Entropy
0/11 completed

- Spontaneous Change
- Entropy and Second Law of Thermodynamics
- Change in Entropy with Temperature
- Change in Entropy of an Ideal Gas
- Change in Entropy at Phase Transition
- Third Law and Boltzmann Formula for Entropy
- Equivalence of Statistical and Thermodynamic Entropy 1
- Equivalence of Statistical and Thermodynamic Entropy 2
- Measuring Entropy as a Function of Temperature
- Standard Molar Entropy
- Standard Reaction Entropy

Global Entropy Change
0/12 completed

Gibbs Free Energy
0/4 completed

{"Free":0,"Sample":1,"Paid":2}

[{"Name":"Entropy","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Spontaneous Change","Duration":"3m 3s","ChapterTopicVideoID":25597,"CourseChapterTopicPlaylistID":244889,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25597.jpeg","UploadDate":"2021-06-02T10:47:25.4370000","DurationForVideoObject":"PT3M3S","Description":null,"MetaTitle":"Spontaneous Change: Video + Workbook | Proprep","MetaDescription":"Thermodynamics Second and Third Law - Entropy. 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/thermodynamics-second-and-third-law/entropy/vid26372","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.560","Text":"In previous videos, we learned about the first law of thermodynamics."},{"Start":"00:04.560 ","End":"00:06.870","Text":"In this video and the ones that follow,"},{"Start":"00:06.870 ","End":"00:11.310","Text":"we\u0027ll learn about entropy and the second law of thermodynamics."},{"Start":"00:11.310 ","End":"00:16.395","Text":"Let us recall what we know about the first law of thermodynamics."},{"Start":"00:16.395 ","End":"00:21.915","Text":"The first law is a version of the law of conservation of energy."},{"Start":"00:21.915 ","End":"00:24.965","Text":"We can write it Delta U."},{"Start":"00:24.965 ","End":"00:29.360","Text":"Where Delta U is the change in internal energy equal to"},{"Start":"00:29.360 ","End":"00:34.115","Text":"q plus w. q is heat and w is work."},{"Start":"00:34.115 ","End":"00:40.170","Text":"This says that the internal energy changes when the system gains or loses heat,"},{"Start":"00:40.170 ","End":"00:41.480","Text":"or does work,"},{"Start":"00:41.480 ","End":"00:44.045","Text":"or has work done on it."},{"Start":"00:44.045 ","End":"00:50.795","Text":"Another way of stating the first law is that the energy of the universe is constant."},{"Start":"00:50.795 ","End":"00:54.185","Text":"Delta U of the universe is equal to zero."},{"Start":"00:54.185 ","End":"01:00.530","Text":"That means that Delta U of the system is equal to minus Delta U of the surroundings."},{"Start":"01:00.530 ","End":"01:04.155","Text":"Delta U is internal energy."},{"Start":"01:04.155 ","End":"01:08.710","Text":"Now, let\u0027s talk about spontaneous change."},{"Start":"01:08.710 ","End":"01:13.280","Text":"A spontaneous change is a change that has a tendency to"},{"Start":"01:13.280 ","End":"01:17.765","Text":"occur without being driven by an external force."},{"Start":"01:17.765 ","End":"01:23.285","Text":"Now, thermodynamics doesn\u0027t tell us how fast a spontaneous change will occur."},{"Start":"01:23.285 ","End":"01:25.534","Text":"To know the rate of change,"},{"Start":"01:25.534 ","End":"01:27.575","Text":"we need to study kinetics,"},{"Start":"01:27.575 ","End":"01:30.620","Text":"and we\u0027ll do that later in this course."},{"Start":"01:30.620 ","End":"01:37.970","Text":"Now, sometimes, a spontaneous process only occurs when it is initiated."},{"Start":"01:37.970 ","End":"01:41.590","Text":"Let\u0027s consider a few examples."},{"Start":"01:41.590 ","End":"01:47.900","Text":"Thermodynamics tells us the diamond will turn into graphite spontaneously,"},{"Start":"01:47.900 ","End":"01:50.990","Text":"but the rate is extremely low,"},{"Start":"01:50.990 ","End":"01:56.000","Text":"so you can keep your diamonds for a while."},{"Start":"01:56.000 ","End":"01:59.480","Text":"Thermodynamics also tells us that a mixture of hydrogen and"},{"Start":"01:59.480 ","End":"02:04.545","Text":"oxygen will react spontaneously to form water."},{"Start":"02:04.545 ","End":"02:11.320","Text":"But this doesn\u0027t happen unless the mixtures first ignited by a spark."},{"Start":"02:11.320 ","End":"02:14.855","Text":"Now, what\u0027s non-spontaneous change?"},{"Start":"02:14.855 ","End":"02:21.475","Text":"A non-spontaneous change can be forced to occur by doing work on the system."},{"Start":"02:21.475 ","End":"02:24.935","Text":"If a reaction is spontaneous in 1 direction,"},{"Start":"02:24.935 ","End":"02:28.684","Text":"it\u0027s non-spontaneous in the reverse direction."},{"Start":"02:28.684 ","End":"02:30.535","Text":"Here are a few examples."},{"Start":"02:30.535 ","End":"02:33.290","Text":"A gas will spontaneously expand into"},{"Start":"02:33.290 ","End":"02:37.430","Text":"a vacuum but will not contract spontaneously into smaller volume."},{"Start":"02:37.430 ","End":"02:40.835","Text":"To contract it, we need to push it into the container."},{"Start":"02:40.835 ","End":"02:44.240","Text":"Now, ice melts spontaneously at room temperature,"},{"Start":"02:44.240 ","End":"02:47.915","Text":"but water doesn\u0027t spontaneously freeze at room temperature."},{"Start":"02:47.915 ","End":"02:52.220","Text":"However, recently, it\u0027s been discovered that a very small sample"},{"Start":"02:52.220 ","End":"02:56.690","Text":"of water can be made to freeze when an electric current is applied,"},{"Start":"02:56.690 ","End":"02:58.580","Text":"but that\u0027s somewhat exceptional."},{"Start":"02:58.580 ","End":"03:03.360","Text":"In this video, we learned about spontaneous change."}],"ID":26372},{"Watched":false,"Name":"Entropy and Second Law of Thermodynamics","Duration":"4m 26s","ChapterTopicVideoID":25604,"CourseChapterTopicPlaylistID":244889,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"In the previous video,"},{"Start":"00:02.145 ","End":"00:05.910","Text":"we learnt about spontaneous or non-spontaneous change."},{"Start":"00:05.910 ","End":"00:11.490","Text":"In this video, we\u0027ll learn about entropy and the second law of thermodynamics."},{"Start":"00:11.490 ","End":"00:15.990","Text":"We\u0027re going to talk about entropy and disorder."},{"Start":"00:15.990 ","End":"00:21.810","Text":"Entropy was first discovered by Clausius in 1854."},{"Start":"00:21.810 ","End":"00:29.580","Text":"Later on, Boltzmann provided a statistical interpretation of entropy, sometimes at 1870s."},{"Start":"00:29.580 ","End":"00:33.250","Text":"We\u0027re going to cover this in great detail in later videos."},{"Start":"00:33.250 ","End":"00:34.790","Text":"Now the entropy,"},{"Start":"00:34.790 ","End":"00:37.190","Text":"which is indicated by the letter S,"},{"Start":"00:37.190 ","End":"00:40.355","Text":"is a measure of disorder."},{"Start":"00:40.355 ","End":"00:43.130","Text":"Low entropy means low disorder."},{"Start":"00:43.130 ","End":"00:46.535","Text":"High entropy means greater disorder."},{"Start":"00:46.535 ","End":"00:49.700","Text":"Let\u0027s take an example."},{"Start":"00:49.700 ","End":"00:54.175","Text":"Supposing we release a gas in one corner of a room,"},{"Start":"00:54.175 ","End":"00:56.990","Text":"it will spread all over the room."},{"Start":"00:56.990 ","End":"01:02.915","Text":"This process is spontaneous and the gas becomes more disordered as it expands."},{"Start":"01:02.915 ","End":"01:06.875","Text":"We say that the entropy increases."},{"Start":"01:06.875 ","End":"01:10.805","Text":"Now, here\u0027s the second law of thermodynamics,"},{"Start":"01:10.805 ","End":"01:14.920","Text":"of which the example we just spoke about is one case."},{"Start":"01:14.920 ","End":"01:21.020","Text":"The entropy of an isolated system increases in a spontaneous process."},{"Start":"01:21.020 ","End":"01:26.840","Text":"The later videos we\u0027ll define very carefully what we mean by an isolated system."},{"Start":"01:26.840 ","End":"01:30.560","Text":"But that\u0027s the definition of the second law of thermodynamics."},{"Start":"01:30.560 ","End":"01:36.905","Text":"The entropy of an isolated system increases in a spontaneous process."},{"Start":"01:36.905 ","End":"01:43.410","Text":"We need a quantitative definition for entropy and here it is."},{"Start":"01:43.410 ","End":"01:45.015","Text":"At constant temperature,"},{"Start":"01:45.015 ","End":"01:50.945","Text":"the change in entropy can be defined as Delta S equal to q_rev,"},{"Start":"01:50.945 ","End":"01:53.570","Text":"q_reversible, divided by T,"},{"Start":"01:53.570 ","End":"01:56.245","Text":"the absolute temperature in Kelvin."},{"Start":"01:56.245 ","End":"01:59.265","Text":"Delta S is the change in entropy,"},{"Start":"01:59.265 ","End":"02:01.565","Text":"T is temperature in Kelvin,"},{"Start":"02:01.565 ","End":"02:07.115","Text":"and q_rev is the energy transferred reversibly as heat."},{"Start":"02:07.115 ","End":"02:10.670","Text":"Now in a reversible process,"},{"Start":"02:10.670 ","End":"02:16.759","Text":"the process can be reversed by making an infinitesimal change in a variable,"},{"Start":"02:16.759 ","End":"02:18.710","Text":"such as the temperature."},{"Start":"02:18.710 ","End":"02:21.800","Text":"That\u0027s a reversible process."},{"Start":"02:21.800 ","End":"02:26.405","Text":"Let\u0027s talk a little bit more about a reversible process."},{"Start":"02:26.405 ","End":"02:29.240","Text":"Heat can flow reversibly between a system and"},{"Start":"02:29.240 ","End":"02:31.880","Text":"its surroundings if the difference in temperature,"},{"Start":"02:31.880 ","End":"02:39.290","Text":"Delta T, between the system and its surroundings is infinitesimal, extremely small."},{"Start":"02:39.290 ","End":"02:42.335","Text":"Here\u0027s a picture, here\u0027s our system."},{"Start":"02:42.335 ","End":"02:46.150","Text":"A temperature of T plus Delta T and the surroundings at"},{"Start":"02:46.150 ","End":"02:51.730","Text":"a slightly lower temperature T. If that\u0027s the case,"},{"Start":"02:51.730 ","End":"02:55.990","Text":"heat will flow out of the system into the surroundings."},{"Start":"02:55.990 ","End":"03:00.580","Text":"Now, if the system has a temperature slightly smaller than"},{"Start":"03:00.580 ","End":"03:05.710","Text":"the surroundings T minus Delta T with the surroundings have a temperature T,"},{"Start":"03:05.710 ","End":"03:10.000","Text":"then heat will flow from the surroundings into the system."},{"Start":"03:10.000 ","End":"03:14.500","Text":"We\u0027ll see later that this combined system of system"},{"Start":"03:14.500 ","End":"03:19.760","Text":"and surroundings is what we call an isolated system."},{"Start":"03:22.380 ","End":"03:26.230","Text":"A few more points about the entropy."},{"Start":"03:26.230 ","End":"03:30.835","Text":"The units are joules per degree Kelvin."},{"Start":"03:30.835 ","End":"03:36.505","Text":"Delta S increases as q_rev increases and T decreases."},{"Start":"03:36.505 ","End":"03:39.910","Text":"We can see that from the equation here,"},{"Start":"03:39.910 ","End":"03:43.630","Text":"Delta S is proportional q_reversible."},{"Start":"03:43.630 ","End":"03:48.250","Text":"As q_rev increases, Delta S will increase,"},{"Start":"03:48.250 ","End":"03:50.890","Text":"and as T increases,"},{"Start":"03:50.890 ","End":"03:54.115","Text":"Delta S will decrease."},{"Start":"03:54.115 ","End":"03:58.603","Text":"Another important point is the entropy like internal energy,"},{"Start":"03:58.603 ","End":"04:01.010","Text":"remember Delta U,"},{"Start":"04:01.350 ","End":"04:08.054","Text":"and enthalpy Delta H is a state function."},{"Start":"04:08.054 ","End":"04:10.835","Text":"Entropy is a state function."},{"Start":"04:10.835 ","End":"04:13.475","Text":"The path is unimportant."},{"Start":"04:13.475 ","End":"04:16.470","Text":"How we go from A to B is unimportant."},{"Start":"04:16.470 ","End":"04:18.170","Text":"Just A and B themselves,"},{"Start":"04:18.170 ","End":"04:21.600","Text":"the beginning and the end are important."},{"Start":"04:21.730 ","End":"04:27.210","Text":"In this video, we introduced the concept of entropy."}],"ID":26379},{"Watched":false,"Name":"Change in Entropy with Temperature","Duration":"4m 56s","ChapterTopicVideoID":25603,"CourseChapterTopicPlaylistID":244889,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.420","Text":"In the previous video we introduced entropy."},{"Start":"00:03.420 ","End":"00:08.835","Text":"In this video we\u0027ll discuss the change in entropy on heating or cooling."},{"Start":"00:08.835 ","End":"00:13.425","Text":"We\u0027re going to talk about the change in entropy with temperature."},{"Start":"00:13.425 ","End":"00:17.310","Text":"This is the expression we want to obtain, Delta S,"},{"Start":"00:17.310 ","End":"00:22.094","Text":"the entropy is equal to c which is the heat capacity of the system,"},{"Start":"00:22.094 ","End":"00:25.545","Text":"times ln t_2 divided by t_1,"},{"Start":"00:25.545 ","End":"00:29.970","Text":"where t_1 is initial temperature and t_2 is the final temperature."},{"Start":"00:29.970 ","End":"00:32.370","Text":"Now if we\u0027ve at constant pressure,"},{"Start":"00:32.370 ","End":"00:34.095","Text":"C is C_p,"},{"Start":"00:34.095 ","End":"00:39.390","Text":"and if we have constant volume is C_v. Here\u0027s the proof."},{"Start":"00:39.390 ","End":"00:46.970","Text":"We learned that the definition of Delta S is q reversible divided by t,"},{"Start":"00:46.970 ","End":"00:49.190","Text":"the absolute temperature in Kelvin."},{"Start":"00:49.190 ","End":"00:54.740","Text":"Now for small change we can write this as ds equal to dq_rev"},{"Start":"00:54.740 ","End":"01:01.145","Text":"divided by t. That\u0027s for an infinitesimal reversible transfer of heat."},{"Start":"01:01.145 ","End":"01:08.930","Text":"Now we know from our discussion of heat transfer that q reversible is equal to C,"},{"Start":"01:08.930 ","End":"01:11.390","Text":"that\u0027s heat capacity times Delta t,"},{"Start":"01:11.390 ","End":"01:13.265","Text":"the change in temperature."},{"Start":"01:13.265 ","End":"01:16.455","Text":"For a small value of q,"},{"Start":"01:16.455 ","End":"01:21.130","Text":"this is dq_rev equal to Cdt."},{"Start":"01:21.130 ","End":"01:24.425","Text":"That\u0027s for an infinitesimal change in temperature."},{"Start":"01:24.425 ","End":"01:28.670","Text":"We\u0027re going to assume that C is constant."},{"Start":"01:28.670 ","End":"01:32.465","Text":"Now we\u0027re going to combine these 2 equations."},{"Start":"01:32.465 ","End":"01:36.920","Text":"Instead of writing dq in this equation,"},{"Start":"01:36.920 ","End":"01:40.060","Text":"we\u0027re going to replace it by Cdt."},{"Start":"01:40.060 ","End":"01:48.979","Text":"We get ds is equal to Cdt divided by t. We can integrate this equation."},{"Start":"01:48.979 ","End":"01:54.320","Text":"We integrate the left hand side we get Delta S. That\u0027s the difference in"},{"Start":"01:54.320 ","End":"02:00.300","Text":"entropy between the temperature T_1 and T_2 equal to C,"},{"Start":"02:00.300 ","End":"02:01.680","Text":"we\u0027re assuming C is a constant.,"},{"Start":"02:01.680 ","End":"02:03.465","Text":"so it goes outside the integral."},{"Start":"02:03.465 ","End":"02:06.255","Text":"The integral from T_1 to T_2,"},{"Start":"02:06.255 ","End":"02:10.505","Text":"dT divided by t. Now we know"},{"Start":"02:10.505 ","End":"02:15.965","Text":"that the integral of dx divided by x is ln x plus a constant."},{"Start":"02:15.965 ","End":"02:20.495","Text":"We should really write another constant not to confuse it with the heat capacity,"},{"Start":"02:20.495 ","End":"02:25.665","Text":"say k. Then this integral is equal to C ln T_2,"},{"Start":"02:25.665 ","End":"02:29.970","Text":"that\u0027s the upper limit T_2, minus ln T_1."},{"Start":"02:29.970 ","End":"02:33.370","Text":"We know that ln a minus ln b,"},{"Start":"02:33.370 ","End":"02:36.875","Text":"is equal to ln of a divided by b."},{"Start":"02:36.875 ","End":"02:40.545","Text":"That\u0027s the expression we wanted to prove."},{"Start":"02:40.545 ","End":"02:44.045","Text":"Now, if T_2 is greater than T_1,"},{"Start":"02:44.045 ","End":"02:46.560","Text":"ratio is greater than 1,"},{"Start":"02:46.560 ","End":"02:52.575","Text":"then the ln is positive and Delta S is also positive,"},{"Start":"02:52.575 ","End":"02:54.800","Text":"so we get Delta S positive."},{"Start":"02:54.800 ","End":"02:57.155","Text":"If we raise the temperature,"},{"Start":"02:57.155 ","End":"02:58.850","Text":"the entropy increases,"},{"Start":"02:58.850 ","End":"03:01.550","Text":"and if we decrease the temperature,"},{"Start":"03:01.550 ","End":"03:03.425","Text":"T_2 is smaller than T_1,"},{"Start":"03:03.425 ","End":"03:06.830","Text":"then the entropy decreases."},{"Start":"03:06.830 ","End":"03:08.945","Text":"Here\u0027s an example."},{"Start":"03:08.945 ","End":"03:15.050","Text":"Calculate the entropy change when the temperature of 3 moles of methane is increased from"},{"Start":"03:15.050 ","End":"03:21.370","Text":"298 Kelvin to 350 kelvin at a constant pressure of 1 bar."},{"Start":"03:21.370 ","End":"03:29.155","Text":"We\u0027re given the information that C_p is equal to 35.7 joules per mole per Kelvin."},{"Start":"03:29.155 ","End":"03:35.224","Text":"Delta s is equal to C_p because we\u0027re working at constant pressure,"},{"Start":"03:35.224 ","End":"03:38.510","Text":"ln T_2 divided by T_1."},{"Start":"03:38.510 ","End":"03:45.875","Text":"Now we see that C_p has units of joules per mole per Kelvin."},{"Start":"03:45.875 ","End":"03:48.500","Text":"We have to multiply by the number of moles."},{"Start":"03:48.500 ","End":"03:51.305","Text":"This is the volume for 1 mole only."},{"Start":"03:51.305 ","End":"03:56.190","Text":"Delta S is equal to C_p ln T_2 divide by T_1."},{"Start":"03:56.190 ","End":"04:02.375","Text":"We could write as 3 moles times 35.7 joules per mole per Kelvin."},{"Start":"04:02.375 ","End":"04:05.010","Text":"That\u0027s the value of C_p,"},{"Start":"04:05.800 ","End":"04:10.980","Text":"ln T_2 divide by T_1."},{"Start":"04:11.320 ","End":"04:16.945","Text":"The initial temperature is 298 and the final temperature is 350."},{"Start":"04:16.945 ","End":"04:21.705","Text":"Now, 3 times 35.7 is 107.1,"},{"Start":"04:21.705 ","End":"04:28.440","Text":"and the units are joules per Kelvin because the mole cancels out,"},{"Start":"04:28.440 ","End":"04:35.255","Text":"and ln of 350 over 298 is 0.1608."},{"Start":"04:35.255 ","End":"04:40.235","Text":"We multiply that we get 17.2 joules per Kelvin."},{"Start":"04:40.235 ","End":"04:42.935","Text":"We can see that as expected,"},{"Start":"04:42.935 ","End":"04:44.990","Text":"by increasing the temperature,"},{"Start":"04:44.990 ","End":"04:47.600","Text":"Delta S is positive,"},{"Start":"04:47.600 ","End":"04:50.014","Text":"we increase the entropy."},{"Start":"04:50.014 ","End":"04:56.430","Text":"In this video we learned about the change in entropy on heating or cooling."}],"ID":26378},{"Watched":false,"Name":"Change in Entropy of an Ideal Gas","Duration":"7m 34s","ChapterTopicVideoID":25602,"CourseChapterTopicPlaylistID":244889,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.635","Text":"In the previous video,"},{"Start":"00:01.635 ","End":"00:06.045","Text":"we discussed the change in entropy when the system is heated or cooled."},{"Start":"00:06.045 ","End":"00:11.355","Text":"In this video, discuss the Change in Entropy of an Ideal gas."},{"Start":"00:11.355 ","End":"00:14.550","Text":"We\u0027re going to talk about the change in entropy of"},{"Start":"00:14.550 ","End":"00:18.405","Text":"an isothermal ideal gas with volume or pressure."},{"Start":"00:18.405 ","End":"00:21.420","Text":"Isothermal means constant temperature,"},{"Start":"00:21.420 ","End":"00:25.845","Text":"T. These are the expressions we\u0027re going to prove."},{"Start":"00:25.845 ","End":"00:29.040","Text":"Delta S, a change in entropy is equal to n,"},{"Start":"00:29.040 ","End":"00:30.540","Text":"the number of moles times R,"},{"Start":"00:30.540 ","End":"00:36.810","Text":"the gas constant lnV_2 divided by V_1 where V_1 is initial volume,"},{"Start":"00:36.810 ","End":"00:39.160","Text":"V_2 is the final volume."},{"Start":"00:39.160 ","End":"00:45.500","Text":"Or alternatively, Delta S is equal to n_RInP_1 divided by P_2."},{"Start":"00:45.500 ","End":"00:49.175","Text":"P_1 is initial pressure and P_2 is a final pressure."},{"Start":"00:49.175 ","End":"00:54.800","Text":"Here\u0027s the proof. We start off by saying that Delta U,"},{"Start":"00:54.800 ","End":"00:58.865","Text":"the change in internal energy is 0 for an ideal gas."},{"Start":"00:58.865 ","End":"01:01.145","Text":"How do we know that?"},{"Start":"01:01.145 ","End":"01:03.770","Text":"We know that the internal energy"},{"Start":"01:03.770 ","End":"01:07.300","Text":"is the sum of the kinetic energy and the potential energy."},{"Start":"01:07.300 ","End":"01:12.070","Text":"Now, the average kinetic energy of an ideal gas is 3/2 RT."},{"Start":"01:12.070 ","End":"01:14.980","Text":"It\u0027s a constant when T is constant."},{"Start":"01:14.980 ","End":"01:17.156","Text":"We don\u0027t change the temperature,"},{"Start":"01:17.156 ","End":"01:18.985","Text":"this is a constant."},{"Start":"01:18.985 ","End":"01:20.660","Text":"For an ideal gas,"},{"Start":"01:20.660 ","End":"01:25.850","Text":"we know that there are no intermolecular forces or at least there are negligible."},{"Start":"01:25.850 ","End":"01:29.270","Text":"The potential energy is 0."},{"Start":"01:29.270 ","End":"01:33.440","Text":"There\u0027s no change in internal energy because there\u0027s no change in"},{"Start":"01:33.440 ","End":"01:38.155","Text":"temperature and there are no intermolecular forces."},{"Start":"01:38.155 ","End":"01:45.530","Text":"Now, Delta U is equal to q plus w. That we know from the first law of thermodynamics,"},{"Start":"01:45.530 ","End":"01:48.725","Text":"q is the heat, w is work."},{"Start":"01:48.725 ","End":"01:51.380","Text":"If Delta U is equal to 0,"},{"Start":"01:51.380 ","End":"01:56.690","Text":"q is equal to minus w. If we\u0027re talking about reversible processes,"},{"Start":"01:56.690 ","End":"02:01.240","Text":"q reversible is equal to minus w reversible."},{"Start":"02:01.240 ","End":"02:04.940","Text":"Now if the work is pressure volume work,"},{"Start":"02:04.940 ","End":"02:08.825","Text":"then w the work is equal to minus P external,"},{"Start":"02:08.825 ","End":"02:11.479","Text":"external pressure on the system,"},{"Start":"02:11.479 ","End":"02:14.870","Text":"Delta V, the change in volume of the system,"},{"Start":"02:14.870 ","End":"02:24.005","Text":"and that can be written for small changes and volume as dw equal to minus P external dV."},{"Start":"02:24.005 ","End":"02:32.135","Text":"Now if we\u0027re talking about reversible processes that P external and P are equal."},{"Start":"02:32.135 ","End":"02:35.750","Text":"The pressure inside and outside the system,"},{"Start":"02:35.750 ","End":"02:38.645","Text":"outside the gas are equal."},{"Start":"02:38.645 ","End":"02:42.275","Text":"We can write instead of P external P,"},{"Start":"02:42.275 ","End":"02:46.895","Text":"so now we have dw reversible is equal to minus PdV."},{"Start":"02:46.895 ","End":"02:53.730","Text":"We remember from the gas law, that PV=nRT."},{"Start":"02:53.730 ","End":"03:01.505","Text":"P=nRT divided by V. Here we have it,"},{"Start":"03:01.505 ","End":"03:07.485","Text":"nRT divided by V. Now we can integrate this expression,"},{"Start":"03:07.485 ","End":"03:09.050","Text":"and on the left-hand side,"},{"Start":"03:09.050 ","End":"03:15.515","Text":"we have w reversible equal to the integral from V_1 to V_2, d_w reversible."},{"Start":"03:15.515 ","End":"03:20.450","Text":"The right-hand side, we have minus nRT outside the integral because these are"},{"Start":"03:20.450 ","End":"03:26.680","Text":"constants times the integral from V_1 to V_2, dV/V."},{"Start":"03:26.680 ","End":"03:30.700","Text":"Now the integral of 1/V is lnV."},{"Start":"03:30.700 ","End":"03:38.844","Text":"Because we know the integral of dx/x is lnx plus a constant."},{"Start":"03:38.844 ","End":"03:45.560","Text":"We know that ln a minus ln b is equal to ln a/b,"},{"Start":"03:45.560 ","End":"03:53.880","Text":"so ln V_2 minus lnV_1 is equal to ln V_2 over V_1."},{"Start":"03:53.880 ","End":"03:56.155","Text":"That\u0027s our answer here."},{"Start":"03:56.155 ","End":"04:03.000","Text":"W reversible is equal to minus nRT ln V_2 over V_1."},{"Start":"04:03.200 ","End":"04:09.070","Text":"Now, before we said that q reversible is equal to minus w reversible."},{"Start":"04:09.070 ","End":"04:14.015","Text":"Here q reversible is not equal to the negative of this expression,"},{"Start":"04:14.015 ","End":"04:19.500","Text":"so it\u0027s nRT plus nRTlnV_2 over V_1."},{"Start":"04:20.210 ","End":"04:27.920","Text":"Delta S, the change in entropy is q reversible divided by T. We have to divide"},{"Start":"04:27.920 ","End":"04:38.720","Text":"this expression by T. The T cancels and we get nRlnV_2 over V_1."},{"Start":"04:38.720 ","End":"04:40.820","Text":"Here\u0027s our expression."},{"Start":"04:40.820 ","End":"04:50.270","Text":"Delta S is equal to nRlnV_2 over V_1 and that\u0027s equivalent to nRln P_1 over P_2."},{"Start":"04:50.270 ","End":"04:52.640","Text":"Because we know from Boyle\u0027s law,"},{"Start":"04:52.640 ","End":"04:53.930","Text":"from the gas law,"},{"Start":"04:53.930 ","End":"04:56.990","Text":"that P_1V_1 is equal to P_2V_2,"},{"Start":"04:56.990 ","End":"05:01.120","Text":"V_2 over V_1 is equal to P_1 over P_2."},{"Start":"05:01.120 ","End":"05:03.240","Text":"Instead of writing V_2 over V_1,"},{"Start":"05:03.240 ","End":"05:06.550","Text":"we could write P_1 over P_2."},{"Start":"05:06.550 ","End":"05:10.670","Text":"Now there\u0027s a more general equation that I haven\u0027t proven in"},{"Start":"05:10.670 ","End":"05:14.535","Text":"this video that can be proven."},{"Start":"05:14.535 ","End":"05:19.270","Text":"That\u0027s the Delta S is equal to C_P lnT_2 over T_1."},{"Start":"05:19.270 ","End":"05:23.600","Text":"That\u0027s expression we got before for the changes in temperature minus"},{"Start":"05:23.600 ","End":"05:28.985","Text":"nR ln P_2 over P_1. Here\u0027s an example."},{"Start":"05:28.985 ","End":"05:34.235","Text":"Calculate the entropy change with a temperature of 3 moles of methane,"},{"Start":"05:34.235 ","End":"05:39.905","Text":"CH_4, is increased from 298 Kelvin to 350 Kelvin."},{"Start":"05:39.905 ","End":"05:45.260","Text":"Then the gas pressure is increased to 1 bar to 3 bars and we\u0027re"},{"Start":"05:45.260 ","End":"05:50.740","Text":"given the value of C as 35.7 joules per mole per Kelvin."},{"Start":"05:50.740 ","End":"05:56.520","Text":"Delta S is equal to C_P ln T_2 divided by T_1."},{"Start":"05:56.520 ","End":"06:04.500","Text":"We worked out the change from 298 to 350 Kelvin in the previous video,"},{"Start":"06:04.500 ","End":"06:11.010","Text":"so we\u0027ll just repeat that here and we got 17.2 joules per Kelvin."},{"Start":"06:11.010 ","End":"06:20.075","Text":"Now the new part minus nRlnP_2 divided by P_1 is given by n=3 moles."},{"Start":"06:20.075 ","End":"06:24.740","Text":"The gas constant is 8.314 joules per mole per Kelvin,"},{"Start":"06:24.740 ","End":"06:28.240","Text":"so mole goes with mole to the power of minus 1,"},{"Start":"06:28.240 ","End":"06:32.850","Text":"and the initial pressure is 1 bar,"},{"Start":"06:32.850 ","End":"06:37.290","Text":"the final pressure is 3 bars so that\u0027s a ratio of 3 to 1,"},{"Start":"06:37.290 ","End":"06:40.530","Text":"so we have times ln3."},{"Start":"06:40.530 ","End":"06:43.125","Text":"We work that out,"},{"Start":"06:43.125 ","End":"06:47.525","Text":"we get minus 27.4."},{"Start":"06:47.525 ","End":"06:51.800","Text":"We have 17.2 from the change in temperature,"},{"Start":"06:51.800 ","End":"06:58.590","Text":"a positive change in entropy and then a negative change from changing"},{"Start":"06:58.590 ","End":"07:06.530","Text":"the pressure and so the 2 going together give us minus 10.2 joules per Kelvin."},{"Start":"07:06.530 ","End":"07:10.255","Text":"We see that in this case,"},{"Start":"07:10.255 ","End":"07:12.710","Text":"when we increase the pressure,"},{"Start":"07:12.710 ","End":"07:20.510","Text":"we compress the gas that gave a greater change in entropy than the change in temperature."},{"Start":"07:20.510 ","End":"07:24.610","Text":"Overall we get a negative change in entropy,"},{"Start":"07:24.610 ","End":"07:27.880","Text":"a decrease in entropy."},{"Start":"07:28.240 ","End":"07:34.650","Text":"In this video, we learnt about the Change in Entropy of an Ideal Gas."}],"ID":26377},{"Watched":false,"Name":"Change in Entropy at Phase Transition","Duration":"6m 25s","ChapterTopicVideoID":25601,"CourseChapterTopicPlaylistID":244889,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.110","Text":"In the previous two videos,"},{"Start":"00:02.110 ","End":"00:04.465","Text":"we talked about changes in entropy."},{"Start":"00:04.465 ","End":"00:09.985","Text":"In this video, we\u0027ll learn about the change in entropy at a phase transition."},{"Start":"00:09.985 ","End":"00:14.155","Text":"Let\u0027s recall what we know about phase transitions."},{"Start":"00:14.155 ","End":"00:17.320","Text":"We know that the normal boiling point T_b,"},{"Start":"00:17.320 ","End":"00:22.420","Text":"is the temperature in Kelvin at which the vapor pressure of"},{"Start":"00:22.420 ","End":"00:28.435","Text":"the liquid is equal to atmospheric pressure of 1 atmosphere."},{"Start":"00:28.435 ","End":"00:32.125","Text":"We know that the normal melting point,"},{"Start":"00:32.125 ","End":"00:34.485","Text":"or fusion point of fusion,"},{"Start":"00:34.485 ","End":"00:36.700","Text":"fusion is in other words for melting,"},{"Start":"00:36.700 ","End":"00:42.445","Text":"T_f is the temperature in Kelvin at which a substance melts,"},{"Start":"00:42.445 ","End":"00:45.405","Text":"at pressure of 1 atmosphere."},{"Start":"00:45.405 ","End":"00:48.620","Text":"Now, at the transition temperature,"},{"Start":"00:48.620 ","End":"00:55.085","Text":"the temperature of the material remains constant until the phase transition is complete."},{"Start":"00:55.085 ","End":"00:57.080","Text":"If we melt something,"},{"Start":"00:57.080 ","End":"01:00.830","Text":"it gets to the melting point and then"},{"Start":"01:00.830 ","End":"01:05.525","Text":"stays at that temperature until the process is complete."},{"Start":"01:05.525 ","End":"01:08.240","Text":"Now, at the transition temperature,"},{"Start":"01:08.240 ","End":"01:10.760","Text":"the transfer of heat is reversible."},{"Start":"01:10.760 ","End":"01:13.340","Text":"Remember we had all these curves,"},{"Start":"01:13.340 ","End":"01:15.695","Text":"one side is a gas,"},{"Start":"01:15.695 ","End":"01:17.720","Text":"say the other side is a liquid."},{"Start":"01:17.720 ","End":"01:21.470","Text":"A tiny change in temperature will go from one to the other,"},{"Start":"01:21.470 ","End":"01:22.985","Text":"from liquid to vapor,"},{"Start":"01:22.985 ","End":"01:24.835","Text":"from vapor to liquid."},{"Start":"01:24.835 ","End":"01:27.875","Text":"The transfer of heat is reversible."},{"Start":"01:27.875 ","End":"01:29.630","Text":"The pressure is constant,"},{"Start":"01:29.630 ","End":"01:32.360","Text":"usually open air pressure is constant."},{"Start":"01:32.360 ","End":"01:34.820","Text":"Q_rev is equal to Delta H,"},{"Start":"01:34.820 ","End":"01:37.160","Text":"the change in enthalpy."},{"Start":"01:37.160 ","End":"01:42.800","Text":"Now, we know that a liquid is more disordered than the solid."},{"Start":"01:42.800 ","End":"01:47.480","Text":"Delta H of fusion is greater than 0."},{"Start":"01:47.480 ","End":"01:49.580","Text":"We\u0027re going from a solid to a liquid,"},{"Start":"01:49.580 ","End":"01:52.100","Text":"so Delta S is greater than 0."},{"Start":"01:52.100 ","End":"01:54.740","Text":"We go from a solid to a liquid."},{"Start":"01:54.740 ","End":"01:58.760","Text":"Now, also a gas is more disordered than a liquid."},{"Start":"01:58.760 ","End":"02:02.850","Text":"Delta S of vaporization is greater than 0,"},{"Start":"02:02.850 ","End":"02:06.025","Text":"when we go from a liquid to a gas."},{"Start":"02:06.025 ","End":"02:11.120","Text":"Of course, recall that Delta S vaporization is larger than"},{"Start":"02:11.120 ","End":"02:16.475","Text":"Delta S of fusion because the gas is very much disordered."},{"Start":"02:16.475 ","End":"02:22.720","Text":"Delta S_vap is greater than Delta S fusion."},{"Start":"02:22.720 ","End":"02:27.410","Text":"Now what above the entropy of vaporization?"},{"Start":"02:27.410 ","End":"02:32.090","Text":"At the boiling point of Delta S,"},{"Start":"02:32.090 ","End":"02:36.380","Text":"which is the definition of Clausius for Delta S,"},{"Start":"02:36.380 ","End":"02:41.480","Text":"becomes Delta S of vaporization equal to Delta H of"},{"Start":"02:41.480 ","End":"02:48.390","Text":"vaporization divided by T_b and that is greater than 0."},{"Start":"02:48.390 ","End":"02:52.250","Text":"Because to vaporize something you have to put energy in."},{"Start":"02:52.250 ","End":"02:55.295","Text":"It\u0027s an endothermic process."},{"Start":"02:55.295 ","End":"03:00.149","Text":"Delta H vaporization over T_b is positive,"},{"Start":"03:00.149 ","End":"03:04.400","Text":"so Delta S of vaporization is positive."},{"Start":"03:04.400 ","End":"03:08.780","Text":"When the liquid and vapor in standard state at one bar,"},{"Start":"03:08.780 ","End":"03:14.615","Text":"then Delta S of vaporization can be written as Delta S vaporization little 0 here."},{"Start":"03:14.615 ","End":"03:17.855","Text":"It\u0027s a standard entropy of vaporization,"},{"Start":"03:17.855 ","End":"03:22.105","Text":"and it\u0027s usually at 25 degrees Celsius."},{"Start":"03:22.105 ","End":"03:27.015","Text":"Now we\u0027re going to learn a rule called Trouton\u0027s rule."},{"Start":"03:27.015 ","End":"03:30.210","Text":"Now, how did we get to Trouton\u0027s rule?"},{"Start":"03:30.210 ","End":"03:33.740","Text":"We know that strong intermolecular forces increase"},{"Start":"03:33.740 ","End":"03:38.300","Text":"both the enthalpy of vaporization and the boiling point."},{"Start":"03:38.300 ","End":"03:45.950","Text":"If both increase, then approximately the ratio of them is approximately constant."},{"Start":"03:45.950 ","End":"03:51.830","Text":"Delta S vaporization is equal to Delta H of vaporization divided"},{"Start":"03:51.830 ","End":"03:57.830","Text":"by T. These are both for standard states and that\u0027s approximately constant."},{"Start":"03:57.830 ","End":"04:05.135","Text":"For many materials, it\u0027s close to 85-88 joules per Kelvin per mole."},{"Start":"04:05.135 ","End":"04:08.765","Text":"Now, if hydrogen bonding is important,"},{"Start":"04:08.765 ","End":"04:11.284","Text":"then the entropy is even larger."},{"Start":"04:11.284 ","End":"04:17.510","Text":"Delta S vaporization is greater than 85-88 joules per Kelvin per mole."},{"Start":"04:17.510 ","End":"04:19.475","Text":"Here\u0027s some examples."},{"Start":"04:19.475 ","End":"04:24.875","Text":"Acetone, it\u0027s 88.3 and benzene 87.2."},{"Start":"04:24.875 ","End":"04:26.720","Text":"But water and methanol,"},{"Start":"04:26.720 ","End":"04:28.325","Text":"which have hydrogen bonding,"},{"Start":"04:28.325 ","End":"04:34.880","Text":"water is a 109 and for methanol 105 and all this is in joules per Kelvin per mole."},{"Start":"04:34.880 ","End":"04:38.340","Text":"Here this hydrogen bonding, and here this hydrogen bonding."},{"Start":"04:39.370 ","End":"04:43.855","Text":"Now what about the entropy of melting or fusion?"},{"Start":"04:43.855 ","End":"04:48.495","Text":"You can write that at the melting point Delta S^0 of fusion,"},{"Start":"04:48.495 ","End":"04:53.375","Text":"the standard entropy of fusion is Delta H fusion,"},{"Start":"04:53.375 ","End":"04:56.210","Text":"that\u0027s the standard enthalpy of fusion,"},{"Start":"04:56.210 ","End":"04:58.235","Text":"divided by T_f,"},{"Start":"04:58.235 ","End":"05:01.740","Text":"the temperature of melting."},{"Start":"05:01.740 ","End":"05:06.350","Text":"That, because Delta H of fusion is positive,"},{"Start":"05:06.350 ","End":"05:09.320","Text":"we have to put in energy to melt something."},{"Start":"05:09.320 ","End":"05:10.910","Text":"This is positive."},{"Start":"05:10.910 ","End":"05:15.335","Text":"Delta S of fusion is positive."},{"Start":"05:15.335 ","End":"05:17.180","Text":"As we said before,"},{"Start":"05:17.180 ","End":"05:23.450","Text":"the values of Delta S of fusion are lower than the values of Delta S of vaporization."},{"Start":"05:23.450 ","End":"05:25.489","Text":"Let\u0027s take an example."},{"Start":"05:25.489 ","End":"05:30.820","Text":"What is the standard entropy change for the melting of ice at its freezing point?"},{"Start":"05:30.820 ","End":"05:33.740","Text":"We\u0027re given the information that Delta H,"},{"Start":"05:33.740 ","End":"05:36.800","Text":"the standard enthalpy change,"},{"Start":"05:36.800 ","End":"05:41.470","Text":"is 6.02 kilojoules per mole."},{"Start":"05:41.470 ","End":"05:43.770","Text":"Now here we should remember,"},{"Start":"05:43.770 ","End":"05:45.380","Text":"that this is kilojoules,"},{"Start":"05:45.380 ","End":"05:49.580","Text":"whereas entropy is usually in joules."},{"Start":"05:49.580 ","End":"05:55.410","Text":"We have Delta S of fusion equal to Delta H of fusion divided by"},{"Start":"05:55.410 ","End":"06:00.890","Text":"T_f 6.02 times 10^3 joules per mole."},{"Start":"06:00.890 ","End":"06:08.570","Text":"We\u0027ve converted the kilojoules to joules divided by 273.15 Kelvin."},{"Start":"06:08.570 ","End":"06:14.085","Text":"We divide these two we get 22.0 joules per mole per Kelvin."},{"Start":"06:14.085 ","End":"06:18.499","Text":"That\u0027s the standard entropy of fusion."},{"Start":"06:18.499 ","End":"06:25.650","Text":"In this video, we learned about the change in entropy that accompanies phase transitions."}],"ID":26376},{"Watched":false,"Name":"Third Law and Boltzmann Formula for Entropy","Duration":"8m 2s","ChapterTopicVideoID":25600,"CourseChapterTopicPlaylistID":244889,"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 previous videos,"},{"Start":"00:02.310 ","End":"00:04.950","Text":"we spoke about changes in entropy."},{"Start":"00:04.950 ","End":"00:08.205","Text":"In this video we\u0027ll learn about entropy itself,"},{"Start":"00:08.205 ","End":"00:10.425","Text":"the third law of thermodynamics,"},{"Start":"00:10.425 ","End":"00:14.259","Text":"and the statistical interpretation of entropy."},{"Start":"00:14.360 ","End":"00:18.780","Text":"Let\u0027s begin with 0 entropy."},{"Start":"00:18.780 ","End":"00:23.565","Text":"As we know, entropy is a measure of disorder."},{"Start":"00:23.565 ","End":"00:26.610","Text":"Now if we have a perfectly ordered crystal,"},{"Start":"00:26.610 ","End":"00:31.050","Text":"that means all the molecules pointing in the exact same direction."},{"Start":"00:31.050 ","End":"00:34.095","Text":"There\u0027ll be no positional disorder."},{"Start":"00:34.095 ","End":"00:36.450","Text":"Positional disorder is when say,"},{"Start":"00:36.450 ","End":"00:42.480","Text":"1 molecule is pointing one way and another molecule is pointing the opposite way."},{"Start":"00:42.480 ","End":"00:48.365","Text":"Now in addition, when the temperature approaches 0 Kelvin,"},{"Start":"00:48.365 ","End":"00:52.595","Text":"we never can absolutely get to 0 Kelvin but we can approach it."},{"Start":"00:52.595 ","End":"00:55.625","Text":"The thermal disorder approaches 0."},{"Start":"00:55.625 ","End":"01:03.160","Text":"There\u0027s no kinetic energy to move the molecule so thermal disorder approaches 0."},{"Start":"01:03.160 ","End":"01:07.870","Text":"We can say that as T approaches 0 Kelvin,"},{"Start":"01:07.870 ","End":"01:13.700","Text":"S entropy also approaches 0 for a perfect crystal."},{"Start":"01:13.700 ","End":"01:17.690","Text":"Now here\u0027s the third law of thermodynamics,"},{"Start":"01:17.690 ","End":"01:20.060","Text":"which says that the entropy of"},{"Start":"01:20.060 ","End":"01:26.045","Text":"a perfect crystal approaches 0 as the absolute temperature approaches 0."},{"Start":"01:26.045 ","End":"01:27.890","Text":"That\u0027s what we wrote here."},{"Start":"01:27.890 ","End":"01:30.335","Text":"As the temperature approaches 0,"},{"Start":"01:30.335 ","End":"01:35.300","Text":"the entropy approaches 0 and that\u0027s only for a perfect crystal."},{"Start":"01:35.300 ","End":"01:42.450","Text":"Now Boltzmann wrote his formula for entropy in the 1870s."},{"Start":"01:42.520 ","End":"01:47.390","Text":"He thought it was so important compared to"},{"Start":"01:47.390 ","End":"01:53.810","Text":"all his other great achievements that he had it written on his tombstone."},{"Start":"01:53.810 ","End":"02:01.190","Text":"It\u0027s written S equal to k log W. Nowadays we write it"},{"Start":"02:01.190 ","End":"02:09.075","Text":"as S equal to K ln W. In those days I didn\u0027t distinguish ln and log."},{"Start":"02:09.075 ","End":"02:13.770","Text":"Now k is the Boltzmann constant equal to R,"},{"Start":"02:13.770 ","End":"02:18.895","Text":"the gas constant divided by Avogadro\u0027s number."},{"Start":"02:18.895 ","End":"02:20.970","Text":"This way we divide the 2,"},{"Start":"02:20.970 ","End":"02:25.625","Text":"we get 1.381 times 10 to the power minus 23"},{"Start":"02:25.625 ","End":"02:31.655","Text":"Joules per degree Kelvin and it\u0027s called the Boltzmann constant."},{"Start":"02:31.655 ","End":"02:33.560","Text":"Now what about this W?"},{"Start":"02:33.560 ","End":"02:36.680","Text":"It\u0027s a very famous expression."},{"Start":"02:36.680 ","End":"02:42.486","Text":"W is the number of microstates of a system at a particular energy."},{"Start":"02:42.486 ","End":"02:46.715","Text":"What\u0027s the meaning of W?"},{"Start":"02:46.715 ","End":"02:52.010","Text":"W is the number of ways the atoms or molecules in a sample can"},{"Start":"02:52.010 ","End":"02:57.545","Text":"be arranged while still giving the same total energy."},{"Start":"02:57.545 ","End":"03:04.430","Text":"Now, each arrangement of the atoms or molecules is called a microstate."},{"Start":"03:04.430 ","End":"03:08.270","Text":"For example, if there\u0027s only 1 possible arrangement,"},{"Start":"03:08.270 ","End":"03:10.625","Text":"then W would be equal to 1."},{"Start":"03:10.625 ","End":"03:16.264","Text":"That\u0027s the number of microstates will be 1 and S will be equal to 0,"},{"Start":"03:16.264 ","End":"03:20.420","Text":"because S is k ln W,"},{"Start":"03:20.420 ","End":"03:24.865","Text":"the ln of 1 is 0 so S would be 0."},{"Start":"03:24.865 ","End":"03:28.065","Text":"Obviously, if there\u0027s more than 1 arrangement,"},{"Start":"03:28.065 ","End":"03:38.195","Text":"W will be greater than 1 so ln W will be greater than 0 and S will be greater than 0."},{"Start":"03:38.195 ","End":"03:42.240","Text":"Now what\u0027s the importance of this formula?"},{"Start":"03:42.460 ","End":"03:45.680","Text":"The Boltzmann formula sometimes called"},{"Start":"03:45.680 ","End":"03:50.210","Text":"the Boltzmann Planck formula for statistical entropy will often"},{"Start":"03:50.210 ","End":"03:52.250","Text":"consider Boltzmann entropy will cause"},{"Start":"03:52.250 ","End":"03:57.635","Text":"statistical entropy helps us to interpret entropy at the molecular level."},{"Start":"03:57.635 ","End":"04:01.730","Text":"The expression of Clausius doesn\u0027t really explain the origin of"},{"Start":"04:01.730 ","End":"04:07.415","Text":"entropy whereas Boltzmann can help to explain its origin."},{"Start":"04:07.415 ","End":"04:11.824","Text":"We can use this expression to calculate the entropy."},{"Start":"04:11.824 ","End":"04:16.610","Text":"It\u0027s way beyond this course to actually calculate the entropy using"},{"Start":"04:16.610 ","End":"04:22.820","Text":"statistical mechanics or thermodynamics but it\u0027s very interesting and very nice to do it."},{"Start":"04:22.820 ","End":"04:25.070","Text":"One needs quite a lot of knowledge of"},{"Start":"04:25.070 ","End":"04:29.915","Text":"quantum mechanics to understand statistical mechanics."},{"Start":"04:29.915 ","End":"04:32.434","Text":"Let\u0027s take an example."},{"Start":"04:32.434 ","End":"04:38.255","Text":"What is the statistical entropy of 1 mole of carbon monoxide CO,"},{"Start":"04:38.255 ","End":"04:43.900","Text":"assuming each molecule can be oriented in 1 of 2 ways."},{"Start":"04:43.900 ","End":"04:48.660","Text":"What is the statistical entropy of 1 mole of CO"},{"Start":"04:48.660 ","End":"04:54.025","Text":"assuming each molecule can be oriented in 1 of 2 ways?"},{"Start":"04:54.025 ","End":"04:57.875","Text":"It can be either CO or OC."},{"Start":"04:57.875 ","End":"05:02.855","Text":"It gives us a hint to help us with our calculations,"},{"Start":"05:02.855 ","End":"05:12.715","Text":"ln of x to the power a is a ln of x. Supposing we have just 1 molecule."},{"Start":"05:12.715 ","End":"05:15.575","Text":"It could be either like this or this."},{"Start":"05:15.575 ","End":"05:17.795","Text":"If we have 2 molecules,"},{"Start":"05:17.795 ","End":"05:20.545","Text":"they could be arranged in 4 different ways,"},{"Start":"05:20.545 ","End":"05:23.115","Text":"CO, CO,"},{"Start":"05:23.115 ","End":"05:25.800","Text":"OC, OC,"},{"Start":"05:25.800 ","End":"05:31.890","Text":"and CO, OC and OC, CO."},{"Start":"05:31.890 ","End":"05:34.365","Text":"If we have 2 molecules,"},{"Start":"05:34.365 ","End":"05:42.155","Text":"they could be arranged in 4 different ways and if we have a whole mole of molecules,"},{"Start":"05:42.155 ","End":"05:47.470","Text":"we have 2 to the power of Avogadro\u0027s number,"},{"Start":"05:47.470 ","End":"05:52.640","Text":"2 to the power of 6.02 times 10 to the power 23."},{"Start":"05:52.640 ","End":"05:56.835","Text":"We saw that ln of x to the power a is"},{"Start":"05:56.835 ","End":"06:03.045","Text":"a ln x so here we could write S is equal to K, Boltzmann constant,"},{"Start":"06:03.045 ","End":"06:09.120","Text":"the ln of all this and that can be written as k"},{"Start":"06:09.120 ","End":"06:17.580","Text":"1.381 times 10 to the power minus 23 Joules per Kelvin times Avogadro\u0027s number,"},{"Start":"06:17.580 ","End":"06:23.620","Text":"6.02 times 10 to the power 23 times the ln of 2."},{"Start":"06:23.620 ","End":"06:31.830","Text":"If we work that out, we get 5.76 Joules per Kelvin. Here are the units."},{"Start":"06:31.830 ","End":"06:41.900","Text":"Here\u0027s our value of W. Now if we actually measure the entropy at very low temperatures,"},{"Start":"06:41.900 ","End":"06:45.895","Text":"we discover it\u0027s 4.6 Joules per Kelvin."},{"Start":"06:45.895 ","End":"06:50.270","Text":"Now we know that volume molecules had the same orientation."},{"Start":"06:50.270 ","End":"06:58.350","Text":"They could only be arranged 1 way and S would be equal to 0 because W is 1."},{"Start":"07:00.160 ","End":"07:08.015","Text":"The deviation of 0 entropy near 0 Kelvin is called residual entropy."},{"Start":"07:08.015 ","End":"07:14.160","Text":"That\u0027s deviation because not all the molecules are in the same orientation."},{"Start":"07:14.620 ","End":"07:19.355","Text":"We see that it\u0027s a similarity between the 2 values,"},{"Start":"07:19.355 ","End":"07:24.020","Text":"between the value we calculated assuming that"},{"Start":"07:24.020 ","End":"07:30.395","Text":"each molecule could be in 2 possible ways and the actual experimental volume,"},{"Start":"07:30.395 ","End":"07:32.990","Text":"just 4.6 Joules per Kelvin."},{"Start":"07:32.990 ","End":"07:40.055","Text":"We can conclude that this suggests that the CO molecules are arranged nearly randomly."},{"Start":"07:40.055 ","End":"07:42.260","Text":"They\u0027re not exactly equal."},{"Start":"07:42.260 ","End":"07:45.480","Text":"What we calculated was slightly greater than what\u0027s"},{"Start":"07:45.480 ","End":"07:50.135","Text":"measured so they\u0027re not arranged completely randomly,"},{"Start":"07:50.135 ","End":"07:53.990","Text":"but there\u0027s a very large measure of randomness."},{"Start":"07:53.990 ","End":"07:57.800","Text":"In this video, we introduced the third law of"},{"Start":"07:57.800 ","End":"08:02.159","Text":"thermodynamics and the Boltzmann formula for entropy."}],"ID":26375},{"Watched":false,"Name":"Equivalence of Statistical and Thermodynamic Entropy 1","Duration":"7m 21s","ChapterTopicVideoID":25605,"CourseChapterTopicPlaylistID":244889,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:02.610","Text":"In the previous videos,"},{"Start":"00:02.610 ","End":"00:06.870","Text":"we introduced the thermodynamic and statistical definitions of entropy."},{"Start":"00:06.870 ","End":"00:08.760","Text":"In this video and the next,"},{"Start":"00:08.760 ","End":"00:13.740","Text":"we will show that the 2 definitions of entropy are equivalent."},{"Start":"00:13.740 ","End":"00:16.740","Text":"Here are the 2 definitions of entropy."},{"Start":"00:16.740 ","End":"00:22.230","Text":"The thermodynamic entropy is Delta S equal to q_rev divided by capital T and"},{"Start":"00:22.230 ","End":"00:27.930","Text":"the statistical entropy is Delta S equal to k ln W. Now,"},{"Start":"00:27.930 ","End":"00:32.285","Text":"in the previous video, we talked about the volume dependence of an ideal gas."},{"Start":"00:32.285 ","End":"00:39.800","Text":"Here, the aim is to prove that Delta S equal to nR ln V_2 divided by V_1,"},{"Start":"00:39.800 ","End":"00:41.180","Text":"which we showed using"},{"Start":"00:41.180 ","End":"00:48.270","Text":"the thermodynamic approach is also true when we use the statistical approach."},{"Start":"00:48.340 ","End":"00:54.740","Text":"To do this, we\u0027re going to model an ideal gas by a 1D particle in the box."},{"Start":"00:54.740 ","End":"01:01.985","Text":"This is a model that we discussed when we talked about electrons in atoms."},{"Start":"01:01.985 ","End":"01:04.700","Text":"According to the particle in a box,"},{"Start":"01:04.700 ","End":"01:06.680","Text":"the energy levels E_n,"},{"Start":"01:06.680 ","End":"01:07.850","Text":"n is an index,"},{"Start":"01:07.850 ","End":"01:09.530","Text":"runs from 1,"},{"Start":"01:09.530 ","End":"01:11.260","Text":"2 and so on."},{"Start":"01:11.260 ","End":"01:14.985","Text":"E_n is equal to n^2h^2,"},{"Start":"01:14.985 ","End":"01:17.134","Text":"h is Planck\u0027s constant,"},{"Start":"01:17.134 ","End":"01:20.370","Text":"divided by 8mL^2,"},{"Start":"01:20.370 ","End":"01:23.435","Text":"m is the mass of the particle in the box,"},{"Start":"01:23.435 ","End":"01:26.390","Text":"and L is the length of the box."},{"Start":"01:26.390 ","End":"01:31.819","Text":"We\u0027ll see that when the length of the box increases at constant temperature,"},{"Start":"01:31.819 ","End":"01:34.010","Text":"the energy levels become closer,"},{"Start":"01:34.010 ","End":"01:36.875","Text":"so more energy levels can be populated."},{"Start":"01:36.875 ","End":"01:41.540","Text":"We can see here that E_n is proportional to 1/L^2."},{"Start":"01:41.540 ","End":"01:51.100","Text":"L increases, E_n decreases."},{"Start":"01:51.550 ","End":"01:53.680","Text":"On the left-hand side,"},{"Start":"01:53.680 ","End":"01:59.125","Text":"we have a box which has the length of 1 unit."},{"Start":"01:59.125 ","End":"02:01.920","Text":"X runs from 0-1."},{"Start":"02:01.920 ","End":"02:03.290","Text":"On the right-hand side,"},{"Start":"02:03.290 ","End":"02:07.535","Text":"we have a box where X runs from 0-1.5."},{"Start":"02:07.535 ","End":"02:12.735","Text":"The right-hand box is 1.5 times the length of the left-hand one."},{"Start":"02:12.735 ","End":"02:15.409","Text":"Here are the energy levels."},{"Start":"02:15.409 ","End":"02:17.995","Text":"This is n equal to 1,"},{"Start":"02:17.995 ","End":"02:21.435","Text":"2, 3, and 4."},{"Start":"02:21.435 ","End":"02:28.850","Text":"The black line gives us the amount of energy available and it\u0027s the same,"},{"Start":"02:28.850 ","End":"02:30.710","Text":"the left and the right."},{"Start":"02:30.710 ","End":"02:32.550","Text":"Now, what\u0027s the difference?"},{"Start":"02:32.550 ","End":"02:35.030","Text":"In the right, the box is longer,"},{"Start":"02:35.030 ","End":"02:38.480","Text":"is bigger, and the energy levels are closer."},{"Start":"02:38.480 ","End":"02:41.110","Text":"We have n equal to 1, 2,"},{"Start":"02:41.110 ","End":"02:43.065","Text":"3, 4,"},{"Start":"02:43.065 ","End":"02:45.820","Text":"5, and 6."},{"Start":"02:45.820 ","End":"02:52.790","Text":"I\u0027ve distributed the molecules or atoms amongst these energy levels."},{"Start":"02:52.790 ","End":"02:53.930","Text":"Of course, in reality,"},{"Start":"02:53.930 ","End":"02:56.810","Text":"we\u0027d have many more particles."},{"Start":"02:56.810 ","End":"02:59.165","Text":"Here, we\u0027ve just got a few particles."},{"Start":"02:59.165 ","End":"03:03.170","Text":"On the right-hand side, we can see that the particles are"},{"Start":"03:03.170 ","End":"03:06.920","Text":"distributed amongst many more energy levels,"},{"Start":"03:06.920 ","End":"03:10.560","Text":"1-6 as opposed to 1-4."},{"Start":"03:10.600 ","End":"03:12.670","Text":"On the right-hand side,"},{"Start":"03:12.670 ","End":"03:15.280","Text":"there are many more ways of distributing"},{"Start":"03:15.280 ","End":"03:22.070","Text":"the atoms or molecules than there are in the left-hand side."},{"Start":"03:22.380 ","End":"03:28.345","Text":"We\u0027ve shown that as the box increases in length,"},{"Start":"03:28.345 ","End":"03:32.005","Text":"W increases, there\u0027re more microstates."},{"Start":"03:32.005 ","End":"03:37.340","Text":"W increases as the length of the box increases."},{"Start":"03:37.670 ","End":"03:41.665","Text":"Now, if we go over to a 3-dimensional model,"},{"Start":"03:41.665 ","End":"03:44.680","Text":"then we can assume that the number of microstates available to"},{"Start":"03:44.680 ","End":"03:47.920","Text":"a single molecule depends on the volume V. Before,"},{"Start":"03:47.920 ","End":"03:49.300","Text":"it depended on the length, now,"},{"Start":"03:49.300 ","End":"03:54.260","Text":"it depends on the volume because this is a 3D model."},{"Start":"03:56.000 ","End":"04:00.440","Text":"Now, W, the number of microstates is equal to"},{"Start":"04:00.440 ","End":"04:04.285","Text":"cV is proportional to the volume for 1 particle."},{"Start":"04:04.285 ","End":"04:09.810","Text":"For N particles, it (cV)^N."},{"Start":"04:09.860 ","End":"04:13.575","Text":"S is equal to k ln W,"},{"Start":"04:13.575 ","End":"04:18.960","Text":"current Boltzmann, that\u0027s k ln (cV)^N."},{"Start":"04:18.960 ","End":"04:27.910","Text":"Now, what we want is Delta S. Delta S is k ln (cV_2)^N."},{"Start":"04:27.910 ","End":"04:33.800","Text":"That\u0027s the final volume minus ln (cv_1)^N,"},{"Start":"04:33.800 ","End":"04:35.930","Text":"that\u0027s the initial volume."},{"Start":"04:35.930 ","End":"04:43.995","Text":"Now, we know that ln of x^a is a times ln x."},{"Start":"04:43.995 ","End":"04:50.190","Text":"We can write ln (cV_2)^N as"},{"Start":"04:50.190 ","End":"04:56.730","Text":"N ln cV_2 and the same thing for cV_1."},{"Start":"04:56.730 ","End":"05:06.377","Text":"We have kN times ln (cV_2) minus ln (cV_1)."},{"Start":"05:06.377 ","End":"05:08.440","Text":"That\u0027s equal to kN."},{"Start":"05:08.440 ","End":"05:16.705","Text":"Now, we know that the difference between ln a and ln b is ln of the ratio."},{"Start":"05:16.705 ","End":"05:20.610","Text":"We can write ln (cV_2) minus ln"},{"Start":"05:20.610 ","End":"05:28.090","Text":"(cV_1) as ln (cV_2) divided by (cV_1) and we can cancel the Cs."},{"Start":"05:28.280 ","End":"05:34.050","Text":"That is equal to kN ln V_2/V_1."},{"Start":"05:34.050 ","End":"05:42.540","Text":"Now,f we want to rewrite kN and show that it\u0027s equal to nR."},{"Start":"05:42.950 ","End":"05:47.360","Text":"We can do that by noting that k, Boltzmann\u0027s constant,"},{"Start":"05:47.360 ","End":"05:53.485","Text":"is equal to the gas constant R divided by N_A Avogadro\u0027s number."},{"Start":"05:53.485 ","End":"05:57.800","Text":"N, the number of moles is equal to N, capital N,"},{"Start":"05:57.800 ","End":"06:02.210","Text":"the number of particles divided by Avogadro\u0027s number."},{"Start":"06:02.210 ","End":"06:03.840","Text":"In this case here,"},{"Start":"06:03.840 ","End":"06:08.285","Text":"we can write the capital N equal nN_A,"},{"Start":"06:08.285 ","End":"06:12.320","Text":"the number of moles times Avogadro\u0027s number."},{"Start":"06:12.320 ","End":"06:16.250","Text":"Now, we have k times N. K"},{"Start":"06:16.250 ","End":"06:24.420","Text":"is R divided by N_A."},{"Start":"06:24.420 ","End":"06:26.595","Text":"This is kN,"},{"Start":"06:26.595 ","End":"06:31.800","Text":"k capital N is n times N_A multiplied by R divided by N_A."},{"Start":"06:31.800 ","End":"06:37.395","Text":"The N_A cancels and we get kN is equal to nR."},{"Start":"06:37.395 ","End":"06:43.715","Text":"Now, we have the final expression."},{"Start":"06:43.715 ","End":"06:50.460","Text":"Delta S is equal to nR ln V_2/V_1."},{"Start":"06:50.620 ","End":"06:53.630","Text":"That\u0027s what we wanted to prove."},{"Start":"06:53.630 ","End":"07:03.590","Text":"We\u0027ve proved that Delta S is equal to nR ln V_2/V_1 using a statistical approach."},{"Start":"07:03.590 ","End":"07:08.195","Text":"That means that at least in this respect, Clausius,"},{"Start":"07:08.195 ","End":"07:13.610","Text":"the thermodynamic approach is equivalent to the statistical approach."},{"Start":"07:13.610 ","End":"07:16.820","Text":"In this video, we showed that both definitions of"},{"Start":"07:16.820 ","End":"07:21.120","Text":"entropy gives the same result for an ideal gas."}],"ID":26380},{"Watched":false,"Name":"Equivalence of Statistical and Thermodynamic Entropy 2","Duration":"4m 30s","ChapterTopicVideoID":25606,"CourseChapterTopicPlaylistID":244889,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.190","Text":"In the previous video,"},{"Start":"00:02.190 ","End":"00:07.139","Text":"we showed the equivalence of the statistical thermodynamic entropy for ideal gases."},{"Start":"00:07.139 ","End":"00:13.110","Text":"In this video, we\u0027ll talk about more ways in which the definitions are equivalent."},{"Start":"00:13.110 ","End":"00:19.020","Text":"We\u0027re going to begin with the change in thermodynamic entropy with temperature."},{"Start":"00:19.020 ","End":"00:22.470","Text":"We showed in a previous video that Delta S,"},{"Start":"00:22.470 ","End":"00:25.170","Text":"the change in entropy is equal to C,"},{"Start":"00:25.170 ","End":"00:30.420","Text":"the heat capacity times ln of T_2 divided by T_1."},{"Start":"00:30.420 ","End":"00:40.945","Text":"Now, this shows that Delta S is positive if T_2 is greater than T_1."},{"Start":"00:40.945 ","End":"00:47.615","Text":"We want to see if we can reach the same conclusion using this statistical entropy."},{"Start":"00:47.615 ","End":"00:52.925","Text":"We\u0027re going to talk about changes statistical entropy with temperature."},{"Start":"00:52.925 ","End":"00:55.160","Text":"Now on the left-hand side,"},{"Start":"00:55.160 ","End":"00:57.890","Text":"we have a particle in a box."},{"Start":"00:57.890 ","End":"01:01.820","Text":"They\u0027re energy levels are particle in a box at low temperature."},{"Start":"01:01.820 ","End":"01:05.165","Text":"On the right-hand side we have the same energy levels,"},{"Start":"01:05.165 ","End":"01:07.925","Text":"but the temperature is higher."},{"Start":"01:07.925 ","End":"01:09.920","Text":"As the temperature increases,"},{"Start":"01:09.920 ","End":"01:14.420","Text":"more energy levels are populated here just goes up to this purple one,"},{"Start":"01:14.420 ","End":"01:17.600","Text":"and here it goes right up to this green one."},{"Start":"01:17.870 ","End":"01:23.585","Text":"As a result, there are more ways of arranging the particles at higher temperature."},{"Start":"01:23.585 ","End":"01:25.820","Text":"That means there are more microstates."},{"Start":"01:25.820 ","End":"01:30.740","Text":"W will be higher and the entropy will also be higher."},{"Start":"01:30.740 ","End":"01:34.100","Text":"This turns out to be true for all states of matter,"},{"Start":"01:34.100 ","End":"01:36.755","Text":"not just ideal gases."},{"Start":"01:36.755 ","End":"01:45.810","Text":"We\u0027ve got agreement with thermodynamic entropy that S increases as temperature increases."},{"Start":"01:50.090 ","End":"01:54.830","Text":"Here we can write it then entropy increases with increasing temperature."},{"Start":"01:54.830 ","End":"01:56.929","Text":"Very, very important conclusion."},{"Start":"01:56.929 ","End":"02:02.075","Text":"Now there are more equivalent properties of thermodynamics and statistical entropy."},{"Start":"02:02.075 ","End":"02:05.705","Text":"The first thing to note is that both are state functions."},{"Start":"02:05.705 ","End":"02:09.710","Text":"We solve this when we talked about the thermodynamic definition."},{"Start":"02:09.710 ","End":"02:13.670","Text":"Statistical entropy depends on the current state of the system,"},{"Start":"02:13.670 ","End":"02:15.950","Text":"not on its past history."},{"Start":"02:15.950 ","End":"02:19.240","Text":"It also is a state function."},{"Start":"02:19.240 ","End":"02:22.280","Text":"Now, both are extensive properties."},{"Start":"02:22.280 ","End":"02:25.505","Text":"That means they\u0027re proportional to the number of molecules."},{"Start":"02:25.505 ","End":"02:29.305","Text":"Let\u0027s remember what we found in the last video."},{"Start":"02:29.305 ","End":"02:34.485","Text":"We said that S_1 is equal to klnW."},{"Start":"02:34.485 ","End":"02:41.600","Text":"KInW is proportional to the kln(cV). V is the volume."},{"Start":"02:41.600 ","End":"02:44.990","Text":"C is just a proportionality constant to the power N,"},{"Start":"02:44.990 ","End":"02:46.864","Text":"where N is the number of atoms."},{"Start":"02:46.864 ","End":"02:51.815","Text":"We wrote that as kNln(cV)."},{"Start":"02:51.815 ","End":"02:54.680","Text":"Now supposing we double the number of molecules,"},{"Start":"02:54.680 ","End":"02:56.870","Text":"instead of having N molecules,"},{"Start":"02:56.870 ","End":"03:00.060","Text":"we have 2N molecules."},{"Start":"03:00.140 ","End":"03:02.570","Text":"Now we have S_2,"},{"Start":"03:02.570 ","End":"03:04.250","Text":"that\u0027s the entropy in this case,"},{"Start":"03:04.250 ","End":"03:08.915","Text":"is equal to kln(cV)^2N."},{"Start":"03:08.915 ","End":"03:13.165","Text":"We can write that as 2kNln(cV)."},{"Start":"03:13.165 ","End":"03:19.268","Text":"That\u0027s twice the previous entropy, twice S_1,"},{"Start":"03:19.268 ","End":"03:23.150","Text":"so we can see that when we doubled the number of molecules,"},{"Start":"03:23.150 ","End":"03:25.310","Text":"we got double the entropy."},{"Start":"03:25.310 ","End":"03:30.720","Text":"That means that the entropy is proportional to the number of molecules."},{"Start":"03:34.310 ","End":"03:39.125","Text":"Now, both the thermodynamic entropy"},{"Start":"03:39.125 ","End":"03:43.640","Text":"and the statistical entropy increase in a spontaneous change."},{"Start":"03:43.640 ","End":"03:45.545","Text":"According to the second law,"},{"Start":"03:45.545 ","End":"03:50.599","Text":"the entropy of an isolated system increases in a spontaneous process."},{"Start":"03:50.599 ","End":"03:53.030","Text":"We know that in a spontaneous change,"},{"Start":"03:53.030 ","End":"03:55.550","Text":"the system becomes more disordered,"},{"Start":"03:55.550 ","End":"03:58.295","Text":"so the W increases."},{"Start":"03:58.295 ","End":"04:04.280","Text":"Then the consequence of w increase in lnW increases and S increases."},{"Start":"04:04.280 ","End":"04:08.390","Text":"We can see that statistical entries also"},{"Start":"04:08.390 ","End":"04:12.590","Text":"in agreement with the second law of thermodynamics."},{"Start":"04:12.590 ","End":"04:15.920","Text":"It increases in a spontaneous change in"},{"Start":"04:15.920 ","End":"04:20.180","Text":"the same way as the thermodynamic entropy increases."},{"Start":"04:20.180 ","End":"04:24.030","Text":"In this video we showed that the definition of thermodynamic and"},{"Start":"04:24.030 ","End":"04:29.850","Text":"statistical entropy gives the same results, so they\u0027re consistent."}],"ID":26381},{"Watched":false,"Name":"Measuring Entropy as a Function of Temperature","Duration":"6m 21s","ChapterTopicVideoID":25607,"CourseChapterTopicPlaylistID":244889,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.560 ","End":"00:03.165","Text":"In the previous videos,"},{"Start":"00:03.165 ","End":"00:06.630","Text":"we learned about thermodynamic and statistical entropy."},{"Start":"00:06.630 ","End":"00:12.910","Text":"In this video, we\u0027ll discuss how to measure the entropy as a function of temperature."},{"Start":"00:13.460 ","End":"00:19.910","Text":"We\u0027re interested in measuring the entropy at a temperature T. Now,"},{"Start":"00:19.910 ","End":"00:23.300","Text":"it\u0027s difficult to calculate the entropy for Boltzmann\u0027s definition,"},{"Start":"00:23.300 ","End":"00:26.970","Text":"you need to know quite a lot of quantum mechanics to do so."},{"Start":"00:27.010 ","End":"00:30.560","Text":"But we can combine the thermodynamic definition,"},{"Start":"00:30.560 ","End":"00:35.750","Text":"the third law to measure entropy as a function of temperature."},{"Start":"00:35.750 ","End":"00:39.320","Text":"We know that according to the third law of thermodynamics,"},{"Start":"00:39.320 ","End":"00:43.415","Text":"S temperature T equals 0 is equal to 0."},{"Start":"00:43.415 ","End":"00:48.260","Text":"We can write S temperature T is equal S at temperature"},{"Start":"00:48.260 ","End":"00:54.420","Text":"0 plus S temperature T minus S temperature 0."},{"Start":"00:54.530 ","End":"01:02.345","Text":"Of course, these are equal because S at temperature 0 minus S at temperature 0 is just 0."},{"Start":"01:02.345 ","End":"01:06.919","Text":"This gives us S temperature 0 is equal to 0"},{"Start":"01:06.919 ","End":"01:12.823","Text":"and S temperature T minus S temperature 0 is Delta S,"},{"Start":"01:12.823 ","End":"01:20.070","Text":"so we get the S temperature T is equal to Delta S. That means heating from 0 to T,"},{"Start":"01:20.070 ","End":"01:25.425","Text":"and we need to know the difference in entropy between these two temperatures."},{"Start":"01:25.425 ","End":"01:27.385","Text":"Now in the previous video,"},{"Start":"01:27.385 ","End":"01:29.665","Text":"we calculated Delta S,"},{"Start":"01:29.665 ","End":"01:33.085","Text":"assuming that the heat capacity C was constant."},{"Start":"01:33.085 ","End":"01:38.664","Text":"Then we wrote Delta S is equal to C outside the integral,"},{"Start":"01:38.664 ","End":"01:41.865","Text":"the integral from T_1 to T_2,"},{"Start":"01:41.865 ","End":"01:46.815","Text":"dT divided by capital T. We worked that out,"},{"Start":"01:46.815 ","End":"01:50.205","Text":"we got C In T_2 over T_1."},{"Start":"01:50.205 ","End":"01:56.660","Text":"Now that assumes C is constant over the temperature range T_1 to T_2."},{"Start":"01:56.660 ","End":"02:01.405","Text":"That\u0027s generally only true for quite short ranges."},{"Start":"02:01.405 ","End":"02:08.465","Text":"In this video, we\u0027ll see how to calculate this Delta S when C is not a constant,"},{"Start":"02:08.465 ","End":"02:15.200","Text":"and that allows us to go over a far larger range of temperatures."},{"Start":"02:15.480 ","End":"02:18.310","Text":"Now if C does change with temperature,"},{"Start":"02:18.310 ","End":"02:21.955","Text":"we can write S at temperature T is equal to Delta S,"},{"Start":"02:21.955 ","End":"02:25.210","Text":"the integral from 0 to T,"},{"Start":"02:25.210 ","End":"02:28.530","Text":"C_p dT over T. That\u0027s"},{"Start":"02:28.530 ","End":"02:32.850","Text":"a constant pressure so we have the heat capacity at constant pressure."},{"Start":"02:33.760 ","End":"02:41.840","Text":"Here\u0027s how we can measure the entropy using this expression integral from 0 to T of"},{"Start":"02:41.840 ","End":"02:44.180","Text":"C_p dT divided by"},{"Start":"02:44.180 ","End":"02:51.529","Text":"T. The first thing to do is to measure C_p as a function of temperature."},{"Start":"02:51.529 ","End":"02:58.090","Text":"We\u0027ll see that C_p becomes smaller as the temperature decreases."},{"Start":"02:58.090 ","End":"03:06.800","Text":"Here is C_p increases as temperature increases first rapidly and then more slowly."},{"Start":"03:06.800 ","End":"03:11.730","Text":"Then we plot C_p divided by T as"},{"Start":"03:11.730 ","End":"03:17.349","Text":"a function of T. Now we see that C_p increases with temperature,"},{"Start":"03:17.349 ","End":"03:20.025","Text":"1 over T decreases with temperature,"},{"Start":"03:20.025 ","End":"03:23.369","Text":"so some point we\u0027re going to have a maximum."},{"Start":"03:24.230 ","End":"03:28.740","Text":"We have C_p divided by T as a function of T,"},{"Start":"03:28.740 ","End":"03:33.220","Text":"rises, gets to a maximum, and then decreases."},{"Start":"03:33.400 ","End":"03:38.090","Text":"Supposing we\u0027re interested at this particular temperature."},{"Start":"03:38.090 ","End":"03:41.780","Text":"Now, we need to calculate the integral,"},{"Start":"03:41.780 ","End":"03:47.165","Text":"this area underneath this curve at this particular temperature."},{"Start":"03:47.165 ","End":"03:50.490","Text":"We\u0027re calculating this area here,"},{"Start":"03:51.860 ","End":"03:58.420","Text":"then we can plot the area at that particular temperature,"},{"Start":"03:58.420 ","End":"04:00.980","Text":"so we get this point here."},{"Start":"04:01.460 ","End":"04:06.715","Text":"Now, we can repeat the process and at every temperature,"},{"Start":"04:06.715 ","End":"04:10.315","Text":"calculate the area underneath the curve."},{"Start":"04:10.315 ","End":"04:18.175","Text":"The area will increase with temperature as we proceed to higher and higher temperatures."},{"Start":"04:18.175 ","End":"04:20.245","Text":"It looks something like this."},{"Start":"04:20.245 ","End":"04:23.770","Text":"Of course, all these graphs are rather approximate."},{"Start":"04:23.770 ","End":"04:30.649","Text":"Here is our entropy at this particular temperature."},{"Start":"04:31.370 ","End":"04:36.275","Text":"So far we haven\u0027t included phase transitions."},{"Start":"04:36.275 ","End":"04:44.030","Text":"We\u0027re talking about water and we\u0027re going from minus 20-120 degrees Celsius,"},{"Start":"04:44.030 ","End":"04:49.280","Text":"then the ice will melt at"},{"Start":"04:49.280 ","End":"04:54.650","Text":"zero degrees Celsius and then as we heat further,"},{"Start":"04:54.650 ","End":"04:58.160","Text":"it will vaporize at 100 degrees Celsius."},{"Start":"04:58.160 ","End":"05:01.594","Text":"We have two phase transitions involved."},{"Start":"05:01.594 ","End":"05:04.115","Text":"Let\u0027s try to draw this."},{"Start":"05:04.115 ","End":"05:11.004","Text":"S is a function of T will increase for the solid."},{"Start":"05:11.004 ","End":"05:15.365","Text":"Then we get a rapid change in S,"},{"Start":"05:15.365 ","End":"05:18.720","Text":"where we go from a solid to a liquid."},{"Start":"05:19.750 ","End":"05:25.720","Text":"Then again, an increase as the liquid is heated."},{"Start":"05:25.720 ","End":"05:34.120","Text":"Then another sudden change when we boil the liquid."},{"Start":"05:34.220 ","End":"05:37.180","Text":"Then of course we form the gas,"},{"Start":"05:37.180 ","End":"05:40.130","Text":"and there\u0027ll be another more gradual change."},{"Start":"05:40.130 ","End":"05:45.010","Text":"Here\u0027s the liquid to gas and here\u0027s gas."},{"Start":"05:45.010 ","End":"05:47.475","Text":"Had I drawn it better,"},{"Start":"05:47.475 ","End":"05:50.445","Text":"this would be longer than this one."},{"Start":"05:50.445 ","End":"05:57.520","Text":"This red bit should be longer than this one."},{"Start":"05:58.370 ","End":"06:03.694","Text":"We can say that the entropy increases sharply when there\u0027s a phase change,"},{"Start":"06:03.694 ","End":"06:07.670","Text":"and it increases more at the boiling point than the melting point."},{"Start":"06:07.670 ","End":"06:13.690","Text":"Here\u0027s melting, and here\u0027s boiling."},{"Start":"06:15.340 ","End":"06:21.210","Text":"In this video, we discussed entropy as a function of temperature."}],"ID":26382},{"Watched":false,"Name":"Standard Molar Entropy","Duration":"5m 31s","ChapterTopicVideoID":25598,"CourseChapterTopicPlaylistID":244889,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.990","Text":"In the previous video,"},{"Start":"00:01.990 ","End":"00:05.979","Text":"we saw how to measure the entropy at a particular temperature."},{"Start":"00:05.979 ","End":"00:10.540","Text":"In this video, we\u0027ll discuss standard molar entropy."},{"Start":"00:10.540 ","End":"00:14.095","Text":"Let\u0027s define the standard molar entropy."},{"Start":"00:14.095 ","End":"00:17.950","Text":"The standard molar entropy is the entropy of 1 mole"},{"Start":"00:17.950 ","End":"00:21.610","Text":"of a substance in its standard state and it\u0027s often"},{"Start":"00:21.610 ","End":"00:29.455","Text":"written as S with a little 0 or O at the top and then m at the bottom."},{"Start":"00:29.455 ","End":"00:34.810","Text":"Many books miss out the m and just write S^0."},{"Start":"00:34.810 ","End":"00:39.670","Text":"Now, just to remind you what we mean by a standard state."},{"Start":"00:39.670 ","End":"00:42.790","Text":"If we\u0027re talking about liquid or solid,"},{"Start":"00:42.790 ","End":"00:48.590","Text":"it\u0027s a pure element or compound at a pressure of 1 bar and specified temperature,"},{"Start":"00:48.590 ","End":"00:52.330","Text":"and it\u0027s usually 25 degrees Celsius."},{"Start":"00:52.330 ","End":"00:55.995","Text":"Most tables of entropies,"},{"Start":"00:55.995 ","End":"01:00.430","Text":"write the entropy at 25 degrees Celsius."},{"Start":"01:00.430 ","End":"01:02.960","Text":"If we\u0027re talking about a gas,"},{"Start":"01:02.960 ","End":"01:06.410","Text":"it\u0027s a pure gas behaving as an ideal gas and"},{"Start":"01:06.410 ","End":"01:10.085","Text":"the pressure of 1 bar and a specified temperature,"},{"Start":"01:10.085 ","End":"01:13.715","Text":"again, usually 25 degrees Celsius."},{"Start":"01:13.715 ","End":"01:19.100","Text":"Now, here\u0027s some generalizations about standard molar entropy."},{"Start":"01:19.100 ","End":"01:23.060","Text":"If we compare gases of different molar masses,"},{"Start":"01:23.060 ","End":"01:26.495","Text":"we find that the entropy increases."},{"Start":"01:26.495 ","End":"01:28.340","Text":"As the mass increases,"},{"Start":"01:28.340 ","End":"01:34.245","Text":"will see that the entropy increases. Why is this so?"},{"Start":"01:34.245 ","End":"01:38.164","Text":"According to the particle in a box model,"},{"Start":"01:38.164 ","End":"01:39.590","Text":"the greater the mass,"},{"Start":"01:39.590 ","End":"01:42.275","Text":"the closer the energy levels."},{"Start":"01:42.275 ","End":"01:45.170","Text":"E_n is proportional to 1 over m,"},{"Start":"01:45.170 ","End":"01:46.525","Text":"1 over mass,"},{"Start":"01:46.525 ","End":"01:49.130","Text":"so the closer the energy levels,"},{"Start":"01:49.130 ","End":"01:52.070","Text":"the more energy levels that are occupied,"},{"Start":"01:52.070 ","End":"01:58.055","Text":"and the more ways there are distributing the molecules amongst the energy levels,"},{"Start":"01:58.055 ","End":"02:00.745","Text":"so the entropy increases,"},{"Start":"02:00.745 ","End":"02:05.560","Text":"so as molar mass increases, the entropy increases."},{"Start":"02:05.560 ","End":"02:07.375","Text":"Here\u0027s some examples."},{"Start":"02:07.375 ","End":"02:11.275","Text":"We\u0027re going to go from hydrogen to nitrogen to oxygen."},{"Start":"02:11.275 ","End":"02:15.565","Text":"We\u0027re in going in the direction of increasing molar mass,"},{"Start":"02:15.565 ","End":"02:18.445","Text":"and we see that the entropy increases."},{"Start":"02:18.445 ","End":"02:24.550","Text":"We go from 130.7 to 291.6 to 205.1,"},{"Start":"02:24.550 ","End":"02:30.180","Text":"all in units of joule per mole per Kelvin at 25 degrees Celsius,"},{"Start":"02:30.180 ","End":"02:34.115","Text":"so as the mass increases, S^0 increases."},{"Start":"02:34.115 ","End":"02:38.920","Text":"Now, what happens when we increase the number of atoms in a molecule?"},{"Start":"02:38.920 ","End":"02:44.005","Text":"We\u0027ll see that as we increase the number of atoms, the entropy increases."},{"Start":"02:44.005 ","End":"02:45.970","Text":"Number of atoms goes up,"},{"Start":"02:45.970 ","End":"02:50.240","Text":"the entropy goes up and the reason for"},{"Start":"02:50.240 ","End":"02:54.860","Text":"this is larger molecules are more ways of vibrating."},{"Start":"02:54.860 ","End":"02:57.800","Text":"We say they have more vibrational degrees of freedom"},{"Start":"02:57.800 ","End":"03:01.790","Text":"and the number of vibrational degrees of freedom is 3N,"},{"Start":"03:01.790 ","End":"03:04.965","Text":"N is the number of atoms minus 6."},{"Start":"03:04.965 ","End":"03:06.755","Text":"So as N increases,"},{"Start":"03:06.755 ","End":"03:09.800","Text":"the entropy increases."},{"Start":"03:09.800 ","End":"03:12.635","Text":"Here\u0027s some examples."},{"Start":"03:12.635 ","End":"03:15.395","Text":"Water, which has 3 atoms,"},{"Start":"03:15.395 ","End":"03:18.170","Text":"the entropy is 69.9."},{"Start":"03:18.170 ","End":"03:21.524","Text":"Methanol, which has 9 atoms,"},{"Start":"03:21.524 ","End":"03:25.605","Text":"is a 160.7, so it\u0027s increased."},{"Start":"03:25.605 ","End":"03:31.100","Text":"Again, this is in joules per mole per Kelvin at 25 degrees Celsius,"},{"Start":"03:31.100 ","End":"03:36.955","Text":"so N increases, causes S to increase."},{"Start":"03:36.955 ","End":"03:43.145","Text":"Now the thing we need to recall is that liquids are more disordered than solids."},{"Start":"03:43.145 ","End":"03:47.210","Text":"Liquids have a higher entropy than solids."},{"Start":"03:47.210 ","End":"03:51.905","Text":"In liquids, the atoms and molecules can move about to some extent."},{"Start":"03:51.905 ","End":"03:54.860","Text":"Whereas in solids, they are fairly rigid."},{"Start":"03:54.860 ","End":"03:59.630","Text":"Liquids have greater entropies than solids."},{"Start":"03:59.630 ","End":"04:04.745","Text":"Here\u0027s an example, liquid water at 69.9,"},{"Start":"04:04.745 ","End":"04:07.570","Text":"whereas diamond is just 2.4,"},{"Start":"04:07.570 ","End":"04:17.525","Text":"so a solid here typified by diamond has a lower entropy than a liquid typified by water."},{"Start":"04:17.525 ","End":"04:22.580","Text":"Now gasses are even much more disordered than liquids."},{"Start":"04:22.580 ","End":"04:26.480","Text":"In gases, the atoms or molecules can move randomly,"},{"Start":"04:26.480 ","End":"04:30.170","Text":"whereas in liquids they can only move to some extent."},{"Start":"04:30.170 ","End":"04:33.740","Text":"Here we have water which is a liquid at"},{"Start":"04:33.740 ","End":"04:38.045","Text":"room temperature was 69.9 joules per mole per Kelvin."},{"Start":"04:38.045 ","End":"04:45.545","Text":"Whereas oxygen, which is a gas at room temperature has an entropy of 205.1."},{"Start":"04:45.545 ","End":"04:50.330","Text":"For the gas, it\u0027s much greater than for the liquid."},{"Start":"04:52.340 ","End":"04:56.120","Text":"We should recall, as we saw in previous videos,"},{"Start":"04:56.120 ","End":"05:02.005","Text":"that Delta S of vaporization is greater than Delta S of fusion."},{"Start":"05:02.005 ","End":"05:05.780","Text":"For example, Delta S of fusion of ice is"},{"Start":"05:05.780 ","End":"05:10.900","Text":"22.5 joules per mole per Kelvin at 0 degrees Celsius."},{"Start":"05:10.900 ","End":"05:14.145","Text":"Whereas Delta S is vaporization of water is"},{"Start":"05:14.145 ","End":"05:18.545","Text":"109 joules per mole per Kelvin at 100 degrees Celsius."},{"Start":"05:18.545 ","End":"05:24.900","Text":"Delta S for vaporization is large and Delta S of fusion."},{"Start":"05:25.870 ","End":"05:31.560","Text":"In this video, we learned about standard molar entropy."}],"ID":26373},{"Watched":false,"Name":"Standard Reaction Entropy","Duration":"7m 32s","ChapterTopicVideoID":25599,"CourseChapterTopicPlaylistID":244889,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"In the previous video,"},{"Start":"00:01.800 ","End":"00:04.530","Text":"we discussed standard molar entropies."},{"Start":"00:04.530 ","End":"00:08.685","Text":"And in this video, we\u0027ll learn about the standard entropy of reactions,"},{"Start":"00:08.685 ","End":"00:12.015","Text":"especially those that involve gases."},{"Start":"00:12.015 ","End":"00:16.560","Text":"We\u0027re going to talk about the standard entropy of reactions."},{"Start":"00:16.560 ","End":"00:19.950","Text":"The standard entropy of reaction is the difference between"},{"Start":"00:19.950 ","End":"00:24.270","Text":"the standard molar entropies of the products and those of the reactants."},{"Start":"00:24.270 ","End":"00:27.810","Text":"We\u0027ll calculate the total standard molar entropies of"},{"Start":"00:27.810 ","End":"00:32.640","Text":"the products and those of the reactants and subtract one from the other."},{"Start":"00:32.640 ","End":"00:36.935","Text":"Now a few points we need to remember before we get on with this,"},{"Start":"00:36.935 ","End":"00:41.390","Text":"the standard molar entropies of gases are higher than those of liquids or solids,"},{"Start":"00:41.390 ","End":"00:44.870","Text":"and that\u0027s because the gases are more disordered."},{"Start":"00:44.870 ","End":"00:48.589","Text":"As a result, if a reaction involves gases,"},{"Start":"00:48.589 ","End":"00:51.755","Text":"these will dominate the reaction entropy."},{"Start":"00:51.755 ","End":"00:57.305","Text":"Let\u0027s look at the ideal gas law, PV=nRT."},{"Start":"00:57.305 ","End":"01:02.185","Text":"We can see from this that the volume is proportional to the number of moles."},{"Start":"01:02.185 ","End":"01:05.010","Text":"If we have a reaction involving gases,"},{"Start":"01:05.010 ","End":"01:08.750","Text":"the number of moles will give us the volume."},{"Start":"01:08.750 ","End":"01:12.860","Text":"Now we saw a few videos ago that the change in"},{"Start":"01:12.860 ","End":"01:17.240","Text":"entropy Delta S of an ideal gas is equal to n,"},{"Start":"01:17.240 ","End":"01:19.700","Text":"the number of moles times R,"},{"Start":"01:19.700 ","End":"01:22.320","Text":"the gas constant,"},{"Start":"01:22.320 ","End":"01:26.910","Text":"ln of the ratio V_2 compared to V_1."},{"Start":"01:26.910 ","End":"01:32.600","Text":"If we identify V_2 as the products in a reaction,"},{"Start":"01:32.600 ","End":"01:34.945","Text":"here reactants to products,"},{"Start":"01:34.945 ","End":"01:39.440","Text":"and V_2 would be associated with the products,"},{"Start":"01:39.440 ","End":"01:41.705","Text":"we can also write that Vp,"},{"Start":"01:41.705 ","End":"01:43.970","Text":"and V_1 with the reactants,"},{"Start":"01:43.970 ","End":"01:45.875","Text":"we\u0027ll call that Vr."},{"Start":"01:45.875 ","End":"01:51.410","Text":"We can see that if the number of moles of products is greater than number of moles of"},{"Start":"01:51.410 ","End":"01:54.140","Text":"reactants so that the volume of the gas"},{"Start":"01:54.140 ","End":"01:57.620","Text":"in the products is greater than that in the reactants,"},{"Start":"01:57.620 ","End":"02:01.460","Text":"we will have Delta S greater than 0,"},{"Start":"02:01.460 ","End":"02:04.070","Text":"because the Vp is greater than Vr."},{"Start":"02:04.070 ","End":"02:08.599","Text":"This ratio will be greater than 1 and the ln will be positive."},{"Start":"02:08.599 ","End":"02:11.615","Text":"So we\u0027ll have Delta S positive."},{"Start":"02:11.615 ","End":"02:14.810","Text":"If on the other hand the number of moles of"},{"Start":"02:14.810 ","End":"02:17.959","Text":"products is less than number of moles in reactants,"},{"Start":"02:17.959 ","End":"02:20.810","Text":"Delta S will be negative."},{"Start":"02:20.810 ","End":"02:25.220","Text":"Now let\u0027s define the standard reaction enthalpy."},{"Start":"02:25.220 ","End":"02:32.420","Text":"Let\u0027s return to the standard reaction enthalpy that we learned about in thermochemistry."},{"Start":"02:32.420 ","End":"02:39.170","Text":"We defined the standard enthalpy of reaction to be the sum over all the products,"},{"Start":"02:39.170 ","End":"02:46.535","Text":"where we multiply the number of moles of each product with its enthalpy of formation."},{"Start":"02:46.535 ","End":"02:48.725","Text":"Then we subtract from this,"},{"Start":"02:48.725 ","End":"02:54.605","Text":"the sum over all the reactants and multiply in each case for each reactant,"},{"Start":"02:54.605 ","End":"02:59.450","Text":"the number of moles times the enthalpy of formation."},{"Start":"02:59.450 ","End":"03:02.990","Text":"That\u0027s what we learned when we talked about enthalpy."},{"Start":"03:02.990 ","End":"03:11.170","Text":"Now, we can write a similar expression for the entropy Delta S reaction,"},{"Start":"03:11.170 ","End":"03:16.250","Text":"the standard reaction entropy,"},{"Start":"03:16.250 ","End":"03:20.165","Text":"is equal to the sum over all the products,"},{"Start":"03:20.165 ","End":"03:22.430","Text":"where we multiply for each product,"},{"Start":"03:22.430 ","End":"03:27.665","Text":"the number of moles times the standard entropy."},{"Start":"03:27.665 ","End":"03:33.140","Text":"We do the same thing for the reactions with sum of all the reactants I\u0027ll multiply in"},{"Start":"03:33.140 ","End":"03:39.250","Text":"each case the number of moles times the standard entropy."},{"Start":"03:39.250 ","End":"03:40.730","Text":"The m, if you recall,"},{"Start":"03:40.730 ","End":"03:44.575","Text":"is just for moles, so molar volume."},{"Start":"03:44.575 ","End":"03:49.190","Text":"It\u0027s different from the enthalpy in that the enthalpy we only know the difference,"},{"Start":"03:49.190 ","End":"03:50.930","Text":"we only know Delta H,"},{"Start":"03:50.930 ","End":"03:57.240","Text":"whereas for the entropy we can know the actual volume of S. Here\u0027s an example."},{"Start":"03:57.240 ","End":"04:05.540","Text":"Estimate the sign of Delta S reaction without doing any calculations for the reaction N2,"},{"Start":"04:05.540 ","End":"04:09.695","Text":"the 3H2 to give 2NH3."},{"Start":"04:09.695 ","End":"04:11.600","Text":"That\u0027s the reaction between nitrogen,"},{"Start":"04:11.600 ","End":"04:13.885","Text":"hydrogen to give ammonia."},{"Start":"04:13.885 ","End":"04:16.640","Text":"After we\u0027ve estimated the sign,"},{"Start":"04:16.640 ","End":"04:18.665","Text":"we\u0027ve worked out what the sign should be,"},{"Start":"04:18.665 ","End":"04:25.789","Text":"we\u0027ll actually calculate Delta S for the reaction using standard molar entropies,"},{"Start":"04:25.789 ","End":"04:30.030","Text":"those which are given at the tables and here are the other values."},{"Start":"04:32.670 ","End":"04:35.945","Text":"Now, let\u0027s look at this reaction."},{"Start":"04:35.945 ","End":"04:40.415","Text":"There\u0027s 1 mole of N2 and 3 moles of hydrogen."},{"Start":"04:40.415 ","End":"04:42.335","Text":"That\u0027s 4 in the beginning."},{"Start":"04:42.335 ","End":"04:48.680","Text":"The reactants have 4 moles of gas and the products only have moles of gas."},{"Start":"04:48.680 ","End":"04:55.630","Text":"We could write Delta n equal to minus 2."},{"Start":"05:02.000 ","End":"05:06.160","Text":"As our 4 moles of reactants and only 2 of the products,"},{"Start":"05:06.160 ","End":"05:08.200","Text":"there\u0027s a decrease in the number of moles."},{"Start":"05:08.200 ","End":"05:10.245","Text":"Delta n is minus 2."},{"Start":"05:10.245 ","End":"05:13.545","Text":"There\u0027s a decrease in the volume of gas."},{"Start":"05:13.545 ","End":"05:16.795","Text":"Since there\u0027s a decrease in the volume of gas,"},{"Start":"05:16.795 ","End":"05:22.585","Text":"we expect that Delta S for the reaction will be negative."},{"Start":"05:22.585 ","End":"05:26.120","Text":"We\u0027re going to see whether this in fact true."},{"Start":"05:27.890 ","End":"05:32.305","Text":"Now, we\u0027re going to do the total calculation,"},{"Start":"05:32.305 ","End":"05:36.230","Text":"and we\u0027ll see if it\u0027s indeed Delta S is negative."},{"Start":"05:36.230 ","End":"05:39.475","Text":"We can write according to the formula we had above,"},{"Start":"05:39.475 ","End":"05:47.330","Text":"Delta S for the reaction is equal to 2 times S_0 for ammonia,"},{"Start":"05:47.330 ","End":"05:51.190","Text":"because we have 2 moles of ammonia."},{"Start":"05:51.230 ","End":"05:57.694","Text":"We\u0027re going to subtract 1 mole times"},{"Start":"05:57.694 ","End":"06:04.415","Text":"the entropy of nitrogen plus 3 moles times the entropy of hydrogen."},{"Start":"06:04.415 ","End":"06:08.015","Text":"So moles for the 2 and the 1 and the 3,"},{"Start":"06:08.015 ","End":"06:15.180","Text":"and joules per degree Kelvin per mole for the entropies."},{"Start":"06:16.070 ","End":"06:18.690","Text":"When we substitute the numbers,"},{"Start":"06:18.690 ","End":"06:21.825","Text":"we get 2 times 192.4."},{"Start":"06:21.825 ","End":"06:28.190","Text":"That\u0027s the value of the entropy for ammonia minus 1 times,"},{"Start":"06:28.190 ","End":"06:35.315","Text":"now the volume for nitrogen is 191.6 here,"},{"Start":"06:35.315 ","End":"06:38.150","Text":"plus 3 times the volume for hydrogen."},{"Start":"06:38.150 ","End":"06:40.705","Text":"The volume of hydrogen is 130.7."},{"Start":"06:40.705 ","End":"06:46.505","Text":"If we work all that out, we get -198.9."},{"Start":"06:46.505 ","End":"06:54.355","Text":"The units are joules per kelvin because moles times moles to the power -1, just give 1."},{"Start":"06:54.355 ","End":"06:56.385","Text":"Here\u0027s our answer,"},{"Start":"06:56.385 ","End":"07:03.000","Text":"minus 198.9 joules per Kelvin."},{"Start":"07:03.000 ","End":"07:06.155","Text":"As we estimated before,"},{"Start":"07:06.155 ","End":"07:09.090","Text":"this is indeed negative."},{"Start":"07:09.940 ","End":"07:15.605","Text":"We can easily find out whether the sign of Delta S,"},{"Start":"07:15.605 ","End":"07:20.640","Text":"if we have gases involved in a reaction."},{"Start":"07:20.950 ","End":"07:25.040","Text":"Here we confirmed it by doing the actual calculation."},{"Start":"07:25.040 ","End":"07:30.480","Text":"In this video we talked about the standard entropy of reactions."}],"ID":26374}],"Thumbnail":null,"ID":244889},{"Name":"Global Entropy Change","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Change in Entropy of Surroundings","Duration":"7m 58s","ChapterTopicVideoID":25612,"CourseChapterTopicPlaylistID":244890,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.200 ","End":"00:02.475","Text":"In a previous video,"},{"Start":"00:02.475 ","End":"00:07.230","Text":"we talked about spontaneous processes and the second law of thermodynamics."},{"Start":"00:07.230 ","End":"00:10.740","Text":"In this video, we\u0027ll talk about the change in total entropy,"},{"Start":"00:10.740 ","End":"00:13.575","Text":"and the entropy of surroundings."},{"Start":"00:13.575 ","End":"00:17.175","Text":"Let\u0027s recall the second law of thermodynamics."},{"Start":"00:17.175 ","End":"00:25.065","Text":"It says that the entropy of an isolated system increases in a spontaneous process."},{"Start":"00:25.065 ","End":"00:29.205","Text":"What do we mean by an isolated system?"},{"Start":"00:29.205 ","End":"00:33.390","Text":"We can see there\u0027s an isolated system consists of"},{"Start":"00:33.390 ","End":"00:37.110","Text":"the system of interest, and its surroundings."},{"Start":"00:37.110 ","End":"00:40.979","Text":"The 2 together make an isolated system,"},{"Start":"00:40.979 ","End":"00:46.560","Text":"which is not in contact with the rest of the universe."},{"Start":"00:47.200 ","End":"00:50.150","Text":"Here\u0027s our system."},{"Start":"00:50.150 ","End":"00:52.580","Text":"Here are the surroundings,"},{"Start":"00:52.580 ","End":"00:56.040","Text":"which we\u0027re going to assume are large."},{"Start":"00:56.650 ","End":"01:00.335","Text":"We can write that Delta S total."},{"Start":"01:00.335 ","End":"01:03.830","Text":"That means the entropy of the total system,"},{"Start":"01:03.830 ","End":"01:05.330","Text":"system plus surroundings,"},{"Start":"01:05.330 ","End":"01:11.810","Text":"is equal to delta S. Delta S is the change in entropy for the system itself."},{"Start":"01:11.810 ","End":"01:15.620","Text":"We could write the Delta S_sys or just without anything."},{"Start":"01:15.620 ","End":"01:19.535","Text":"Delta S plus Delta S of the surroundings."},{"Start":"01:19.535 ","End":"01:24.957","Text":"This is our total system to make the surroundings as large as we want."},{"Start":"01:24.957 ","End":"01:30.905","Text":"Now, let\u0027s discuss the entropy of the surroundings in a qualitative manner."},{"Start":"01:30.905 ","End":"01:35.315","Text":"Now if the process occurring in the system is exothermic,"},{"Start":"01:35.315 ","End":"01:37.745","Text":"that means it\u0027s releasing heat."},{"Start":"01:37.745 ","End":"01:41.900","Text":"The surroundings will increase in temperature so that the entropy of"},{"Start":"01:41.900 ","End":"01:47.780","Text":"the surroundings will increase and Delta S of the surroundings will be positive."},{"Start":"01:47.780 ","End":"01:52.745","Text":"If the process occurring in system is endothermic on the other hand,"},{"Start":"01:52.745 ","End":"01:59.615","Text":"that means it\u0027s drawing in energy from the surroundings."},{"Start":"01:59.615 ","End":"02:02.555","Text":"The surroundings will decrease in temperature."},{"Start":"02:02.555 ","End":"02:04.718","Text":"When they decrease in temperature,"},{"Start":"02:04.718 ","End":"02:08.360","Text":"the change in entropy will be negative."},{"Start":"02:08.360 ","End":"02:10.790","Text":"Now we want to try and find"},{"Start":"02:10.790 ","End":"02:16.655","Text":"a quantitative expression that will agree with these predictions."},{"Start":"02:16.655 ","End":"02:21.350","Text":"The expression is that at constant temperature and pressure,"},{"Start":"02:21.350 ","End":"02:25.685","Text":"Delta S of surroundings is equal to minus Delta H divided by"},{"Start":"02:25.685 ","End":"02:31.355","Text":"T. Delta H is the enthalpy change of the system."},{"Start":"02:31.355 ","End":"02:36.650","Text":"If Delta H is negative,"},{"Start":"02:36.650 ","End":"02:43.100","Text":"that means we\u0027re talking about an exothermic reaction or process,"},{"Start":"02:43.100 ","End":"02:50.635","Text":"then Delta S will be positive."},{"Start":"02:50.635 ","End":"02:56.245","Text":"It tells us also that if Delta H is positive,"},{"Start":"02:56.245 ","End":"03:03.970","Text":"Delta S will be negative and that\u0027s an agreement with what we said before."},{"Start":"03:03.970 ","End":"03:08.510","Text":"Now we want to prove this expression."},{"Start":"03:08.700 ","End":"03:14.155","Text":"We\u0027re going to assume that the temperature of the surroundings is constant."},{"Start":"03:14.155 ","End":"03:17.150","Text":"Delta S of the surroundings is equal to q_surroundings,"},{"Start":"03:17.150 ","End":"03:22.100","Text":"reversible divided by T."},{"Start":"03:22.350 ","End":"03:28.615","Text":"Now we know from the first law of thermodynamics that for an isolated system,"},{"Start":"03:28.615 ","End":"03:32.263","Text":"Delta U_total is 0."},{"Start":"03:32.263 ","End":"03:38.780","Text":"That\u0027s implies that Delta U of the system plus Delta U of the surroundings is 0."},{"Start":"03:38.780 ","End":"03:43.505","Text":"Delta U of the system is equal to minus Delta U of the surroundings."},{"Start":"03:43.505 ","End":"03:45.590","Text":"If we\u0027re not talking about work,"},{"Start":"03:45.590 ","End":"03:46.940","Text":"there\u0027s no work involved,"},{"Start":"03:46.940 ","End":"03:53.480","Text":"q of the system will be equal to minus q of the surroundings."},{"Start":"03:53.480 ","End":"03:56.585","Text":"Now we know that at constant pressure,"},{"Start":"03:56.585 ","End":"04:03.170","Text":"the heat exchanged is equal to Delta H, the enthalpy change."},{"Start":"04:03.170 ","End":"04:13.280","Text":"So q of the surroundings is at opposite sign to q of the system is minus Delta H. Now,"},{"Start":"04:13.280 ","End":"04:19.160","Text":"the transfer of heat into the surroundings is assumed to be"},{"Start":"04:19.160 ","End":"04:26.180","Text":"reversible because the surroundings are so large that the temperature is hardly altered."},{"Start":"04:26.180 ","End":"04:27.596","Text":"We can write that q_surroundings,"},{"Start":"04:27.596 ","End":"04:36.520","Text":"reversible is equal to minus Delta H. Then we can substitute q_surroundings,"},{"Start":"04:36.520 ","End":"04:43.160","Text":"reversible by minus delta H in this expression,"},{"Start":"04:43.160 ","End":"04:45.583","Text":"Delta S_surrounding equal to q_surroundings,"},{"Start":"04:45.583 ","End":"04:48.840","Text":"reversible divided by T."},{"Start":"04:48.840 ","End":"04:55.070","Text":"Then we get Delta S_surroundings is equal to q_surroundings, reversible."},{"Start":"04:55.070 ","End":"04:57.695","Text":"We\u0027ll divide by T, replace that,"},{"Start":"04:57.695 ","End":"05:01.235","Text":"and we get minus Delta H divided by T,"},{"Start":"05:01.235 ","End":"05:04.625","Text":"which is the expression we wanted to prove."},{"Start":"05:04.625 ","End":"05:07.880","Text":"Let\u0027s take an example."},{"Start":"05:07.880 ","End":"05:14.120","Text":"Show that Ice freezes spontaneously at minus 5 degrees Celsius."},{"Start":"05:14.120 ","End":"05:17.075","Text":"Assuming that Delta H of fusion,"},{"Start":"05:17.075 ","End":"05:21.635","Text":"and Delta S of fusion have the same values as 0 degrees Celsius,"},{"Start":"05:21.635 ","End":"05:26.360","Text":"calculate Delta S of the surroundings, and Delta S_total."},{"Start":"05:26.360 ","End":"05:28.114","Text":"Here are the values."},{"Start":"05:28.114 ","End":"05:32.212","Text":"Delta H of fusion is 6.01 kilojoules per mole and"},{"Start":"05:32.212 ","End":"05:37.650","Text":"Delta S of fusion is 22.0 joules per mole."},{"Start":"05:37.930 ","End":"05:40.760","Text":"If we\u0027re talking not about fusion,"},{"Start":"05:40.760 ","End":"05:41.870","Text":"not about melting,"},{"Start":"05:41.870 ","End":"05:45.799","Text":"but about freezing because the question asked about freezing,"},{"Start":"05:45.799 ","End":"05:50.675","Text":"we need Delta H_freezing is equal to minus Delta H of fusion."},{"Start":"05:50.675 ","End":"05:55.880","Text":"It\u0027s equal to minus 6.01 kilojoules per mole."},{"Start":"05:55.880 ","End":"06:01.250","Text":"Now we can substitute and we get Delta S of surroundings is equal"},{"Start":"06:01.250 ","End":"06:06.260","Text":"to 6.01 minus 1,000 joules per"},{"Start":"06:06.260 ","End":"06:12.245","Text":"mole because if you remember the expression was Delta S_surroundings"},{"Start":"06:12.245 ","End":"06:19.160","Text":"equal to minus Delta H over T. Delta H was negative."},{"Start":"06:19.160 ","End":"06:23.825","Text":"Now we have a positive value, so 6.01."},{"Start":"06:23.825 ","End":"06:29.390","Text":"We need it in joules, not in kilojoules."},{"Start":"06:29.390 ","End":"06:34.910","Text":"We multiply by 1,000 and then we divide by the temperature and"},{"Start":"06:34.910 ","End":"06:41.047","Text":"the temperature is minus 5 degrees Celsius, that\u0027s 268.15 Kelvin."},{"Start":"06:41.047 ","End":"06:42.350","Text":"If we do the calculation,"},{"Start":"06:42.350 ","End":"06:47.495","Text":"we get 22.41 joules per mole per Kelvin."},{"Start":"06:47.495 ","End":"06:52.685","Text":"We\u0027ve calculated Delta S of the surroundings."},{"Start":"06:52.685 ","End":"06:55.670","Text":"Now let\u0027s look at Delta S_total."},{"Start":"06:55.670 ","End":"07:00.830","Text":"Delta S_total is Delta S_freezing plus Delta S of surroundings."},{"Start":"07:00.830 ","End":"07:07.580","Text":"We see that\u0027s minus Delta S of fusion plus Delta S of surroundings."},{"Start":"07:07.580 ","End":"07:12.650","Text":"Delta S of fusion we saw here is 22.0 joules per mole,"},{"Start":"07:12.650 ","End":"07:17.270","Text":"so we have minus 22.0 joules per mole."},{"Start":"07:17.270 ","End":"07:20.294","Text":"The Delta S of the surroundings we already calculated,"},{"Start":"07:20.294 ","End":"07:25.145","Text":"it\u0027ll be 22.41 joules per mole per Kelvin."},{"Start":"07:25.145 ","End":"07:34.070","Text":"Then we add these 2 and we get a small net Delta S. Delta S is positive,"},{"Start":"07:34.070 ","End":"07:37.235","Text":"it\u0027s small, but it\u0027s positive."},{"Start":"07:37.235 ","End":"07:40.325","Text":"If Delta S is positive,"},{"Start":"07:40.325 ","End":"07:43.970","Text":"it means this is a spontaneous process."},{"Start":"07:43.970 ","End":"07:51.290","Text":"We\u0027ve shown that ice freezes spontaneously at minus 5 degrees Celsius."},{"Start":"07:51.290 ","End":"07:58.380","Text":"In this video, we talked about the total entropy and the entropy of the surroundings."}],"ID":30052},{"Watched":false,"Name":"Overall Change in Entropy","Duration":"9m 6s","ChapterTopicVideoID":25613,"CourseChapterTopicPlaylistID":244890,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:02.100","Text":"In the previous video,"},{"Start":"00:02.100 ","End":"00:04.830","Text":"we learned about the entropy change of the surroundings."},{"Start":"00:04.830 ","End":"00:08.985","Text":"In this video, we\u0027ll learn about the overall change in entropy."},{"Start":"00:08.985 ","End":"00:12.930","Text":"We\u0027re going to talk about the overall change in entropy."},{"Start":"00:12.930 ","End":"00:15.000","Text":"Now, we know from the second law of"},{"Start":"00:15.000 ","End":"00:18.645","Text":"thermodynamics that if the overall change in entropy is positive,"},{"Start":"00:18.645 ","End":"00:21.384","Text":"the process is spontaneous."},{"Start":"00:21.384 ","End":"00:22.800","Text":"The opposite is true."},{"Start":"00:22.800 ","End":"00:25.140","Text":"If the overall change in entropy is negative,"},{"Start":"00:25.140 ","End":"00:27.720","Text":"the process is nonspontaneous."},{"Start":"00:27.720 ","End":"00:31.755","Text":"If the overall change is neither positive nor negative, but it\u0027s 0,"},{"Start":"00:31.755 ","End":"00:33.570","Text":"the process is reversible,"},{"Start":"00:33.570 ","End":"00:35.175","Text":"can go in either direction,"},{"Start":"00:35.175 ","End":"00:37.995","Text":"and the system is an equilibrium."},{"Start":"00:37.995 ","End":"00:42.815","Text":"Now, you\u0027ll see all sorts of equivalent nomenclature for overall change."},{"Start":"00:42.815 ","End":"00:46.310","Text":"Sometimes you\u0027ll see overall entropy change,"},{"Start":"00:46.310 ","End":"00:47.975","Text":"or total entropy change,"},{"Start":"00:47.975 ","End":"00:49.670","Text":"or global entropy change,"},{"Start":"00:49.670 ","End":"00:51.755","Text":"or even entropy of the universe."},{"Start":"00:51.755 ","End":"00:53.720","Text":"They\u0027re all equivalent."},{"Start":"00:53.720 ","End":"00:56.480","Text":"Now we\u0027re going to see how the entropy of the system and"},{"Start":"00:56.480 ","End":"01:00.500","Text":"the surroundings inferences the total entropy."},{"Start":"01:00.500 ","End":"01:03.410","Text":"We can write that Delta S total is equal to"},{"Start":"01:03.410 ","End":"01:06.635","Text":"Delta S of the system plus Delta S of the surroundings."},{"Start":"01:06.635 ","End":"01:08.720","Text":"I\u0027ve drawn this in green, red,"},{"Start":"01:08.720 ","End":"01:13.495","Text":"and blue because these are the colors I\u0027m going to use in the diagrams."},{"Start":"01:13.495 ","End":"01:16.800","Text":"Here are 4 cases."},{"Start":"01:16.800 ","End":"01:20.415","Text":"The first case, Delta S of the system,"},{"Start":"01:20.415 ","End":"01:22.640","Text":"Delta S of the surroundings are both positive,"},{"Start":"01:22.640 ","End":"01:24.620","Text":"so there\u0027s some positive."},{"Start":"01:24.620 ","End":"01:28.470","Text":"Delta S total is positive."},{"Start":"01:34.330 ","End":"01:36.410","Text":"We can write that if"},{"Start":"01:36.410 ","End":"01:39.979","Text":"both Delta S of the system and Delta S of the surrounding is a positive,"},{"Start":"01:39.979 ","End":"01:42.170","Text":"the process is spontaneous,"},{"Start":"01:42.170 ","End":"01:45.025","Text":"because Delta S total is positive."},{"Start":"01:45.025 ","End":"01:49.545","Text":"The second case, Delta S of the system is negative."},{"Start":"01:49.545 ","End":"01:51.605","Text":"As Delta S surroundings,"},{"Start":"01:51.605 ","End":"01:54.620","Text":"so that the sum of these 2 is also negative."},{"Start":"01:54.620 ","End":"01:58.050","Text":"Here Delta S total is negative."},{"Start":"02:00.890 ","End":"02:04.250","Text":"We can write that if both Delta S of the system,"},{"Start":"02:04.250 ","End":"02:06.320","Text":"Delta S of surrounding is a negative,"},{"Start":"02:06.320 ","End":"02:11.810","Text":"the process is non-spontaneous because Delta S total is negative."},{"Start":"02:11.810 ","End":"02:16.340","Text":"Now, here we have 2 cases in which Delta S of the system,"},{"Start":"02:16.340 ","End":"02:19.580","Text":"Delta S of the surroundings have opposite signs."},{"Start":"02:19.580 ","End":"02:25.355","Text":"In this case, the system is positive, surroundings is negative."},{"Start":"02:25.355 ","End":"02:29.825","Text":"In this case the system is negative, surroundings is positive."},{"Start":"02:29.825 ","End":"02:32.720","Text":"Let\u0027s look at these 2 cases."},{"Start":"02:32.720 ","End":"02:37.069","Text":"In this case, Delta S of the system is positive,"},{"Start":"02:37.069 ","End":"02:40.310","Text":"and Delta S of the surroundings is negative,"},{"Start":"02:40.310 ","End":"02:42.690","Text":"but Delta S of the system is longer,"},{"Start":"02:42.690 ","End":"02:47.910","Text":"it\u0027s larger than minus Delta S of the surroundings."},{"Start":"02:48.110 ","End":"02:52.550","Text":"The sum of these 2 vectors is positive."},{"Start":"02:52.550 ","End":"02:57.840","Text":"Here we have Delta S, total is positive."},{"Start":"02:58.060 ","End":"03:00.120","Text":"We can sum that up,"},{"Start":"03:00.120 ","End":"03:01.430","Text":"if Delta S is positive,"},{"Start":"03:01.430 ","End":"03:03.245","Text":"Delta S surroundings is negative,"},{"Start":"03:03.245 ","End":"03:07.445","Text":"but Delta S is greater than minus Delta S of the surroundings."},{"Start":"03:07.445 ","End":"03:09.883","Text":"The process is spontaneous,"},{"Start":"03:09.883 ","End":"03:12.590","Text":"and that\u0027s because Delta S total is positive."},{"Start":"03:12.590 ","End":"03:14.385","Text":"Now, in the fourth case,"},{"Start":"03:14.385 ","End":"03:20.605","Text":"Delta S of the system is negative and Delta S of the surroundings is positive."},{"Start":"03:20.605 ","End":"03:23.670","Text":"The sum of the 2 is still positive,"},{"Start":"03:23.670 ","End":"03:32.090","Text":"because Delta S of the surroundings is greater than minus Delta S of the system."},{"Start":"03:32.090 ","End":"03:37.100","Text":"The sum of the red and blue vectors is a green vector which is positive."},{"Start":"03:37.100 ","End":"03:40.505","Text":"Once again, Delta S total is positive."},{"Start":"03:40.505 ","End":"03:42.740","Text":"You write that,"},{"Start":"03:42.740 ","End":"03:44.210","Text":"if Delta S is negative,"},{"Start":"03:44.210 ","End":"03:46.100","Text":"Delta S surroundings is positive,"},{"Start":"03:46.100 ","End":"03:50.315","Text":"but Delta S surroundings is greater than minus Delta S of the system."},{"Start":"03:50.315 ","End":"03:53.785","Text":"The process is again spontaneous."},{"Start":"03:53.785 ","End":"03:56.655","Text":"Now we can answer the question."},{"Start":"03:56.655 ","End":"03:59.801","Text":"Can an endothermic reaction be spontaneous?"},{"Start":"03:59.801 ","End":"04:04.175","Text":"At one time, it was thought that only exothermic reactions was spontaneous."},{"Start":"04:04.175 ","End":"04:09.190","Text":"But we\u0027ll see now that\u0027s an endothermic reaction can indeed be spontaneous."},{"Start":"04:09.190 ","End":"04:12.050","Text":"Now, for an endothermic reaction to be spontaneous,"},{"Start":"04:12.050 ","End":"04:15.980","Text":"we need Delta S total is greater than 0."},{"Start":"04:15.980 ","End":"04:17.705","Text":"We know that already."},{"Start":"04:17.705 ","End":"04:19.640","Text":"For an endothermic reaction,"},{"Start":"04:19.640 ","End":"04:22.235","Text":"Delta H is positive."},{"Start":"04:22.235 ","End":"04:28.505","Text":"Delta H is positive means that it\u0027s taking in heat from the surroundings."},{"Start":"04:28.505 ","End":"04:33.125","Text":"Delta S surroundings will be negative."},{"Start":"04:33.125 ","End":"04:38.625","Text":"Delta H positive gives us Delta H of the surroundings is negative."},{"Start":"04:38.625 ","End":"04:42.735","Text":"Now, that is this case here,"},{"Start":"04:42.735 ","End":"04:44.760","Text":"Delta S of the system,"},{"Start":"04:44.760 ","End":"04:48.695","Text":"positive, Delta S is the surroundings negative."},{"Start":"04:48.695 ","End":"04:56.315","Text":"We see that the sum can be positive provided the red one is longer than the blue vector."},{"Start":"04:56.315 ","End":"04:59.600","Text":"For the reaction to occur spontaneously,"},{"Start":"04:59.600 ","End":"05:03.830","Text":"Delta S is greater than minus Delta S surroundings."},{"Start":"05:03.830 ","End":"05:06.450","Text":"That\u0027s this third case."},{"Start":"05:08.180 ","End":"05:12.590","Text":"Now, we\u0027re going to talk about the Clausius inequality."},{"Start":"05:12.590 ","End":"05:19.370","Text":"Clausius inequality says Delta S is greater or equal to q over T,"},{"Start":"05:19.370 ","End":"05:23.660","Text":"q is the heat and T is the temperature."},{"Start":"05:23.660 ","End":"05:28.220","Text":"For reversible process, we know that Delta S is equal to q reversible"},{"Start":"05:28.220 ","End":"05:32.255","Text":"divided by T. That means that\u0027s the equality here,"},{"Start":"05:32.255 ","End":"05:38.405","Text":"Delta S equal to q over T. We\u0027ll see in a few moments that for an irreversible process,"},{"Start":"05:38.405 ","End":"05:45.530","Text":"Delta S is greater than q reversible over T. That\u0027s the greater than sign."},{"Start":"05:45.530 ","End":"05:48.160","Text":"Here\u0027s our proof."},{"Start":"05:48.160 ","End":"05:52.580","Text":"Now it can be shown that a process produces maximum work,"},{"Start":"05:52.580 ","End":"05:54.170","Text":"the most negative work,"},{"Start":"05:54.170 ","End":"05:56.540","Text":"if it occurs reversibly."},{"Start":"05:56.540 ","End":"06:02.060","Text":"Minus w reversible is always greater than minus w irreversible."},{"Start":"06:02.060 ","End":"06:07.200","Text":"Here\u0027s w reversible, and here w irreversible."},{"Start":"06:11.620 ","End":"06:15.410","Text":"We see that the length of"},{"Start":"06:15.410 ","End":"06:22.350","Text":"the vector w reversible is greater than that of the w irreversible vector."},{"Start":"06:22.690 ","End":"06:26.705","Text":"Now recall that Delta U is a state function."},{"Start":"06:26.705 ","End":"06:30.575","Text":"Change in internal energy is a state function."},{"Start":"06:30.575 ","End":"06:33.835","Text":"Delta U is equal to q plus w,"},{"Start":"06:33.835 ","End":"06:36.710","Text":"q is heat and w is work."},{"Start":"06:36.710 ","End":"06:39.110","Text":"Now, since it\u0027s a state function,"},{"Start":"06:39.110 ","End":"06:46.055","Text":"it doesn\u0027t matter whether we\u0027re talking about reversible or irreversible process,"},{"Start":"06:46.055 ","End":"06:53.100","Text":"q reversible plus w reversible will be equal to q irreversible plus w irreversible."},{"Start":"06:53.890 ","End":"06:57.145","Text":"This is illustrated here."},{"Start":"06:57.145 ","End":"07:01.090","Text":"Here we have w reversible,"},{"Start":"07:01.400 ","End":"07:05.620","Text":"and here we have w irreversible."},{"Start":"07:06.460 ","End":"07:13.325","Text":"The other arrows refer to the heat q, q reversible."},{"Start":"07:13.325 ","End":"07:16.680","Text":"This is q irreversible."},{"Start":"07:16.940 ","End":"07:22.440","Text":"In order for the sum of q_rev and w_rev to be equal to the sum of"},{"Start":"07:22.440 ","End":"07:28.205","Text":"q_irrev and w_irrev we need that q_rev will be greater,"},{"Start":"07:28.205 ","End":"07:34.130","Text":"will be longer than q_irrev switch in here in equations."},{"Start":"07:34.130 ","End":"07:38.570","Text":"If minus w_rev is greater than minus w_irrev,"},{"Start":"07:38.570 ","End":"07:41.750","Text":"then q_rev would be greater than q_irrev."},{"Start":"07:41.750 ","End":"07:43.610","Text":"That\u0027s implies that Delta S,"},{"Start":"07:43.610 ","End":"07:47.930","Text":"which is defined as q_rev divided by T will be greater than"},{"Start":"07:47.930 ","End":"07:53.485","Text":"q_irrev divided by T. Delta S,"},{"Start":"07:53.485 ","End":"07:55.845","Text":"which is equal q_rev over T,"},{"Start":"07:55.845 ","End":"08:01.810","Text":"is also greater than q_rev over T. That tells us that"},{"Start":"08:01.810 ","End":"08:10.240","Text":"Delta S is greater or equal to q over T. That\u0027s our inequality."},{"Start":"08:14.000 ","End":"08:18.535","Text":"Now, let\u0027s go back to the second law of thermodynamics."},{"Start":"08:18.535 ","End":"08:21.565","Text":"Recall for an isolated system,"},{"Start":"08:21.565 ","End":"08:24.430","Text":"q total is equal to 0."},{"Start":"08:24.430 ","End":"08:29.680","Text":"That\u0027s implies that Delta S total is greater or equal to 0."},{"Start":"08:29.680 ","End":"08:32.050","Text":"From here, if q is 0,"},{"Start":"08:32.050 ","End":"08:36.500","Text":"we get Delta S greater or equal to 0."},{"Start":"08:36.890 ","End":"08:41.195","Text":"This tells us that the entropy of an isolated system,"},{"Start":"08:41.195 ","End":"08:44.840","Text":"or sometimes it\u0027s written entropy of the universe cannot decrease."},{"Start":"08:44.840 ","End":"08:47.900","Text":"It\u0027s always greater or equal to 0."},{"Start":"08:47.900 ","End":"08:51.365","Text":"The entropy of the universe is steadily increasing."},{"Start":"08:51.365 ","End":"08:56.150","Text":"That\u0027s another way people express the second law of thermodynamics."},{"Start":"08:56.150 ","End":"09:01.835","Text":"We see we went from the Clausius inequality to the second law of thermodynamics."},{"Start":"09:01.835 ","End":"09:06.270","Text":"In this video, we talked about the overall entropy."}],"ID":30053},{"Watched":false,"Name":"Exercise 1","Duration":"4m 6s","ChapterTopicVideoID":28533,"CourseChapterTopicPlaylistID":244890,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:03.645","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.645 ","End":"00:06.240","Text":"Would you expect the change in entropy to be"},{"Start":"00:06.240 ","End":"00:09.164","Text":"positive or negative for the following reactions?"},{"Start":"00:09.164 ","End":"00:11.235","Text":"Explain your answer."},{"Start":"00:11.235 ","End":"00:13.785","Text":"Before we take a look at a,"},{"Start":"00:13.785 ","End":"00:15.675","Text":"and the rest of the equations,"},{"Start":"00:15.675 ","End":"00:19.065","Text":"let\u0027s talk a little bit about entropy."},{"Start":"00:19.065 ","End":"00:24.675","Text":"If we have a solid and a liquid and a gas,"},{"Start":"00:24.675 ","End":"00:27.434","Text":"comparing the liquid to the solid,"},{"Start":"00:27.434 ","End":"00:29.990","Text":"the liquid is more disordered than the solid,"},{"Start":"00:29.990 ","End":"00:33.195","Text":"and our particles will have more freedom of motion."},{"Start":"00:33.195 ","End":"00:40.425","Text":"Therefore, when we go from a solid to liquid, our entropy increases."},{"Start":"00:40.425 ","End":"00:46.895","Text":"The change in entropy is going to be positive when we go from a solid to liquid."},{"Start":"00:46.895 ","End":"00:49.369","Text":"Now when we go from a liquid to a gas,"},{"Start":"00:49.369 ","End":"00:52.295","Text":"our particles have even more freedom of motion,"},{"Start":"00:52.295 ","End":"00:54.035","Text":"and the energy is more dispersed."},{"Start":"00:54.035 ","End":"00:57.200","Text":"Therefore, again, the entropy increases and"},{"Start":"00:57.200 ","End":"01:01.090","Text":"the change in entropy is going to be positive."},{"Start":"01:01.090 ","End":"01:04.715","Text":"Obviously, when we go from a solid to a gas,"},{"Start":"01:04.715 ","End":"01:08.045","Text":"the change in entropy is also going to be positive."},{"Start":"01:08.045 ","End":"01:11.840","Text":"When we go the other way around from a gas to liquid,"},{"Start":"01:11.840 ","End":"01:14.165","Text":"we\u0027re going to expect our entropy to"},{"Start":"01:14.165 ","End":"01:18.800","Text":"decrease since our particles will have less freedom of motion."},{"Start":"01:18.800 ","End":"01:21.530","Text":"The same will happen when we go from a liquid to solid,"},{"Start":"01:21.530 ","End":"01:23.510","Text":"our particles will become more fixed."},{"Start":"01:23.510 ","End":"01:25.220","Text":"They will have less freedom of motion,"},{"Start":"01:25.220 ","End":"01:28.440","Text":"and therefore our entropy is going to decrease,"},{"Start":"01:28.440 ","End":"01:34.475","Text":"and the change in entropy is going to be smaller than 0 or negative."},{"Start":"01:34.475 ","End":"01:37.925","Text":"Now, in general, when we look at a chemical equation,"},{"Start":"01:37.925 ","End":"01:41.675","Text":"we can just look at the number of moles of gas"},{"Start":"01:41.675 ","End":"01:47.450","Text":"in the reactant side and the number of moles of gas in the product side and compare."},{"Start":"01:47.450 ","End":"01:53.695","Text":"If we have more moles of gas in the product side compared to the reactant side,"},{"Start":"01:53.695 ","End":"01:55.820","Text":"again, the number of moles of"},{"Start":"01:55.820 ","End":"01:59.345","Text":"gas increases when we go from the reactant side to the product side."},{"Start":"01:59.345 ","End":"02:05.470","Text":"It means that the entropy also increases since we have more moles of gas."},{"Start":"02:05.470 ","End":"02:08.030","Text":"If we go from the reactant side to the product side"},{"Start":"02:08.030 ","End":"02:10.564","Text":"and the number of moles of gas decrease,"},{"Start":"02:10.564 ","End":"02:14.420","Text":"we expect our entropy to decrease,"},{"Start":"02:14.420 ","End":"02:17.855","Text":"and the change in entropy will be negative."},{"Start":"02:17.855 ","End":"02:20.030","Text":"Let\u0027s take a look at a."},{"Start":"02:20.030 ","End":"02:21.650","Text":"In a in our reactant side,"},{"Start":"02:21.650 ","End":"02:23.060","Text":"we have 1 mole of a solid,"},{"Start":"02:23.060 ","End":"02:27.040","Text":"and in our product side, we have 1 mole of a solid and we have 1 mole of gas."},{"Start":"02:27.040 ","End":"02:29.210","Text":"If we compare the number of moles of gas,"},{"Start":"02:29.210 ","End":"02:30.755","Text":"in our reactant side we have none,"},{"Start":"02:30.755 ","End":"02:33.065","Text":"in our product side we have 1."},{"Start":"02:33.065 ","End":"02:36.035","Text":"The number of moles of gas increases,"},{"Start":"02:36.035 ","End":"02:38.150","Text":"therefore, our entropy increases."},{"Start":"02:38.150 ","End":"02:43.894","Text":"Therefore, the change in entropy in a is going to be positive."},{"Start":"02:43.894 ","End":"02:47.530","Text":"In b, we can see that in our reactant side we have 1 mole of liquid,"},{"Start":"02:47.530 ","End":"02:50.870","Text":"and in our product side we have 2 moles of gas."},{"Start":"02:50.870 ","End":"02:54.850","Text":"Again, we go from no gas moles to 2 moles of gas,"},{"Start":"02:54.850 ","End":"02:59.529","Text":"meaning the change in entropy is going to be positive."},{"Start":"02:59.529 ","End":"03:03.870","Text":"In c, we begin with 4 moles of solid and 4 moles of gas,"},{"Start":"03:03.870 ","End":"03:06.840","Text":"and in our product side we have 1 mole of a gas."},{"Start":"03:06.840 ","End":"03:08.560","Text":"Again, if we look at our gases,"},{"Start":"03:08.560 ","End":"03:11.840","Text":"we can see that in our reactant side we have 4 moles of gas,"},{"Start":"03:11.840 ","End":"03:14.010","Text":"and in our product side, only 1 mole of gas."},{"Start":"03:14.010 ","End":"03:18.360","Text":"The number of moles of gas decrease, therefore,"},{"Start":"03:18.360 ","End":"03:24.435","Text":"the change in entropy is going to be smaller than 0 or negative."},{"Start":"03:24.435 ","End":"03:27.480","Text":"Let\u0027s take a look at d. In d,"},{"Start":"03:27.480 ","End":"03:31.380","Text":"we have 2 moles of gas in our reactant side,"},{"Start":"03:31.380 ","End":"03:34.670","Text":"in our product side, we only have 1 mole of liquid."},{"Start":"03:34.670 ","End":"03:39.920","Text":"Again, 2 moles of gas on our reactant side and no moles of gas in our product side."},{"Start":"03:39.920 ","End":"03:42.515","Text":"The number of moles of gas decrease,"},{"Start":"03:42.515 ","End":"03:46.325","Text":"therefore, the change in entropy is negative."},{"Start":"03:46.325 ","End":"03:49.505","Text":"Again, for a, we found that the change in entropy is positive."},{"Start":"03:49.505 ","End":"03:53.042","Text":"In b, we found that the change in entropy is positive."},{"Start":"03:53.042 ","End":"03:55.910","Text":"C, we found that the change in entropy is negative."},{"Start":"03:55.910 ","End":"03:59.180","Text":"In d, we found that the change in entropy is negative based on"},{"Start":"03:59.180 ","End":"04:02.915","Text":"the number of moles of gas on our reactant side and on our product side."},{"Start":"04:02.915 ","End":"04:04.220","Text":"That is our final answer."},{"Start":"04:04.220 ","End":"04:06.720","Text":"Thank you very much for watching."}],"ID":30054},{"Watched":false,"Name":"Exercise 2","Duration":"3m 34s","ChapterTopicVideoID":28534,"CourseChapterTopicPlaylistID":244890,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.954","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.954 ","End":"00:05.400","Text":"Would you expect the change in entropy to be"},{"Start":"00:05.400 ","End":"00:08.505","Text":"positive or negative for the following processes?"},{"Start":"00:08.505 ","End":"00:10.245","Text":"Let\u0027s look at a."},{"Start":"00:10.245 ","End":"00:14.910","Text":"In a, we have solid copper at 350 degrees Celsius and 2.5"},{"Start":"00:14.910 ","End":"00:21.140","Text":"atmosphere and then we have solid copper at 450 degrees Celsius and 2.5 atmosphere."},{"Start":"00:21.140 ","End":"00:23.180","Text":"Meaning, if we look at our pressure,"},{"Start":"00:23.180 ","End":"00:25.340","Text":"we have 2.5 atmosphere in both cases,"},{"Start":"00:25.340 ","End":"00:26.750","Text":"so the pressure stays the same,"},{"Start":"00:26.750 ","End":"00:32.970","Text":"doesn\u0027t change, and our temperature increases, meaning we\u0027re heating."},{"Start":"00:32.970 ","End":"00:36.320","Text":"We go from 350 degrees Celsius to 450 degrees"},{"Start":"00:36.320 ","End":"00:40.195","Text":"Celsius and we want to know what happens to the change in entropy."},{"Start":"00:40.195 ","End":"00:42.845","Text":"When the temperature increases,"},{"Start":"00:42.845 ","End":"00:46.535","Text":"the particles get more kinetic energy"},{"Start":"00:46.535 ","End":"00:50.630","Text":"and therefore they have more freedom of motion and therefore we"},{"Start":"00:50.630 ","End":"01:00.500","Text":"expect our entropy to increase and the change in entropy is going to be positive."},{"Start":"01:00.500 ","End":"01:06.635","Text":"Again, since our temperature increases and we\u0027re adding kinetic energy,"},{"Start":"01:06.635 ","End":"01:10.580","Text":"the particles will have more freedom of motion and therefore,"},{"Start":"01:10.580 ","End":"01:15.244","Text":"our entropy is going to increase and the change in entropy is going to be positive."},{"Start":"01:15.244 ","End":"01:17.240","Text":"Now let\u0027s take a look at b. In b,"},{"Start":"01:17.240 ","End":"01:24.530","Text":"we have carbon dioxide gas and we have it at 25 degrees Celsius in both cases and"},{"Start":"01:24.530 ","End":"01:28.160","Text":"what changes here is the pressure because we started at 1 atmosphere and then"},{"Start":"01:28.160 ","End":"01:32.435","Text":"the pressure is 10 millimeters of mercury."},{"Start":"01:32.435 ","End":"01:40.425","Text":"Remember that 1 atmosphere equals 760 millimeters of mercury."},{"Start":"01:40.425 ","End":"01:48.350","Text":"We begin with 760 millimeters of mercury and then we go to 10 millimeters of mercury,"},{"Start":"01:48.350 ","End":"01:54.380","Text":"meaning our pressure decreases when our temperature stays the same."},{"Start":"01:54.380 ","End":"01:58.250","Text":"Now remember that PV equals nRT."},{"Start":"01:58.250 ","End":"02:00.515","Text":"P is the pressure, V is the volume,"},{"Start":"02:00.515 ","End":"02:01.880","Text":"n is the number of moles,"},{"Start":"02:01.880 ","End":"02:03.080","Text":"R is the gas constant,"},{"Start":"02:03.080 ","End":"02:04.370","Text":"and T is the temperature."},{"Start":"02:04.370 ","End":"02:06.590","Text":"Remember, temperature stays constant,"},{"Start":"02:06.590 ","End":"02:09.470","Text":"and so does our number of moles,"},{"Start":"02:09.470 ","End":"02:13.780","Text":"because we\u0027re not adding any moles of gas or taking any out."},{"Start":"02:13.780 ","End":"02:16.400","Text":"When we decrease the pressure,"},{"Start":"02:16.400 ","End":"02:20.405","Text":"what happens is our volume increases."},{"Start":"02:20.405 ","End":"02:23.630","Text":"Now when we have an increase of volume,"},{"Start":"02:23.630 ","End":"02:27.770","Text":"it means that the particles have more freedom of motion."},{"Start":"02:27.770 ","End":"02:31.010","Text":"They can be more dispersed and therefore,"},{"Start":"02:31.010 ","End":"02:33.965","Text":"when the pressure is decreased,"},{"Start":"02:33.965 ","End":"02:40.865","Text":"the volume increases and therefore the change in entropy also increases."},{"Start":"02:40.865 ","End":"02:44.000","Text":"Now, let\u0027s take a look at c. In c,"},{"Start":"02:44.000 ","End":"02:46.175","Text":"we have oxygen gas,"},{"Start":"02:46.175 ","End":"02:51.170","Text":"again at 25 degrees Celsius and it stays at 25 degrees Celsius,"},{"Start":"02:51.170 ","End":"02:55.075","Text":"meaning the temperature is constant, doesn\u0027t change."},{"Start":"02:55.075 ","End":"02:58.350","Text":"This time, we\u0027re given the volume."},{"Start":"02:58.350 ","End":"03:03.380","Text":"We know that we start with 1 liter of oxygen gas and we go to 0.1 liter,"},{"Start":"03:03.380 ","End":"03:06.185","Text":"meaning that volume decreases."},{"Start":"03:06.185 ","End":"03:07.474","Text":"The volume decreases."},{"Start":"03:07.474 ","End":"03:09.335","Text":"It\u0027s the opposite of what we had in b."},{"Start":"03:09.335 ","End":"03:11.360","Text":"The particles will have less room,"},{"Start":"03:11.360 ","End":"03:12.560","Text":"less freedom of motion,"},{"Start":"03:12.560 ","End":"03:15.560","Text":"and therefore our entropy will go down and the change in"},{"Start":"03:15.560 ","End":"03:19.880","Text":"entropy is going to be smaller than 0 or negative."},{"Start":"03:19.880 ","End":"03:23.839","Text":"Again, for a, we found that the change in entropy is positive."},{"Start":"03:23.839 ","End":"03:27.335","Text":"For b we also found that the change in entropy is positive,"},{"Start":"03:27.335 ","End":"03:30.500","Text":"and c, we found that the change in entropy is negative."},{"Start":"03:30.500 ","End":"03:31.970","Text":"That is our final answer."},{"Start":"03:31.970 ","End":"03:34.590","Text":"Thank you very much for watching."}],"ID":30055},{"Watched":false,"Name":"Exercise 3","Duration":"4m 26s","ChapterTopicVideoID":28535,"CourseChapterTopicPlaylistID":244890,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.789","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.789 ","End":"00:06.510","Text":"Which substance would you expect to have the higher molar entropy?"},{"Start":"00:06.510 ","End":"00:07.740","Text":"Before we look at a, b,"},{"Start":"00:07.740 ","End":"00:09.690","Text":"and c. In general,"},{"Start":"00:09.690 ","End":"00:12.330","Text":"the differences in molar entropy are based on"},{"Start":"00:12.330 ","End":"00:15.555","Text":"the molar mass and the molecular complexity. Let\u0027s take a look at a."},{"Start":"00:15.555 ","End":"00:16.710","Text":"In a, we have methanol,"},{"Start":"00:16.710 ","End":"00:20.295","Text":"liquid methanol, and we have liquid propanol."},{"Start":"00:20.295 ","End":"00:22.895","Text":"Now, if we take a look at our molar mass,"},{"Start":"00:22.895 ","End":"00:28.270","Text":"obviously the propanol has a higher molar mass than our methanol."},{"Start":"00:28.270 ","End":"00:32.435","Text":"Now, the Ethanol and propanol are very similar molecules."},{"Start":"00:32.435 ","End":"00:35.405","Text":"However, the propanol has a longer chain."},{"Start":"00:35.405 ","End":"00:38.315","Text":"Now as our chain length increases,"},{"Start":"00:38.315 ","End":"00:41.090","Text":"our entropy increases because the long molecule can"},{"Start":"00:41.090 ","End":"00:44.120","Text":"rotate and vibrate in more ways than a short one."},{"Start":"00:44.120 ","End":"00:45.875","Text":"Therefore our propanol,"},{"Start":"00:45.875 ","End":"00:50.540","Text":"which also has the higher molar mass and also has the higher chain length,"},{"Start":"00:50.540 ","End":"00:52.985","Text":"is going to have the higher molar entropy."},{"Start":"00:52.985 ","End":"01:02.520","Text":"For a, the molar entropy of propanol is higher than the molar entropy of methanol."},{"Start":"01:02.530 ","End":"01:05.285","Text":"Now let\u0027s take a look at b."},{"Start":"01:05.285 ","End":"01:12.785","Text":"In b, we have 1 mole of nitric oxide gas or 1 mole of dinitrogen tetroxide gas."},{"Start":"01:12.785 ","End":"01:15.860","Text":"Now again, we\u0027re looking at very similar molecules, obviously,"},{"Start":"01:15.860 ","End":"01:19.175","Text":"the dinitrogen tetroxide has a higher molar mass,"},{"Start":"01:19.175 ","End":"01:23.225","Text":"so we would expect it to have a higher molar entropy."},{"Start":"01:23.225 ","End":"01:29.030","Text":"In addition, when we have similar molecules and one of the molecules has more atoms,"},{"Start":"01:29.030 ","End":"01:31.250","Text":"when we have an increase in the number of atoms,"},{"Start":"01:31.250 ","End":"01:35.419","Text":"this also increases the number of different vibrational motions."},{"Start":"01:35.419 ","End":"01:39.109","Text":"Therefore, dinitrogen tetroxide,"},{"Start":"01:39.109 ","End":"01:42.890","Text":"which again also has the higher molar mass compared to nitric oxide,"},{"Start":"01:42.890 ","End":"01:44.960","Text":"but also has more atoms,"},{"Start":"01:44.960 ","End":"01:48.350","Text":"which gives us more movement and more freedom,"},{"Start":"01:48.350 ","End":"01:50.360","Text":"is going to have the higher molar entropy."},{"Start":"01:50.360 ","End":"01:54.755","Text":"The entropy of dinitrogen tetroxide"},{"Start":"01:54.755 ","End":"01:59.255","Text":"is expected to be higher than the entropy of nitric oxide."},{"Start":"01:59.255 ","End":"02:01.430","Text":"Now let\u0027s take a look at c. In c,"},{"Start":"02:01.430 ","End":"02:05.775","Text":"we have n-pentene and we have cyclopentane."},{"Start":"02:05.775 ","End":"02:07.605","Text":"Let\u0027s take a look at n-pentene."},{"Start":"02:07.605 ","End":"02:10.880","Text":"We have a double bond and pentene,"},{"Start":"02:10.880 ","End":"02:14.580","Text":"meaning we have 5 carbons, so that\u0027s 1,2,3,4,5."},{"Start":"02:16.310 ","End":"02:20.405","Text":"That\u0027s n-pentene and we also have cyclopentane."},{"Start":"02:20.405 ","End":"02:24.350","Text":"Cyclopentane, meaning again, we have 5 carbons."},{"Start":"02:24.350 ","End":"02:27.140","Text":"However, we have a cyclic compound,"},{"Start":"02:27.140 ","End":"02:29.335","Text":"so it\u0027s going to be a 5-carbon ring."},{"Start":"02:29.335 ","End":"02:33.350","Text":"Both of them are C_5 and cyclopentane is going to have"},{"Start":"02:33.350 ","End":"02:38.510","Text":"2 hydrogens and each carbon because we have room left for 2 hydrogens."},{"Start":"02:38.510 ","End":"02:40.870","Text":"We\u0027re going to have 10 hydrogens and all,"},{"Start":"02:40.870 ","End":"02:42.365","Text":"so that\u0027s C_5, H_10."},{"Start":"02:42.365 ","End":"02:44.540","Text":"Again, in each carbon,"},{"Start":"02:44.540 ","End":"02:47.210","Text":"we can add 2 hydrogens and we have 5 carbons,"},{"Start":"02:47.210 ","End":"02:52.230","Text":"so that\u0027s 5 times 2 and that gives us 10 hydrogens, so C_5H_10."},{"Start":"02:52.250 ","End":"02:55.200","Text":"Now in our n-pentene,"},{"Start":"02:55.200 ","End":"02:57.435","Text":"let\u0027s see how many hydrogens we have."},{"Start":"02:57.435 ","End":"03:03.035","Text":"In our terminal carbon here we have 3 hydrogens. You know what?"},{"Start":"03:03.035 ","End":"03:04.250","Text":"We\u0027re just going to write them down."},{"Start":"03:04.250 ","End":"03:05.480","Text":"This will be easier, 3."},{"Start":"03:05.480 ","End":"03:07.955","Text":"The one next to the carbon next to it,"},{"Start":"03:07.955 ","End":"03:09.745","Text":"we have 2 hydrogens,"},{"Start":"03:09.745 ","End":"03:12.800","Text":"the one next to that, we also have 2 hydrogens."},{"Start":"03:12.800 ","End":"03:17.779","Text":"Now the carbon of the double bond only has 1 room for a hydrogen,"},{"Start":"03:17.779 ","End":"03:21.695","Text":"and the other terminal carbon has 2 rooms for hydrogens."},{"Start":"03:21.695 ","End":"03:26.825","Text":"If we count the number of hydrogen here, it\u0027s 1,2,3,4,5,6,7,8,9,10."},{"Start":"03:26.825 ","End":"03:29.360","Text":"In this case, it\u0027s also C_5H_10."},{"Start":"03:29.360 ","End":"03:33.950","Text":"If we compare the molar mass of both compounds,"},{"Start":"03:33.950 ","End":"03:35.885","Text":"the molar mass is equal."},{"Start":"03:35.885 ","End":"03:39.770","Text":"We have to look at the structure of the compounds."},{"Start":"03:39.770 ","End":"03:43.280","Text":"Now the ring compound obviously is more"},{"Start":"03:43.280 ","End":"03:47.075","Text":"restricted and therefore there\u0027s less freedom of motion."},{"Start":"03:47.075 ","End":"03:50.210","Text":"Therefore, the molar entropy of a ring compound is going to"},{"Start":"03:50.210 ","End":"03:54.425","Text":"be lower than the molar entropy of n-pentene."},{"Start":"03:54.425 ","End":"04:02.570","Text":"In c, the molar entropy of n-pentene is higher than the molar entropy of cyclopentane."},{"Start":"04:02.570 ","End":"04:04.970","Text":"Again, so in a, we found that the molar entropy of"},{"Start":"04:04.970 ","End":"04:08.630","Text":"propanol is higher than the molar entropy of methanol."},{"Start":"04:08.630 ","End":"04:11.705","Text":"In b, we found that the molar entropy of"},{"Start":"04:11.705 ","End":"04:16.855","Text":"dinitrogen tetroxide is higher than the molar entropy of nitric oxide."},{"Start":"04:16.855 ","End":"04:20.360","Text":"In c, we found that the molar entropy of n-pentene is"},{"Start":"04:20.360 ","End":"04:23.630","Text":"higher than the molar entropy of cyclopentane."},{"Start":"04:23.630 ","End":"04:24.980","Text":"That is our final answer."},{"Start":"04:24.980 ","End":"04:27.510","Text":"Thank you very much for watching."}],"ID":30056},{"Watched":false,"Name":"Exercise 4","Duration":"4m 28s","ChapterTopicVideoID":28536,"CourseChapterTopicPlaylistID":244890,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.375","Text":"How we\u0027re going to solve the following exercise?"},{"Start":"00:03.375 ","End":"00:05.730","Text":"Which substance would you expect to have"},{"Start":"00:05.730 ","End":"00:09.435","Text":"the higher molar entropy? Let\u0027s take a look at a."},{"Start":"00:09.435 ","End":"00:16.360","Text":"In a, we have carbon tetrachloride and 1 is a liquid and 1 is a gas."},{"Start":"00:19.640 ","End":"00:24.480","Text":"Remember that a gas has higher entropy compared to a liquid."},{"Start":"00:26.330 ","End":"00:29.765","Text":"First of all, the gas is much more disordered."},{"Start":"00:29.765 ","End":"00:33.260","Text":"The particles have a lot more room between them."},{"Start":"00:33.260 ","End":"00:36.080","Text":"Therefore, their freedom of motion increases."},{"Start":"00:36.080 ","End":"00:41.300","Text":"Therefore, we would expect the entropy of"},{"Start":"00:41.300 ","End":"00:48.080","Text":"the gas to be greater than the entropy of the liquid, carbon tetrachloride."},{"Start":"00:48.080 ","End":"00:50.540","Text":"Now let\u0027s take a look at b."},{"Start":"00:50.540 ","End":"00:54.885","Text":"In b we have graphite and we have diamond."},{"Start":"00:54.885 ","End":"00:59.027","Text":"Both graphite and diamond are composed of carbon,"},{"Start":"00:59.027 ","End":"01:01.040","Text":"but if we look at the structure,"},{"Start":"01:01.040 ","End":"01:05.100","Text":"graphite has 2-dimensional sheets."},{"Start":"01:05.100 ","End":"01:08.950","Text":"In these 2-dimensional sheets we have covalent bonds."},{"Start":"01:12.200 ","End":"01:14.210","Text":"We can look at it like this."},{"Start":"01:14.210 ","End":"01:15.349","Text":"We have 2 sheets."},{"Start":"01:15.349 ","End":"01:17.940","Text":"In the sheets,"},{"Start":"01:17.940 ","End":"01:19.910","Text":"we have strong covalent bonds."},{"Start":"01:19.910 ","End":"01:22.235","Text":"However, between these sheets,"},{"Start":"01:22.235 ","End":"01:24.170","Text":"the forces are not very strong."},{"Start":"01:24.170 ","End":"01:28.505","Text":"Therefore, the sheets can slide 1 past the other fairly easily."},{"Start":"01:28.505 ","End":"01:30.470","Text":"Now, if we take diamond,"},{"Start":"01:30.470 ","End":"01:33.995","Text":"in diamond, we have covalent bonds in 3-dimensions."},{"Start":"01:33.995 ","End":"01:38.825","Text":"Therefore, the diamond has a lot more restriction of movement."},{"Start":"01:38.825 ","End":"01:42.500","Text":"Therefore we would expect the entropy of"},{"Start":"01:42.500 ","End":"01:49.542","Text":"the graphite to be higher than the entropy of the diamond."},{"Start":"01:49.542 ","End":"01:51.214","Text":"Again, because in diamond,"},{"Start":"01:51.214 ","End":"01:55.160","Text":"we have covalent bonds in 3-dimensions and it\u0027s much more restricted."},{"Start":"01:55.160 ","End":"01:58.370","Text":"Now, obviously, the graphite sheets don\u0027t look like this."},{"Start":"01:58.370 ","End":"02:00.020","Text":"I just drew in general sheets,"},{"Start":"02:00.020 ","End":"02:01.670","Text":"so you can get the idea."},{"Start":"02:01.670 ","End":"02:06.540","Text":"Now we\u0027re going to go on to c. In c,"},{"Start":"02:06.540 ","End":"02:08.695","Text":"we have methane gas,"},{"Start":"02:08.695 ","End":"02:12.574","Text":"and we also have liquid methanol."},{"Start":"02:12.574 ","End":"02:15.520","Text":"Now, if you look, first of all,"},{"Start":"02:15.520 ","End":"02:16.910","Text":"at the molar mass,"},{"Start":"02:16.910 ","End":"02:24.120","Text":"obviously the methanol has a higher molar mass compare it to the methane."},{"Start":"02:24.860 ","End":"02:30.905","Text":"However, the methane is a gas and the methanol is a liquid,"},{"Start":"02:30.905 ","End":"02:36.125","Text":"and remember that the entropy of the gas is a lot higher than the entropy of the liquid."},{"Start":"02:36.125 ","End":"02:39.800","Text":"Therefore, the fact that we have a gas and a liquid"},{"Start":"02:39.800 ","End":"02:43.475","Text":"is more significant than the molar mass difference."},{"Start":"02:43.475 ","End":"02:47.270","Text":"Therefore, the entropy of our gas,"},{"Start":"02:47.270 ","End":"02:54.425","Text":"our methane gas is going to be higher than the entropy of our liquid."},{"Start":"02:54.425 ","End":"03:02.680","Text":"So the entropy of our methane gas is higher than the entropy of our liquid methanol."},{"Start":"03:02.680 ","End":"03:07.070","Text":"Again, in a, we had carbon tetrachloride and we have a liquid and a gas."},{"Start":"03:07.070 ","End":"03:09.110","Text":"Obviously, the entropy of the gas is"},{"Start":"03:09.110 ","End":"03:11.210","Text":"going to be greater than the entropy of the liquid since"},{"Start":"03:11.210 ","End":"03:15.800","Text":"the gas is much more disordered and the particles have more freedom of movement."},{"Start":"03:15.800 ","End":"03:18.410","Text":"In b, we have graphite versus diamond."},{"Start":"03:18.410 ","End":"03:21.635","Text":"Again, graphite is built from 2-dimensional sheets,"},{"Start":"03:21.635 ","End":"03:23.585","Text":"which have covalent bonds in them."},{"Start":"03:23.585 ","End":"03:27.480","Text":"However, the sheets don\u0027t have strong forces between them."},{"Start":"03:27.480 ","End":"03:32.390","Text":"Therefore, they can slide 1 past the other rather easily."},{"Start":"03:32.390 ","End":"03:33.980","Text":"As opposed to diamond,"},{"Start":"03:33.980 ","End":"03:38.025","Text":"which has covalent bonds in 3-dimensions."},{"Start":"03:38.025 ","End":"03:44.223","Text":"Therefore, the entropy of graphite is higher than the entropy of diamond."},{"Start":"03:44.223 ","End":"03:46.440","Text":"Again, because in diamond,"},{"Start":"03:46.480 ","End":"03:50.390","Text":"we have covalent bonds in 3-dimensions,"},{"Start":"03:50.390 ","End":"03:53.720","Text":"meaning there is more restriction of movement."},{"Start":"03:53.720 ","End":"03:57.005","Text":"Since we have more freedom in the graphite,"},{"Start":"03:57.005 ","End":"04:01.160","Text":"the entropy of graphite is greater than the entropy of the diamond."},{"Start":"04:01.160 ","End":"04:05.990","Text":"In c, we have methane gas and liquid methanol."},{"Start":"04:05.990 ","End":"04:10.370","Text":"Now we said that the methanol has a higher molar mass compared to the methane."},{"Start":"04:10.370 ","End":"04:13.580","Text":"However, the fact that the methane is a gas"},{"Start":"04:13.580 ","End":"04:17.605","Text":"and the methanol is liquid is more significant."},{"Start":"04:17.605 ","End":"04:19.580","Text":"Therefore, the entropy of"},{"Start":"04:19.580 ","End":"04:24.335","Text":"our methane gas is higher than the entropy of our liquid methanol."},{"Start":"04:24.335 ","End":"04:25.820","Text":"That is our final answer."},{"Start":"04:25.820 ","End":"04:28.440","Text":"Thank you very much for watching."}],"ID":30057},{"Watched":false,"Name":"Exercise 5","Duration":"4m 17s","ChapterTopicVideoID":28537,"CourseChapterTopicPlaylistID":244890,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.440 ","End":"00:03.915","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.915 ","End":"00:09.135","Text":"Calculate the standard change in entropy for the following reaction."},{"Start":"00:09.135 ","End":"00:15.525","Text":"We have 2 times potassium chloride gives us 2 potassium chloride plus 3 oxygen."},{"Start":"00:15.525 ","End":"00:21.210","Text":"We have standard molar entropy values of potassium chloride,"},{"Start":"00:21.210 ","End":"00:24.195","Text":"potassium chloride, and oxygen."},{"Start":"00:24.195 ","End":"00:30.180","Text":"The standard entropy of the reaction is what we need to calculate."},{"Start":"00:30.180 ","End":"00:34.515","Text":"This equals the sum"},{"Start":"00:34.515 ","End":"00:40.005","Text":"of the number of moles of products times the entropy of the products,"},{"Start":"00:40.005 ","End":"00:47.140","Text":"minus the sum of the number of moles of the reactants times the entropy of the reactants."},{"Start":"00:47.140 ","End":"00:49.920","Text":"Let\u0027s take a look. First,"},{"Start":"00:49.920 ","End":"00:51.405","Text":"we\u0027ll start with the products."},{"Start":"00:51.405 ","End":"00:58.115","Text":"The sum of the number of moles of products times the entropy of the products equals."},{"Start":"00:58.115 ","End":"01:03.695","Text":"We can see that we have 2 moles of potassium chloride and we have 3 moles of oxygen gas."},{"Start":"01:03.695 ","End":"01:05.600","Text":"We\u0027re going to start with a potassium chloride."},{"Start":"01:05.600 ","End":"01:09.330","Text":"That\u0027s 2 moles of potassium chloride and it\u0027s times"},{"Start":"01:09.330 ","End":"01:15.185","Text":"the standard molar entropy of potassium chloride."},{"Start":"01:15.185 ","End":"01:20.030","Text":"We need to add to this the number of moles of oxygen gas,"},{"Start":"01:20.030 ","End":"01:23.270","Text":"which is 3 moles of oxygen gas,"},{"Start":"01:23.400 ","End":"01:30.710","Text":"times the molar entropy and standard molar entropy of the oxygen gas."},{"Start":"01:30.710 ","End":"01:36.650","Text":"This equals again, we have 2 moles times the entropy of the potassium chloride,"},{"Start":"01:36.650 ","End":"01:41.705","Text":"which we can say equals 82.59 joule per mole Kelvin."},{"Start":"01:41.705 ","End":"01:47.130","Text":"We can see that our moles cancel out right away and we\u0027re"},{"Start":"01:47.130 ","End":"01:52.295","Text":"going to add 3 moles times the molar entropy of the oxygen."},{"Start":"01:52.295 ","End":"01:55.610","Text":"The oxygen is 205.1,"},{"Start":"01:55.610 ","End":"02:00.885","Text":"so that\u0027s times 205.1 joule per mole Kelvin."},{"Start":"02:00.885 ","End":"02:03.720","Text":"Again, our moles cancel out."},{"Start":"02:03.720 ","End":"02:11.435","Text":"This equals 780.48 joule per Kelvin, and that\u0027s our products."},{"Start":"02:11.435 ","End":"02:14.689","Text":"Now remember, we need to calculate our reactants."},{"Start":"02:14.689 ","End":"02:21.950","Text":"The sum of the number of moles of reactants times the entropy of the reactants equals."},{"Start":"02:21.950 ","End":"02:25.880","Text":"We can see in our reactants we have 2 moles of potassium chloride."},{"Start":"02:25.880 ","End":"02:35.004","Text":"That\u0027s 2 moles times the standard entropy of potassium chlorate."},{"Start":"02:35.004 ","End":"02:41.285","Text":"This equals 2 moles times the molar entropy of potassium chloride,"},{"Start":"02:41.285 ","End":"02:48.975","Text":"which equals 143.1 joule per mole Kelvin."},{"Start":"02:48.975 ","End":"02:59.325","Text":"Our moles cancel out and this equals 286.2 joule per kelvin."},{"Start":"02:59.325 ","End":"03:02.370","Text":"That\u0027s for our reactants."},{"Start":"03:02.370 ","End":"03:08.600","Text":"The standard entropy of the reaction,"},{"Start":"03:08.600 ","End":"03:12.820","Text":"remember, equals the sum of the products minus the sum of the reactants."},{"Start":"03:12.820 ","End":"03:21.700","Text":"The sum of products equals 780.48 joule per Kelvin minus our reactants,"},{"Start":"03:21.700 ","End":"03:29.650","Text":"which equals 286.2 joule per Kelvin and this"},{"Start":"03:29.650 ","End":"03:38.705","Text":"equals 494.28 joule per Kelvin."},{"Start":"03:38.705 ","End":"03:44.510","Text":"The entropy of the reaction that we found equals 494.28 joule per Kelvin."},{"Start":"03:44.510 ","End":"03:46.670","Text":"Now, if we take a look at our equation again,"},{"Start":"03:46.670 ","End":"03:50.735","Text":"we can see that on the reactants side we only have 2 moles of solid,"},{"Start":"03:50.735 ","End":"03:54.020","Text":"in our product side we have 2 moles of solid and 3 moles of gas."},{"Start":"03:54.020 ","End":"03:58.475","Text":"We have no gas in our reactant side and 3 moles of gas in our product side."},{"Start":"03:58.475 ","End":"04:01.910","Text":"Remember when the number of moles of gas increases,"},{"Start":"04:01.910 ","End":"04:06.979","Text":"we expect the change in entropy of the reaction to be positive."},{"Start":"04:06.979 ","End":"04:13.620","Text":"Here we can really see that we got a positive change in entropy of reaction."},{"Start":"04:13.620 ","End":"04:15.230","Text":"That is our final answer."},{"Start":"04:15.230 ","End":"04:17.850","Text":"Thank you very much for watching."}],"ID":30058},{"Watched":false,"Name":"Exercise 6","Duration":"3m 29s","ChapterTopicVideoID":28528,"CourseChapterTopicPlaylistID":244890,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.090","Text":"Hi, we are we going to solve the following exercise."},{"Start":"00:03.090 ","End":"00:06.885","Text":"Calculate the standard change in entropy for the following reaction."},{"Start":"00:06.885 ","End":"00:11.400","Text":"We have sulfur plus oxygen gas gives us sulfur dioxide gas."},{"Start":"00:11.400 ","End":"00:16.410","Text":"We also know the standard entropy values for sulfur,"},{"Start":"00:16.410 ","End":"00:19.290","Text":"oxygen and sulfur dioxide."},{"Start":"00:19.290 ","End":"00:21.690","Text":"The standard entropy of the reaction,"},{"Start":"00:21.690 ","End":"00:25.470","Text":"which is what we need to calculate equals"},{"Start":"00:25.470 ","End":"00:33.289","Text":"the sum of the number of moles of products times the entropy of the products,"},{"Start":"00:33.289 ","End":"00:41.155","Text":"minus the sum of the number of moles of the reactants times the entropy of the reactants."},{"Start":"00:41.155 ","End":"00:43.945","Text":"We\u0027re going to start with the products."},{"Start":"00:43.945 ","End":"00:47.510","Text":"Again, the sum of the moles of products times"},{"Start":"00:47.510 ","End":"00:50.855","Text":"the entropy of the products equals, let\u0027s see."},{"Start":"00:50.855 ","End":"00:56.580","Text":"In our products we only have sulfur dioxide then we have 1 mole."},{"Start":"00:56.580 ","End":"01:03.080","Text":"That\u0027s 1 mole times the entropy of sulfur dioxide."},{"Start":"01:03.080 ","End":"01:11.210","Text":"So again, this equals 1 mole times the entropy of sulfur dioxide, which equals 248.2."},{"Start":"01:13.600 ","End":"01:16.830","Text":"The units are joule per mole-kelvin."},{"Start":"01:19.860 ","End":"01:22.630","Text":"Now the moles cancel out,"},{"Start":"01:22.630 ","End":"01:31.115","Text":"and this equals 1 times 248.2 so this equals 248.2 joule per kelvin."},{"Start":"01:31.115 ","End":"01:34.120","Text":"That\u0027s the products. Now we\u0027re going to look at our reactants."},{"Start":"01:34.120 ","End":"01:41.080","Text":"The sum of the moles of the reactants times the entropy of the reactants equals."},{"Start":"01:41.080 ","End":"01:42.190","Text":"If we look at our reactants,"},{"Start":"01:42.190 ","End":"01:46.255","Text":"we have 1 mole of sulfur and 1 mole of oxygen gas."},{"Start":"01:46.255 ","End":"01:52.695","Text":"That\u0027s 1 mole times the entropy of sulfur."},{"Start":"01:52.695 ","End":"01:56.375","Text":"The standard molar entropy of sulfur plus"},{"Start":"01:56.375 ","End":"02:03.125","Text":"1 mole times the standard entropy of oxygen gas."},{"Start":"02:03.125 ","End":"02:08.180","Text":"This equals 1 mole times the entropy of sulfur,"},{"Start":"02:08.180 ","End":"02:14.160","Text":"which equals 31.8 joule per mole-kelvin."},{"Start":"02:14.160 ","End":"02:23.160","Text":"The moles cancel out plus 1 mole times the entropy of oxygen gas."},{"Start":"02:23.160 ","End":"02:32.615","Text":"The entropy of oxygen gas equals 205.1 joule per mole-kelvin. So the moles cancel out."},{"Start":"02:32.615 ","End":"02:40.310","Text":"This comes to 236.9 joule per kelvin."},{"Start":"02:40.310 ","End":"02:45.035","Text":"Again, the standard entropy of the reaction,"},{"Start":"02:45.035 ","End":"02:49.820","Text":"equals the sum of the moles times the entropy"},{"Start":"02:49.820 ","End":"02:54.200","Text":"of the products minus the sum of the moles times the entropy of the reactants."},{"Start":"02:54.200 ","End":"03:00.560","Text":"The most of the products times the entropy of the products we found to be 248.2"},{"Start":"03:00.560 ","End":"03:07.565","Text":"joule per kelvin and this is minus what we found for the reactants."},{"Start":"03:07.565 ","End":"03:15.260","Text":"For the reactants, we found 236.9 joule per kelvin."},{"Start":"03:15.260 ","End":"03:21.380","Text":"This equals 11.3 joule per kelvin."},{"Start":"03:21.380 ","End":"03:26.600","Text":"The entropy of the reaction that we found equals 11.3 joule per kelvin."},{"Start":"03:26.600 ","End":"03:27.770","Text":"That is our final answer."},{"Start":"03:27.770 ","End":"03:30.300","Text":"Thank you very much for watching."}],"ID":30059},{"Watched":false,"Name":"Exercise 7","Duration":"3m 13s","ChapterTopicVideoID":28529,"CourseChapterTopicPlaylistID":244890,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.165","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.165 ","End":"00:06.135","Text":"Explain why the following reaction is spontaneous?"},{"Start":"00:06.135 ","End":"00:09.375","Text":"We have the oxidation of glucose."},{"Start":"00:09.375 ","End":"00:12.030","Text":"To tell if a reaction is spontaneous,"},{"Start":"00:12.030 ","End":"00:18.180","Text":"there are a number of ways and here we\u0027re going to talk about entropy."},{"Start":"00:18.180 ","End":"00:22.080","Text":"Regarding entropy for a reaction to be spontaneous,"},{"Start":"00:22.080 ","End":"00:27.825","Text":"we have the change in entropy of the universe."},{"Start":"00:27.825 ","End":"00:32.130","Text":"Now the change in entropy of the universe equals the change in entropy of"},{"Start":"00:32.130 ","End":"00:39.165","Text":"the system plus the change in entropy of the surroundings."},{"Start":"00:39.165 ","End":"00:42.255","Text":"Now for a reaction to be spontaneous,"},{"Start":"00:42.255 ","End":"00:48.720","Text":"the change in entropy of the universe needs to be positive or greater than 0."},{"Start":"00:48.720 ","End":"00:50.680","Text":"Now when we have our reaction,"},{"Start":"00:50.680 ","End":"00:54.595","Text":"the change in entropy of the system is the change in entropy of the reaction."},{"Start":"00:54.595 ","End":"00:59.045","Text":"The change in entropy of the surroundings is the change in entropy of the surroundings."},{"Start":"00:59.045 ","End":"01:00.560","Text":"Let\u0027s take a look at our reaction."},{"Start":"01:00.560 ","End":"01:02.915","Text":"We have the oxidation of glucose."},{"Start":"01:02.915 ","End":"01:07.025","Text":"Now, first of all, let\u0027s talk about the change in entropy of our reaction,"},{"Start":"01:07.025 ","End":"01:10.714","Text":"which is also the change in entropy of the system."},{"Start":"01:10.714 ","End":"01:14.260","Text":"First of all, let\u0027s take a look at the number of moles of gas we have on each side."},{"Start":"01:14.260 ","End":"01:17.960","Text":"On the reactant side you see we have 6 moles of oxygen gas,"},{"Start":"01:17.960 ","End":"01:19.520","Text":"so 6 moles of gas."},{"Start":"01:19.520 ","End":"01:25.310","Text":"On the product side, we have 6 moles of carbon dioxide gas plus 6 moles of water,"},{"Start":"01:25.310 ","End":"01:26.540","Text":"which is also in gas form."},{"Start":"01:26.540 ","End":"01:29.240","Text":"That means we have 12 moles of gas."},{"Start":"01:29.240 ","End":"01:30.650","Text":"Again, in the reactant side,"},{"Start":"01:30.650 ","End":"01:32.090","Text":"we have 6 moles of gas,"},{"Start":"01:32.090 ","End":"01:34.460","Text":"and in the product side we have 12 moles of gas."},{"Start":"01:34.460 ","End":"01:37.580","Text":"The number of moles of gas increases,"},{"Start":"01:37.580 ","End":"01:39.720","Text":"meaning the entropy increases."},{"Start":"01:39.720 ","End":"01:45.455","Text":"The change in entropy of the system or of the reaction is going to be greater than 0."},{"Start":"01:45.455 ","End":"01:48.965","Text":"It\u0027s going to be positive. That\u0027s 1."},{"Start":"01:48.965 ","End":"01:53.330","Text":"That\u0027s the change in entropy of the reaction or the change in entropy of the system."},{"Start":"01:53.330 ","End":"01:57.440","Text":"Now we need to look at the change in entropy of the surroundings."},{"Start":"01:57.440 ","End":"01:59.390","Text":"Now again, if we take a look at our reaction,"},{"Start":"01:59.390 ","End":"02:02.185","Text":"in our product side, we see that we have heat."},{"Start":"02:02.185 ","End":"02:05.850","Text":"Meaning that this is an exothermic reaction."},{"Start":"02:05.850 ","End":"02:08.404","Text":"Heat is given off to the surroundings."},{"Start":"02:08.404 ","End":"02:10.250","Text":"Now the heat which is given off to"},{"Start":"02:10.250 ","End":"02:14.000","Text":"the surroundings increases the entropy of the surroundings."},{"Start":"02:14.000 ","End":"02:18.900","Text":"The change in entropy of the surroundings will also be positive."},{"Start":"02:19.210 ","End":"02:26.125","Text":"Again, we have an exothermic reaction since we see heat is given off in the products."},{"Start":"02:26.125 ","End":"02:28.480","Text":"This heat is given off to the surroundings"},{"Start":"02:28.480 ","End":"02:31.805","Text":"and it increases the entropy of the surroundings,"},{"Start":"02:31.805 ","End":"02:34.400","Text":"because it increases the freedom of motion and"},{"Start":"02:34.400 ","End":"02:37.085","Text":"the dispersal of energy in the surroundings."},{"Start":"02:37.085 ","End":"02:40.010","Text":"Now we know that the change in entropy of the system is"},{"Start":"02:40.010 ","End":"02:43.525","Text":"positive and the change in entropy of the surroundings is positive."},{"Start":"02:43.525 ","End":"02:47.475","Text":"Obviously, the change in entropy of the universe,"},{"Start":"02:47.475 ","End":"02:51.020","Text":"which equals the change in entropy of the system,"},{"Start":"02:51.020 ","End":"02:57.410","Text":"which is positive, plus the change in entropy of the surroundings,"},{"Start":"02:57.410 ","End":"02:59.165","Text":"which is also positive,"},{"Start":"02:59.165 ","End":"03:02.164","Text":"is also going to be positive."},{"Start":"03:02.164 ","End":"03:07.414","Text":"Since the change in entropy of the universe is a positive value,"},{"Start":"03:07.414 ","End":"03:10.550","Text":"we know that our reaction is spontaneous."},{"Start":"03:10.550 ","End":"03:12.020","Text":"That is our final answer."},{"Start":"03:12.020 ","End":"03:14.400","Text":"Thank you very much for watching."}],"ID":30060},{"Watched":false,"Name":"Exercise 8","Duration":"3m 11s","ChapterTopicVideoID":28530,"CourseChapterTopicPlaylistID":244890,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:03.479","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.479 ","End":"00:05.565","Text":"The following reaction is spontaneous,"},{"Start":"00:05.565 ","End":"00:10.965","Text":"even though the change in entropy of the reaction is smaller than 0, is negative."},{"Start":"00:10.965 ","End":"00:12.660","Text":"Explain how this can be."},{"Start":"00:12.660 ","End":"00:18.780","Text":"The reaction is calcium oxide plus carbon dioxide gives us calcium carbonate plus heat."},{"Start":"00:18.780 ","End":"00:24.195","Text":"Remember when we\u0027re dealing with entropy and we want a reaction to be spontaneous,"},{"Start":"00:24.195 ","End":"00:32.385","Text":"we need the total entropy of the universe to be positive, larger than 0."},{"Start":"00:32.385 ","End":"00:39.090","Text":"Now, the total entropy of the universe equals the change in entropy of the system,"},{"Start":"00:39.090 ","End":"00:42.180","Text":"which is the change in entropy of the reaction"},{"Start":"00:42.180 ","End":"00:46.400","Text":"plus the change in entropy of the surroundings."},{"Start":"00:46.400 ","End":"00:51.925","Text":"We know that the change in entropy of our system of a reaction is smaller than 0."},{"Start":"00:51.925 ","End":"00:57.740","Text":"In order for the total entropy of the universe to be larger than 0,"},{"Start":"00:57.740 ","End":"00:59.960","Text":"first of all, we need the change in entropy of"},{"Start":"00:59.960 ","End":"01:03.560","Text":"the surroundings to be positive, larger than 0."},{"Start":"01:03.560 ","End":"01:08.495","Text":"We also need the absolute value to be larger"},{"Start":"01:08.495 ","End":"01:14.780","Text":"than the absolute value of the change in entropy of the system."},{"Start":"01:14.780 ","End":"01:18.780","Text":"Now, first of all, let\u0027s take a look at our reaction."},{"Start":"01:18.780 ","End":"01:20.640","Text":"First of all, we have in our reactant side,"},{"Start":"01:20.640 ","End":"01:22.940","Text":"we have 1 mole of gas and then our product side,"},{"Start":"01:22.940 ","End":"01:24.840","Text":"we don\u0027t have any moles of gas."},{"Start":"01:24.840 ","End":"01:28.820","Text":"We can assume that the entropy is going to decrease and that"},{"Start":"01:28.820 ","End":"01:30.740","Text":"the change in entropy of our reaction is really"},{"Start":"01:30.740 ","End":"01:33.410","Text":"going to be negative, so that makes sense."},{"Start":"01:33.410 ","End":"01:36.860","Text":"Now, if we look at our products,"},{"Start":"01:36.860 ","End":"01:38.330","Text":"we can see that heat is given off,"},{"Start":"01:38.330 ","End":"01:40.170","Text":"meaning this is an exothermic reaction."},{"Start":"01:40.170 ","End":"01:42.110","Text":"Remember when you have an exothermic reaction,"},{"Start":"01:42.110 ","End":"01:43.925","Text":"heat is given off to the surroundings."},{"Start":"01:43.925 ","End":"01:48.035","Text":"The surroundings except the heat and the entropy goes up,"},{"Start":"01:48.035 ","End":"01:52.775","Text":"meaning that the change in entropy of the surroundings is going to be positive."},{"Start":"01:52.775 ","End":"01:56.105","Text":"Now we\u0027re asked to explain how this reaction is spontaneous,"},{"Start":"01:56.105 ","End":"02:00.875","Text":"even though the change in entropy of the reaction or of the system is negative."},{"Start":"02:00.875 ","End":"02:02.860","Text":"Again, this can be."},{"Start":"02:02.860 ","End":"02:05.600","Text":"All we need to know is that the change in entropy of"},{"Start":"02:05.600 ","End":"02:11.190","Text":"the surroundings is positive and we also need"},{"Start":"02:11.980 ","End":"02:18.680","Text":"this absolute value to be greater than the absolute value of the change in"},{"Start":"02:18.680 ","End":"02:21.260","Text":"entropy of the system and that way we\u0027ll still"},{"Start":"02:21.260 ","End":"02:25.280","Text":"have a total change in entropy of the universe,"},{"Start":"02:25.280 ","End":"02:28.890","Text":"which is positive or greater than 0."},{"Start":"02:28.890 ","End":"02:31.415","Text":"One last time the explanation is,"},{"Start":"02:31.415 ","End":"02:36.890","Text":"even though our change in entropy of a reaction is negative,"},{"Start":"02:36.890 ","End":"02:40.285","Text":"the change in entropy of the surroundings is positive."},{"Start":"02:40.285 ","End":"02:43.129","Text":"It has a larger absolute value,"},{"Start":"02:43.129 ","End":"02:47.315","Text":"than the change in entropy of the reaction or of the system."},{"Start":"02:47.315 ","End":"02:50.000","Text":"Therefore, when we add 1 to the other,"},{"Start":"02:50.000 ","End":"02:51.740","Text":"1 is positive and 1 is negative,"},{"Start":"02:51.740 ","End":"02:55.025","Text":"we can say that 1 is more positive than the negative."},{"Start":"02:55.025 ","End":"02:59.420","Text":"We will eventually receive a positive change in entropy of"},{"Start":"02:59.420 ","End":"03:03.980","Text":"the universe and that\u0027s how this reaction can be spontaneous,"},{"Start":"03:03.980 ","End":"03:08.045","Text":"even though the change in entropy of the reaction is negative."},{"Start":"03:08.045 ","End":"03:09.545","Text":"That is our final answer."},{"Start":"03:09.545 ","End":"03:12.120","Text":"Thank you very much for watching."}],"ID":30061},{"Watched":false,"Name":"Exercise 9","Duration":"4m 1s","ChapterTopicVideoID":28531,"CourseChapterTopicPlaylistID":244890,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.380 ","End":"00:04.230","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:04.230 ","End":"00:08.850","Text":"Calculate the standard Gibbs free energy change for the following reaction."},{"Start":"00:08.850 ","End":"00:13.365","Text":"We have nitrogen gas plus oxygen gas gives us 2 nitric oxide gas."},{"Start":"00:13.365 ","End":"00:20.520","Text":"We are also given values of standard Gibbs free energy change of formation."},{"Start":"00:20.520 ","End":"00:28.620","Text":"So the standard change in free energy for a reaction,"},{"Start":"00:28.620 ","End":"00:34.815","Text":"equals the sum of the number of moles of products"},{"Start":"00:34.815 ","End":"00:43.370","Text":"times the standard change in free energy of formation of the products minus,"},{"Start":"00:43.370 ","End":"00:46.580","Text":"and I\u0027m just going to continue on this line here because I don\u0027t have enough space,"},{"Start":"00:46.580 ","End":"00:51.500","Text":"minus the sum of the moles of reactants"},{"Start":"00:51.500 ","End":"00:58.160","Text":"times the standard change in free energy of formation of the reactants."},{"Start":"00:58.160 ","End":"01:01.700","Text":"Now just remember before we continue that the standard change in"},{"Start":"01:01.700 ","End":"01:05.780","Text":"free energy of formation for elements in their reference form equals 0,"},{"Start":"01:05.780 ","End":"01:08.150","Text":"therefore the change in free energy for"},{"Start":"01:08.150 ","End":"01:11.965","Text":"the nitrogen gas and for the oxygen gas both equals 0."},{"Start":"01:11.965 ","End":"01:17.405","Text":"Let\u0027s begin first of all by calculating the sum of the products."},{"Start":"01:17.405 ","End":"01:22.070","Text":"It\u0027s equals the sum of the number of moles of products"},{"Start":"01:22.070 ","End":"01:26.810","Text":"times the change in free energy of the products."},{"Start":"01:26.810 ","End":"01:28.175","Text":"Let\u0027s just write p here."},{"Start":"01:28.175 ","End":"01:29.900","Text":"This equals, if we look at our products,"},{"Start":"01:29.900 ","End":"01:31.910","Text":"we have 2 nitric oxides."},{"Start":"01:31.910 ","End":"01:37.340","Text":"That\u0027s 2 moles times the change in free energy of formation,"},{"Start":"01:37.340 ","End":"01:41.480","Text":"times the standard change in free energy of formation of nitric oxide,"},{"Start":"01:41.480 ","End":"01:43.460","Text":"and that equals 86.55."},{"Start":"01:43.460 ","End":"01:49.760","Text":"This is times 86.55 kilojoules per mole."},{"Start":"01:49.760 ","End":"01:52.321","Text":"The moles cancel out, we can see that right away,"},{"Start":"01:52.321 ","End":"01:55.710","Text":"and this equals 173.1."},{"Start":"01:55.910 ","End":"01:58.745","Text":"We\u0027re left with kilojoules."},{"Start":"01:58.745 ","End":"02:02.110","Text":"Now we\u0027re going to calculate the sum of the reactants."},{"Start":"02:02.110 ","End":"02:06.920","Text":"We have the sum of the number of moles of reactants times"},{"Start":"02:06.920 ","End":"02:12.875","Text":"the standard change in free energy of formation of the reactants, and this equals."},{"Start":"02:12.875 ","End":"02:18.650","Text":"If we look at our reactants, we have 1 mole of the nitrogen gas and 1 mole of oxygen."},{"Start":"02:18.650 ","End":"02:25.760","Text":"We have 1 mole times the change in free energy of the nitrogen and remember it equals 0,"},{"Start":"02:25.760 ","End":"02:28.910","Text":"so that\u0027s times 0 kilojoules per mole."},{"Start":"02:28.910 ","End":"02:30.515","Text":"The moles cancel out."},{"Start":"02:30.515 ","End":"02:35.165","Text":"This is plus, because remember it\u0027s the sum of,"},{"Start":"02:35.165 ","End":"02:36.625","Text":"let\u0027s look at our oxygen."},{"Start":"02:36.625 ","End":"02:38.750","Text":"Again, we have 1 mole of oxygen,"},{"Start":"02:38.750 ","End":"02:41.300","Text":"times the change in free energy of the oxygen,"},{"Start":"02:41.300 ","End":"02:42.430","Text":"which also equals 0,"},{"Start":"02:42.430 ","End":"02:46.940","Text":"so that\u0027s also times 0 kilojoules per mole."},{"Start":"02:46.940 ","End":"02:53.820","Text":"The moles cancel out and summing this up we obviously have 0 kilojoules."},{"Start":"02:57.440 ","End":"03:00.830","Text":"That general change in the reaction,"},{"Start":"03:00.830 ","End":"03:04.880","Text":"the standard free energy change for reaction,"},{"Start":"03:04.880 ","End":"03:07.160","Text":"we said equals the sum,"},{"Start":"03:07.160 ","End":"03:11.375","Text":"again, of the number of moles of products times"},{"Start":"03:11.375 ","End":"03:15.905","Text":"the standard change in energy of formation of the products"},{"Start":"03:15.905 ","End":"03:19.820","Text":"minus the sum of the number of moles of reactants"},{"Start":"03:19.820 ","End":"03:26.060","Text":"times the standard change in energy of formation of the reactants."},{"Start":"03:26.060 ","End":"03:30.830","Text":"I\u0027m just going to write r because it\u0027s shorter and this equal."},{"Start":"03:30.830 ","End":"03:35.060","Text":"We found that the sum of the products equals 173.1"},{"Start":"03:35.060 ","End":"03:42.515","Text":"kilojoules minus some of the reactants and that equals 0 kilojoules."},{"Start":"03:42.515 ","End":"03:50.290","Text":"Obviously, we\u0027re left with 173.1 kilojoules."},{"Start":"03:50.600 ","End":"03:58.970","Text":"The standard change in free energy for the reaction equals 173.1 kilojoules."},{"Start":"03:58.970 ","End":"04:02.880","Text":"That is our final answer. Thank you very much for watching."}],"ID":30062},{"Watched":false,"Name":"Exercise 10","Duration":"4m 10s","ChapterTopicVideoID":28532,"CourseChapterTopicPlaylistID":244890,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.670","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.670 ","End":"00:07.350","Text":"Calculate the standard Gibbs free energy change for the following reaction."},{"Start":"00:07.350 ","End":"00:13.215","Text":"Phosphorus trichloride gas plus chlorine gas gives us phosphorous pentachloride gas."},{"Start":"00:13.215 ","End":"00:21.550","Text":"We\u0027re also given different values of the standard change in free energy of formation."},{"Start":"00:21.670 ","End":"00:30.735","Text":"First of all, the standard change in free energy of a reaction equals"},{"Start":"00:30.735 ","End":"00:34.680","Text":"the sum of the number of moles of products times"},{"Start":"00:34.680 ","End":"00:40.085","Text":"the change in the free energy of formation of the products,"},{"Start":"00:40.085 ","End":"00:43.415","Text":"minus, and we\u0027re just going to write that here so we have enough room,"},{"Start":"00:43.415 ","End":"00:46.790","Text":"the sum of the number of moles of reactants times"},{"Start":"00:46.790 ","End":"00:52.850","Text":"the standard change in free energy of formation of the reactants."},{"Start":"00:52.850 ","End":"00:55.520","Text":"I like to calculate their product first and then"},{"Start":"00:55.520 ","End":"00:58.430","Text":"the reactants and then subtract one from the other."},{"Start":"00:58.430 ","End":"01:01.895","Text":"Obviously, you can do this in 1 stage and you don\u0027t have to divide it into 2."},{"Start":"01:01.895 ","End":"01:03.665","Text":"I\u0027m going to start with the products."},{"Start":"01:03.665 ","End":"01:07.265","Text":"The sum of the number of moles of products"},{"Start":"01:07.265 ","End":"01:12.800","Text":"times the standard change in free energy of formation of the products,"},{"Start":"01:12.800 ","End":"01:15.575","Text":"so I\u0027ll just write p equals,"},{"Start":"01:15.575 ","End":"01:17.060","Text":"so let\u0027s take a look at our product."},{"Start":"01:17.060 ","End":"01:20.030","Text":"In our products, we only have the phosphorus pentachloride,"},{"Start":"01:20.030 ","End":"01:21.560","Text":"and we see that it\u0027s 1,"},{"Start":"01:21.560 ","End":"01:25.535","Text":"so that\u0027s 1 mole times the change"},{"Start":"01:25.535 ","End":"01:29.180","Text":"in free energy of formation of the phosphorus pentachloride,"},{"Start":"01:29.180 ","End":"01:31.930","Text":"which equals minus 305."},{"Start":"01:31.930 ","End":"01:35.085","Text":"That\u0027s minus 305."},{"Start":"01:35.085 ","End":"01:38.585","Text":"Don\u0027t forget your units, kilojoules per mole."},{"Start":"01:38.585 ","End":"01:40.850","Text":"That\u0027s kilojoules per mole,"},{"Start":"01:40.850 ","End":"01:42.625","Text":"the moles cancel out,"},{"Start":"01:42.625 ","End":"01:47.800","Text":"and this equals minus 305 kilojoules."},{"Start":"01:47.800 ","End":"01:50.389","Text":"That\u0027s our products."},{"Start":"01:50.389 ","End":"01:52.070","Text":"Now let\u0027s go onto our reactants."},{"Start":"01:52.070 ","End":"01:55.430","Text":"Our reactants we have the sum of the number of moles of reactants times"},{"Start":"01:55.430 ","End":"02:00.090","Text":"the change in free energy of formation of the reactants."},{"Start":"02:00.500 ","End":"02:08.190","Text":"In our reactants, we have our phosphorus trichloride and we also have our chlorine gas."},{"Start":"02:08.190 ","End":"02:11.835","Text":"Let\u0027s start with our phosphorus trichloride."},{"Start":"02:11.835 ","End":"02:16.970","Text":"First of all, we have 1 mole times our change in free energy,"},{"Start":"02:16.970 ","End":"02:26.125","Text":"which equals minus 267.8 kilojoules per mole."},{"Start":"02:26.125 ","End":"02:28.590","Text":"Our moles cancel out,"},{"Start":"02:28.590 ","End":"02:32.070","Text":"and now we need to add the chlorine gas."},{"Start":"02:32.480 ","End":"02:35.325","Text":"Again we have 1 mole,"},{"Start":"02:35.325 ","End":"02:40.250","Text":"so that\u0027s 1 mole times the change in free energy of the chlorine,"},{"Start":"02:40.250 ","End":"02:42.005","Text":"which equals 0,"},{"Start":"02:42.005 ","End":"02:45.170","Text":"so that\u0027s 0 kilojoules per mole."},{"Start":"02:45.170 ","End":"02:48.275","Text":"Because remember an element and its reference form,"},{"Start":"02:48.275 ","End":"02:51.695","Text":"the change in free energy equals 0."},{"Start":"02:51.695 ","End":"02:53.840","Text":"This is an easy calculation,"},{"Start":"02:53.840 ","End":"03:00.640","Text":"and this equals minus 267.8 kilojoules."},{"Start":"03:00.640 ","End":"03:07.470","Text":"Now let\u0027s go for the standard change in free energy of the reaction."},{"Start":"03:07.610 ","End":"03:13.010","Text":"Remember, it\u0027s the sum of the number of moles of products"},{"Start":"03:13.010 ","End":"03:17.540","Text":"times the change in free energy of formation of"},{"Start":"03:17.540 ","End":"03:22.280","Text":"the products minus the sum of the number of moles of reactants"},{"Start":"03:22.280 ","End":"03:27.545","Text":"times the standard change in free energy of formation of the reactants."},{"Start":"03:27.545 ","End":"03:30.500","Text":"This equals, we\u0027ll start with our products,"},{"Start":"03:30.500 ","End":"03:35.100","Text":"minus 305 kilojoules,"},{"Start":"03:35.360 ","End":"03:43.075","Text":"and our reactants is minus 267.8 kilojoules."},{"Start":"03:43.075 ","End":"03:51.770","Text":"This equals minus 305 kilojoules plus 267.8 kilojoules."},{"Start":"03:51.770 ","End":"03:59.385","Text":"That equals minus 37.2 kilojoules."},{"Start":"03:59.385 ","End":"04:03.845","Text":"The standard change in free energy of the reaction that we found equals"},{"Start":"04:03.845 ","End":"04:07.025","Text":"minus 37.2 kilojoules."},{"Start":"04:07.025 ","End":"04:08.690","Text":"That is our final answer."},{"Start":"04:08.690 ","End":"04:11.190","Text":"Thank you very much for watching."}],"ID":30063}],"Thumbnail":null,"ID":244890},{"Name":"Gibbs Free Energy","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Gibbs Free Energy","Duration":"5m 22s","ChapterTopicVideoID":25608,"CourseChapterTopicPlaylistID":244891,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.570 ","End":"00:03.145","Text":"In the previous videos,"},{"Start":"00:03.145 ","End":"00:06.595","Text":"we talked about global entropy and spontaneity."},{"Start":"00:06.595 ","End":"00:11.740","Text":"In this video, we\u0027ll talk about a criterion based on the system alone."},{"Start":"00:11.740 ","End":"00:14.500","Text":"In order to begin the discussion,"},{"Start":"00:14.500 ","End":"00:18.415","Text":"let\u0027s recall what we\u0027ve learned about entropy."},{"Start":"00:18.415 ","End":"00:24.370","Text":"We learned that if Delta S total or global is greater 0,"},{"Start":"00:24.370 ","End":"00:27.085","Text":"the process is spontaneous."},{"Start":"00:27.085 ","End":"00:31.120","Text":"We wrote that Delta S total is equal to Delta S of"},{"Start":"00:31.120 ","End":"00:35.800","Text":"the system plus Delta S of the surroundings."},{"Start":"00:35.800 ","End":"00:39.205","Text":"That means that in order to know whether process is spontaneous,"},{"Start":"00:39.205 ","End":"00:45.170","Text":"we need to know the entropy change of both the system, and the surroundings."},{"Start":"00:45.170 ","End":"00:48.740","Text":"Now, this is rather inconvenient and it would"},{"Start":"00:48.740 ","End":"00:52.670","Text":"be better to have a criterion for the system alone."},{"Start":"00:52.670 ","End":"00:55.385","Text":"There is such a criterion,"},{"Start":"00:55.385 ","End":"00:59.050","Text":"and it\u0027s called Gibbs free energy."},{"Start":"00:59.050 ","End":"01:03.634","Text":"Now, Gibbs free energy is named after J. Willard Gibbs,"},{"Start":"01:03.634 ","End":"01:06.470","Text":"who lived mainly in the 19th century,"},{"Start":"01:06.470 ","End":"01:13.780","Text":"and was one of the greatest mathematical physicists of his time and indeed of all times."},{"Start":"01:13.780 ","End":"01:17.310","Text":"Now, in order to introduce Gibbs free energy,"},{"Start":"01:17.310 ","End":"01:20.810","Text":"let\u0027s recall some things we learned in previous videos."},{"Start":"01:20.810 ","End":"01:23.270","Text":"At constant T and P,"},{"Start":"01:23.270 ","End":"01:24.740","Text":"temperature and pressure,"},{"Start":"01:24.740 ","End":"01:29.910","Text":"we learned that Delta S of the surroundings can be written as minus Delta H,"},{"Start":"01:29.910 ","End":"01:33.215","Text":"that\u0027s the enthalpy change of the system,"},{"Start":"01:33.215 ","End":"01:36.925","Text":"divided by T, the temperature."},{"Start":"01:36.925 ","End":"01:41.100","Text":"We can write Delta S total is equal to Delta S of"},{"Start":"01:41.100 ","End":"01:44.850","Text":"the system plus Delta S of the surroundings."},{"Start":"01:44.850 ","End":"01:48.900","Text":"Then we can substitute Delta S surrounding is"},{"Start":"01:48.900 ","End":"01:53.430","Text":"equal to minus Delta H divided by T in this equation."},{"Start":"01:53.430 ","End":"01:59.020","Text":"We get Delta S minus Delta H divided by T. Now,"},{"Start":"01:59.020 ","End":"02:04.380","Text":"we\u0027re going to multiply this equation by T,"},{"Start":"02:05.300 ","End":"02:14.565","Text":"and then we get T Delta S total is equal to T times Delta S"},{"Start":"02:14.565 ","End":"02:20.130","Text":"minus Delta H. Here we have minus Delta H. Now we have"},{"Start":"02:20.130 ","End":"02:26.430","Text":"T Delta S total is equal to T Delta S minus Delta H. Now,"},{"Start":"02:26.430 ","End":"02:29.275","Text":"we\u0027re going to change the sign of that."},{"Start":"02:29.275 ","End":"02:32.570","Text":"Multiply by minus sign,"},{"Start":"02:32.570 ","End":"02:34.085","Text":"by minus 1,"},{"Start":"02:34.085 ","End":"02:40.830","Text":"and we\u0027ll get minus T Delta S total is equal to plus"},{"Start":"02:40.830 ","End":"02:50.460","Text":"Delta H minus T Delta S. We need this for the next part."},{"Start":"02:50.860 ","End":"02:54.230","Text":"Now we\u0027re going to define the Gibbs free energy."},{"Start":"02:54.230 ","End":"02:55.696","Text":"It\u0027s defined as G,"},{"Start":"02:55.696 ","End":"03:00.725","Text":"after Gibbs, equal to H minus TS."},{"Start":"03:00.725 ","End":"03:03.800","Text":"If we want the change in Gibbs free energy,"},{"Start":"03:03.800 ","End":"03:08.887","Text":"we can write Delta G is equal to Delta H minus T"},{"Start":"03:08.887 ","End":"03:15.640","Text":"Delta S. We don\u0027t have Delta T because its constant temperature."},{"Start":"03:15.640 ","End":"03:23.455","Text":"Here\u0027s our definition of the change in Gibbs free energy."},{"Start":"03:23.455 ","End":"03:28.539","Text":"The first thing to note is Delta G is a state function"},{"Start":"03:28.539 ","End":"03:32.670","Text":"since Delta H and Delta S are also state functions."},{"Start":"03:32.670 ","End":"03:36.594","Text":"Delta G is a state function."},{"Start":"03:36.594 ","End":"03:42.035","Text":"It\u0027s not dependent on the actual path from the beginning to the end of the process,"},{"Start":"03:42.035 ","End":"03:46.100","Text":"but just on the 2 states at the beginning and end."},{"Start":"03:46.100 ","End":"03:49.590","Text":"Now, if we compare these 2 equations here,"},{"Start":"03:49.590 ","End":"03:53.025","Text":"we see that they\u0027re the same on the right-hand side."},{"Start":"03:53.025 ","End":"03:59.190","Text":"We can write that Delta G is equal to minus T Delta S total."},{"Start":"03:59.190 ","End":"04:07.795","Text":"We have Delta G is equal to minus T Delta S total."},{"Start":"04:07.795 ","End":"04:11.420","Text":"Let\u0027s recall what we learned about Delta S total."},{"Start":"04:11.420 ","End":"04:16.700","Text":"We recall that if Delta S total is positive,"},{"Start":"04:16.700 ","End":"04:21.637","Text":"then our process is spontaneous,"},{"Start":"04:21.637 ","End":"04:23.205","Text":"but if it\u0027s positive,"},{"Start":"04:23.205 ","End":"04:26.290","Text":"then Delta G is negative."},{"Start":"04:26.290 ","End":"04:35.555","Text":"Now we\u0027re in a position to write the criteria for spontaneous change at constant T and P."},{"Start":"04:35.555 ","End":"04:45.335","Text":"Constant P because Delta H is for constant P. We have that if Delta G is negative,"},{"Start":"04:45.335 ","End":"04:47.930","Text":"the process is spontaneous."},{"Start":"04:47.930 ","End":"04:51.755","Text":"This is an extremely important conclusion."},{"Start":"04:51.755 ","End":"04:55.025","Text":"We\u0027ll see that we\u0027ll use it a great deal."},{"Start":"04:55.025 ","End":"04:58.220","Text":"On the other hand, if Delta G is positive,"},{"Start":"04:58.220 ","End":"05:01.700","Text":"the process is non-spontaneous."},{"Start":"05:01.700 ","End":"05:04.425","Text":"If Delta G is 0,"},{"Start":"05:04.425 ","End":"05:06.585","Text":"neither positive nor negative,"},{"Start":"05:06.585 ","End":"05:09.175","Text":"the process is reversible,"},{"Start":"05:09.175 ","End":"05:14.285","Text":"can go in either direction and the system is in equilibrium."},{"Start":"05:14.285 ","End":"05:17.975","Text":"This is a dynamic equilibrium, usually."},{"Start":"05:17.975 ","End":"05:23.040","Text":"In this video, we introduce Gibbs free energy."}],"ID":26383},{"Watched":false,"Name":"Spontaneous and Nonspontaneous Processes","Duration":"8m 56s","ChapterTopicVideoID":25609,"CourseChapterTopicPlaylistID":244891,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.680","Text":"In the previous video,"},{"Start":"00:01.680 ","End":"00:03.900","Text":"we defined Gibbs Free Energy."},{"Start":"00:03.900 ","End":"00:06.390","Text":"This video, we\u0027ll see how it can be used to"},{"Start":"00:06.390 ","End":"00:10.695","Text":"predict whether a process is spontaneous or not."},{"Start":"00:10.695 ","End":"00:14.048","Text":"We\u0027re going to see how the sign of the enthalpy,"},{"Start":"00:14.048 ","End":"00:19.240","Text":"and entropy influence the sign of Gibbs Free Energy."},{"Start":"00:19.280 ","End":"00:24.420","Text":"Now we know that Delta G is equal to Delta H minus T Delta S,"},{"Start":"00:24.420 ","End":"00:26.310","Text":"where Delta G is Gibbs Free Energy,"},{"Start":"00:26.310 ","End":"00:27.945","Text":"Delta H is the enthalpy,"},{"Start":"00:27.945 ","End":"00:30.425","Text":"and Delta S is the entropy."},{"Start":"00:30.425 ","End":"00:32.960","Text":"Now, drawn them in green, red,"},{"Start":"00:32.960 ","End":"00:39.060","Text":"and blue because these are the colors I\u0027ll use in the diagram that follows."},{"Start":"00:39.710 ","End":"00:42.495","Text":"Let\u0027s look at the first case."},{"Start":"00:42.495 ","End":"00:47.794","Text":"Here, the enthalpy is negative and the entropy is positive."},{"Start":"00:47.794 ","End":"00:52.290","Text":"Delta H minus T Delta S will be negative,"},{"Start":"00:52.290 ","End":"00:55.450","Text":"so Delta G is negative."},{"Start":"00:55.880 ","End":"01:00.855","Text":"In this case, if Delta H is negative and Delta S is positive,"},{"Start":"01:00.855 ","End":"01:02.700","Text":"Delta G is negative."},{"Start":"01:02.700 ","End":"01:05.905","Text":"The process is spontaneous."},{"Start":"01:05.905 ","End":"01:08.600","Text":"Let\u0027s look at the second case."},{"Start":"01:08.600 ","End":"01:11.600","Text":"Here Delta H is positive and Delta S is negative,"},{"Start":"01:11.600 ","End":"01:21.050","Text":"so Delta H minus this arrow here will give us a positive result,"},{"Start":"01:21.050 ","End":"01:23.840","Text":"so Delta G is positive."},{"Start":"01:23.840 ","End":"01:25.790","Text":"That\u0027s the second case."},{"Start":"01:25.790 ","End":"01:27.485","Text":"Delta H is positive,"},{"Start":"01:27.485 ","End":"01:29.345","Text":"Delta S is negative."},{"Start":"01:29.345 ","End":"01:32.015","Text":"Delta G will be positive,"},{"Start":"01:32.015 ","End":"01:35.840","Text":"and the process will be non-spontaneous."},{"Start":"01:35.840 ","End":"01:38.750","Text":"Now in the third case,"},{"Start":"01:38.750 ","End":"01:41.330","Text":"there are 2 cases like that."},{"Start":"01:41.330 ","End":"01:45.620","Text":"Delta H is positive and Delta S is positive."},{"Start":"01:45.620 ","End":"01:48.770","Text":"But the difference between them is that in the first case,"},{"Start":"01:48.770 ","End":"01:52.730","Text":"Delta H is greater than T Delta S."},{"Start":"01:52.730 ","End":"01:56.660","Text":"Delta H minus T Delta S will give us"},{"Start":"01:56.660 ","End":"02:01.440","Text":"a positive result for Delta G. Whereas in this case,"},{"Start":"02:01.440 ","End":"02:05.855","Text":"Delta H is smaller than T Delta S,"},{"Start":"02:05.855 ","End":"02:11.075","Text":"so Delta H minus T Delta S will give us a negative result."},{"Start":"02:11.075 ","End":"02:14.100","Text":"Here Delta G is negative."},{"Start":"02:14.480 ","End":"02:16.610","Text":"We can summarize it."},{"Start":"02:16.610 ","End":"02:18.620","Text":"If Delta H is greater than 0,"},{"Start":"02:18.620 ","End":"02:21.905","Text":"Delta H is positive and Delta S is positive,"},{"Start":"02:21.905 ","End":"02:29.015","Text":"Delta G will be positive if Delta H is greater than T Delta S that\u0027s this case,"},{"Start":"02:29.015 ","End":"02:35.700","Text":"and negative if Delta H is less than T Delta S, that\u0027s this case."},{"Start":"02:35.980 ","End":"02:39.470","Text":"If we look at carefully at the first case,"},{"Start":"02:39.470 ","End":"02:41.330","Text":"this is low temperature."},{"Start":"02:41.330 ","End":"02:44.665","Text":"Here, it is high temperature."},{"Start":"02:44.665 ","End":"02:49.160","Text":"We can say that process is spontaneous at"},{"Start":"02:49.160 ","End":"02:53.435","Text":"high temperatures and non spontaneous at low temperatures."},{"Start":"02:53.435 ","End":"02:56.250","Text":"That\u0027s this case 3."},{"Start":"02:56.630 ","End":"02:59.405","Text":"Let\u0027s look at the fourth case."},{"Start":"02:59.405 ","End":"03:06.140","Text":"Here, the enthalpy is negative and the entropy is negative in both of these diagrams."},{"Start":"03:06.140 ","End":"03:10.400","Text":"But in the first one, Delta H is more negative than"},{"Start":"03:10.400 ","End":"03:19.684","Text":"T Delta S. Delta H minus T Delta S is still negative."},{"Start":"03:19.684 ","End":"03:27.240","Text":"Delta H is longer than T Delta S. But in this case,"},{"Start":"03:27.240 ","End":"03:29.629","Text":"Delta H is a shorter vectors,"},{"Start":"03:29.629 ","End":"03:37.540","Text":"a smaller number than T Delta S. Delta H minus T Delta S will be positive."},{"Start":"03:37.540 ","End":"03:42.740","Text":"Here we have Delta G negative and here Delta G positive."},{"Start":"03:42.740 ","End":"03:47.090","Text":"Can summarize that by saying if Delta H is negative,"},{"Start":"03:47.090 ","End":"03:48.650","Text":"and Delta S is negative,"},{"Start":"03:48.650 ","End":"03:53.330","Text":"Delta G will be negative if Delta H is greater than T Delta S."},{"Start":"03:53.330 ","End":"03:58.950","Text":"The absolute value of Delta H is greater than absolute value of T Delta S,"},{"Start":"03:58.950 ","End":"04:04.260","Text":"and will be positive if the absolute value of Delta H is smaller"},{"Start":"04:04.260 ","End":"04:09.995","Text":"than the absolute value of T Delta S. When is this going to occur?"},{"Start":"04:09.995 ","End":"04:14.905","Text":"This will be high temperature and this will be low temperature."},{"Start":"04:14.905 ","End":"04:17.795","Text":"We can see that in this case,"},{"Start":"04:17.795 ","End":"04:24.785","Text":"the process is spontaneous at low temperatures and non spontaneous at high temperatures."},{"Start":"04:24.785 ","End":"04:27.530","Text":"Let\u0027s look at some examples."},{"Start":"04:27.530 ","End":"04:33.515","Text":"Example 1, N_2 turning into nitrogen and oxygen,"},{"Start":"04:33.515 ","End":"04:40.900","Text":"then Delta H is negative here is an exothermic process."},{"Start":"04:43.300 ","End":"04:47.240","Text":"We have more moles of products than reactants."},{"Start":"04:47.240 ","End":"04:56.920","Text":"Delta S will be positive and Delta G is negative."},{"Start":"04:58.760 ","End":"05:04.130","Text":"The second case which relates to number 2 up here,"},{"Start":"05:04.130 ","End":"05:08.990","Text":"oxygen is being converted into ozone."},{"Start":"05:08.990 ","End":"05:14.610","Text":"This process is endothermic."},{"Start":"05:14.920 ","End":"05:18.425","Text":"Delta H is positive."},{"Start":"05:18.425 ","End":"05:21.455","Text":"3 moles goes into 2 moles,"},{"Start":"05:21.455 ","End":"05:29.760","Text":"so Delta S is negative and Delta G will be positive."},{"Start":"05:30.160 ","End":"05:37.370","Text":"The third example, which is ammonia decomposing into nitrogen and hydrogen."},{"Start":"05:37.370 ","End":"05:42.690","Text":"Then here Delta H is positive."},{"Start":"05:44.960 ","End":"05:48.240","Text":"We go from 2 moles to 4 moles."},{"Start":"05:48.240 ","End":"05:50.980","Text":"Delta S is also positive."},{"Start":"05:50.980 ","End":"05:55.865","Text":"This reaction is spontaneous at high temperatures,"},{"Start":"05:55.865 ","End":"05:58.490","Text":"and non spontaneous at low temperatures."},{"Start":"05:58.490 ","End":"06:02.360","Text":"Ammonia at low temperatures doesn\u0027t decompose into nitrogen-hydrogen,"},{"Start":"06:02.360 ","End":"06:05.760","Text":"but it can do at high temperatures."},{"Start":"06:05.950 ","End":"06:12.305","Text":"The fourth example is liquid water turning into ice."},{"Start":"06:12.305 ","End":"06:16.970","Text":"Here, Delta H is"},{"Start":"06:16.970 ","End":"06:24.510","Text":"negative and Delta S is negative."},{"Start":"06:24.510 ","End":"06:27.880","Text":"Then this process is spontaneous at"},{"Start":"06:27.880 ","End":"06:32.695","Text":"low temperatures and non spontaneous at high temperatures."},{"Start":"06:32.695 ","End":"06:38.425","Text":"We all know that water only it turns into ice at low temperatures."},{"Start":"06:38.425 ","End":"06:43.465","Text":"Now we\u0027re interested in the temperature which Gibbs Free Energy changes sign."},{"Start":"06:43.465 ","End":"06:49.720","Text":"At equilibrium, we know that Delta G equal to Delta H minus T Delta S equals 0."},{"Start":"06:49.720 ","End":"06:54.970","Text":"The temperature at which this occurs is T equilibrium equal to Delta H over"},{"Start":"06:54.970 ","End":"06:59.110","Text":"Delta S. Here\u0027s an example for"},{"Start":"06:59.110 ","End":"07:04.005","Text":"the reaction of ammonia decomposing into nitrogen and hydrogen,"},{"Start":"07:04.005 ","End":"07:11.060","Text":"Delta S is positive plus a 198.9 joules per Kelvin."},{"Start":"07:11.060 ","End":"07:13.670","Text":"Delta H is also positive,"},{"Start":"07:13.670 ","End":"07:19.850","Text":"92.22 kilo joules, and all this is 25 degrees Celsius."},{"Start":"07:19.850 ","End":"07:23.143","Text":"We want to calculate Delta G,"},{"Start":"07:23.143 ","End":"07:29.350","Text":"and then decide at which temperatures reaction will be spontaneous."},{"Start":"07:29.350 ","End":"07:33.140","Text":"Delta G is equal to Delta H minus T Delta S. We"},{"Start":"07:33.140 ","End":"07:37.790","Text":"can substitute the numbers for Delta H. We have 92.22."},{"Start":"07:37.790 ","End":"07:41.735","Text":"For T Delta S, we have 298.15,"},{"Start":"07:41.735 ","End":"07:47.100","Text":"that\u0027s T. Then Delta S is 198.9."},{"Start":"07:47.100 ","End":"07:48.960","Text":"Now here it\u0027s joules,"},{"Start":"07:48.960 ","End":"07:52.800","Text":"so we have to convert it into kilojoules by dividing by a 1000."},{"Start":"07:52.800 ","End":"07:54.330","Text":"Work that all out,"},{"Start":"07:54.330 ","End":"08:03.700","Text":"we get 92.22 minus 59.30 and that\u0027s equal to 32.90 kilojoules."},{"Start":"08:04.340 ","End":"08:09.135","Text":"This temperature, 25 degrees Celsius,"},{"Start":"08:09.135 ","End":"08:11.330","Text":"Delta G is positive."},{"Start":"08:11.330 ","End":"08:15.330","Text":"The reaction is non-spontaneous."},{"Start":"08:16.060 ","End":"08:20.960","Text":"We can work out the temperature which becomes spontaneous."},{"Start":"08:20.960 ","End":"08:27.830","Text":"That\u0027s Delta H divided by Delta S as 92.22 divided by 0.1989."},{"Start":"08:27.830 ","End":"08:31.210","Text":"We had to convert that from joules to kilojoules."},{"Start":"08:31.210 ","End":"08:36.670","Text":"We get the answer 463.7 Kelvin."},{"Start":"08:37.390 ","End":"08:42.845","Text":"We know that when the temperature is greater than 463.7,"},{"Start":"08:42.845 ","End":"08:46.055","Text":"the reaction will be spontaneous."},{"Start":"08:46.055 ","End":"08:48.755","Text":"This of course is in Kelvin."},{"Start":"08:48.755 ","End":"08:56.460","Text":"In this video, we use Gibbs Free Energy to calculate whether a process is spontaneous."}],"ID":26384},{"Watched":false,"Name":"Gibbs Free Energy of Formation and Reaction","Duration":"6m 43s","ChapterTopicVideoID":25611,"CourseChapterTopicPlaylistID":244891,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.530","Text":"In previous videos,"},{"Start":"00:01.530 ","End":"00:04.860","Text":"we talked about the standard enthalpy and entropy of reactions."},{"Start":"00:04.860 ","End":"00:05.940","Text":"In this video,"},{"Start":"00:05.940 ","End":"00:10.365","Text":"we\u0027ll talk about the standard Gibbs free energy of a reaction."},{"Start":"00:10.365 ","End":"00:16.530","Text":"Before we start that,"},{"Start":"00:16.530 ","End":"00:21.345","Text":"we\u0027ll talk about the standard Gibbs free energy of formation."},{"Start":"00:21.345 ","End":"00:27.345","Text":"Now the standard Gibbs free energy of formation, Delta G_f,"},{"Start":"00:27.345 ","End":"00:30.325","Text":"with a little o,"},{"Start":"00:30.325 ","End":"00:36.455","Text":"is defined in the same way as the standard enthalpy of formation, Delta H_f."},{"Start":"00:36.455 ","End":"00:41.780","Text":"The standard Gibbs free energy of formation is the change in the Gibbs free energy per"},{"Start":"00:41.780 ","End":"00:47.850","Text":"mole when a substance is formed from its elements in their standard states."},{"Start":"00:48.020 ","End":"00:55.775","Text":"Now, Delta G formation is 0 for an element in its reference form."},{"Start":"00:55.775 ","End":"00:57.929","Text":"For most elements,"},{"Start":"00:57.929 ","End":"01:03.410","Text":"that\u0027s simply what you expect to obtain at 25 degrees Celsius,"},{"Start":"01:03.410 ","End":"01:06.095","Text":"like a gas or hydrogen,"},{"Start":"01:06.095 ","End":"01:10.355","Text":"or oxygen or so on."},{"Start":"01:10.355 ","End":"01:16.765","Text":"There are a few exceptions which we discussed when we learned about thermochemistry."},{"Start":"01:16.765 ","End":"01:21.125","Text":"Now, Delta G_f standard,"},{"Start":"01:21.125 ","End":"01:25.000","Text":"the standard volume of Gibbs free energy of formation is"},{"Start":"01:25.000 ","End":"01:29.545","Text":"listed in tables together with Delta H_f^0 and S^0."},{"Start":"01:29.545 ","End":"01:34.880","Text":"You should find them altogether and they usually at 25 degrees Celsius."},{"Start":"01:35.520 ","End":"01:39.945","Text":"Delta G_f^0 cannot be measured directly."},{"Start":"01:39.945 ","End":"01:48.250","Text":"But you can calculate from Delta G_f^0 equal to Delta H_f^0 minus T Delta S_f^0."},{"Start":"01:48.290 ","End":"01:50.430","Text":"If we know Delta H,"},{"Start":"01:50.430 ","End":"01:52.005","Text":"we know Delta S,"},{"Start":"01:52.005 ","End":"01:59.220","Text":"we can calculate Delta G. The units are kilojoules per mole."},{"Start":"01:59.620 ","End":"02:04.555","Text":"I\u0027m going to say a few words about thermodynamic stability."},{"Start":"02:04.555 ","End":"02:08.150","Text":"Now, if Delta G_f is negative,"},{"Start":"02:08.150 ","End":"02:13.835","Text":"we say the compound is thermodynamically stable and if Delta G_f is positive,"},{"Start":"02:13.835 ","End":"02:17.740","Text":"the compound is thermodynamically unstable."},{"Start":"02:17.740 ","End":"02:22.250","Text":"For example, Delta G_f for benzene is plus"},{"Start":"02:22.250 ","End":"02:26.915","Text":"124 kilojoules per mole at 25 degrees Celsius,"},{"Start":"02:26.915 ","End":"02:31.650","Text":"so benzene is less stable than its elements."},{"Start":"02:32.290 ","End":"02:37.100","Text":"However, we all know that benzene can be kept in"},{"Start":"02:37.100 ","End":"02:42.190","Text":"the laboratory for quite some period of time and it does not decompose."},{"Start":"02:42.190 ","End":"02:50.120","Text":"It\u0027s thermodynamically unstable, but it decomposes extremely slowly."},{"Start":"02:50.120 ","End":"02:52.765","Text":"We say it\u0027s non-labile."},{"Start":"02:52.765 ","End":"02:58.985","Text":"So thermodynamically unstable, but kinetically decays very slowly."},{"Start":"02:58.985 ","End":"03:01.385","Text":"We say it\u0027s non-labile."},{"Start":"03:01.385 ","End":"03:07.195","Text":"Whereas unstable compounds that decompose rapidly are said to be labile."},{"Start":"03:07.195 ","End":"03:12.490","Text":"Now that we know about the standard Gibbs free energy of formation,"},{"Start":"03:12.490 ","End":"03:16.225","Text":"we can define the standard Gibbs free energy of reaction."},{"Start":"03:16.225 ","End":"03:22.440","Text":"This is the same as the definition of Delta H of reaction."},{"Start":"03:22.440 ","End":"03:28.660","Text":"Delta G of reaction is equal to the sum over all the products where for each product,"},{"Start":"03:28.660 ","End":"03:33.519","Text":"we multiply the number of moles times Delta G of formation."},{"Start":"03:33.519 ","End":"03:35.950","Text":"Then we subtract from that,"},{"Start":"03:35.950 ","End":"03:38.229","Text":"the sum over all the reactants,"},{"Start":"03:38.229 ","End":"03:40.660","Text":"where we multiply the number of moles for"},{"Start":"03:40.660 ","End":"03:47.345","Text":"each reactant times Delta G of formation for that particular reactant."},{"Start":"03:47.345 ","End":"03:49.814","Text":"Let\u0027s take an example."},{"Start":"03:49.814 ","End":"03:57.394","Text":"Calculate Delta G of reaction for the reaction 2SO_3 gas,"},{"Start":"03:57.394 ","End":"04:02.640","Text":"giving us 2SO_2 gas plus oxygen gas."},{"Start":"04:02.640 ","End":"04:07.470","Text":"So SO_3 decomposes to SO_2 and oxygen."},{"Start":"04:07.470 ","End":"04:14.335","Text":"We\u0027re given the Delta G of formation of SO_2 is minus 300.2 kilojoules per mole."},{"Start":"04:14.335 ","End":"04:25.735","Text":"For SO_3 is minus 371.1 kilojoules per mole and Delta G_f for oxygen is of course 0,"},{"Start":"04:25.735 ","End":"04:35.685","Text":"because we\u0027re talking about the standard Gibbs free energy and oxygen is an element."},{"Start":"04:35.685 ","End":"04:38.055","Text":"Now, we can do the calculation."},{"Start":"04:38.055 ","End":"04:41.025","Text":"Delta G reaction is equal to,"},{"Start":"04:41.025 ","End":"04:44.265","Text":"first of all, we look at the products."},{"Start":"04:44.265 ","End":"04:48.915","Text":"We have twice, we have 2 moles times"},{"Start":"04:48.915 ","End":"04:57.335","Text":"Delta G_f^0 for SO_2 plus 1 times Delta G_f^0 of O_2,"},{"Start":"04:57.335 ","End":"05:05.965","Text":"which of course is 0 minus 2 times Delta G_f^0 of SO_3 that\u0027s the reactants."},{"Start":"05:05.965 ","End":"05:12.810","Text":"The units are moles for these 2,1 and 2 and kilojoules per"},{"Start":"05:12.810 ","End":"05:19.590","Text":"mole for the Delta G values and mole times moles apart minus 1 is 1 of course,"},{"Start":"05:19.590 ","End":"05:21.885","Text":"so finally, we get kilojoules."},{"Start":"05:21.885 ","End":"05:25.275","Text":"Now, we can substitute these values,"},{"Start":"05:25.275 ","End":"05:31.590","Text":"2 times minus 300.2 plus 0 minus 2 times"},{"Start":"05:31.590 ","End":"05:37.820","Text":"minus 371.1 and that\u0027s in kilojoules and we work all this out,"},{"Start":"05:37.820 ","End":"05:43.175","Text":"we get 141.8 kilojoules."},{"Start":"05:43.175 ","End":"05:46.479","Text":"We\u0027re getting Delta G positive,"},{"Start":"05:46.479 ","End":"05:53.310","Text":"which tells us that SO_3 will"},{"Start":"05:53.310 ","End":"06:01.510","Text":"not decompose spontaneously to SO_2 plus oxygen at 25 degrees Celsius."},{"Start":"06:01.510 ","End":"06:05.105","Text":"Now, if we go to sufficiently high temperatures,"},{"Start":"06:05.105 ","End":"06:11.180","Text":"SO_3 can decompose to SO_2 and oxygen and the reason for"},{"Start":"06:11.180 ","End":"06:18.210","Text":"this is that Delta H is positive,"},{"Start":"06:18.210 ","End":"06:21.810","Text":"but Delta S is also positive."},{"Start":"06:21.810 ","End":"06:29.820","Text":"At some point, T Delta S becomes greater than Delta H,"},{"Start":"06:29.820 ","End":"06:36.380","Text":"and then the SO_3 will decompose to SO_2 and oxygen."},{"Start":"06:36.380 ","End":"06:43.200","Text":"In this video, we learned about Gibbs free energy of formation and reaction."}],"ID":26386},{"Watched":false,"Name":"Gibbs Free Energy and Non-Expansion Work","Duration":"7m 5s","ChapterTopicVideoID":25610,"CourseChapterTopicPlaylistID":244891,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:02.400","Text":"In previous videos,"},{"Start":"00:02.400 ","End":"00:04.559","Text":"we talked about Gibbs free energy."},{"Start":"00:04.559 ","End":"00:05.640","Text":"In this video,"},{"Start":"00:05.640 ","End":"00:09.100","Text":"we\u0027ll learn why it\u0027s called free energy."},{"Start":"00:09.380 ","End":"00:16.120","Text":"We\u0027re going to talk about Gibbs free energy and non-expansion work."},{"Start":"00:16.460 ","End":"00:19.980","Text":"The Gibbs free energy is a measure of the energy of"},{"Start":"00:19.980 ","End":"00:24.345","Text":"a system that is free to do non-expansion work."},{"Start":"00:24.345 ","End":"00:27.120","Text":"Now what\u0027s non-expansion work?"},{"Start":"00:27.120 ","End":"00:30.180","Text":"Non-expansion work, which we\u0027ll call w_e,"},{"Start":"00:30.180 ","End":"00:31.410","Text":"e is for extra,"},{"Start":"00:31.410 ","End":"00:35.930","Text":"is any type of work apart from expansion against an external pressure,"},{"Start":"00:35.930 ","End":"00:38.735","Text":"what we\u0027ll call PV work."},{"Start":"00:38.735 ","End":"00:41.600","Text":"Now this includes electrical work, for example,"},{"Start":"00:41.600 ","End":"00:45.200","Text":"pushing electrons through electrical circuit and mechanical work,"},{"Start":"00:45.200 ","End":"00:49.805","Text":"for example, stretching a spring or pushing a weight up a slope."},{"Start":"00:49.805 ","End":"00:54.800","Text":"It also includes biological work such as sending nerve signals through neurons,"},{"Start":"00:54.800 ","End":"00:57.830","Text":"and it\u0027s important in bioenergetics,"},{"Start":"00:57.830 ","End":"01:01.130","Text":"the study of energy in living cells."},{"Start":"01:01.130 ","End":"01:05.060","Text":"What we\u0027re going to prove is that the change in Gibbs free energy is"},{"Start":"01:05.060 ","End":"01:09.275","Text":"the maximum non-expansion work a system can do."},{"Start":"01:09.275 ","End":"01:14.215","Text":"In other words, Delta G is equal to w_e maximum."},{"Start":"01:14.215 ","End":"01:20.700","Text":"Now the proof is rather long and involved and you may not need it in your course."},{"Start":"01:20.700 ","End":"01:23.775","Text":"You can skip this bit if you don\u0027t need it."},{"Start":"01:23.775 ","End":"01:29.905","Text":"It starts off by saying that for an infinitesimal change at constant temperature,"},{"Start":"01:29.905 ","End":"01:32.435","Text":"G=H minus TS,"},{"Start":"01:32.435 ","End":"01:34.735","Text":"which is the definition of Gibbs free energy,"},{"Start":"01:34.735 ","End":"01:39.640","Text":"implies dG=dH minus TdS."},{"Start":"01:39.640 ","End":"01:41.290","Text":"The temperature is constant."},{"Start":"01:41.290 ","End":"01:43.090","Text":"T is constant."},{"Start":"01:43.090 ","End":"01:46.255","Text":"If the pressure is also constant,"},{"Start":"01:46.255 ","End":"01:53.810","Text":"H=U plus PV implies that dH=dU plus PdV."},{"Start":"01:53.810 ","End":"01:58.740","Text":"Now, we\u0027re going to substitute equation 2 into equation 1."},{"Start":"01:58.740 ","End":"02:04.980","Text":"Then we get dG=dU plus PdV."},{"Start":"02:04.980 ","End":"02:08.460","Text":"We\u0027ve written dH=dU plus PdV,"},{"Start":"02:08.460 ","End":"02:13.545","Text":"so we have dU plus PdV minus TdS."},{"Start":"02:13.545 ","End":"02:16.620","Text":"We know there\u0027s a constant temperature and pressure."},{"Start":"02:16.620 ","End":"02:17.990","Text":"U, the internal energy,"},{"Start":"02:17.990 ","End":"02:20.285","Text":"is equal to q plus w,"},{"Start":"02:20.285 ","End":"02:26.270","Text":"and for a small change that\u0027s dU=dq plus dw."},{"Start":"02:26.270 ","End":"02:31.175","Text":"Now if we substitute equation 4 into equation 3,"},{"Start":"02:31.175 ","End":"02:37.580","Text":"we get dG instead of dU we\u0027re writing dq plus dw from here,"},{"Start":"02:37.580 ","End":"02:41.435","Text":"plus PdV minus TdS."},{"Start":"02:41.435 ","End":"02:44.690","Text":"Now if we\u0027re talking about a reversible change,"},{"Start":"02:44.690 ","End":"02:55.540","Text":"equation 5 becomes dG=dq_rev plus dw_rev plus PdV minus TdS."},{"Start":"02:55.540 ","End":"03:00.390","Text":"Now we know that dS=dq_reversible divided by"},{"Start":"03:00.390 ","End":"03:07.300","Text":"T. We can rewrite that dq_rev=TdS."},{"Start":"03:07.300 ","End":"03:12.825","Text":"Now if we substitute equation 7 into 6,"},{"Start":"03:12.825 ","End":"03:17.040","Text":"we\u0027ll get dG is equal to instead of dq_rev,"},{"Start":"03:17.040 ","End":"03:27.405","Text":"we write TdS plus dw_rev plus PdV minus TdS and then TdS cancels."},{"Start":"03:27.405 ","End":"03:33.550","Text":"We\u0027re left with dG=dw_rev plus PdV."},{"Start":"03:33.550 ","End":"03:38.060","Text":"Now we\u0027re going to divide the work into expansion in non-expansion work."},{"Start":"03:38.060 ","End":"03:47.420","Text":"dw_rev=dw_rev,e, let\u0027s say non-expansion work, does dw_rev expansion."},{"Start":"03:47.420 ","End":"03:53.690","Text":"We know that dw_expansion is equal to minus P_ex dV."},{"Start":"03:53.690 ","End":"03:57.670","Text":"The external pressure is P_ex."},{"Start":"03:57.670 ","End":"04:01.925","Text":"If we\u0027re interested in the reversible version of this,"},{"Start":"04:01.925 ","End":"04:05.360","Text":"it\u0027s dw_rev,expansion is equal to minus"},{"Start":"04:05.360 ","End":"04:12.040","Text":"PdV because at every point we are equating P external to P internal."},{"Start":"04:12.040 ","End":"04:15.010","Text":"P_ex=P."},{"Start":"04:18.280 ","End":"04:23.550","Text":"Now we can substitute equation 10 into 9."},{"Start":"04:23.550 ","End":"04:28.285","Text":"We get dw_rev=dw_rev,e,"},{"Start":"04:28.285 ","End":"04:31.350","Text":"and instead of the dw_rev,expansion,"},{"Start":"04:31.350 ","End":"04:33.285","Text":"we write minus PdV."},{"Start":"04:33.285 ","End":"04:39.700","Text":"We have dw_rev=dw_rev,e minus PdV."},{"Start":"04:40.130 ","End":"04:42.395","Text":"Then it\u0027s a long story."},{"Start":"04:42.395 ","End":"04:46.175","Text":"We can substitute equation 11 into 8,"},{"Start":"04:46.175 ","End":"04:51.050","Text":"so this expression for dw_rev into equation 8,"},{"Start":"04:51.050 ","End":"04:54.860","Text":"we get dG is equal to, instead of dw_rev,"},{"Start":"04:54.860 ","End":"04:58.910","Text":"we\u0027re writing dw_rev,e minus PdV."},{"Start":"04:58.910 ","End":"05:01.220","Text":"Then we have plus PdV."},{"Start":"05:01.220 ","End":"05:05.790","Text":"Now minus PdV and plus PdV give us"},{"Start":"05:05.790 ","End":"05:12.270","Text":"0 and we have left with dG=dw_rev,e."},{"Start":"05:12.270 ","End":"05:22.070","Text":"Now dw_rev,e is the maximum work the system could do"},{"Start":"05:22.070 ","End":"05:25.880","Text":"at constant T and P so that"},{"Start":"05:25.880 ","End":"05:35.325","Text":"dG=dw_e,max and that\u0027s implies that Delta G=w_e,max."},{"Start":"05:35.325 ","End":"05:38.380","Text":"We\u0027ve proved what we wanted to prove."},{"Start":"05:40.100 ","End":"05:43.315","Text":"Now a word about the signs."},{"Start":"05:43.315 ","End":"05:47.345","Text":"We know that when a system does work, w is negative."},{"Start":"05:47.345 ","End":"05:49.055","Text":"If w is negative,"},{"Start":"05:49.055 ","End":"05:51.695","Text":"Delta G must also be negative."},{"Start":"05:51.695 ","End":"05:55.770","Text":"We\u0027re talking about spontaneous processes."},{"Start":"05:55.790 ","End":"05:58.010","Text":"Let\u0027s take an example,"},{"Start":"05:58.010 ","End":"06:00.665","Text":"the oxidation of glucose."},{"Start":"06:00.665 ","End":"06:05.520","Text":"Here\u0027s the equation for the oxidation of glucose."},{"Start":"06:07.510 ","End":"06:11.035","Text":"C_6H_12O_6, it\u0027s a solid,"},{"Start":"06:11.035 ","End":"06:14.750","Text":"reacting with 6 moles of oxygens and gas, of course,"},{"Start":"06:14.750 ","End":"06:16.760","Text":"to give us 6 moles of carbon dioxide,"},{"Start":"06:16.760 ","End":"06:18.620","Text":"6 moles of water."},{"Start":"06:18.620 ","End":"06:27.020","Text":"Delta G^0 for this process is minus 2879 kilojoules."},{"Start":"06:27.020 ","End":"06:32.560","Text":"The maximum work obtained from oxidation of 1 mole of glucose,"},{"Start":"06:32.560 ","End":"06:37.890","Text":"oxidation or burning, is 2879 kilojoules."},{"Start":"06:37.890 ","End":"06:42.500","Text":"Now take 17 kilojoules to make 1 mole of links between amino acids,"},{"Start":"06:42.500 ","End":"06:44.080","Text":"we call these peptide links."},{"Start":"06:44.080 ","End":"06:50.210","Text":"The maximum number of links is 2879 divided by 17,"},{"Start":"06:50.210 ","End":"06:53.045","Text":"that means 170 peptide links."},{"Start":"06:53.045 ","End":"06:58.820","Text":"This is absolute maximum and in practice not so many are formed."},{"Start":"06:58.820 ","End":"07:05.100","Text":"In this video, we discuss the maximum work that a system can perform."}],"ID":26385}],"Thumbnail":null,"ID":244891}]

[{"ID":244889,"Videos":[26372,26379,26378,26377,26376,26375,26380,26381,26382,26373,26374]},{"ID":244890,"Videos":[30052,30053,30054,30055,30056,30057,30058,30059,30060,30061,30062,30063]},{"ID":244891,"Videos":[26383,26384,26386,26385]}];

[26372,26379,26378,26377,26376,26375,26380,26381,26382,26373,26374];

1

5

Get unlimited access to **1500 subjects** including **personalised modules**

Start your free trial
We couldn't find any results for

Upload your syllabus now and our team will create a customised module especially for you!

Alert

and we will create a personalised module (just for you) in less than **48 hours...**