{"Start":"04:56.325 ","End":"05:02.080","Text":"e^2 over 4 pi Epsilon_0 d. Now this e is"},{"Start":"05:02.080 ","End":"05:07.470","Text":"called the Madelung constant and it\u0027s different for different types of crystals."},{"Start":"05:07.470 ","End":"05:09.250","Text":"We\u0027re not going to calculate it here."},{"Start":"05:09.250 ","End":"05:14.450","Text":"Its values can be found in all texts."},{"Start":"05:14.450 ","End":"05:23.950","Text":"For NaCl in the rock salt crystal structure, D= 1.763."},{"Start":"05:24.600 ","End":"05:27.700","Text":"What do we see from this equation?"},{"Start":"05:27.700 ","End":"05:30.610","Text":"We see that V becomes more negative."},{"Start":"05:30.610 ","End":"05:33.850","Text":"That means it has a stronger interaction with"},{"Start":"05:33.850 ","End":"05:39.475","Text":"increasing charge and decreasing inter ionic distance."},{"Start":"05:39.475 ","End":"05:45.790","Text":"As the charges increase the crystal becomes more stable and does"},{"Start":"05:45.790 ","End":"05:52.835","Text":"the distances between the atoms decreases it also becomes more stable."},{"Start":"05:52.835 ","End":"05:55.855","Text":"These are the two important quantities we need."},{"Start":"05:55.855 ","End":"06:04.520","Text":"z_1 times z_2 and d. Let\u0027s take an example."},{"Start":"06:04.550 ","End":"06:10.155","Text":"Assuming that NaCl and KCl form the same type of crystal,"},{"Start":"06:10.155 ","End":"06:14.385","Text":"in which crystal with the ions interact more strongly?"},{"Start":"06:14.385 ","End":"06:18.720","Text":"Which crystal has a more negative lattice energy?"},{"Start":"06:18.720 ","End":"06:22.690","Text":"Greater lattice energy means more negative."},{"Start":"06:22.690 ","End":"06:29.340","Text":"The inter ionic distance is d=283 picometers for NaCl."},{"Start":"06:29.340 ","End":"06:35.085","Text":"We saw that before and 319 picometers for KCl."},{"Start":"06:35.085 ","End":"06:39.885","Text":"KCl is larger than NaCl."},{"Start":"06:39.885 ","End":"06:44.660","Text":"The ions in NaCl interact more strongly."},{"Start":"06:44.660 ","End":"06:49.990","Text":"We expect the lattice energy to be more negative for NaCl."},{"Start":"06:49.990 ","End":"06:54.740","Text":"In this video, we discussed ionic solids and we will"},{"Start":"06:54.740 ","End":"07:00.300","Text":"discuss ionic crystals in greater detail in later videos."}],"ID":24812}],"Thumbnail":null,"ID":112678},{"Name":"Crystal Structures","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Crystal Lattices and Unit Cells","Duration":"5m 4s","ChapterTopicVideoID":23871,"CourseChapterTopicPlaylistID":112679,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:02.460","Text":"In the next few videos,"},{"Start":"00:02.460 ","End":"00:05.220","Text":"we\u0027ll talk about the structure of crystals."},{"Start":"00:05.220 ","End":"00:10.650","Text":"In this video, we\u0027ll talk about crystal lattices and unit cells."},{"Start":"00:10.650 ","End":"00:13.695","Text":"What\u0027s a crystal structure?"},{"Start":"00:13.695 ","End":"00:18.869","Text":"The crystal structure is a description of the ordered arrangement of atoms,"},{"Start":"00:18.869 ","End":"00:24.195","Text":"ions or molecules in a crystalline material."},{"Start":"00:24.195 ","End":"00:30.814","Text":"A very important concept that helps us to describe crystals is the unit cell."},{"Start":"00:30.814 ","End":"00:36.160","Text":"This is the smallest group of atoms or ions or molecules,"},{"Start":"00:36.160 ","End":"00:39.385","Text":"I\u0027ll just say atoms in the rest of this video,"},{"Start":"00:39.385 ","End":"00:42.470","Text":"for which the entire lattice of the crystal can be"},{"Start":"00:42.470 ","End":"00:45.860","Text":"built up by translation in 3 dimensions,"},{"Start":"00:45.860 ","End":"00:49.735","Text":"by moving in 3 dimensions."},{"Start":"00:49.735 ","End":"00:56.140","Text":"Let\u0027s start by looking at 2 dimensional square unit cell."},{"Start":"00:56.860 ","End":"01:01.970","Text":"Here in the center is a square unit cell,"},{"Start":"01:01.970 ","End":"01:05.850","Text":"it\u0027s a very simple shape."},{"Start":"01:06.140 ","End":"01:10.250","Text":"Now, if we move this black square,"},{"Start":"01:10.250 ","End":"01:12.900","Text":"if it\u0027s translated in 2 dimensions,"},{"Start":"01:12.900 ","End":"01:16.515","Text":"that\u0027s means left and right or up and down,"},{"Start":"01:16.515 ","End":"01:18.635","Text":"we can build up the whole pattern."},{"Start":"01:18.635 ","End":"01:23.730","Text":"You can see that we can build up a whole pattern of squares."},{"Start":"01:24.620 ","End":"01:29.895","Text":"The black square is called a 2D unit cell,"},{"Start":"01:29.895 ","End":"01:33.963","Text":"and the points at the corners, 1, 2, 3,"},{"Start":"01:33.963 ","End":"01:39.070","Text":"4 are called lattice points."},{"Start":"01:39.730 ","End":"01:43.115","Text":"Now it need not necessarily be a square,"},{"Start":"01:43.115 ","End":"01:45.260","Text":"it can be another parallelogram,"},{"Start":"01:45.260 ","End":"01:48.610","Text":"you can recall the parallelogram is"},{"Start":"01:48.610 ","End":"01:52.414","Text":"a geometrical shape in which the opposite sides are parallel."},{"Start":"01:52.414 ","End":"01:54.997","Text":"Here\u0027s an example."},{"Start":"01:54.997 ","End":"01:57.975","Text":"What about 3 dimensions?"},{"Start":"01:57.975 ","End":"02:02.640","Text":"Now we\u0027ll first look at the 3 dimensional cubic unit cell."},{"Start":"02:02.640 ","End":"02:06.980","Text":"Here\u0027s our cube. In 3 dimensions,"},{"Start":"02:06.980 ","End":"02:10.340","Text":"we have a cube instead of a square."},{"Start":"02:10.340 ","End":"02:15.140","Text":"This cube could be translated in 3 dimensions: Up and down,"},{"Start":"02:15.140 ","End":"02:19.915","Text":"left and right, backwards and forwards."},{"Start":"02:19.915 ","End":"02:21.660","Text":"If we do this,"},{"Start":"02:21.660 ","End":"02:26.202","Text":"we build up a 3 dimensional cubic lattice."},{"Start":"02:26.202 ","End":"02:30.050","Text":"The unit cell, this is a cubic unit cell,"},{"Start":"02:30.050 ","End":"02:31.940","Text":"need not be cubic,"},{"Start":"02:31.940 ","End":"02:33.620","Text":"it can be other shapes."},{"Start":"02:33.620 ","End":"02:39.680","Text":"We can have 7 possible 3 dimensional parallelepipeds."},{"Start":"02:39.680 ","End":"02:46.260","Text":"A parallelepiped is a 3 dimensional equivalent of a parallelogram."},{"Start":"02:46.260 ","End":"02:53.855","Text":"These 7 possible parallelepipeds are called 7 crystal classes."},{"Start":"02:53.855 ","End":"02:57.905","Text":"Now if a unit cell has the atoms at its corners,"},{"Start":"02:57.905 ","End":"03:00.230","Text":"at each of these 8 corners,"},{"Start":"03:00.230 ","End":"03:05.400","Text":"it\u0027s called a primitive or a simple unit cell."},{"Start":"03:05.810 ","End":"03:11.290","Text":"Here\u0027s a simple primitive cubic unit cell."},{"Start":"03:11.290 ","End":"03:13.070","Text":"I\u0027ve drawn it in 2 forms."},{"Start":"03:13.070 ","End":"03:16.663","Text":"Here, the balls are touching each other,"},{"Start":"03:16.663 ","End":"03:19.470","Text":"and here it\u0027s at the corners of cube,"},{"Start":"03:19.470 ","End":"03:22.160","Text":"we can actually see the cubic unit cell."},{"Start":"03:22.160 ","End":"03:26.210","Text":"This is called simple or primitive cubic."},{"Start":"03:26.210 ","End":"03:28.984","Text":"We can have the same thing,"},{"Start":"03:28.984 ","End":"03:32.070","Text":"but a ball at the center,"},{"Start":"03:32.070 ","End":"03:33.723","Text":"an atom at the center,"},{"Start":"03:33.723 ","End":"03:36.230","Text":"and that\u0027s called body-centered cubic."},{"Start":"03:36.230 ","End":"03:37.774","Text":"Here are 2 forms of it."},{"Start":"03:37.774 ","End":"03:43.175","Text":"Here we can see the atoms touching the center ball,"},{"Start":"03:43.175 ","End":"03:47.360","Text":"and here we just see the positions of them."},{"Start":"03:47.360 ","End":"03:57.460","Text":"Then we can have a face-centered cubic unit cell where we have atoms at the corners,"},{"Start":"03:57.460 ","End":"04:04.605","Text":"8 atoms at the corners and 6 atoms in the centers of the faces."},{"Start":"04:04.605 ","End":"04:06.660","Text":"Here we can see it clearly,"},{"Start":"04:06.660 ","End":"04:09.935","Text":"this red ball is in the center of this face."},{"Start":"04:09.935 ","End":"04:12.530","Text":"This is called face-centered cubic,"},{"Start":"04:12.530 ","End":"04:15.598","Text":"body-centered cubic, simple cubic,"},{"Start":"04:15.598 ","End":"04:23.020","Text":"and we can abbreviate this by SC, BCC, FCC."},{"Start":"04:23.210 ","End":"04:27.800","Text":"To reiterate, if there\u0027s an atom at the center of the cubic unit cell,"},{"Start":"04:27.800 ","End":"04:30.410","Text":"it\u0027s called body-centered cubic."},{"Start":"04:30.410 ","End":"04:33.440","Text":"If there is an atom at the center of the faces,"},{"Start":"04:33.440 ","End":"04:36.115","Text":"it\u0027s called face-centered cubic."},{"Start":"04:36.115 ","End":"04:44.345","Text":"These cubic unit cells are just 3 of the 14 possible lattice types,"},{"Start":"04:44.345 ","End":"04:48.995","Text":"and the 14 possible lattice types are called Bravais lattices."},{"Start":"04:48.995 ","End":"04:51.815","Text":"We\u0027re not going to discuss all the Bravais lattices,"},{"Start":"04:51.815 ","End":"04:58.000","Text":"just the cubic ones in this particular course."},{"Start":"04:58.000 ","End":"05:04.140","Text":"In this video, we learned about crystal lattices and unit cells."}],"ID":24814},{"Watched":false,"Name":"Close-Packed Structures","Duration":"7m 54s","ChapterTopicVideoID":23875,"CourseChapterTopicPlaylistID":112679,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.040 ","End":"00:03.120","Text":"In a previous video,"},{"Start":"00:03.120 ","End":"00:06.180","Text":"we talked about the free-electron model of metals."},{"Start":"00:06.180 ","End":"00:07.980","Text":"In this video, we\u0027ll describe"},{"Start":"00:07.980 ","End":"00:14.730","Text":"the possible close-packed structure of the atoms in metallic solids."},{"Start":"00:14.730 ","End":"00:18.975","Text":"We\u0027re going to talk about close-packed structures."},{"Start":"00:18.975 ","End":"00:24.670","Text":"We\u0027ll first talk about the first 2 layers of the close-packed structure."},{"Start":"00:25.040 ","End":"00:30.060","Text":"First thing to note is that the cations in metals can be"},{"Start":"00:30.060 ","End":"00:35.730","Text":"described as hard spheres that are packed as closely as possible."},{"Start":"00:35.730 ","End":"00:37.800","Text":"The very first layer,"},{"Start":"00:37.800 ","End":"00:39.090","Text":"which we\u0027ll call A,"},{"Start":"00:39.090 ","End":"00:44.410","Text":"each sphere is surrounded by 6 spheres forming a hexagon."},{"Start":"00:44.410 ","End":"00:46.580","Text":"Here\u0027s a picture."},{"Start":"00:46.580 ","End":"00:53.850","Text":"Here\u0027s a central sphere and it\u0027s surrounded by 6 spheres,"},{"Start":"00:53.850 ","End":"00:55.186","Text":"1, 2, 3, 4, 5,"},{"Start":"00:55.186 ","End":"01:01.360","Text":"6 forming a hexagon."},{"Start":"01:01.420 ","End":"01:04.640","Text":"Now you see that there are holes."},{"Start":"01:04.640 ","End":"01:06.589","Text":"If we take these 3 atoms,"},{"Start":"01:06.589 ","End":"01:08.000","Text":"there\u0027s a hole between it,"},{"Start":"01:08.000 ","End":"01:10.595","Text":"which we call a trigonal hole."},{"Start":"01:10.595 ","End":"01:13.250","Text":"Here\u0027s another one, 1,"},{"Start":"01:13.250 ","End":"01:15.800","Text":"2, 3 atoms and here\u0027s a trigonal hole."},{"Start":"01:15.800 ","End":"01:18.370","Text":"Here\u0027s a picture of it and we can see"},{"Start":"01:18.370 ","End":"01:24.360","Text":"the 3 atoms are like a triangle and it\u0027s called a trigonal hole."},{"Start":"01:24.380 ","End":"01:27.900","Text":"Now in the second layer, which we\u0027ll call B,"},{"Start":"01:27.900 ","End":"01:29.115","Text":"this is A,"},{"Start":"01:29.115 ","End":"01:33.825","Text":"and here we have A is red and the blue is B."},{"Start":"01:33.825 ","End":"01:35.975","Text":"In the second layer B,"},{"Start":"01:35.975 ","End":"01:40.970","Text":"the atoms sit over the trigonal holes between the atoms of the first layer."},{"Start":"01:40.970 ","End":"01:46.705","Text":"These blue atoms sit on top of the trigonal holes."},{"Start":"01:46.705 ","End":"01:50.550","Text":"Again, in the second layer B,"},{"Start":"01:50.550 ","End":"01:54.015","Text":"each atom is surrounded by 6 atoms."},{"Start":"01:54.015 ","End":"01:57.540","Text":"Here\u0027s a central one, 1, 2,"},{"Start":"01:57.540 ","End":"02:01.830","Text":"3, 4, 5, 6."},{"Start":"02:01.830 ","End":"02:05.830","Text":"Each atom is surrounded by 6 atoms."},{"Start":"02:06.290 ","End":"02:09.650","Text":"Now if we look at this picture,"},{"Start":"02:09.650 ","End":"02:13.065","Text":"we see that there are 2 holes,"},{"Start":"02:13.065 ","End":"02:16.595","Text":"tetrahedral holes and octahedral holes."},{"Start":"02:16.595 ","End":"02:18.920","Text":"Now the tetrahedral hole,"},{"Start":"02:18.920 ","End":"02:21.200","Text":"which are the red ones,"},{"Start":"02:21.200 ","End":"02:24.205","Text":"are formed from 4 atoms,"},{"Start":"02:24.205 ","End":"02:28.985","Text":"1 in A and 3 in B in the tetrahedral configuration."},{"Start":"02:28.985 ","End":"02:31.230","Text":"Let\u0027s see a picture."},{"Start":"02:31.940 ","End":"02:34.785","Text":"Here\u0027s 1 atom from A,"},{"Start":"02:34.785 ","End":"02:38.260","Text":"3 atoms from B."},{"Start":"02:38.750 ","End":"02:43.550","Text":"Here\u0027s a tetrahedron connecting the various atoms."},{"Start":"02:43.550 ","End":"02:47.735","Text":"Now if we turn it around so the red is on top,"},{"Start":"02:47.735 ","End":"02:51.000","Text":"you can clearly see the tetrahedral."},{"Start":"02:51.250 ","End":"02:54.845","Text":"Now in addition to the tetrahedral hole,"},{"Start":"02:54.845 ","End":"02:56.840","Text":"we also have octahedral holes,"},{"Start":"02:56.840 ","End":"02:59.885","Text":"and these are white in this picture."},{"Start":"02:59.885 ","End":"03:02.135","Text":"These are these white holes."},{"Start":"03:02.135 ","End":"03:04.995","Text":"They\u0027re formed from 6 atoms,"},{"Start":"03:04.995 ","End":"03:07.020","Text":"3 in A and 3 in B."},{"Start":"03:07.020 ","End":"03:09.345","Text":"Here\u0027s a picture, 3 in A,"},{"Start":"03:09.345 ","End":"03:12.510","Text":"the red ones and 3 on top for B."},{"Start":"03:12.510 ","End":"03:15.170","Text":"Here are the holes."},{"Start":"03:15.170 ","End":"03:16.730","Text":"You get a hole from"},{"Start":"03:16.730 ","End":"03:23.405","Text":"the red layer and a hole from the blue layer in the opposite direction."},{"Start":"03:23.405 ","End":"03:27.515","Text":"Now it\u0027s a bit difficult to see this is a octahedron."},{"Start":"03:27.515 ","End":"03:29.615","Text":"But if we turn it round,"},{"Start":"03:29.615 ","End":"03:35.560","Text":"you can clearly see the octahedron connecting the 6 atoms."},{"Start":"03:35.560 ","End":"03:38.160","Text":"This is called an octahedral hole,"},{"Start":"03:38.160 ","End":"03:42.860","Text":"and this is called a tetrahedral hole."},{"Start":"03:42.860 ","End":"03:45.110","Text":"Now when we come to the third layer,"},{"Start":"03:45.110 ","End":"03:51.810","Text":"we can either cover the tetrahedral holes or cover the octahedral holes."},{"Start":"03:54.670 ","End":"04:00.950","Text":"I will first talk about covering the tetrahedral holes."},{"Start":"04:01.130 ","End":"04:07.545","Text":"We\u0027re going to have the third layer on top of these red balls in the first layer."},{"Start":"04:07.545 ","End":"04:13.440","Text":"Exactly on top of the red atom, of course,"},{"Start":"04:13.440 ","End":"04:17.950","Text":"there are blue atoms in the middle and so it\u0027s another layer,"},{"Start":"04:17.950 ","End":"04:21.680","Text":"just like the layer of the red atoms."},{"Start":"04:21.680 ","End":"04:26.140","Text":"The atoms in the third layer sits in the tetrahedral holes,"},{"Start":"04:26.140 ","End":"04:28.375","Text":"there are directly above the first layer."},{"Start":"04:28.375 ","End":"04:30.835","Text":"The structure is ABA,"},{"Start":"04:30.835 ","End":"04:37.740","Text":"and this is called hexagonal close packing or hcp for short."},{"Start":"04:37.740 ","End":"04:43.780","Text":"Here\u0027s a picture, here\u0027s A, B, A."},{"Start":"04:43.780 ","End":"04:46.160","Text":"If we draw the unit cell,"},{"Start":"04:46.160 ","End":"04:47.825","Text":"we\u0027ll see it\u0027s like a hexagon."},{"Start":"04:47.825 ","End":"04:50.945","Text":"Here we have 6 atoms in the first layer."},{"Start":"04:50.945 ","End":"04:53.600","Text":"Remember, that\u0027s like a hexagon."},{"Start":"04:53.600 ","End":"04:56.870","Text":"Each atom surrounded by a hexagon of atoms,"},{"Start":"04:56.870 ","End":"05:00.980","Text":"3 blue ones, 3 in the second layer."},{"Start":"05:00.980 ","End":"05:04.250","Text":"The third layer is like the first layer, a hexagon."},{"Start":"05:04.250 ","End":"05:08.675","Text":"This is called hexagonal close packing."},{"Start":"05:08.675 ","End":"05:12.920","Text":"If we draw the unit cell, here it is."},{"Start":"05:12.920 ","End":"05:16.100","Text":"We have 3 unit cells in this arrangement,"},{"Start":"05:16.100 ","End":"05:18.215","Text":"in this executable close packing,"},{"Start":"05:18.215 ","End":"05:20.075","Text":"we have 3 unit cells."},{"Start":"05:20.075 ","End":"05:27.579","Text":"Each one has an angle here of 60 degrees and here of a 120 degrees."},{"Start":"05:27.579 ","End":"05:32.560","Text":"It\u0027s not cubic, it\u0027s called hexagonal."},{"Start":"05:32.560 ","End":"05:39.485","Text":"Now if we put the third layer to cover the octahedral holes,"},{"Start":"05:39.485 ","End":"05:42.904","Text":"we\u0027re covering these white holes."},{"Start":"05:42.904 ","End":"05:46.965","Text":"Here\u0027s a picture, here\u0027s A, B,"},{"Start":"05:46.965 ","End":"05:52.850","Text":"and C. C is no longer exactly the same as A."},{"Start":"05:52.850 ","End":"05:58.710","Text":"We have A, B, C. Now,"},{"Start":"05:58.710 ","End":"06:02.520","Text":"if the atoms in the third layer cover the octahedral holes,"},{"Start":"06:02.520 ","End":"06:05.795","Text":"they are neither directly above the first layer or the second layer."},{"Start":"06:05.795 ","End":"06:09.155","Text":"So this is called ABC."},{"Start":"06:09.155 ","End":"06:11.420","Text":"Now when we get to the fourth layer,"},{"Start":"06:11.420 ","End":"06:13.055","Text":"it is above the first layer."},{"Start":"06:13.055 ","End":"06:15.890","Text":"So the structure is ABCA."},{"Start":"06:15.890 ","End":"06:17.510","Text":"A, B, C,"},{"Start":"06:17.510 ","End":"06:26.540","Text":"A and this is called cubic close packing or ccp for short."},{"Start":"06:26.540 ","End":"06:31.080","Text":"Now if we take 14 atoms"},{"Start":"06:31.100 ","End":"06:36.360","Text":"out of the crystal so that we have 6 atoms in each of the middle layers,"},{"Start":"06:36.360 ","End":"06:38.979","Text":"here\u0027s 6 atoms and other 6 atoms,"},{"Start":"06:38.979 ","End":"06:42.355","Text":"one on the top and one on the bottom."},{"Start":"06:42.355 ","End":"06:47.590","Text":"We\u0027ve taken a section out of a larger structure."},{"Start":"06:47.590 ","End":"06:53.875","Text":"We have to take a large number of atoms in each layer."},{"Start":"06:53.875 ","End":"06:56.860","Text":"Then if we turn it round,"},{"Start":"06:56.860 ","End":"06:59.900","Text":"we\u0027re rotating it this way."},{"Start":"07:01.400 ","End":"07:09.580","Text":"Now we have something like this and we can clearly see that this is face-centered cubic."},{"Start":"07:09.580 ","End":"07:17.525","Text":"Here\u0027s one cubic face and at the center there is an atom."},{"Start":"07:17.525 ","End":"07:21.290","Text":"This is true of all the faces of the cube."},{"Start":"07:21.290 ","End":"07:28.850","Text":"When rotated, we can see the face-centered cube unit cell, fcc for short."},{"Start":"07:28.850 ","End":"07:35.275","Text":"We can conclude that ccp is equivalent to fcc."},{"Start":"07:35.275 ","End":"07:41.640","Text":"Cubic close packing is the same as face-centered cubic."},{"Start":"07:41.740 ","End":"07:49.535","Text":"In this video, we talked about close-packed ccp and hcp structures."},{"Start":"07:49.535 ","End":"07:54.510","Text":"Cubic close packing and hexagonal close packing."}],"ID":24818},{"Watched":false,"Name":"Coordination Number and Atoms in Unit Cell","Duration":"7m ","ChapterTopicVideoID":23876,"CourseChapterTopicPlaylistID":112679,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.070","Text":"In the previous video,"},{"Start":"00:02.070 ","End":"00:04.695","Text":"we learned about close packed structures."},{"Start":"00:04.695 ","End":"00:10.960","Text":"In this video, we\u0027ll learn about the coordination number and atoms in the unit cell."},{"Start":"00:11.240 ","End":"00:15.930","Text":"Let\u0027s begin with the coordination number."},{"Start":"00:15.930 ","End":"00:23.880","Text":"This is defined as the number of atoms with which each atom is in contact."},{"Start":"00:23.880 ","End":"00:27.965","Text":"Let\u0027s begin with a simple cubic cell."},{"Start":"00:27.965 ","End":"00:33.830","Text":"Here the coordination number is 6 and the example is polonium."},{"Start":"00:33.830 ","End":"00:38.285","Text":"Here\u0027s a picture of the simple cubic cell."},{"Start":"00:38.285 ","End":"00:43.145","Text":"Each atom has 1 to the right, 1 below,"},{"Start":"00:43.145 ","End":"00:44.420","Text":"1 to the left,"},{"Start":"00:44.420 ","End":"00:46.655","Text":"1 above, 1 behind,"},{"Start":"00:46.655 ","End":"00:48.080","Text":"and 1 in front."},{"Start":"00:48.080 ","End":"00:52.580","Text":"Each atom\u0027s in contact with 6 other atoms."},{"Start":"00:52.580 ","End":"00:56.180","Text":"That means that the coordination number is 6."},{"Start":"00:56.180 ","End":"00:59.285","Text":"Let\u0027s look at the body-centered cube."},{"Start":"00:59.285 ","End":"01:04.225","Text":"Here it is. Here the coordination number is 8."},{"Start":"01:04.225 ","End":"01:06.555","Text":"Let\u0027s take the central atom."},{"Start":"01:06.555 ","End":"01:09.210","Text":"It\u0027s in contact with 1, 2, 3,"},{"Start":"01:09.210 ","End":"01:12.480","Text":"4 atoms in front and 1,"},{"Start":"01:12.480 ","End":"01:15.180","Text":"2, 3, 4 atoms behind."},{"Start":"01:15.180 ","End":"01:16.870","Text":"Total of 8."},{"Start":"01:16.870 ","End":"01:18.740","Text":"Examples are sodium,"},{"Start":"01:18.740 ","End":"01:21.800","Text":"potassium, iron, and tungsten."},{"Start":"01:21.800 ","End":"01:26.315","Text":"I just want to make the point that"},{"Start":"01:26.315 ","End":"01:30.439","Text":"usually simple cubic and body-centered cubic structures"},{"Start":"01:30.439 ","End":"01:34.205","Text":"are not considered to be close packed structures."},{"Start":"01:34.205 ","End":"01:38.490","Text":"Or as some books write closest packed."},{"Start":"01:38.660 ","End":"01:43.595","Text":"That\u0027s because the holes in the cubic packed,"},{"Start":"01:43.595 ","End":"01:49.340","Text":"these holes, are larger than the holes in the close packed structure."},{"Start":"01:49.340 ","End":"01:57.170","Text":"Here is simple trigonal holes and here tetragonal holes."},{"Start":"01:57.170 ","End":"02:01.310","Text":"Let\u0027s look at the close packed structures."},{"Start":"02:01.310 ","End":"02:05.480","Text":"We have the hexagonal close packed and here the coordination numbers"},{"Start":"02:05.480 ","End":"02:09.380","Text":"we\u0027ll see is 12 with examples of cadmium,"},{"Start":"02:09.380 ","End":"02:12.410","Text":"magnesium, zinc, and titanium."},{"Start":"02:12.410 ","End":"02:15.890","Text":"The cubic close packed structure,"},{"Start":"02:15.890 ","End":"02:20.495","Text":"which is equivalent as we saw to the face-centered cubic structure,"},{"Start":"02:20.495 ","End":"02:27.630","Text":"the coordination number is 12 and examples here are silver, copper or lead."},{"Start":"02:27.760 ","End":"02:30.125","Text":"As we saw before,"},{"Start":"02:30.125 ","End":"02:33.710","Text":"both the cubic close packed structure and"},{"Start":"02:33.710 ","End":"02:40.490","Text":"the hexagonal close packed structure have the same first 2 layers, A and B."},{"Start":"02:40.490 ","End":"02:44.960","Text":"The blue one\u0027s B. These are both the same."},{"Start":"02:44.960 ","End":"02:48.590","Text":"If we take a typical atom, here,"},{"Start":"02:48.590 ","End":"02:55.835","Text":"it has 6 atoms around it in the same layer and 3 below,"},{"Start":"02:55.835 ","End":"02:59.200","Text":"making a total here of 9."},{"Start":"02:59.390 ","End":"03:02.700","Text":"In the case of hcp,"},{"Start":"03:02.700 ","End":"03:06.955","Text":"there are 3 atoms covering the tetrahedral holes."},{"Start":"03:06.955 ","End":"03:11.275","Text":"Remember these red holes are the tetrahedral holes."},{"Start":"03:11.275 ","End":"03:16.700","Text":"There are 3 atoms covering the tetrahedral holes so that makes a total of 12."},{"Start":"03:16.700 ","End":"03:18.695","Text":"Then the ccp,"},{"Start":"03:18.695 ","End":"03:20.725","Text":"which is the same as fcc,"},{"Start":"03:20.725 ","End":"03:25.700","Text":"there are 3 atoms covering the octahedral holes, these white ones."},{"Start":"03:25.700 ","End":"03:27.780","Text":"That\u0027s a total of 12."},{"Start":"03:27.780 ","End":"03:30.765","Text":"Both hcp and ccp,"},{"Start":"03:30.765 ","End":"03:34.380","Text":"the coordination number is 12."},{"Start":"03:34.380 ","End":"03:41.140","Text":"We also want to know how many atoms there are in each unit cell."},{"Start":"03:41.140 ","End":"03:45.210","Text":"Here are cubic unit cells."},{"Start":"03:45.210 ","End":"03:49.340","Text":"If we look at an atom in the corner,"},{"Start":"03:49.340 ","End":"03:52.790","Text":"we see that it contributes to 4 unit cells,"},{"Start":"03:52.790 ","End":"03:55.040","Text":"these 1, 2, 3,"},{"Start":"03:55.040 ","End":"03:58.955","Text":"4 and also to 4 above."},{"Start":"03:58.955 ","End":"04:02.630","Text":"It contributes to each unit cells and therefore"},{"Start":"04:02.630 ","End":"04:06.850","Text":"contributes just 1/8 of an atom to each unit cell."},{"Start":"04:06.850 ","End":"04:10.280","Text":"If we look at an atom on a face,"},{"Start":"04:10.280 ","End":"04:12.200","Text":"it contributes to 2 unit cells."},{"Start":"04:12.200 ","End":"04:13.850","Text":"Here\u0027s an example."},{"Start":"04:13.850 ","End":"04:19.790","Text":"The one to the red one on the left and to the black one on the right."},{"Start":"04:19.790 ","End":"04:24.360","Text":"It contributes 1/2 an atom to each unit cell."},{"Start":"04:25.820 ","End":"04:32.455","Text":"An atom in the center of a cubic cell contributes only to that unit cell."},{"Start":"04:32.455 ","End":"04:36.320","Text":"It contributes 1 atom to each unit cell."},{"Start":"04:36.590 ","End":"04:41.000","Text":"We can consider some examples."},{"Start":"04:43.140 ","End":"04:46.615","Text":"If we have a simple cubic cell,"},{"Start":"04:46.615 ","End":"04:51.145","Text":"it has 8 atoms corners, 1,"},{"Start":"04:51.145 ","End":"04:53.280","Text":"2, 3, 4,"},{"Start":"04:53.280 ","End":"04:56.880","Text":"and 4 behind and each one contributes 1/8."},{"Start":"04:56.880 ","End":"05:00.075","Text":"That\u0027s 8 times 1/8, that\u0027s a total of 1."},{"Start":"05:00.075 ","End":"05:04.930","Text":"There\u0027s 1 atom in each cubic cell."},{"Start":"05:06.200 ","End":"05:11.155","Text":"The body-centered cube once again, has 8 corners,"},{"Start":"05:11.155 ","End":"05:13.390","Text":"8 atoms at the corners,"},{"Start":"05:13.390 ","End":"05:14.680","Text":"8 times an 1/8,"},{"Start":"05:14.680 ","End":"05:17.394","Text":"and also has 1 in the center."},{"Start":"05:17.394 ","End":"05:20.000","Text":"That\u0027s a total of 2."},{"Start":"05:20.540 ","End":"05:25.485","Text":"The face centered cube has 8 corners,"},{"Start":"05:25.485 ","End":"05:28.095","Text":"give me 8 times 1/8,"},{"Start":"05:28.095 ","End":"05:32.555","Text":"and 6 on the faces because of course,"},{"Start":"05:32.555 ","End":"05:34.595","Text":"a cube has 6 faces."},{"Start":"05:34.595 ","End":"05:38.345","Text":"That\u0027s a total of 1 plus 3,"},{"Start":"05:38.345 ","End":"05:41.280","Text":"giving 4 in total."},{"Start":"05:43.660 ","End":"05:49.145","Text":"We get to the hexagonal close packed, that\u0027s hcp."},{"Start":"05:49.145 ","End":"05:51.744","Text":"In the hexagonal cell,"},{"Start":"05:51.744 ","End":"05:54.325","Text":"which we have here,"},{"Start":"05:54.325 ","End":"05:57.245","Text":"there are 12 atoms,"},{"Start":"05:57.245 ","End":"06:01.970","Text":"6 on the top and 6 on the bottom at the corners."},{"Start":"06:01.970 ","End":"06:08.240","Text":"Each of these contributes 1/6."},{"Start":"06:08.240 ","End":"06:13.025","Text":"Then there are 2 here and here at the top and the bottom,"},{"Start":"06:13.025 ","End":"06:15.487","Text":"it\u0027s on a face,"},{"Start":"06:15.487 ","End":"06:17.720","Text":"each of these contributes a 1/2."},{"Start":"06:17.720 ","End":"06:19.595","Text":"That\u0027s 2 times 1/2."},{"Start":"06:19.595 ","End":"06:22.610","Text":"Then we have 3 in the center,"},{"Start":"06:22.610 ","End":"06:26.270","Text":"which only contribute to this hexagonal cell."},{"Start":"06:26.270 ","End":"06:29.210","Text":"That makes a total of 6."},{"Start":"06:29.210 ","End":"06:33.405","Text":"2 here, 1, and 3."},{"Start":"06:33.405 ","End":"06:35.440","Text":"That\u0027s a total of 6."},{"Start":"06:35.440 ","End":"06:39.965","Text":"However, in this hexagonal close packed structure,"},{"Start":"06:39.965 ","End":"06:41.960","Text":"there are 3 unit cells,"},{"Start":"06:41.960 ","End":"06:46.650","Text":"so that\u0027s 6 divide by 3, giving us 2."},{"Start":"06:50.240 ","End":"06:52.820","Text":"In this video, we learnt about"},{"Start":"06:52.820 ","End":"06:58.290","Text":"coordination number and the number of atoms in a unit cell."}],"ID":24819},{"Watched":false,"Name":"Volume Occupied in Close-Packed Structures","Duration":"5m 6s","ChapterTopicVideoID":23873,"CourseChapterTopicPlaylistID":112679,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.590 ","End":"00:02.970","Text":"In the last 2 videos,"},{"Start":"00:02.970 ","End":"00:05.310","Text":"we talked about close-packed structures."},{"Start":"00:05.310 ","End":"00:11.620","Text":"In this video, we\u0027ll talk about the volume occupied in close-packed structures."},{"Start":"00:11.930 ","End":"00:17.340","Text":"We\u0027re going to calculate the volume occupied in close-packed structures."},{"Start":"00:17.340 ","End":"00:23.055","Text":"Now the volume actually occupied varies from structure to structure."},{"Start":"00:23.055 ","End":"00:27.885","Text":"We\u0027re going to take the example of body-centered cube."},{"Start":"00:27.885 ","End":"00:31.590","Text":"We have 1 atom in the corner,"},{"Start":"00:31.590 ","End":"00:34.995","Text":"another atom at the diagonal corners."},{"Start":"00:34.995 ","End":"00:36.450","Text":"That\u0027s 1 here,"},{"Start":"00:36.450 ","End":"00:40.735","Text":"and 1 here, and 1 in the center."},{"Start":"00:40.735 ","End":"00:46.970","Text":"If we draw a line from the center of the top one,"},{"Start":"00:46.970 ","End":"00:51.210","Text":"the top atom to the bottom atom."},{"Start":"00:52.880 ","End":"00:57.915","Text":"Now this line has a length of 4r."},{"Start":"00:57.915 ","End":"01:01.355","Text":"This is the radius of the first atom."},{"Start":"01:01.355 ","End":"01:03.665","Text":"Another 2 radii here,"},{"Start":"01:03.665 ","End":"01:05.863","Text":"the diameter of this atom,"},{"Start":"01:05.863 ","End":"01:08.255","Text":"and the radius of the third atom."},{"Start":"01:08.255 ","End":"01:11.135","Text":"We have a total of 4r."},{"Start":"01:11.135 ","End":"01:15.410","Text":"The length of this blue line is 4r."},{"Start":"01:18.260 ","End":"01:22.420","Text":"The blue diagonal has a length of 4r."},{"Start":"01:22.420 ","End":"01:25.640","Text":"Now if we look at the red diagonal and assume that"},{"Start":"01:25.640 ","End":"01:30.140","Text":"the length of a particular side of this cube is l,"},{"Start":"01:30.140 ","End":"01:38.465","Text":"then we can see the red diagonal has the length of square root of l^2 from this side,"},{"Start":"01:38.465 ","End":"01:40.565","Text":"l^2 from this side,"},{"Start":"01:40.565 ","End":"01:44.885","Text":"and the sum of that is 2l^2."},{"Start":"01:44.885 ","End":"01:48.695","Text":"The length of the red diagonal is the square root of 2l^2,"},{"Start":"01:48.695 ","End":"01:51.920","Text":"which is root 2l."},{"Start":"01:51.920 ","End":"01:56.030","Text":"Now if we look at the blue diagonal once again and"},{"Start":"01:56.030 ","End":"02:00.470","Text":"consider it to be the hypotenuse of this triangle,"},{"Start":"02:00.470 ","End":"02:02.640","Text":"with the red line at one side,"},{"Start":"02:02.640 ","End":"02:08.235","Text":"then its length is the square root of l^2 from this side,"},{"Start":"02:08.235 ","End":"02:16.385","Text":"2l^2 from the red diagonal,"},{"Start":"02:16.385 ","End":"02:19.025","Text":"giving a total of 3l^2."},{"Start":"02:19.025 ","End":"02:21.530","Text":"The square root of that is root 3 times"},{"Start":"02:21.530 ","End":"02:26.690","Text":"l. But we know that the blue diagonal also has a length of 4r."},{"Start":"02:26.690 ","End":"02:32.540","Text":"We know that 4r is equal to the square root of"},{"Start":"02:32.540 ","End":"02:39.750","Text":"3 times l. l is 4r divided by the square root of 3."},{"Start":"02:40.160 ","End":"02:43.470","Text":"Now we know that each unit cell, well,"},{"Start":"02:43.470 ","End":"02:47.905","Text":"BCC cell contains 2 atoms."},{"Start":"02:47.905 ","End":"02:53.570","Text":"The volume of each atom is 4/3 times Pi r^3."},{"Start":"02:53.570 ","End":"02:56.930","Text":"The total volume of the atoms is twice that,"},{"Start":"02:56.930 ","End":"03:00.575","Text":"2 times 4 divided by 3 Pi r^3,"},{"Start":"03:00.575 ","End":"03:05.010","Text":"giving us 8 over 3 Pi r^3."},{"Start":"03:05.500 ","End":"03:09.825","Text":"Now the volume of each unit cell is l^3."},{"Start":"03:09.825 ","End":"03:11.580","Text":"We know what l is,"},{"Start":"03:11.580 ","End":"03:16.140","Text":"we know that l is 4r divided by square root of 3."},{"Start":"03:16.140 ","End":"03:19.780","Text":"We have 4r divided by square root of 3^3,"},{"Start":"03:20.060 ","End":"03:23.745","Text":"and that\u0027s equal to 64,"},{"Start":"03:23.745 ","End":"03:29.880","Text":"4^3 is 64r^3 divided by the square root of 3^3."},{"Start":"03:29.880 ","End":"03:33.370","Text":"That\u0027s 3 times root 3."},{"Start":"03:33.570 ","End":"03:38.485","Text":"The fraction of the cell occupied is"},{"Start":"03:38.485 ","End":"03:43.505","Text":"the volume of the 2 atoms divided by the volume of the cell."},{"Start":"03:43.505 ","End":"03:48.525","Text":"If you work that out the r^3 cancels, of each case."},{"Start":"03:48.525 ","End":"03:51.110","Text":"We have 8 here, we have 64 here,"},{"Start":"03:51.110 ","End":"03:53.635","Text":"that gives us 8."},{"Start":"03:53.635 ","End":"03:56.610","Text":"We have 3 here and 3 here,"},{"Start":"03:56.610 ","End":"03:59.960","Text":"so that gives us in the end,"},{"Start":"03:59.960 ","End":"04:04.160","Text":"Pi square root of 3 divided by 8."},{"Start":"04:04.160 ","End":"04:07.990","Text":"If you work that out, that\u0027s 0.68."},{"Start":"04:07.990 ","End":"04:12.250","Text":"It\u0027s equivalent to 68%."},{"Start":"04:12.250 ","End":"04:14.645","Text":"Known as simple cube."},{"Start":"04:14.645 ","End":"04:18.660","Text":"We can work out that it\u0027s just 52%."},{"Start":"04:19.420 ","End":"04:22.235","Text":"Then hexagonal close-packed,"},{"Start":"04:22.235 ","End":"04:25.805","Text":"it turns out to be 74%."},{"Start":"04:25.805 ","End":"04:31.440","Text":"In a face-centered cube it\u0027s also 74%."},{"Start":"04:31.940 ","End":"04:37.675","Text":"We can see that by considering that there are 4 atoms in the unit cell."},{"Start":"04:37.675 ","End":"04:45.055","Text":"In this case, the face-centered cubic 4r is equal to root 2 times"},{"Start":"04:45.055 ","End":"04:50.120","Text":"l. That\u0027s because the diagonal"},{"Start":"04:50.120 ","End":"04:54.450","Text":"we\u0027re considering for the face-centered cube is the red one."},{"Start":"04:54.580 ","End":"04:58.400","Text":"One at this corner, one at this corner, one at the center."},{"Start":"04:58.400 ","End":"05:02.450","Text":"In this video, we learned how to calculate the volume"},{"Start":"05:02.450 ","End":"05:06.780","Text":"occupied in a close-packed structure."}],"ID":24816},{"Watched":false,"Name":"Relating Density of Metal to Atomic Radius","Duration":"4m 23s","ChapterTopicVideoID":23872,"CourseChapterTopicPlaylistID":112679,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.230 ","End":"00:02.340","Text":"In the previous video,"},{"Start":"00:02.340 ","End":"00:06.690","Text":"we calculated the volume occupied in a close packed structure."},{"Start":"00:06.690 ","End":"00:13.000","Text":"In this video, we\u0027ll relate the density of a metal to its atomic radius."},{"Start":"00:13.670 ","End":"00:19.720","Text":"We\u0027re going to relate the density of a metal to its atomic radius."},{"Start":"00:20.510 ","End":"00:25.780","Text":"We\u0027re going to do this by considering an example."},{"Start":"00:26.300 ","End":"00:31.214","Text":"Potassium crystallizes in the bcc structure."},{"Start":"00:31.214 ","End":"00:33.525","Text":"If the density is known,"},{"Start":"00:33.525 ","End":"00:37.005","Text":"0.806 grams per cubic centimeter,"},{"Start":"00:37.005 ","End":"00:40.450","Text":"what is metallic radius?"},{"Start":"00:41.260 ","End":"00:45.830","Text":"Recall that the density is equal to the mass over the volume."},{"Start":"00:45.830 ","End":"00:52.430","Text":"We need to calculate the mass of the unit cell and the volume of the unit cell."},{"Start":"00:52.430 ","End":"00:54.380","Text":"In the previous video,"},{"Start":"00:54.380 ","End":"00:59.480","Text":"we saw the volume of the unit cell in a bcc structure is given by V,"},{"Start":"00:59.480 ","End":"01:01.130","Text":"the volume, equal to l^3."},{"Start":"01:01.130 ","End":"01:05.165","Text":"L is a side of the unit cell,"},{"Start":"01:05.165 ","End":"01:09.575","Text":"and l is equal to 4r over the square root of 3."},{"Start":"01:09.575 ","End":"01:11.945","Text":"We need to cube that."},{"Start":"01:11.945 ","End":"01:19.270","Text":"That\u0027s equal to 64r^3 divided by 3 times the square root of 3."},{"Start":"01:19.270 ","End":"01:21.725","Text":"If we workout that numerically,"},{"Start":"01:21.725 ","End":"01:26.135","Text":"we get 12.317 times r^3."},{"Start":"01:26.135 ","End":"01:29.220","Text":"r is the radius of the atom."},{"Start":"01:29.260 ","End":"01:31.955","Text":"Now we need the mass."},{"Start":"01:31.955 ","End":"01:38.330","Text":"The mass of the potassium atom is 39.098 atomic mass units."},{"Start":"01:38.330 ","End":"01:47.495","Text":"Recall that each atomic mass unit is 1.6605 times 10^minus 24 grams."},{"Start":"01:47.495 ","End":"01:49.655","Text":"If we multiply this out,"},{"Start":"01:49.655 ","End":"01:52.190","Text":"of course, units cancels with unit,"},{"Start":"01:52.190 ","End":"01:54.425","Text":"and we get the answer in grams,"},{"Start":"01:54.425 ","End":"02:01.800","Text":"it\u0027s 6.4922 times 10^minus 23 grams."},{"Start":"02:02.450 ","End":"02:07.260","Text":"But we know that there are 2 atoms in a bcc unit cell,"},{"Start":"02:07.260 ","End":"02:10.900","Text":"so the mass of the unit cell is twice this number."},{"Start":"02:10.900 ","End":"02:18.460","Text":"That\u0027s 1.2984 times 10^minus 22 grams."},{"Start":"02:19.140 ","End":"02:23.155","Text":"Now we\u0027ve calculated the mass and the volume,"},{"Start":"02:23.155 ","End":"02:30.115","Text":"and it\u0027s given in the question that the density is 0.862 grams per cubic centimeter."},{"Start":"02:30.115 ","End":"02:36.100","Text":"We have the mass 1.2984 times 10^minus 22 grams"},{"Start":"02:36.100 ","End":"02:43.025","Text":"and the volume 12.317 times r^3 centimeters cubed."},{"Start":"02:43.025 ","End":"02:45.410","Text":"If we work out this numerically,"},{"Start":"02:45.410 ","End":"02:48.470","Text":"we get 1.0541 times"},{"Start":"02:48.470 ","End":"02:56.700","Text":"10^minus 23 grams divided by r^3 centimeters cubed."},{"Start":"02:57.940 ","End":"03:01.085","Text":"If we rearrange this equation,"},{"Start":"03:01.085 ","End":"03:09.635","Text":"0.862 equal to 1.0541 times 10^minus 23 divided by r^3,"},{"Start":"03:09.635 ","End":"03:13.250","Text":"we can work out what r^3 is."},{"Start":"03:13.250 ","End":"03:22.160","Text":"We get the r^3 is 1.0541 times 10^minus 23 grams divided by the density,"},{"Start":"03:22.160 ","End":"03:26.225","Text":"which is 0.862 grams per cubic centimeter."},{"Start":"03:26.225 ","End":"03:28.375","Text":"Work that out numerically,"},{"Start":"03:28.375 ","End":"03:37.020","Text":"we get it\u0027s 1.2229 times 10^minus 23 centimeters cubed."},{"Start":"03:37.580 ","End":"03:44.115","Text":"From that we can work out the value of r. If r^3 is equal to this."},{"Start":"03:44.115 ","End":"03:51.620","Text":"then r itself is 2.3039 times 10^minus 8 centimeters."},{"Start":"03:51.620 ","End":"03:55.430","Text":"We can work out what that is in meters because centimeters"},{"Start":"03:55.430 ","End":"04:03.990","Text":"10^minus 2 times 10^minus 8 is 10^minus 10 meters."},{"Start":"04:03.990 ","End":"04:07.850","Text":"It\u0027s usual to give these radii in picometers,"},{"Start":"04:07.850 ","End":"04:13.810","Text":"and we can work out that that is 230 picometers."},{"Start":"04:13.810 ","End":"04:22.860","Text":"In this video, we calculated the radius of an atom in a metal from the density."}],"ID":24815},{"Watched":false,"Name":"X-Ray Diffraction","Duration":"7m 36s","ChapterTopicVideoID":23874,"CourseChapterTopicPlaylistID":112679,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.440 ","End":"00:05.010","Text":"In previous videos we learned about crystal structure."},{"Start":"00:05.010 ","End":"00:08.040","Text":"In this video we\u0027ll talk about X-ray diffraction,"},{"Start":"00:08.040 ","End":"00:12.300","Text":"a method used to determine the structure of materials."},{"Start":"00:12.300 ","End":"00:15.600","Text":"We\u0027re going to talk about X-ray diffraction,"},{"Start":"00:15.600 ","End":"00:19.335","Text":"which is often abbreviated as XRD."},{"Start":"00:19.335 ","End":"00:24.630","Text":"Now the diffraction of X-rays by crystals were discovered by Maxwell von Laue,"},{"Start":"00:24.630 ","End":"00:27.870","Text":"and he got Nobel Prize for it in 1914."},{"Start":"00:27.870 ","End":"00:32.130","Text":"We\u0027re talking about a very old technique."},{"Start":"00:32.130 ","End":"00:34.650","Text":"Now, the Bragg father and son,"},{"Start":"00:34.650 ","End":"00:37.580","Text":"William Henry Bragg and William Lawrence Bragg,"},{"Start":"00:37.580 ","End":"00:40.580","Text":"who both got the Nobel Prize in 1915,"},{"Start":"00:40.580 ","End":"00:45.300","Text":"were the first to apply to determine the structure of crystals."},{"Start":"00:45.650 ","End":"00:48.000","Text":"Now, what are X-rays?"},{"Start":"00:48.000 ","End":"00:50.435","Text":"We learned to about this in a previous video."},{"Start":"00:50.435 ","End":"00:54.725","Text":"X-rays are a form of high-energy electromagnetic radiation."},{"Start":"00:54.725 ","End":"01:00.155","Text":"The wavelength Lambda is in the range of 10 picometers to 10 nanometers."},{"Start":"01:00.155 ","End":"01:05.435","Text":"That makes them suitable for measuring the distance between layers of atoms in a crystal."},{"Start":"01:05.435 ","End":"01:10.025","Text":"That distance is typically 100 picometers."},{"Start":"01:10.025 ","End":"01:13.760","Text":"Now when the X-rays irradiate crystals,"},{"Start":"01:13.760 ","End":"01:18.669","Text":"they are scattered or diffracted by the electrons in the atoms."},{"Start":"01:18.669 ","End":"01:20.750","Text":"Now, most directions,"},{"Start":"01:20.750 ","End":"01:24.980","Text":"there\u0027s destructive interference between the scattered waves."},{"Start":"01:24.980 ","End":"01:30.865","Text":"However, in 1 particular direction there is constructive interference."},{"Start":"01:30.865 ","End":"01:33.405","Text":"That\u0027s when the Bragg law holds."},{"Start":"01:33.405 ","End":"01:36.720","Text":"Now we\u0027re going to learn about the Bragg law."},{"Start":"01:37.300 ","End":"01:41.915","Text":"Here we have the layers within the crystal."},{"Start":"01:41.915 ","End":"01:45.590","Text":"The distance between 2 layers is d."},{"Start":"01:45.590 ","End":"01:50.990","Text":"Now the radiation comes down and the angle of incidence,"},{"Start":"01:50.990 ","End":"01:57.235","Text":"the angle which the radiation meets the layers is Theta."},{"Start":"01:57.235 ","End":"02:03.005","Text":"Now, the X-rays are coherent in the sense"},{"Start":"02:03.005 ","End":"02:09.005","Text":"that the Lambda here has the same phase as the Lambda in this ray."},{"Start":"02:09.005 ","End":"02:14.300","Text":"We have the same Lambda and they are in phase."},{"Start":"02:14.300 ","End":"02:18.020","Text":"In phase means 1 wave,"},{"Start":"02:18.020 ","End":"02:23.330","Text":"and the other wave has exactly the same maxima and minima."},{"Start":"02:23.330 ","End":"02:25.910","Text":"They\u0027re in phase."},{"Start":"02:25.910 ","End":"02:30.275","Text":"Top ray meets the top layer,"},{"Start":"02:30.275 ","End":"02:33.665","Text":"and the second 1 meets the second layer."},{"Start":"02:33.665 ","End":"02:38.665","Text":"Here it summarize the wave front propagates towards the crystal."},{"Start":"02:38.665 ","End":"02:42.710","Text":"The incident angle is Theta and the distance in planes is"},{"Start":"02:42.710 ","End":"02:47.930","Text":"d. The first ray"},{"Start":"02:47.930 ","End":"02:52.380","Text":"is distracted from the top layer and the second ray from the second layer."},{"Start":"02:53.930 ","End":"02:58.770","Text":"Now, we can find that the second ray travels"},{"Start":"02:58.770 ","End":"03:03.500","Text":"2d sine Theta more than the first ray. How is that?"},{"Start":"03:03.500 ","End":"03:09.900","Text":"We need to look at the length of this blue section."},{"Start":"03:09.900 ","End":"03:11.595","Text":"Now if this is Theta,"},{"Start":"03:11.595 ","End":"03:14.250","Text":"this angle here is also Theta."},{"Start":"03:14.250 ","End":"03:22.380","Text":"This is d. We can easily see that the length of this section is d sine Theta."},{"Start":"03:24.140 ","End":"03:26.745","Text":"We have 2 of them,"},{"Start":"03:26.745 ","End":"03:30.075","Text":"so it\u0027s 2d sine Theta."},{"Start":"03:30.075 ","End":"03:35.805","Text":"The second ray travels 2d sine Theta more than the first ray."},{"Start":"03:35.805 ","End":"03:42.050","Text":"In order for the 2 diffracted rays to be in phase and to interfere constructively,"},{"Start":"03:42.050 ","End":"03:48.230","Text":"we need a whole number of Lambdas to be equal to 2d sine Theta."},{"Start":"03:48.230 ","End":"03:56.220","Text":"Then we get the situation and diffracted rays that they are instead they\u0027re in phase."},{"Start":"03:58.540 ","End":"04:02.940","Text":"This is called Bragg\u0027s law."},{"Start":"04:03.800 ","End":"04:07.590","Text":"If we know Lambda and Theta,"},{"Start":"04:07.590 ","End":"04:12.920","Text":"we can calculate d. Of course n can take different values,"},{"Start":"04:12.920 ","End":"04:14.705","Text":"1, 2, 3 and so on."},{"Start":"04:14.705 ","End":"04:17.870","Text":"Now, there are 2 main ways"},{"Start":"04:17.870 ","End":"04:23.345","Text":"this equation is used and we\u0027re going to describe them in brief."},{"Start":"04:23.345 ","End":"04:26.480","Text":"I must emphasize this is in brief because these are"},{"Start":"04:26.480 ","End":"04:30.860","Text":"very complicated techniques and a great deal of"},{"Start":"04:30.860 ","End":"04:35.645","Text":"mathematics is required in order to get the solutions and of course,"},{"Start":"04:35.645 ","End":"04:37.955","Text":"a great deal of computer power."},{"Start":"04:37.955 ","End":"04:43.520","Text":"First 1 is a single crystal diffraction, often called crystallography."},{"Start":"04:43.520 ","End":"04:45.410","Text":"First, you have to grow a crystal,"},{"Start":"04:45.410 ","End":"04:49.730","Text":"a tiny single crystal is grown and this can be very,"},{"Start":"04:49.730 ","End":"04:52.400","Text":"very difficult for biological molecules."},{"Start":"04:52.400 ","End":"04:57.725","Text":"Sometimes it takes many years and there are experts in growing crystals."},{"Start":"04:57.725 ","End":"05:01.490","Text":"The crystals are exposed to a beam of white radiation."},{"Start":"05:01.490 ","End":"05:06.005","Text":"White radiation means that we have many different values of Lambda."},{"Start":"05:06.005 ","End":"05:09.780","Text":"All the colors are mixed up so it looks white."},{"Start":"05:09.980 ","End":"05:13.795","Text":"The crystal detector are rotated."},{"Start":"05:13.795 ","End":"05:16.610","Text":"Now each set of lattice planes,"},{"Start":"05:16.610 ","End":"05:20.525","Text":"as we rotate we get different sets of lattice planes."},{"Start":"05:20.525 ","End":"05:25.920","Text":"Each 1 will choose the appropriate Lambda to give Bragg diffraction."},{"Start":"05:26.960 ","End":"05:29.840","Text":"Now, we use computer analysis of"},{"Start":"05:29.840 ","End":"05:34.475","Text":"diffraction pattern and this gives the position of all the atoms,"},{"Start":"05:34.475 ","End":"05:35.750","Text":"the bond lengths,"},{"Start":"05:35.750 ","End":"05:37.910","Text":"and the bond angles."},{"Start":"05:37.910 ","End":"05:41.675","Text":"Nowadays, there are databases available of"},{"Start":"05:41.675 ","End":"05:47.405","Text":"all the molecules that have been analyzed in this way."},{"Start":"05:47.405 ","End":"05:53.310","Text":"Here\u0027s a picture of a diffraction spectrum."},{"Start":"05:53.780 ","End":"05:57.555","Text":"This is the diffraction pattern from ZnS,"},{"Start":"05:57.555 ","End":"06:04.670","Text":"zinc sulfide recorded by Laue on a photographic plate in the early 20th century."},{"Start":"06:04.760 ","End":"06:11.125","Text":"Now, the second method I\u0027m going to talk about is the powder method."},{"Start":"06:11.125 ","End":"06:19.000","Text":"Now the powder method is used routinely in material science to characterize materials."},{"Start":"06:19.000 ","End":"06:22.870","Text":"In this method, monochromatic radiation,"},{"Start":"06:22.870 ","End":"06:25.720","Text":"that means radiation with a single value of Lambda,"},{"Start":"06:25.720 ","End":"06:31.209","Text":"is used to bombard a finely powdered of fine grain sample of the material."},{"Start":"06:31.209 ","End":"06:33.205","Text":"You put the material on a support."},{"Start":"06:33.205 ","End":"06:37.430","Text":"We no longer have a single crystal, but a powder."},{"Start":"06:37.430 ","End":"06:41.540","Text":"The angle Theta is varied and planes in"},{"Start":"06:41.540 ","End":"06:46.135","Text":"the crystal with the appropriate orientation give Bragg diffraction."},{"Start":"06:46.135 ","End":"06:50.805","Text":"Here\u0027s an example of an XRD spectrum."},{"Start":"06:50.805 ","End":"06:54.545","Text":"It\u0027s for asparagine, which is a nucleic acid"},{"Start":"06:54.545 ","End":"06:59.970","Text":"crystallized in water ethanol system. Here\u0027s a spectrum."},{"Start":"06:59.970 ","End":"07:02.150","Text":"We can analyze the various peaks."},{"Start":"07:02.150 ","End":"07:06.540","Text":"We see there\u0027s a very high peak here between 15 and 20."},{"Start":"07:06.770 ","End":"07:12.640","Text":"This allows us to get information about our material."},{"Start":"07:12.640 ","End":"07:17.990","Text":"It\u0027s not as detailed as what is obtained from single crystals,"},{"Start":"07:17.990 ","End":"07:20.974","Text":"but it\u0027s very, very useful and used routinely."},{"Start":"07:20.974 ","End":"07:25.470","Text":"We\u0027re plotting intensity versus 2 Theta."},{"Start":"07:26.960 ","End":"07:35.790","Text":"This video, we gave a glimpse into the vast field of XR diffraction."}],"ID":24817}],"Thumbnail":null,"ID":112679},{"Name":"Ionic Crystals","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Ionic Crystals Radius Ratio Rules","Duration":"8m ","ChapterTopicVideoID":23878,"CourseChapterTopicPlaylistID":113861,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.010 ","End":"00:03.180","Text":"In previous videos,"},{"Start":"00:03.180 ","End":"00:05.565","Text":"we talked about close-packed structures."},{"Start":"00:05.565 ","End":"00:08.070","Text":"In this video, we\u0027ll learn about the rules that"},{"Start":"00:08.070 ","End":"00:11.680","Text":"determine the structure of ionic crystals."},{"Start":"00:11.750 ","End":"00:16.005","Text":"We\u0027re going to learn about ionic crystals."},{"Start":"00:16.005 ","End":"00:21.945","Text":"Now what\u0027s the difference between atomic crystals and ionic crystals?"},{"Start":"00:21.945 ","End":"00:25.140","Text":"The first thing is that the ions are of opposite charges,"},{"Start":"00:25.140 ","End":"00:29.925","Text":"we have like charges that repel and opposite charges that attract."},{"Start":"00:29.925 ","End":"00:32.340","Text":"We have to take that into account."},{"Start":"00:32.340 ","End":"00:36.240","Text":"The second thing is that the ions have different sizes."},{"Start":"00:36.240 ","End":"00:38.625","Text":"Usually anions are greater,"},{"Start":"00:38.625 ","End":"00:42.630","Text":"are larger than cations."},{"Start":"00:42.630 ","End":"00:46.460","Text":"Now some ionic crystals are modifications of"},{"Start":"00:46.460 ","End":"00:51.510","Text":"closed-packed arrangements and these are the ones we\u0027re going to talk about here."},{"Start":"00:51.980 ","End":"00:58.279","Text":"In this case, the larger anions form approximately close-packed structures,"},{"Start":"00:58.279 ","End":"01:04.195","Text":"and the smaller cations fill the holes in the closed-packed structure."},{"Start":"01:04.195 ","End":"01:09.625","Text":"Let\u0027s talk first about face-centered cubic arrangement."},{"Start":"01:09.625 ","End":"01:11.555","Text":"Now as we learned before,"},{"Start":"01:11.555 ","End":"01:14.795","Text":"there are 3 types of holes, there\u0027s trigonal,"},{"Start":"01:14.795 ","End":"01:21.420","Text":"tetrahedral, and octahedral in face-centered cubic arrangement."},{"Start":"01:21.520 ","End":"01:27.575","Text":"Here\u0027s a picture. Here\u0027s the trigonal hole between these 3 atoms."},{"Start":"01:27.575 ","End":"01:31.880","Text":"Here\u0027s a tetrahedral hole between these 4 atoms,"},{"Start":"01:31.880 ","End":"01:33.215","Text":"1, 2, 3,"},{"Start":"01:33.215 ","End":"01:34.910","Text":"and 1 above,"},{"Start":"01:34.910 ","End":"01:40.500","Text":"that forms a tetrahedral and the hole somewhere in the middle,"},{"Start":"01:40.500 ","End":"01:42.630","Text":"and an octahedron,"},{"Start":"01:42.630 ","End":"01:45.225","Text":"in which we have 6 atoms,"},{"Start":"01:45.225 ","End":"01:48.200","Text":"1 at the top, 4 in the center, and 1 at the bottom."},{"Start":"01:48.200 ","End":"01:49.490","Text":"The hole is in the middle,"},{"Start":"01:49.490 ","End":"01:51.750","Text":"it\u0027s an octahedral hole."},{"Start":"01:52.150 ","End":"01:59.590","Text":"Let\u0027s calculate the radius of a cation that fits into an octahedral hole."},{"Start":"01:59.590 ","End":"02:03.785","Text":"Here is the central plane of this octahedron,"},{"Start":"02:03.785 ","End":"02:06.815","Text":"these red anions,"},{"Start":"02:06.815 ","End":"02:10.280","Text":"and here\u0027s a small cation in the center."},{"Start":"02:10.280 ","End":"02:17.135","Text":"Let\u0027s call the radius of the anion r minus and the radius of the cation r plus."},{"Start":"02:17.135 ","End":"02:21.530","Text":"Now we want to calculate the length of this diagonal"},{"Start":"02:21.530 ","End":"02:27.559","Text":"d. We have one side that has length 2r minus,"},{"Start":"02:27.559 ","End":"02:30.770","Text":"another side of the same length 2r minus."},{"Start":"02:30.770 ","End":"02:37.200","Text":"The length of d is the square root of 4r minus squared plus 4r minus squared,"},{"Start":"02:37.200 ","End":"02:40.050","Text":"and that gives us 2 root 2r minus."},{"Start":"02:40.050 ","End":"02:41.700","Text":"But in addition,"},{"Start":"02:41.700 ","End":"02:44.690","Text":"let\u0027s look at the length of the diagonal."},{"Start":"02:44.690 ","End":"02:49.790","Text":"It\u0027s like the length of the anion r minus,"},{"Start":"02:49.790 ","End":"02:51.695","Text":"the radius of the anion,"},{"Start":"02:51.695 ","End":"02:54.630","Text":"twice the radius of the cation,"},{"Start":"02:55.090 ","End":"02:59.940","Text":"plus the radius of the anion."},{"Start":"03:00.070 ","End":"03:04.920","Text":"That gives us 2r minus, plus 2r plus."},{"Start":"03:04.920 ","End":"03:06.975","Text":"If we equate these two,"},{"Start":"03:06.975 ","End":"03:13.370","Text":"we get r plus is equal to the square root of 2 minus 1 times r minus."},{"Start":"03:13.370 ","End":"03:18.040","Text":"The ratio, what we call the radius ratio,"},{"Start":"03:18.740 ","End":"03:24.070","Text":"is equal to 0.414."},{"Start":"03:24.560 ","End":"03:28.510","Text":"Now if we did a similar calculation for a tetrahedral hole,"},{"Start":"03:28.510 ","End":"03:37.160","Text":"we discover that the ratio is 0.225 and for a trigonal hole, 0.155."},{"Start":"03:40.290 ","End":"03:44.244","Text":"Now let\u0027s look at another possible arrangement."},{"Start":"03:44.244 ","End":"03:46.190","Text":"A simple cubic arrangement,"},{"Start":"03:46.190 ","End":"03:49.420","Text":"is not exactly closed-packed if you remember,"},{"Start":"03:49.420 ","End":"03:54.335","Text":"it\u0027s a little less packed than a closed-packed arrangement."},{"Start":"03:54.335 ","End":"04:01.740","Text":"Here\u0027s our cube and we have 8 atoms in the cube."},{"Start":"04:01.740 ","End":"04:09.960","Text":"Let\u0027s consider the atoms that we have here and here,"},{"Start":"04:09.960 ","End":"04:15.150","Text":"and the atoms, what have here and here."},{"Start":"04:15.150 ","End":"04:17.760","Text":"Here are the 4 atoms in red."},{"Start":"04:17.760 ","End":"04:20.735","Text":"The 4 anions."},{"Start":"04:20.735 ","End":"04:27.260","Text":"We\u0027re interested in the little cation that fits into this central hole."},{"Start":"04:28.170 ","End":"04:31.780","Text":"Now if we look at the side of a cube,"},{"Start":"04:31.780 ","End":"04:36.265","Text":"we can see that the side of the cube is this distance,"},{"Start":"04:36.265 ","End":"04:39.590","Text":"which is 2r minus."},{"Start":"04:39.620 ","End":"04:48.129","Text":"The length of the red diagonal is square root of 4r minus squared plus 4r minus squared,"},{"Start":"04:48.129 ","End":"04:53.979","Text":"because the length here is 2r minus and the length here is 2r minus,"},{"Start":"04:53.979 ","End":"04:58.255","Text":"and that gives us 2 times the square root of r minus."},{"Start":"04:58.255 ","End":"05:01.295","Text":"That\u0027s the length of the red diagonal."},{"Start":"05:01.295 ","End":"05:04.520","Text":"d, which is this length here,"},{"Start":"05:04.520 ","End":"05:06.470","Text":"it\u0027s equivalent to this one here,"},{"Start":"05:06.470 ","End":"05:10.670","Text":"this is d, and in this picture, this is also d,"},{"Start":"05:10.670 ","End":"05:15.604","Text":"is equal to square root of 8r minus squared,"},{"Start":"05:15.604 ","End":"05:17.695","Text":"that\u0027s the red diagonal,"},{"Start":"05:17.695 ","End":"05:23.070","Text":"plus this side which has a length 2r minus."},{"Start":"05:23.070 ","End":"05:27.600","Text":"It\u0027s square root of 8r minus squared plus 4r minus squared."},{"Start":"05:27.600 ","End":"05:30.405","Text":"That gives us 12r minus squared,"},{"Start":"05:30.405 ","End":"05:36.320","Text":"and the square root of that is 2 root 3r minus."},{"Start":"05:36.320 ","End":"05:40.540","Text":"But in addition, it has the length of the anion"},{"Start":"05:40.540 ","End":"05:47.695","Text":"twice the radius of the cation and plus the radius of the anion."},{"Start":"05:47.695 ","End":"05:52.205","Text":"That gives us a total of 2r minus plus 2r plus."},{"Start":"05:52.205 ","End":"05:55.900","Text":"We can equate these two and get the r plus is"},{"Start":"05:55.900 ","End":"06:00.440","Text":"equal to the square root of 3 minus 1 times r minus,"},{"Start":"06:00.440 ","End":"06:04.320","Text":"and that\u0027s 0.732r minus."},{"Start":"06:04.320 ","End":"06:10.930","Text":"The ratio r plus divided by r minus is 0.732."},{"Start":"06:14.570 ","End":"06:19.129","Text":"Now we can form the radius ratio of rules."},{"Start":"06:19.129 ","End":"06:20.660","Text":"How do we do this?"},{"Start":"06:20.660 ","End":"06:27.985","Text":"We first note that the cation occupies a hole slightly smaller than its actual size."},{"Start":"06:27.985 ","End":"06:32.345","Text":"The reason for this is that it pushes the anions apart,"},{"Start":"06:32.345 ","End":"06:35.420","Text":"minimizing repulsion between them."},{"Start":"06:35.420 ","End":"06:38.990","Text":"In addition, the cation and the anion are in"},{"Start":"06:38.990 ","End":"06:43.330","Text":"contact and that maximizes attraction between them."},{"Start":"06:43.330 ","End":"06:45.210","Text":"Here are the 3 rules."},{"Start":"06:45.210 ","End":"06:54.580","Text":"The first one is that if r plus divided by r minus lies between 0.225,"},{"Start":"06:54.730 ","End":"06:59.390","Text":"which remember was for a tetrahedral hole,"},{"Start":"06:59.390 ","End":"07:05.495","Text":"and the result 0.414 for an octahedral hole,"},{"Start":"07:05.495 ","End":"07:08.585","Text":"if it lies between these two numbers,"},{"Start":"07:08.585 ","End":"07:13.500","Text":"the cation occupies a tetrahedral hole."},{"Start":"07:13.720 ","End":"07:20.855","Text":"If it\u0027s between the result that we got for a tetrahedral hole, 0.414,"},{"Start":"07:20.855 ","End":"07:24.130","Text":"and that we got for cubic hole,"},{"Start":"07:24.130 ","End":"07:31.840","Text":"0.732, then we say that the cation occupies an octahedral hole."},{"Start":"07:31.840 ","End":"07:35.925","Text":"If the ratio is greater than 0.732,"},{"Start":"07:35.925 ","End":"07:39.425","Text":"the cation occupies a cubic hole."},{"Start":"07:39.425 ","End":"07:43.040","Text":"Now these are very approximate rules."},{"Start":"07:43.040 ","End":"07:46.340","Text":"They often hold, but occasionally they don\u0027t."},{"Start":"07:46.340 ","End":"07:51.034","Text":"In the next video, we\u0027ll show some examples of where they hold."},{"Start":"07:51.034 ","End":"07:53.480","Text":"In this video, we learned about"},{"Start":"07:53.480 ","End":"07:58.470","Text":"ionic crystals and the rules that determine the structure."}],"ID":24821},{"Watched":false,"Name":"Examples of Ionic Crystals","Duration":"8m 29s","ChapterTopicVideoID":23877,"CourseChapterTopicPlaylistID":113861,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.950 ","End":"00:03.090","Text":"In the previous video,"},{"Start":"00:03.090 ","End":"00:06.750","Text":"we learned about ionic crystals and the radius ratio rules."},{"Start":"00:06.750 ","End":"00:11.619","Text":"In this video, we\u0027ll show how to use these rules to find the crystal structure."},{"Start":"00:11.660 ","End":"00:15.495","Text":"The first example is NaCl."},{"Start":"00:15.495 ","End":"00:20.730","Text":"Here the ratio of the radii is 0.55,"},{"Start":"00:20.730 ","End":"00:26.130","Text":"which is intermediate between octahedral and cubic."},{"Start":"00:26.130 ","End":"00:33.570","Text":"The Na plus cations squeeze into the octahedral holes"},{"Start":"00:33.570 ","End":"00:42.285","Text":"which are smaller than the ratio in our fcc or ccp array of Cl minus anions."},{"Start":"00:42.285 ","End":"00:44.520","Text":"Here\u0027s the familiar picture,"},{"Start":"00:44.520 ","End":"00:49.085","Text":"the Cl minus anions form a face-centered cube,"},{"Start":"00:49.085 ","End":"00:54.065","Text":"and Na plus cations fill the octahedral holes."},{"Start":"00:54.065 ","End":"00:56.890","Text":"You can see it clearly with the central one."},{"Start":"00:56.890 ","End":"01:03.305","Text":"Now the number of Na plus cations in unit cell is 12 times 1/4"},{"Start":"01:03.305 ","End":"01:09.845","Text":"because we have 12 Na plus ions on the edges,"},{"Start":"01:09.845 ","End":"01:13.145","Text":"and each edge contributes 1/4,"},{"Start":"01:13.145 ","End":"01:16.400","Text":"and 1 in the center giving a total of 4."},{"Start":"01:16.400 ","End":"01:21.560","Text":"A number of Cl minus anions is 8 times an 1/8 because we\u0027ve 8 at"},{"Start":"01:21.560 ","End":"01:27.915","Text":"the corners and 6 times 1/2 because we have 6 on the faces,"},{"Start":"01:27.915 ","End":"01:30.435","Text":"that gives a total of 4."},{"Start":"01:30.435 ","End":"01:32.310","Text":"The ratio is 4 to 4,"},{"Start":"01:32.310 ","End":"01:34.155","Text":"which is equivalent to 1 to 1,"},{"Start":"01:34.155 ","End":"01:38.600","Text":"and that\u0027s an agreement with the formula NaCl."},{"Start":"01:38.600 ","End":"01:42.680","Text":"The coordination numbers, that\u0027s the number of nearest neighbors"},{"Start":"01:42.680 ","End":"01:46.295","Text":"must be the same for Na plus and Cl minus,"},{"Start":"01:46.295 ","End":"01:53.495","Text":"because they\u0027re in a 1 to 1 ratio and it works out to be 6 for both Na plus and Cl minus."},{"Start":"01:53.495 ","End":"02:00.895","Text":"You can see that from the green one center has 6 Cl minus anions around it."},{"Start":"02:00.895 ","End":"02:04.930","Text":"The next example, CsCl."},{"Start":"02:04.930 ","End":"02:09.020","Text":"Here the ratio is 0.934,"},{"Start":"02:09.020 ","End":"02:14.370","Text":"which is larger than cubic, cubic is 0.732."},{"Start":"02:16.730 ","End":"02:25.505","Text":"The cesium plus cations occupy cubic holes in the simple cubic array of Cl minus anions."},{"Start":"02:25.505 ","End":"02:27.725","Text":"Here\u0027s a picture,"},{"Start":"02:27.725 ","End":"02:33.615","Text":"we have 8 Cl minus anions forming a cube,"},{"Start":"02:33.615 ","End":"02:38.095","Text":"and we have 1 Cs plus cation in the center."},{"Start":"02:38.095 ","End":"02:43.720","Text":"Now the number of cations in the unit cell is just 1, is red 1,"},{"Start":"02:43.720 ","End":"02:47.255","Text":"and number of Cl minus anions is 8,"},{"Start":"02:47.255 ","End":"02:49.050","Text":"we have 1, 2, 3, 4,"},{"Start":"02:49.050 ","End":"02:50.445","Text":"5, 6, 7, 8,"},{"Start":"02:50.445 ","End":"02:53.130","Text":"and each contributes an 1/8 because of corners,"},{"Start":"02:53.130 ","End":"02:54.525","Text":"that\u0027s a total of 1."},{"Start":"02:54.525 ","End":"02:59.810","Text":"The ratio is 1 to 1 in agreement with the formula CsCl."},{"Start":"02:59.810 ","End":"03:06.290","Text":"The coordination numbers which must be the same for both ions is our 8."},{"Start":"03:06.290 ","End":"03:13.210","Text":"You can see that Cs plus in the center surrounded by 8 Cl minus anions,"},{"Start":"03:13.210 ","End":"03:14.940","Text":"so it\u0027s 8,"},{"Start":"03:14.940 ","End":"03:17.680","Text":"and the ratio of 1 to 1."},{"Start":"03:17.950 ","End":"03:22.640","Text":"Now we\u0027re going to talk about more complicated crystal structures,"},{"Start":"03:22.640 ","End":"03:28.240","Text":"some of which are in agreement with the rules and some of which deviate from them."},{"Start":"03:28.240 ","End":"03:30.825","Text":"We have 3 structures here,"},{"Start":"03:30.825 ","End":"03:34.635","Text":"zinc blende and ZnS is typical of it."},{"Start":"03:34.635 ","End":"03:39.900","Text":"Fluorite, CaF_2, rutile, TiO_2."},{"Start":"03:39.900 ","End":"03:43.745","Text":"We\u0027re going to begin with zinc blende structure."},{"Start":"03:43.745 ","End":"03:49.835","Text":"Zn2 plus is in mauve and the S2 minus is in yellow."},{"Start":"03:49.835 ","End":"03:53.220","Text":"The ratio of 0.408,"},{"Start":"03:53.830 ","End":"03:58.960","Text":"which is intermediate between tetrahedral and octahedral."},{"Start":"03:58.960 ","End":"04:04.355","Text":"The S2 minus anions form a face-centered cube and"},{"Start":"04:04.355 ","End":"04:09.520","Text":"Zn2 plus cations fill half the tetrahedral holes."},{"Start":"04:09.520 ","End":"04:12.285","Text":"They squeeze into tetrahedral holes,"},{"Start":"04:12.285 ","End":"04:17.310","Text":"but there are only half as many Zn2 pluses as there are holes,"},{"Start":"04:17.310 ","End":"04:19.355","Text":"so only fills half of them."},{"Start":"04:19.355 ","End":"04:22.040","Text":"You can see that clearly in this picture."},{"Start":"04:22.040 ","End":"04:24.110","Text":"Here is a tetrahedral hole,"},{"Start":"04:24.110 ","End":"04:29.025","Text":"you can see it has 4 S2 minus anions around it."},{"Start":"04:29.025 ","End":"04:33.960","Text":"Now the number of Zn2 plus ions in the unit cell is"},{"Start":"04:33.960 ","End":"04:39.675","Text":"4 and the number of S2 minus ions in the unit cell is also 4."},{"Start":"04:39.675 ","End":"04:41.580","Text":"The Zn2 plus,"},{"Start":"04:41.580 ","End":"04:43.161","Text":"we can see clearly 1,"},{"Start":"04:43.161 ","End":"04:44.625","Text":"2, 3, 4,"},{"Start":"04:44.625 ","End":"04:50.045","Text":"and the S2 minus is face-centered cubic cell,"},{"Start":"04:50.045 ","End":"04:52.795","Text":"so it\u0027s 4 as we saw before."},{"Start":"04:52.795 ","End":"04:54.990","Text":"The ratio is 4 to 4,"},{"Start":"04:54.990 ","End":"04:56.280","Text":"that\u0027s 1 to 1."},{"Start":"04:56.280 ","End":"04:59.370","Text":"That\u0027s in agreement with ZnS,"},{"Start":"04:59.370 ","End":"05:01.995","Text":"which is the ratio of 1 to 1."},{"Start":"05:01.995 ","End":"05:06.060","Text":"The coordination number is 4 for each ion."},{"Start":"05:06.060 ","End":"05:09.300","Text":"You can see it clearly here,"},{"Start":"05:09.300 ","End":"05:16.300","Text":"Zn2 plus has 4 S2 minus ions around it."},{"Start":"05:16.300 ","End":"05:20.180","Text":"Let\u0027s go on to the fluorite structure."},{"Start":"05:20.180 ","End":"05:24.050","Text":"Here the ratio is 0.752,"},{"Start":"05:24.050 ","End":"05:30.710","Text":"which is slightly greater than octahedral, octahedral is 0.732."},{"Start":"05:30.710 ","End":"05:34.730","Text":"We expect it to fit into cubic holes,"},{"Start":"05:34.730 ","End":"05:38.610","Text":"but in fact it fits into tetrahedral holes,"},{"Start":"05:38.610 ","End":"05:41.480","Text":"so it\u0027s not an agreement with the rules."},{"Start":"05:41.480 ","End":"05:46.025","Text":"We get a face-centered cube of Ca2 plus,"},{"Start":"05:46.025 ","End":"05:48.520","Text":"there is a picture there in gray."},{"Start":"05:48.520 ","End":"05:52.265","Text":"The F minus anions,"},{"Start":"05:52.265 ","End":"05:57.635","Text":"which are in green fill all the tetrahedral holes, here."},{"Start":"05:57.635 ","End":"05:59.630","Text":"They\u0027re filling all the tetrahedral holes."},{"Start":"05:59.630 ","End":"06:02.435","Text":"You can clearly see it\u0027s a tetrahedral."},{"Start":"06:02.435 ","End":"06:09.865","Text":"Now the number of calcium ions in the unit cell is 4."},{"Start":"06:09.865 ","End":"06:14.585","Text":"That\u0027s because the calcium 2 plus just are either face-centered cube,"},{"Start":"06:14.585 ","End":"06:16.760","Text":"so it\u0027s 4 as before."},{"Start":"06:16.760 ","End":"06:21.985","Text":"Number F minus ions in the unit cell is 8, we can count them."},{"Start":"06:21.985 ","End":"06:26.000","Text":"They in green, we can count that there are 8 of them."},{"Start":"06:26.000 ","End":"06:28.490","Text":"The ratio is 4 to 8,"},{"Start":"06:28.490 ","End":"06:29.885","Text":"that\u0027s 1 to 2,"},{"Start":"06:29.885 ","End":"06:35.560","Text":"and that\u0027s in agreement with a formula CaF_2, 1 to 2."},{"Start":"06:35.560 ","End":"06:38.370","Text":"We look at the coordination numbers,"},{"Start":"06:38.370 ","End":"06:41.465","Text":"they also have to be in agreement of 1 to 2,"},{"Start":"06:41.465 ","End":"06:44.120","Text":"so Ca2 plus it\u0027s 8,"},{"Start":"06:44.120 ","End":"06:46.495","Text":"and for F minus is 4."},{"Start":"06:46.495 ","End":"06:51.680","Text":"If we look at one of the green F minus ions,"},{"Start":"06:51.680 ","End":"06:57.865","Text":"we can see clearly that there are 4 gray ions around it."},{"Start":"06:57.865 ","End":"07:01.320","Text":"The last structure is TiO_2,"},{"Start":"07:01.320 ","End":"07:04.260","Text":"that\u0027s called the rutile structure."},{"Start":"07:04.260 ","End":"07:10.680","Text":"Ti4 plus is in gray and O2 minus in red, this picture."},{"Start":"07:10.680 ","End":"07:14.520","Text":"Here, it\u0027s nothing to do with a cubic structure,"},{"Start":"07:14.520 ","End":"07:16.725","Text":"in fact, it\u0027s not cubic at all."},{"Start":"07:16.725 ","End":"07:23.240","Text":"The number of Ti4 plus ions in the unit cell is equal to 8 times an 1/8,"},{"Start":"07:23.240 ","End":"07:25.685","Text":"8 corners and 1 in the center."},{"Start":"07:25.685 ","End":"07:29.790","Text":"Here we have 8 of the corners and 1 in the center."},{"Start":"07:29.790 ","End":"07:34.050","Text":"That\u0027s 8 times an 1/8 plus 1 giving a total of 2."},{"Start":"07:34.050 ","End":"07:40.346","Text":"The number O2 minus ions in unit cell is 4 times 1/2, 1, 2, 3,"},{"Start":"07:40.346 ","End":"07:41.925","Text":"4 on the faces,"},{"Start":"07:41.925 ","End":"07:43.935","Text":"and another 2,"},{"Start":"07:43.935 ","End":"07:46.635","Text":"so that\u0027s 4 times 1/2 plus 2,"},{"Start":"07:46.635 ","End":"07:48.165","Text":"giving a total of 4."},{"Start":"07:48.165 ","End":"07:50.280","Text":"The ratio is 2 to 4,"},{"Start":"07:50.280 ","End":"07:51.660","Text":"that\u0027s 1 to 2,"},{"Start":"07:51.660 ","End":"07:54.045","Text":"in agreement with TiO2,"},{"Start":"07:54.045 ","End":"07:56.530","Text":"which is the ratio of 1 to 2."},{"Start":"07:56.530 ","End":"07:59.375","Text":"Now let\u0027s look at the coordination number,"},{"Start":"07:59.375 ","End":"08:02.675","Text":"the coordination of Ti4 plus is 6,"},{"Start":"08:02.675 ","End":"08:06.000","Text":"and for O2 minus is 3."},{"Start":"08:06.130 ","End":"08:08.420","Text":"If we look in the center,"},{"Start":"08:08.420 ","End":"08:10.068","Text":"we can see 1, 2,"},{"Start":"08:10.068 ","End":"08:11.638","Text":"3, 4, 5,"},{"Start":"08:11.638 ","End":"08:13.775","Text":"6 surrounding it,"},{"Start":"08:13.775 ","End":"08:17.090","Text":"so that\u0027s clearly that the coordination number of"},{"Start":"08:17.090 ","End":"08:21.260","Text":"Ti4 plus is 6 and we can count what it is for O2 minus,"},{"Start":"08:21.260 ","End":"08:24.290","Text":"it works out to 3 and the ratio is correct."},{"Start":"08:24.290 ","End":"08:29.100","Text":"In this video we talked about examples of crystal structure."}],"ID":24820},{"Watched":false,"Name":"Lattice Energy Born-Fajer-Haber Cycle","Duration":"7m 32s","ChapterTopicVideoID":23879,"CourseChapterTopicPlaylistID":113861,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.095","Text":"In previous videos we talked about ionic crystals."},{"Start":"00:04.095 ","End":"00:07.500","Text":"In this video we\u0027ll discuss an indirect way to"},{"Start":"00:07.500 ","End":"00:11.969","Text":"calculate the lattice energy of such a crystal."},{"Start":"00:11.969 ","End":"00:16.665","Text":"Let\u0027s recall what we mean by lattice energy."},{"Start":"00:16.665 ","End":"00:20.880","Text":"The lattice energy Delta H lattice of a crystal is"},{"Start":"00:20.880 ","End":"00:25.620","Text":"the enthalpy change when a crystal is formed from its ions in the gas phase."},{"Start":"00:25.620 ","End":"00:27.540","Text":"Now this is just a definition."},{"Start":"00:27.540 ","End":"00:33.600","Text":"You don\u0027t actually ever form a crystal directly from ions in the gas phase."},{"Start":"00:33.600 ","End":"00:37.260","Text":"Now the lattice energy is negative since the formation"},{"Start":"00:37.260 ","End":"00:41.130","Text":"of a crystal is an exothermic process."},{"Start":"00:41.130 ","End":"00:44.715","Text":"Here\u0027s an example, NaCl."},{"Start":"00:44.715 ","End":"00:49.700","Text":"Here\u0027s Na+, it\u0027s the ion of sodium in its gas phase,"},{"Start":"00:49.700 ","End":"00:54.715","Text":"plus Cl-, the ion of chlorine in its gas phase."},{"Start":"00:54.715 ","End":"00:59.165","Text":"And when they form a solid, NaCl(s),"},{"Start":"00:59.165 ","End":"01:04.580","Text":"Delta H lattice is -787 kilo joules per mole."},{"Start":"01:04.580 ","End":"01:08.085","Text":"So it\u0027s a very strong crystal."},{"Start":"01:08.085 ","End":"01:14.454","Text":"Now, we\u0027re going to talk about the Born-Fajan-Haber cycle,"},{"Start":"01:14.454 ","End":"01:18.820","Text":"often just called the Born-Haber cycle."},{"Start":"01:18.820 ","End":"01:22.910","Text":"Now, it\u0027s difficult to calculate the lattice energy directly,"},{"Start":"01:22.910 ","End":"01:26.150","Text":"but it can be calculated indirectly by applying"},{"Start":"01:26.150 ","End":"01:30.860","Text":"Hess\u0027s Law to a number of steps of known enthalpy."},{"Start":"01:30.860 ","End":"01:34.860","Text":"The method is called the Born-Fajan or Fajan-Haber"},{"Start":"01:34.860 ","End":"01:39.980","Text":"cycle or Born-Haber cycle after people proposed it."},{"Start":"01:39.980 ","End":"01:44.195","Text":"Max Born got the Nobel Prize in 1954."},{"Start":"01:44.195 ","End":"01:49.835","Text":"Kasimir Fajans and Fritz Haber got the Nobel Prize in 1918."},{"Start":"01:49.835 ","End":"01:54.560","Text":"I should point out that none of them got the Nobel Prize for this cycle,"},{"Start":"01:54.560 ","End":"01:57.120","Text":"they got it for other things."},{"Start":"01:57.290 ","End":"02:01.485","Text":"So let\u0027s look at NaCl."},{"Start":"02:01.485 ","End":"02:06.625","Text":"The first thing we\u0027re going to do is to sublimate solid sodium."},{"Start":"02:06.625 ","End":"02:09.175","Text":"We\u0027re going to form a gas,"},{"Start":"02:09.175 ","End":"02:12.595","Text":"sodium gas, from sodium solid."},{"Start":"02:12.595 ","End":"02:18.745","Text":"This number is known delta H of sublimation is a 107 kilo joules per mole."},{"Start":"02:18.745 ","End":"02:21.460","Text":"The next process we\u0027re going to do,"},{"Start":"02:21.460 ","End":"02:24.804","Text":"is to dissociate the chlorine molecule."},{"Start":"02:24.804 ","End":"02:27.820","Text":"We only want one chlorine atom,"},{"Start":"02:27.820 ","End":"02:30.400","Text":"so it\u0027s half of a chlorine molecule in"},{"Start":"02:30.400 ","End":"02:36.055","Text":"the gas phase dissociating to give one chlorine atom."},{"Start":"02:36.055 ","End":"02:39.130","Text":"That\u0027s half of the dissociation energy."},{"Start":"02:39.130 ","End":"02:44.025","Text":"Of course, the dissociation energy is to dissociate the CO2 molecule."},{"Start":"02:44.025 ","End":"02:46.850","Text":"This is half of the dissociation energy,"},{"Start":"02:46.850 ","End":"02:50.670","Text":"that\u0027s a 122 kilo joules per mole."},{"Start":"02:50.780 ","End":"02:54.470","Text":"Now that we\u0027ve got sodium in the gas phase,"},{"Start":"02:54.470 ","End":"02:56.245","Text":"we can form the ion."},{"Start":"02:56.245 ","End":"03:00.710","Text":"To form the ion, is a process of ionization."},{"Start":"03:00.710 ","End":"03:04.580","Text":"So sodium gas going to sodium plus in the gas phase,"},{"Start":"03:04.580 ","End":"03:10.250","Text":"plus an electron, that\u0027s equivalent to ionization."},{"Start":"03:10.250 ","End":"03:14.285","Text":"Delta H of ionization is the same as I1."},{"Start":"03:14.285 ","End":"03:18.980","Text":"We talked about this when we talked about the periodic table and"},{"Start":"03:18.980 ","End":"03:24.965","Text":"the various properties of the atoms of the elements and the ionization."},{"Start":"03:24.965 ","End":"03:31.415","Text":"The first ionization of sodium is 496 kilo joules per mole."},{"Start":"03:31.415 ","End":"03:34.610","Text":"Having formed the sodium ion,"},{"Start":"03:34.610 ","End":"03:37.055","Text":"we can now form the chlorine ions."},{"Start":"03:37.055 ","End":"03:42.020","Text":"We have chlorine in the gas phase, Cl gas phase,"},{"Start":"03:42.020 ","End":"03:45.650","Text":"and we add an electron to this to get Cl- and"},{"Start":"03:45.650 ","End":"03:50.300","Text":"the gas phase and this process is called electron affinity."},{"Start":"03:50.300 ","End":"03:56.885","Text":"Delta H of electron affinity is minus 349 kilo joules per mole."},{"Start":"03:56.885 ","End":"03:59.704","Text":"This is electron affinity."},{"Start":"03:59.704 ","End":"04:08.705","Text":"It\u0027s an exothermic process when we form the Cl- anion from the Cl atom."},{"Start":"04:08.705 ","End":"04:15.560","Text":"If we take the Na+ that we formed and the Cl- that we\u0027ve formed,"},{"Start":"04:15.560 ","End":"04:22.475","Text":"put them together and form the crystal NaCl in the solid phase,"},{"Start":"04:22.475 ","End":"04:28.250","Text":"that\u0027s the delta H of lattice that we want to find."},{"Start":"04:28.250 ","End":"04:30.465","Text":"This is what we don\u0027t know."},{"Start":"04:30.465 ","End":"04:33.915","Text":"We know all these, we don\u0027t know this."},{"Start":"04:33.915 ","End":"04:38.645","Text":"If we add up all these equations,"},{"Start":"04:38.645 ","End":"04:42.935","Text":"we can cancel out certain quantities."},{"Start":"04:42.935 ","End":"04:47.645","Text":"For example, Na gas cancels with Na gas."},{"Start":"04:47.645 ","End":"04:52.400","Text":"Cl gas cancels with Cl gas."},{"Start":"04:52.400 ","End":"04:56.445","Text":"Na+ cancels with Na+."},{"Start":"04:56.445 ","End":"05:01.230","Text":"Cl- cancels with Cl-."},{"Start":"05:01.230 ","End":"05:09.405","Text":"We\u0027re left with Na solid plus half Cl2 gas,"},{"Start":"05:09.405 ","End":"05:12.889","Text":"to give us NaCl solid."},{"Start":"05:12.889 ","End":"05:19.145","Text":"This is a process for which we know the enthalpy."},{"Start":"05:19.145 ","End":"05:26.420","Text":"So this is the enthalpy of formation of NaCl in the solid phase,"},{"Start":"05:26.420 ","End":"05:34.700","Text":"from its elements, Na solid and Cl2 in the gas phase."},{"Start":"05:34.700 ","End":"05:37.460","Text":"This is called Delta H of formation."},{"Start":"05:37.460 ","End":"05:40.640","Text":"We talked about that when we talked about enthalpy and that\u0027s"},{"Start":"05:40.640 ","End":"05:44.420","Text":"minus 411 kilo joules per mole."},{"Start":"05:44.420 ","End":"05:47.515","Text":"This is another quantity we know."},{"Start":"05:47.515 ","End":"05:55.145","Text":"We can write that Delta H of formation is the sum of all these processes."},{"Start":"05:55.145 ","End":"05:59.210","Text":"So it\u0027s Delta H of sublimation plus half of Delta H of"},{"Start":"05:59.210 ","End":"06:03.830","Text":"dissociation plus ionization energy,"},{"Start":"06:03.830 ","End":"06:06.825","Text":"plus the electron affinity,"},{"Start":"06:06.825 ","End":"06:09.510","Text":"plus the lattice energy."},{"Start":"06:09.510 ","End":"06:11.870","Text":"The one we want to know, of course,"},{"Start":"06:11.870 ","End":"06:13.835","Text":"is Delta H lattice."},{"Start":"06:13.835 ","End":"06:15.500","Text":"Here I\u0027ve written LATT,"},{"Start":"06:15.500 ","End":"06:19.595","Text":"sometimes it\u0027s LATT, Sometimes it\u0027s LATTICE."},{"Start":"06:19.595 ","End":"06:29.630","Text":"Same thing. Delta H lattice is delta H of formation minus the sum of"},{"Start":"06:29.630 ","End":"06:33.815","Text":"delta H of sublimation plus half Delta H of dissociation plus"},{"Start":"06:33.815 ","End":"06:41.465","Text":"I1 ionization energy plus Delta H EA electron affinity."},{"Start":"06:41.465 ","End":"06:45.470","Text":"And if we substitute all the values above,"},{"Start":"06:45.470 ","End":"06:48.890","Text":"we get -411 for the formation,"},{"Start":"06:48.890 ","End":"06:53.210","Text":"the sum is 376,"},{"Start":"06:53.210 ","End":"06:57.395","Text":"and then we get -411 -(376),"},{"Start":"06:57.395 ","End":"07:01.790","Text":"and that\u0027s -787 kilo joules per mole."},{"Start":"07:01.790 ","End":"07:05.435","Text":"That\u0027s exactly the value we\u0027re seeking."},{"Start":"07:05.435 ","End":"07:11.600","Text":"Of course, if we know the enthalpy change for any 5 of the 6 reactions here,"},{"Start":"07:11.600 ","End":"07:13.400","Text":"we can find the sixth one."},{"Start":"07:13.400 ","End":"07:15.380","Text":"For example, if we knew the lattice energy,"},{"Start":"07:15.380 ","End":"07:16.550","Text":"we knew all the others,"},{"Start":"07:16.550 ","End":"07:20.500","Text":"we could find Delta H of formation."},{"Start":"07:20.500 ","End":"07:23.510","Text":"In this video, we learned about calculating"},{"Start":"07:23.510 ","End":"07:29.219","Text":"the lattice energy using the Born-Fajan-Haber cycle."}],"ID":24822}],"Thumbnail":null,"ID":113861}]

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