Reactions at Equilibrium
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Tools For Equilibrium Calculations
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- Equilibrium Constant Using Gas Concentrations
- Exercise 1
- Exercise 2
- Exercise 3
- Equilibrium Calculations 1
- Equilibrium Calculations 2
- Equilibrium Calculations with Approximation 1
- Equilibrium Calculations with Approximations 2
- Exercise 4
- Exercise 5
- Exercise 6
- Combining Equilibrium Constants
- Exercise 7

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[{"Name":"Reactions at Equilibrium","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Equilibrium Constant and Law of Mass Action","Duration":"10m 47s","ChapterTopicVideoID":26272,"CourseChapterTopicPlaylistID":253355,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/26272.jpeg","UploadDate":"2021-08-27T13:29:01.6630000","DurationForVideoObject":"PT10M47S","Description":null,"MetaTitle":"Equilibrium Constant and Law of Mass Action: Video + Workbook | Proprep","MetaDescription":"Chemical Equilibria - Reactions at Equilibrium. 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/chemical-equilibria/reactions-at-equilibrium/vid30453","VideoComments":[],"Subtitles":[{"Start":"00:00.470 ","End":"00:02.640","Text":"In the previous video,"},{"Start":"00:02.640 ","End":"00:06.420","Text":"we talked about dynamic equilibrium in chemical reactions."},{"Start":"00:06.420 ","End":"00:12.795","Text":"In this video, we\u0027ll talk about the equilibrium constant and the law of mass action."},{"Start":"00:12.795 ","End":"00:16.725","Text":"First, we\u0027re going to talk about the equilibrium constant."},{"Start":"00:16.725 ","End":"00:24.105","Text":"Now in 1864, Guldberg and Waage noticed that for every reaction at equilibrium,"},{"Start":"00:24.105 ","End":"00:27.060","Text":"one can define an equilibrium constant,"},{"Start":"00:27.060 ","End":"00:29.010","Text":"which they called capital K,"},{"Start":"00:29.010 ","End":"00:35.355","Text":"that doesn\u0027t depend on the initial concentration of the reactants and products."},{"Start":"00:35.355 ","End":"00:37.910","Text":"Let\u0027s take an example."},{"Start":"00:37.910 ","End":"00:41.765","Text":"Supposing we have the reaction SO_2,"},{"Start":"00:41.765 ","End":"00:46.115","Text":"the gas phase, reacting with oxygen to form SO_3,"},{"Start":"00:46.115 ","End":"00:53.888","Text":"sulfur dioxide plus oxygen to give a sulfur trioxide,"},{"Start":"00:53.888 ","End":"00:58.075","Text":"and this happens at 1,000 degrees Kelvin."},{"Start":"00:58.075 ","End":"01:06.545","Text":"Now, they found that if you write K is equal to the pressure of SO_3 divided by p^0,"},{"Start":"01:06.545 ","End":"01:09.715","Text":"where p^0 is 1 bar,"},{"Start":"01:09.715 ","End":"01:14.222","Text":"and you square that and you divide that by the pressure of"},{"Start":"01:14.222 ","End":"01:20.255","Text":"SO_2 divided by p^0 squared times p,"},{"Start":"01:20.255 ","End":"01:23.575","Text":"the pressure of oxygen, divided by p^0."},{"Start":"01:23.575 ","End":"01:26.515","Text":"If you work out that K,"},{"Start":"01:26.515 ","End":"01:31.595","Text":"it\u0027s approximately the same regardless of the initial composition."},{"Start":"01:31.595 ","End":"01:36.360","Text":"This is a constant at a particular temperature."},{"Start":"01:36.530 ","End":"01:39.225","Text":"So here P_A,"},{"Start":"01:39.225 ","End":"01:43.420","Text":"like P_SO_2 or P_SO_3 or P_O_2,"},{"Start":"01:43.420 ","End":"01:47.070","Text":"is the partial pressure A at equilibrium and"},{"Start":"01:47.070 ","End":"01:52.600","Text":"p^0 is 1 bar and that\u0027s the standard pressure."},{"Start":"01:53.360 ","End":"02:01.360","Text":"Now because we\u0027re dividing pressure by pressure, K is dimensionless."},{"Start":"02:01.360 ","End":"02:04.955","Text":"It has no dimensions at all."},{"Start":"02:04.955 ","End":"02:11.750","Text":"For simplicity, sometimes we write K is equal to the partial pressure of SO_3"},{"Start":"02:11.750 ","End":"02:15.290","Text":"squared divided by the partial pressure of SO_2 squared"},{"Start":"02:15.290 ","End":"02:19.325","Text":"times the partial pressure of oxygen,"},{"Start":"02:19.325 ","End":"02:23.860","Text":"where now P_A is the numerical value of the pressure in bars."},{"Start":"02:23.860 ","End":"02:28.330","Text":"In other words, these pressures have no units."},{"Start":"02:28.330 ","End":"02:30.715","Text":"So if the pressures have no units,"},{"Start":"02:30.715 ","End":"02:33.310","Text":"K has no units."},{"Start":"02:33.310 ","End":"02:37.580","Text":"Now, we need the law of mass action."},{"Start":"02:38.400 ","End":"02:43.105","Text":"Now the law of mass action says that at equilibrium,"},{"Start":"02:43.105 ","End":"02:46.120","Text":"the composition of the reaction mixture can be"},{"Start":"02:46.120 ","End":"02:50.750","Text":"expressed in terms of an equilibrium constant."},{"Start":"02:50.750 ","End":"02:55.720","Text":"It is a formal way of telling us that every reaction that"},{"Start":"02:55.720 ","End":"03:00.890","Text":"goes to equilibrium has an equilibrium constant."},{"Start":"03:02.600 ","End":"03:07.015","Text":"Up to now we\u0027ve talked about a specific reaction,"},{"Start":"03:07.015 ","End":"03:11.530","Text":"but we need a general expression for the equilibrium constant."},{"Start":"03:11.530 ","End":"03:15.280","Text":"Supposing we have a general reaction."},{"Start":"03:15.280 ","End":"03:20.590","Text":"a moles of A say in the gas phase plus b moles of B in"},{"Start":"03:20.590 ","End":"03:27.130","Text":"the gas phase to give c moles of C in the gas phase plus d moles of D in the gas phase."},{"Start":"03:27.130 ","End":"03:30.385","Text":"We treat all the gases as ideal."},{"Start":"03:30.385 ","End":"03:36.790","Text":"Then K can be defined as the pressure of C. Again,"},{"Start":"03:36.790 ","End":"03:40.659","Text":"we\u0027re talking about the pressure in bars,"},{"Start":"03:40.659 ","End":"03:44.035","Text":"but without any units being written."},{"Start":"03:44.035 ","End":"03:45.970","Text":"If it\u0027s 5 bars,"},{"Start":"03:45.970 ","End":"03:47.945","Text":"we\u0027ll just write 5."},{"Start":"03:47.945 ","End":"03:55.090","Text":"That\u0027s taken to the power C. The power C here, the number of moles,"},{"Start":"03:55.090 ","End":"03:59.950","Text":"times the pressure of D. That\u0027s another product to the power"},{"Start":"03:59.950 ","End":"04:05.950","Text":"D. Then in the numerator we have the products,"},{"Start":"04:06.560 ","End":"04:11.275","Text":"then in the denominator we have the reactants."},{"Start":"04:11.275 ","End":"04:16.085","Text":"In the denominator, we have the pressure of A to the power a,"},{"Start":"04:16.085 ","End":"04:18.595","Text":"because we have A moles here,"},{"Start":"04:18.595 ","End":"04:23.905","Text":"times the pressure of B to the power b because we have b moles of B."},{"Start":"04:23.905 ","End":"04:27.380","Text":"This is a general expression."},{"Start":"04:27.500 ","End":"04:31.210","Text":"Sometimes people write K_P,"},{"Start":"04:31.210 ","End":"04:32.960","Text":"but if you define it like this,"},{"Start":"04:32.960 ","End":"04:40.850","Text":"you don\u0027t really need to write the P. What we did above was for gases."},{"Start":"04:40.850 ","End":"04:43.610","Text":"Now, we\u0027ll do it for ideal solutions or write"},{"Start":"04:43.610 ","End":"04:47.450","Text":"the equilibrium constant for ideal solutions."},{"Start":"04:47.450 ","End":"04:49.670","Text":"Instead of the pressure,"},{"Start":"04:49.670 ","End":"04:56.900","Text":"we have the molar concentration A divided by the C^0."},{"Start":"04:56.900 ","End":"04:59.225","Text":"C^0 is 1 mole per liter,"},{"Start":"04:59.225 ","End":"05:07.065","Text":"which that means that we have the concentration in moles per liter."},{"Start":"05:07.065 ","End":"05:08.850","Text":"The molar concentration,"},{"Start":"05:08.850 ","End":"05:11.990","Text":"the molarity without any units."},{"Start":"05:11.990 ","End":"05:17.250","Text":"It\u0027s the concentration A divided by C^0."},{"Start":"05:17.250 ","End":"05:23.135","Text":"It\u0027s the molar concentration relative to the standard concentration of C^0,"},{"Start":"05:23.135 ","End":"05:24.845","Text":"which is 1 mole per liter."},{"Start":"05:24.845 ","End":"05:29.580","Text":"In other words, just the molar concentration without any units."},{"Start":"05:30.380 ","End":"05:35.270","Text":"Often A is written alone and the units of K are ignored."},{"Start":"05:35.270 ","End":"05:39.145","Text":"You can just ignore the units of K. It\u0027s supposed to be dimensionless."},{"Start":"05:39.145 ","End":"05:44.270","Text":"Now, pure liquids and solids are not included in the expression for"},{"Start":"05:44.270 ","End":"05:50.965","Text":"K. Now we\u0027re going to define the activity."},{"Start":"05:50.965 ","End":"05:55.285","Text":"Assuming we have ideal gases of very dilute solutions."},{"Start":"05:55.285 ","End":"06:02.860","Text":"The activity is in general use and deviates from the values we\u0027re going to give"},{"Start":"06:02.860 ","End":"06:10.630","Text":"now when we have non-ideal gases or concentrated solutions but in this course,"},{"Start":"06:10.630 ","End":"06:16.550","Text":"I\u0027m just going to assume we have ideal gases of very dilute solutions."},{"Start":"06:16.550 ","End":"06:19.740","Text":"Now for an ideal gas a,"},{"Start":"06:19.740 ","End":"06:26.335","Text":"the activity of A is the pressure of A divided by p^0."},{"Start":"06:26.335 ","End":"06:30.860","Text":"Or for simplicity, a is equal to P_A."},{"Start":"06:31.220 ","End":"06:34.665","Text":"For a solute in a dilute solution,"},{"Start":"06:34.665 ","End":"06:41.200","Text":"the activity is just the concentration A divided by C^0 or A,"},{"Start":"06:41.200 ","End":"06:45.290","Text":"the concentration of a, for simplicity."},{"Start":"06:45.320 ","End":"06:48.430","Text":"For a pure solid or liquid,"},{"Start":"06:48.430 ","End":"06:51.790","Text":"the activity is just 1."},{"Start":"06:51.790 ","End":"06:56.545","Text":"Note that the activities are unitless, dimensionless."},{"Start":"06:56.545 ","End":"06:59.215","Text":"So if K is defined in terms of them,"},{"Start":"06:59.215 ","End":"07:05.060","Text":"K will also be dimensionless or unitless."},{"Start":"07:08.660 ","End":"07:12.400","Text":"Now, we\u0027re going to write a general expression for the equilibrium"},{"Start":"07:12.400 ","End":"07:15.800","Text":"constant in terms of activities."},{"Start":"07:15.800 ","End":"07:22.185","Text":"So for the reaction a A plus b B to c C plus d D,"},{"Start":"07:22.185 ","End":"07:28.130","Text":"K is going to be the activity of C to the power of c times the activity of D to power d,"},{"Start":"07:28.130 ","End":"07:31.403","Text":"that\u0027s the products,"},{"Start":"07:31.403 ","End":"07:35.610","Text":"divided by a A to the power a times a B to the power of b,"},{"Start":"07:35.610 ","End":"07:38.110","Text":"and that\u0027s the reactants."},{"Start":"07:38.780 ","End":"07:44.045","Text":"Now, we can define homogeneous and heterogeneous equilibrium."},{"Start":"07:44.045 ","End":"07:47.660","Text":"Now if all the reactants and products are in the same phase,"},{"Start":"07:47.660 ","End":"07:50.045","Text":"they\u0027re all gas or solid or liquid,"},{"Start":"07:50.045 ","End":"07:52.820","Text":"the equilibrium is homogeneous."},{"Start":"07:52.820 ","End":"07:56.150","Text":"If the reactants and products are in more than 1 phase,"},{"Start":"07:56.150 ","End":"07:58.805","Text":"the equilibrium is called heterogeneous."},{"Start":"07:58.805 ","End":"08:01.745","Text":"Now, we\u0027re going to consider 2 examples."},{"Start":"08:01.745 ","End":"08:04.910","Text":"Write the equilibrium constants for the reactions."},{"Start":"08:04.910 ","End":"08:11.100","Text":"Magnesium hydroxide, that\u0027s a solid in equilibrium with magnesium^2+,"},{"Start":"08:11.100 ","End":"08:15.620","Text":"that\u0027s the magnesium cation in the aqueous phase,"},{"Start":"08:15.620 ","End":"08:22.610","Text":"plus 2 OH- hydroxide anion in the aqueous phase."},{"Start":"08:22.610 ","End":"08:30.140","Text":"That\u0027s the first one. We can write K for this first reaction as, first of all,"},{"Start":"08:30.140 ","End":"08:34.530","Text":"we consider the products a of Mg^2+,"},{"Start":"08:35.530 ","End":"08:41.550","Text":"and that\u0027s just to the power of 1 because it\u0027s one mole,"},{"Start":"08:41.550 ","End":"08:44.668","Text":"times a^2 of OH-,"},{"Start":"08:44.668 ","End":"08:48.455","Text":"and that\u0027s squared because of the 2 here,"},{"Start":"08:48.455 ","End":"08:52.955","Text":"and then it\u0027s divided by a_Mg(OH)_2,"},{"Start":"08:52.955 ","End":"08:55.890","Text":"and that\u0027s for the reactants."},{"Start":"08:55.990 ","End":"09:02.265","Text":"Now, we have to write the values of these activities."},{"Start":"09:02.265 ","End":"09:06.620","Text":"So for Mg^2+,"},{"Start":"09:06.620 ","End":"09:09.840","Text":"it\u0027s just the concentration."},{"Start":"09:10.160 ","End":"09:17.000","Text":"For OH-, it\u0027s the concentration of OH- and that\u0027s squared."},{"Start":"09:17.000 ","End":"09:20.105","Text":"We have the squared here and a squared here."},{"Start":"09:20.105 ","End":"09:27.679","Text":"Now, the numerator here is just 1 because this is a solid."},{"Start":"09:27.679 ","End":"09:30.660","Text":"The numerator is 1."},{"Start":"09:30.660 ","End":"09:37.110","Text":"We\u0027re left with [Mg2+][OH-]^2."},{"Start":"09:37.110 ","End":"09:39.410","Text":"Now let\u0027s look at the second example."},{"Start":"09:39.410 ","End":"09:44.780","Text":"Calcium carbonate, which is a solid in equilibrium with calcium,"},{"Start":"09:44.780 ","End":"09:48.575","Text":"which is also solid, and CO_2 gas."},{"Start":"09:48.575 ","End":"09:56.705","Text":"We can write K_2 is equal to the activity of Ca times the activity of CO_2."},{"Start":"09:56.705 ","End":"10:02.570","Text":"Now these are not raised to any power because we just have a 1 here and a 1 here,"},{"Start":"10:02.570 ","End":"10:06.730","Text":"divided by activity of CaCO_3,"},{"Start":"10:06.730 ","End":"10:08.825","Text":"and again it\u0027s just one mole."},{"Start":"10:08.825 ","End":"10:12.140","Text":"Now, Ca is a solid,"},{"Start":"10:12.140 ","End":"10:14.850","Text":"so the activity is just one."},{"Start":"10:15.050 ","End":"10:17.640","Text":"CaCO_3 is also a solid,"},{"Start":"10:17.640 ","End":"10:23.000","Text":"so the activity is one and we\u0027re left just with CO_2, and that\u0027s a gas."},{"Start":"10:23.000 ","End":"10:30.290","Text":"K_2 is just the pressure of CO_2 because a for calcium,"},{"Start":"10:30.290 ","End":"10:33.230","Text":"and a for CaCO_3 are just one."},{"Start":"10:33.230 ","End":"10:35.375","Text":"We\u0027re left with a for CO_2,"},{"Start":"10:35.375 ","End":"10:40.060","Text":"and that\u0027s the pressure of CO_2 in dimensionless units."},{"Start":"10:40.060 ","End":"10:46.680","Text":"In this video, we learned about equilibrium constants and the law of mass action."}],"ID":30453},{"Watched":false,"Name":"Exercise 1 Part 1","Duration":"5m 35s","ChapterTopicVideoID":28914,"CourseChapterTopicPlaylistID":253355,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.925","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.925 ","End":"00:07.425","Text":"Write the equilibrium constant expression Kc for the following reactions."},{"Start":"00:07.425 ","End":"00:11.310","Text":"In a, we have carbon dioxide plus hydrogen gives us"},{"Start":"00:11.310 ","End":"00:16.575","Text":"carbon monoxide plus water. We\u0027re going to start with a."},{"Start":"00:16.575 ","End":"00:23.265","Text":"First of all, the equilibrium constant expression K in general, first of all,"},{"Start":"00:23.265 ","End":"00:30.675","Text":"equals the activities of the products divided by the activities of the reactants."},{"Start":"00:30.675 ","End":"00:32.430","Text":"We\u0027re going to start now."},{"Start":"00:32.430 ","End":"00:34.740","Text":"Our first product is carbon monoxide."},{"Start":"00:34.740 ","End":"00:38.100","Text":"It\u0027s the activity of carbon monoxide and this is"},{"Start":"00:38.100 ","End":"00:42.210","Text":"raised to the power of the coefficient of the product."},{"Start":"00:42.210 ","End":"00:44.000","Text":"Our coefficient is 1, therefore,"},{"Start":"00:44.000 ","End":"00:46.520","Text":"it\u0027s going to be raised to the power of 1."},{"Start":"00:46.520 ","End":"00:48.320","Text":"We\u0027re not going to write anything,"},{"Start":"00:48.320 ","End":"00:50.815","Text":"but in this case, since is 1."},{"Start":"00:50.815 ","End":"00:53.360","Text":"Now we\u0027re going to multiply it by the next product,"},{"Start":"00:53.360 ","End":"00:56.705","Text":"which is the activity of water."},{"Start":"00:56.705 ","End":"01:01.735","Text":"Again, our coefficient is 1, so it\u0027s going to be raised to the power of 1."},{"Start":"01:01.735 ","End":"01:05.355","Text":"You can write them if you want to, you don\u0027t have to."},{"Start":"01:05.355 ","End":"01:08.760","Text":"Divided by the activities of the reactants."},{"Start":"01:08.760 ","End":"01:10.575","Text":"We have carbon dioxide,"},{"Start":"01:10.575 ","End":"01:12.650","Text":"the activity of carbon dioxide."},{"Start":"01:12.650 ","End":"01:15.830","Text":"Again, our coefficient is 1 so it\u0027s raised to"},{"Start":"01:15.830 ","End":"01:20.350","Text":"the power of 1 times the activity of hydrogen,"},{"Start":"01:20.350 ","End":"01:24.370","Text":"and also raised this time to the power of 1."},{"Start":"01:24.370 ","End":"01:28.120","Text":"That\u0027s the general equilibrium constant expression."},{"Start":"01:28.120 ","End":"01:31.060","Text":"Again, it\u0027s the activities of the products raised to"},{"Start":"01:31.060 ","End":"01:33.550","Text":"their coefficients divided by"},{"Start":"01:33.550 ","End":"01:37.060","Text":"the activities of the reactants raised to their coefficients."},{"Start":"01:37.060 ","End":"01:40.015","Text":"If we have ideal gases or dilute solutions,"},{"Start":"01:40.015 ","End":"01:42.940","Text":"the activity, we\u0027re talking in general,"},{"Start":"01:42.940 ","End":"01:48.985","Text":"the activity for example of a equals the concentration of a,"},{"Start":"01:48.985 ","End":"01:51.040","Text":"the molar concentration of a,"},{"Start":"01:51.040 ","End":"01:58.170","Text":"divided by the standard molar concentration."},{"Start":"01:58.170 ","End":"02:08.685","Text":"Now the standard molar concentration equals 1 mole per liter."},{"Start":"02:08.685 ","End":"02:14.975","Text":"Again, the activity equals the molar concentration of a certain product or reactant,"},{"Start":"02:14.975 ","End":"02:22.070","Text":"divided by or relative to the standard molar concentration,"},{"Start":"02:22.070 ","End":"02:25.085","Text":"which equals 1 mole per liter."},{"Start":"02:25.085 ","End":"02:28.880","Text":"Now this is in case we\u0027re talking about concentrations."},{"Start":"02:28.880 ","End":"02:35.685","Text":"The equilibrium constant expression can also be written in terms of pressures,"},{"Start":"02:35.685 ","End":"02:45.310","Text":"and then the activity of a equals the pressure of a divided by the standard pressure."},{"Start":"02:45.310 ","End":"02:51.885","Text":"In this case, the standard pressure equals 1 bar."},{"Start":"02:51.885 ","End":"02:54.605","Text":"Again, the equilibrium constant expression,"},{"Start":"02:54.605 ","End":"02:56.195","Text":"and we will see this in a minute,"},{"Start":"02:56.195 ","End":"02:58.850","Text":"can either be written in terms of concentrations."},{"Start":"02:58.850 ","End":"03:00.980","Text":"In this case, the activity of a equals"},{"Start":"03:00.980 ","End":"03:03.740","Text":"the molar concentration of a divided by a relative to"},{"Start":"03:03.740 ","End":"03:06.985","Text":"the standard concentration or"},{"Start":"03:06.985 ","End":"03:12.355","Text":"the equilibrium constant expression can be written in terms of pressures."},{"Start":"03:12.355 ","End":"03:16.355","Text":"There the activity of a equals the pressure of"},{"Start":"03:16.355 ","End":"03:21.815","Text":"a relative to the standard pressure and the standard pressure equals 1 bar in this case."},{"Start":"03:21.815 ","End":"03:25.030","Text":"Now, in this question, we\u0027re asked to find Kc."},{"Start":"03:25.030 ","End":"03:35.505","Text":"Kc is equilibrium constant expression in terms of concentrations."},{"Start":"03:35.505 ","End":"03:37.455","Text":"First of all, we know,"},{"Start":"03:37.455 ","End":"03:39.750","Text":"as we\u0027ve written that it equals"},{"Start":"03:39.750 ","End":"03:42.920","Text":"the activities again of the products raised to the coefficient of each product"},{"Start":"03:42.920 ","End":"03:48.360","Text":"divided by the activities of the reactants raised to the power of each reactant."},{"Start":"03:48.360 ","End":"03:50.190","Text":"Let\u0027s just continue here."},{"Start":"03:50.190 ","End":"03:54.280","Text":"Kc equals, so we have the activity of the carbon monoxide."},{"Start":"03:54.280 ","End":"04:02.810","Text":"It\u0027s going to be the concentration of carbon monoxide divided by or relative to"},{"Start":"04:02.810 ","End":"04:06.590","Text":"the standard concentration raised to the power of"},{"Start":"04:06.590 ","End":"04:10.130","Text":"1 times the concentration of"},{"Start":"04:10.130 ","End":"04:15.494","Text":"the water divided by again the relative to the standard concentration,"},{"Start":"04:15.494 ","End":"04:17.250","Text":"again raised to the power of 1."},{"Start":"04:17.250 ","End":"04:26.600","Text":"We\u0027re going to divide this by the concentration of the carbon dioxide relative to"},{"Start":"04:26.600 ","End":"04:33.250","Text":"the standard concentration to the first power times the concentration of"},{"Start":"04:33.250 ","End":"04:43.280","Text":"hydrogen relative to the standard concentration and again raised to the power of 1."},{"Start":"04:43.280 ","End":"04:46.500","Text":"The standard concentration equals 1 mole per liter."},{"Start":"04:46.500 ","End":"04:48.310","Text":"We\u0027re dealing with molar concentrations,"},{"Start":"04:48.310 ","End":"04:51.475","Text":"so our units are going to just cancel out and then Kc,"},{"Start":"04:51.475 ","End":"04:54.970","Text":"the equilibrium constant equals the concentration of"},{"Start":"04:54.970 ","End":"04:58.930","Text":"the carbon monoxide times the concentration of water,"},{"Start":"04:58.930 ","End":"05:06.135","Text":"divided by the concentration of carbon dioxide times the concentration of hydrogen."},{"Start":"05:06.135 ","End":"05:09.565","Text":"Now, you may see some cases that for simplicity,"},{"Start":"05:09.565 ","End":"05:12.150","Text":"the concentrations are given right away"},{"Start":"05:12.150 ","End":"05:16.385","Text":"and you won\u0027t even see the activities in some places."},{"Start":"05:16.385 ","End":"05:19.340","Text":"Again, the equilibrium constant expression that we found"},{"Start":"05:19.340 ","End":"05:22.790","Text":"for a equals the concentration of"},{"Start":"05:22.790 ","End":"05:26.270","Text":"carbon monoxide times the concentration of water divided by"},{"Start":"05:26.270 ","End":"05:31.400","Text":"the concentration of the carbon dioxide times the concentration of hydrogen."},{"Start":"05:31.400 ","End":"05:33.365","Text":"That is our final answer for a."},{"Start":"05:33.365 ","End":"05:35.790","Text":"Now we\u0027re going to go on to b."}],"ID":30454},{"Watched":false,"Name":"Exercise 1 Part 2","Duration":"4m 47s","ChapterTopicVideoID":28912,"CourseChapterTopicPlaylistID":253355,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.000","Text":"Now we\u0027re going to take a look at b."},{"Start":"00:03.000 ","End":"00:14.260","Text":"In b we have chromium plus 3 times the silver cation gives us the chromium plus 3 silver."},{"Start":"00:14.420 ","End":"00:21.825","Text":"We\u0027re going to start the same way kc equals the activity of the products,"},{"Start":"00:21.825 ","End":"00:24.750","Text":"which we have the chromium cation."},{"Start":"00:24.750 ","End":"00:27.060","Text":"Again, its coefficient is 1,"},{"Start":"00:27.060 ","End":"00:30.390","Text":"so we\u0027re just not going to raise to the power of anything since it"},{"Start":"00:30.390 ","End":"00:34.155","Text":"equals 1 times the activity"},{"Start":"00:34.155 ","End":"00:41.750","Text":"of the silver raised to the power of the coefficient of the silver,"},{"Start":"00:41.750 ","End":"00:50.610","Text":"which equals 3 and this is divided by the activities of the reactants."},{"Start":"00:50.610 ","End":"00:54.710","Text":"This is divided by the activity of chromium,"},{"Start":"00:54.710 ","End":"00:57.115","Text":"again raised to the power of 1,"},{"Start":"00:57.115 ","End":"01:03.025","Text":"times the activity of the silver cation,"},{"Start":"01:03.025 ","End":"01:07.105","Text":"raised to the third power."},{"Start":"01:07.105 ","End":"01:11.690","Text":"Now there\u0027s something that I didn\u0027t mention when we solved a,"},{"Start":"01:11.690 ","End":"01:17.350","Text":"and that is that the activities of pure solids and liquids equals 1."},{"Start":"01:17.350 ","End":"01:19.925","Text":"If we look at our reaction,"},{"Start":"01:19.925 ","End":"01:24.220","Text":"our chromium is a solid and we also have solid silver."},{"Start":"01:24.220 ","End":"01:29.550","Text":"Therefore, since the activities of the solids equal 1,"},{"Start":"01:29.710 ","End":"01:37.550","Text":"we can just take off the activity of the silver and the activity of the chromium."},{"Start":"01:37.550 ","End":"01:42.260","Text":"You will see again that there are places that they just don\u0027t include"},{"Start":"01:42.260 ","End":"01:48.105","Text":"the pure solids or pure liquids in expression from the beginning."},{"Start":"01:48.105 ","End":"01:50.480","Text":"Now we\u0027ll continue again."},{"Start":"01:50.480 ","End":"01:55.055","Text":"Remember that the activity equals the concentration of,"},{"Start":"01:55.055 ","End":"02:04.410","Text":"in our case that chromium cation divided by or relative to the standard concentration."},{"Start":"02:04.870 ","End":"02:08.300","Text":"This is divided by, again,"},{"Start":"02:08.300 ","End":"02:10.850","Text":"we have the activity of the silver cations,"},{"Start":"02:10.850 ","End":"02:14.745","Text":"so it\u0027s the concentration of"},{"Start":"02:14.745 ","End":"02:22.425","Text":"silver cation divided by the standard concentration."},{"Start":"02:22.425 ","End":"02:25.965","Text":"This is to the power of 3."},{"Start":"02:25.965 ","End":"02:32.035","Text":"Again, the standard concentration equals 1 mole per liter."},{"Start":"02:32.035 ","End":"02:35.240","Text":"What happens is our units just cancel out,"},{"Start":"02:35.240 ","End":"02:41.615","Text":"so we\u0027re left with the concentration of the chromium cation"},{"Start":"02:41.615 ","End":"02:50.960","Text":"divided by the concentration of the silver cation to the third power."},{"Start":"02:50.960 ","End":"02:58.295","Text":"The equilibrium constant expression that we found from b kc equals"},{"Start":"02:58.295 ","End":"03:02.750","Text":"the concentration of the chromium cation divided by"},{"Start":"03:02.750 ","End":"03:09.240","Text":"the concentration of the silver cation to the power of 3."},{"Start":"03:09.290 ","End":"03:11.625","Text":"That is our answer for b."},{"Start":"03:11.625 ","End":"03:18.440","Text":"Now let\u0027s take a look at c. In c we have calcium carbonate and on the product side,"},{"Start":"03:18.440 ","End":"03:22.680","Text":"we have calcium oxide plus oxygen."},{"Start":"03:23.930 ","End":"03:25.970","Text":"First of all, let\u0027s take a look."},{"Start":"03:25.970 ","End":"03:30.200","Text":"We have the calcium carbonate and calcium oxide are both solids."},{"Start":"03:30.200 ","End":"03:33.440","Text":"As I said, we can either write"},{"Start":"03:33.440 ","End":"03:36.950","Text":"them in the beginning and then just take them off as we did in B."},{"Start":"03:36.950 ","End":"03:41.180","Text":"Or we can just not take them into consideration from the beginning."},{"Start":"03:41.180 ","End":"03:48.440","Text":"We know that the activities of the calcium oxide and calcium carbonate equal 1."},{"Start":"03:48.440 ","End":"03:51.725","Text":"Let\u0027s just not take it into consideration from the beginning."},{"Start":"03:51.725 ","End":"03:57.930","Text":"KC is going to equal the activity only of oxygen."},{"Start":"03:57.940 ","End":"04:01.460","Text":"Again, since it\u0027s supposed to equal the activity of"},{"Start":"04:01.460 ","End":"04:04.450","Text":"oxygen times the activity of the calcium oxide."},{"Start":"04:04.450 ","End":"04:06.875","Text":"However, the activity of the calcium oxide equals"},{"Start":"04:06.875 ","End":"04:09.590","Text":"1 divided by the activity of the calcium carbonate,"},{"Start":"04:09.590 ","End":"04:14.270","Text":"which also equals 1, so Kc equals the activity of oxygen,"},{"Start":"04:14.270 ","End":"04:21.650","Text":"which equals the molar concentration of oxygen relative to the standard concentration,"},{"Start":"04:21.650 ","End":"04:26.150","Text":"which equals 1 mole per liter and that cancels our units for us."},{"Start":"04:26.150 ","End":"04:32.720","Text":"Therefore this equals the molar concentration of oxygen."},{"Start":"04:32.720 ","End":"04:37.040","Text":"For c, Kc,"},{"Start":"04:37.040 ","End":"04:43.265","Text":"equilibrium constant expression equals the molar concentration of oxygen."},{"Start":"04:43.265 ","End":"04:45.140","Text":"That is our final answer."},{"Start":"04:45.140 ","End":"04:47.820","Text":"Thank you very much for watching."}],"ID":30455},{"Watched":false,"Name":"Exercise 2","Duration":"6m 17s","ChapterTopicVideoID":28913,"CourseChapterTopicPlaylistID":253355,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.800 ","End":"00:03.870","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.870 ","End":"00:08.460","Text":"Write the equilibrium constant expressions K_c and K_p for the following reaction."},{"Start":"00:08.460 ","End":"00:12.615","Text":"On the reactant side we have carbon disulfide and 4 hydrogen,"},{"Start":"00:12.615 ","End":"00:16.545","Text":"and on the product side we have methane and 2 hydrogen sulfide."},{"Start":"00:16.545 ","End":"00:18.375","Text":"Now as we can see first of all,"},{"Start":"00:18.375 ","End":"00:21.113","Text":"they are all gases."},{"Start":"00:21.113 ","End":"00:22.635","Text":"Now we\u0027re going to start."},{"Start":"00:22.635 ","End":"00:28.425","Text":"Remember that equilibrium constant expression K, in general,"},{"Start":"00:28.425 ","End":"00:32.400","Text":"equals the activity of the products raised to"},{"Start":"00:32.400 ","End":"00:36.870","Text":"the power of each coefficient of each product."},{"Start":"00:36.870 ","End":"00:41.230","Text":"We have methane in our products and we have hydrogen sulfide,"},{"Start":"00:41.230 ","End":"00:42.635","Text":"so it\u0027s going to be the activity of"},{"Start":"00:42.635 ","End":"00:47.810","Text":"methane raised to the power of 1 since the coefficient of methane is 1,"},{"Start":"00:47.810 ","End":"00:51.469","Text":"times the activity of hydrogen sulfide."},{"Start":"00:51.469 ","End":"00:55.190","Text":"However, this time it\u0027s going to be raised to the power of 2 because"},{"Start":"00:55.190 ","End":"00:58.635","Text":"we have 2 in our coefficient,"},{"Start":"00:58.635 ","End":"01:02.870","Text":"and that\u0027s divided by the activities of our reactants."},{"Start":"01:02.870 ","End":"01:04.669","Text":"We have carbon disulfide,"},{"Start":"01:04.669 ","End":"01:06.620","Text":"so that\u0027s the activity of carbon disulfide."},{"Start":"01:06.620 ","End":"01:10.950","Text":"Again, the coefficient is 1, therefore,"},{"Start":"01:10.950 ","End":"01:13.245","Text":"this is raised to 1,"},{"Start":"01:13.245 ","End":"01:17.364","Text":"times the activity of hydrogen."},{"Start":"01:17.364 ","End":"01:20.180","Text":"This is raised to the power of 4."},{"Start":"01:20.180 ","End":"01:23.420","Text":"Now, I just want to comment that if you see in"},{"Start":"01:23.420 ","End":"01:27.860","Text":"the middle 2 arrows pointing at both of the directions and not 1 arrow,"},{"Start":"01:27.860 ","End":"01:30.840","Text":"it means that we\u0027re in a state of equilibrium."},{"Start":"01:33.020 ","End":"01:38.525","Text":"This is our general equilibrium constant expression with the activities."},{"Start":"01:38.525 ","End":"01:47.640","Text":"Now, K_c refers to the constant in terms of concentration."},{"Start":"01:47.640 ","End":"01:50.690","Text":"Remember that the activity of,"},{"Start":"01:50.690 ","End":"01:55.115","Text":"let\u0027s just say a general molecule a,"},{"Start":"01:55.115 ","End":"01:58.730","Text":"equals the molar concentration of a"},{"Start":"01:58.730 ","End":"02:02.968","Text":"divided by or relative to a standard concentration c,"},{"Start":"02:02.968 ","End":"02:07.480","Text":"and c equals 1 mole per liter."},{"Start":"02:07.480 ","End":"02:10.999","Text":"In our case, again, since these are all gases,"},{"Start":"02:10.999 ","End":"02:15.350","Text":"because remember that the activities of pure solids and liquids equals 1."},{"Start":"02:15.350 ","End":"02:17.360","Text":"But here we have all gases."},{"Start":"02:17.360 ","End":"02:23.609","Text":"Therefore, we have the activity of methane, CH_4,"},{"Start":"02:23.609 ","End":"02:31.205","Text":"equals the concentration of methane divided by or relative to the standard concentration,"},{"Start":"02:31.205 ","End":"02:34.145","Text":"multiplied by the concentration of"},{"Start":"02:34.145 ","End":"02:41.000","Text":"hydrogen sulfide relative to the standard concentration."},{"Start":"02:41.000 ","End":"02:51.380","Text":"This is the power of 2 divided by the concentration of carbon disulfide."},{"Start":"02:51.380 ","End":"03:00.305","Text":"Again, relative to the standard concentration times the concentration of hydrogen."},{"Start":"03:00.305 ","End":"03:06.883","Text":"Again, relative to the standard concentration,"},{"Start":"03:06.883 ","End":"03:10.135","Text":"and this is to the fourth power,"},{"Start":"03:10.135 ","End":"03:14.120","Text":"since the coefficient of hydrogen in our reaction is 4."},{"Start":"03:14.120 ","End":"03:18.890","Text":"Remember, standard concentration equals 1 mole per liter and therefore,"},{"Start":"03:18.890 ","End":"03:24.585","Text":"our units cancel out and the equilibrium constant expression is unitless."},{"Start":"03:24.585 ","End":"03:30.190","Text":"K_c comes out to the concentration of the methane times"},{"Start":"03:30.190 ","End":"03:36.790","Text":"the concentration of hydrogen sulfide to the second power,"},{"Start":"03:36.790 ","End":"03:41.230","Text":"divided by the concentration of"},{"Start":"03:41.230 ","End":"03:48.375","Text":"the carbon disulfide times the concentration of hydrogen to the fourth power."},{"Start":"03:48.375 ","End":"03:51.220","Text":"Again, K_c equals the concentration of methane times"},{"Start":"03:51.220 ","End":"03:54.849","Text":"the concentration of hydrogen sulfide squared,"},{"Start":"03:54.849 ","End":"03:57.835","Text":"divided by the concentration of"},{"Start":"03:57.835 ","End":"04:04.305","Text":"carbon disulfide times hydrogen to the fourth, that\u0027s K_c."},{"Start":"04:04.305 ","End":"04:06.585","Text":"Now let\u0027s take a look at K_p."},{"Start":"04:06.585 ","End":"04:11.170","Text":"Now we want to write out K_p,"},{"Start":"04:11.170 ","End":"04:16.640","Text":"meaning the equilibrium constant expression in terms of pressures."},{"Start":"04:16.640 ","End":"04:22.970","Text":"Remember that the activity a of a certain product or reactant A"},{"Start":"04:22.970 ","End":"04:29.083","Text":"equals the pressure of A divided by the standard pressure,"},{"Start":"04:29.083 ","End":"04:32.360","Text":"and the standard pressure equals 1 bar."},{"Start":"04:32.360 ","End":"04:38.675","Text":"This equals the activity of the methane which is the pressure of methane,"},{"Start":"04:38.675 ","End":"04:43.475","Text":"divided by or relative to the standard pressure,"},{"Start":"04:43.475 ","End":"04:46.805","Text":"times the pressure of hydrogen sulfide,"},{"Start":"04:46.805 ","End":"04:49.850","Text":"divided by the standard pressure,"},{"Start":"04:49.850 ","End":"04:51.910","Text":"and this is squared."},{"Start":"04:51.910 ","End":"04:56.290","Text":"This is divided by the pressure of"},{"Start":"04:56.290 ","End":"05:05.424","Text":"carbon disulfide relative to the standard pressure,"},{"Start":"05:05.424 ","End":"05:13.013","Text":"times the pressure of hydrogen relative to the standard pressure,"},{"Start":"05:13.013 ","End":"05:15.440","Text":"and this is to the fourth power."},{"Start":"05:15.440 ","End":"05:18.670","Text":"Now remember, the standard pressure is 1 bar,"},{"Start":"05:18.670 ","End":"05:21.445","Text":"and these pressures will also be in bars,"},{"Start":"05:21.445 ","End":"05:27.025","Text":"and therefore the units are going to cancel out and K_p will be without units."},{"Start":"05:27.025 ","End":"05:32.719","Text":"K_p equals, now we can take out the standard pressure."},{"Start":"05:32.719 ","End":"05:34.980","Text":"It equals the pressure of"},{"Start":"05:34.980 ","End":"05:44.645","Text":"the methane times the pressure of hydrogen sulfide squared,"},{"Start":"05:44.645 ","End":"05:54.135","Text":"divided by the pressure of carbon disulfide times the pressure of hydrogen to the fourth."},{"Start":"05:54.135 ","End":"05:57.215","Text":"Again, K_p is going to be equal to"},{"Start":"05:57.215 ","End":"06:01.310","Text":"the pressure of methane times the pressure of hydrogen sulfide squared,"},{"Start":"06:01.310 ","End":"06:04.025","Text":"divided by the pressure of"},{"Start":"06:04.025 ","End":"06:12.785","Text":"carbon disulfide times the pressure of hydrogen to the fourth."},{"Start":"06:12.785 ","End":"06:14.480","Text":"That is our final answer."},{"Start":"06:14.480 ","End":"06:17.010","Text":"Thank you very much for watching."}],"ID":30456},{"Watched":false,"Name":"Extent of Reaction","Duration":"6m 4s","ChapterTopicVideoID":26274,"CourseChapterTopicPlaylistID":253355,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:02.369","Text":"In previous videos,"},{"Start":"00:02.369 ","End":"00:04.950","Text":"we talked about the equilibrium constant."},{"Start":"00:04.950 ","End":"00:07.980","Text":"In this video, we\u0027ll talk about the relation between"},{"Start":"00:07.980 ","End":"00:12.850","Text":"the equilibrium constant and the extent of reaction."},{"Start":"00:12.920 ","End":"00:17.430","Text":"What did we learn about the equilibrium constant?"},{"Start":"00:17.430 ","End":"00:23.340","Text":"We saw that for the reaction a moles of A plus b of B in"},{"Start":"00:23.340 ","End":"00:28.740","Text":"equilibrium with c moles of C and d moles of D. K,"},{"Start":"00:28.740 ","End":"00:32.850","Text":"the equilibrium constant is equal to the activity of C^c,"},{"Start":"00:32.850 ","End":"00:34.815","Text":"that\u0027s a small c, this one,"},{"Start":"00:34.815 ","End":"00:37.980","Text":"times the activity of D^d,"},{"Start":"00:37.980 ","End":"00:41.940","Text":"divided by the activity of A^a,"},{"Start":"00:41.940 ","End":"00:45.360","Text":"and the activity of B^b."},{"Start":"00:45.360 ","End":"00:52.100","Text":"On the top, we have the products and on the bottom,"},{"Start":"00:52.100 ","End":"00:54.450","Text":"we have the reactants."},{"Start":"00:54.580 ","End":"00:58.490","Text":"We also learned that there\u0027s a relationship between"},{"Start":"00:58.490 ","End":"01:03.230","Text":"the Gibbs free energy reaction and the equilibrium constant."},{"Start":"01:03.230 ","End":"01:12.605","Text":"We saw that the standard Gibbs free energy of reaction is equal to minus RTlnK."},{"Start":"01:12.605 ","End":"01:17.125","Text":"That means this Delta G is positive,"},{"Start":"01:17.125 ","End":"01:20.235","Text":"lnK will be negative."},{"Start":"01:20.235 ","End":"01:26.055","Text":"The negative log means K is less than 1."},{"Start":"01:26.055 ","End":"01:27.975","Text":"Now, if we look at this,"},{"Start":"01:27.975 ","End":"01:30.525","Text":"we\u0027ll see the trip is less than 1,"},{"Start":"01:30.525 ","End":"01:32.900","Text":"the reactants must be more important,"},{"Start":"01:32.900 ","End":"01:36.695","Text":"must have a greater pressure than the products so"},{"Start":"01:36.695 ","End":"01:41.835","Text":"the reactants will be favored at equilibrium."},{"Start":"01:41.835 ","End":"01:46.680","Text":"If Delta G reaction is negative,"},{"Start":"01:46.680 ","End":"01:50.860","Text":"lnK will be positive and K will be greater than 1,"},{"Start":"01:50.860 ","End":"01:54.485","Text":"that means that the products will be favored."},{"Start":"01:54.485 ","End":"01:56.720","Text":"Products are in the numerator,"},{"Start":"01:56.720 ","End":"01:59.300","Text":"whereas reactants are in the denominator."},{"Start":"01:59.300 ","End":"02:03.150","Text":"The products will be favored at equilibrium."},{"Start":"02:03.410 ","End":"02:06.910","Text":"Now we\u0027re going to talk about the extent of reaction,"},{"Start":"02:06.910 ","End":"02:09.280","Text":"to which extent reaction occurs,"},{"Start":"02:09.280 ","End":"02:11.560","Text":"has it barely started,"},{"Start":"02:11.560 ","End":"02:14.945","Text":"or has it proceeded to completion?"},{"Start":"02:14.945 ","End":"02:21.390","Text":"Supposing Delta G of reaction is very much greater than 0,"},{"Start":"02:21.390 ","End":"02:24.895","Text":"is a large number, large positive number,"},{"Start":"02:24.895 ","End":"02:28.480","Text":"then K will be much smaller than 1,"},{"Start":"02:28.480 ","End":"02:31.120","Text":"and the reaction will hardly occur."},{"Start":"02:31.120 ","End":"02:35.275","Text":"The reactions will be much more important than the product."},{"Start":"02:35.275 ","End":"02:38.200","Text":"The reaction will hardly occur."},{"Start":"02:38.200 ","End":"02:46.095","Text":"If Delta G of reaction is much less than 0, very negative."},{"Start":"02:46.095 ","End":"02:49.460","Text":"K will be much greater than 1."},{"Start":"02:49.460 ","End":"02:53.335","Text":"Then the reaction goes almost to completion."},{"Start":"02:53.335 ","End":"02:57.620","Text":"We have many more products than we have reactants."},{"Start":"02:57.740 ","End":"03:01.280","Text":"Let\u0027s look at 2 examples."},{"Start":"03:01.280 ","End":"03:06.205","Text":"Here we have the fluorine molecule F2 in the gas phase,"},{"Start":"03:06.205 ","End":"03:11.635","Text":"dissociating to 2 fluorine atoms also in the gas phase."},{"Start":"03:11.635 ","End":"03:14.770","Text":"The K for this is the pressure fluorine"},{"Start":"03:14.770 ","End":"03:19.225","Text":"squared divided by the pressure of the fluorine molecules."},{"Start":"03:19.225 ","End":"03:21.740","Text":"Squares because of the 2 here."},{"Start":"03:21.740 ","End":"03:29.190","Text":"That turns out to be 3 times 10^-11 at 500 Kelvin."},{"Start":"03:29.190 ","End":"03:33.644","Text":"That means it\u0027s a very small number."},{"Start":"03:33.644 ","End":"03:36.715","Text":"If it\u0027s a very small number,"},{"Start":"03:36.715 ","End":"03:39.180","Text":"reaction will hardly occur,"},{"Start":"03:39.180 ","End":"03:44.875","Text":"that means the very little of a fluorine molecule dissociate to atoms."},{"Start":"03:44.875 ","End":"03:47.170","Text":"There will be many more molecules,"},{"Start":"03:47.170 ","End":"03:50.810","Text":"that will be much more important than atoms."},{"Start":"03:53.700 ","End":"03:57.165","Text":"On the other hand, let\u0027s look at this reaction,"},{"Start":"03:57.165 ","End":"04:00.075","Text":"2 SO_2 in the gas phase,"},{"Start":"04:00.075 ","End":"04:01.890","Text":"reacting with oxygen,"},{"Start":"04:01.890 ","End":"04:05.820","Text":"the gas phase to give 2 SO_3 in the gas phase."},{"Start":"04:05.820 ","End":"04:11.535","Text":"Now, here the K is written as the pressure SO_3 all squared."},{"Start":"04:11.535 ","End":"04:15.995","Text":"That\u0027s because this 2 divided by the pressure of"},{"Start":"04:15.995 ","End":"04:21.025","Text":"SO_2 squared because of this 2 times the pressure focused."},{"Start":"04:21.025 ","End":"04:24.275","Text":"This turns out to be very large."},{"Start":"04:24.275 ","End":"04:29.135","Text":"K is equal to 4.0 times 10^24."},{"Start":"04:29.135 ","End":"04:32.450","Text":"That\u0027s 298 Kelvin."},{"Start":"04:32.450 ","End":"04:38.900","Text":"That\u0027s a room temperature. What does that mean?"},{"Start":"04:38.900 ","End":"04:44.225","Text":"It means that the reaction will go to completion."},{"Start":"04:44.225 ","End":"04:48.710","Text":"The products will be much more important than the reactants."},{"Start":"04:48.710 ","End":"04:50.210","Text":"This will go to completion."},{"Start":"04:50.210 ","End":"04:54.890","Text":"That means there will be almost no SO_2 or oxygen left,"},{"Start":"04:54.890 ","End":"04:59.010","Text":"and they\u0027ll all be converted into SO_3."},{"Start":"04:59.050 ","End":"05:05.180","Text":"Here are a few rules of thumb that help us to understand qualitatively,"},{"Start":"05:05.180 ","End":"05:08.630","Text":"to get a qualitative feel for the values of"},{"Start":"05:08.630 ","End":"05:15.475","Text":"K. For large values of K greater than about 10^3,"},{"Start":"05:15.475 ","End":"05:19.000","Text":"the products are favored at equilibrium."},{"Start":"05:19.000 ","End":"05:21.465","Text":"For small values of K,"},{"Start":"05:21.465 ","End":"05:24.490","Text":"smaller than 10^-3,"},{"Start":"05:24.490 ","End":"05:28.150","Text":"the reactants are favored at equilibrium."},{"Start":"05:28.150 ","End":"05:34.820","Text":"Now, if the values are in the range between 10^-3 and 10^3,"},{"Start":"05:34.820 ","End":"05:39.410","Text":"neither reactants nor products are strongly favored."},{"Start":"05:39.410 ","End":"05:43.490","Text":"Here, when K is very large,"},{"Start":"05:43.490 ","End":"05:45.634","Text":"the products are favored."},{"Start":"05:45.634 ","End":"05:47.465","Text":"When K is very small,"},{"Start":"05:47.465 ","End":"05:50.885","Text":"the reactants are favored and if it\u0027s intermediate,"},{"Start":"05:50.885 ","End":"05:53.000","Text":"we can\u0027t know without doing"},{"Start":"05:53.000 ","End":"05:59.195","Text":"detailed calculations whether the reactants or products are more important."},{"Start":"05:59.195 ","End":"06:03.900","Text":"In this video, we talked about the extent of reaction."}],"ID":30457},{"Watched":false,"Name":"Equilibrium Constant Enthalpy and Entropy","Duration":"5m 24s","ChapterTopicVideoID":26273,"CourseChapterTopicPlaylistID":253355,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.700","Text":"In the previous 2 videos,"},{"Start":"00:02.700 ","End":"00:05.340","Text":"we talked about the relation between Gibbs free energy of"},{"Start":"00:05.340 ","End":"00:08.295","Text":"reaction and the reaction composition."},{"Start":"00:08.295 ","End":"00:11.549","Text":"In this video, we\u0027ll relate the equilibrium"},{"Start":"00:11.549 ","End":"00:15.850","Text":"constant to the reaction enthalpy and entropy."},{"Start":"00:16.250 ","End":"00:19.800","Text":"Let\u0027s recall what we learned about the equilibrium"},{"Start":"00:19.800 ","End":"00:23.805","Text":"constant and the Gibbs free energy of reaction."},{"Start":"00:23.805 ","End":"00:29.850","Text":"We showed that Delta G0 of reaction is equal to minus RT Ln K,"},{"Start":"00:29.850 ","End":"00:32.145","Text":"K is a equilibrium constant."},{"Start":"00:32.145 ","End":"00:37.920","Text":"We can rearrange this equation and write that Ln K is equal"},{"Start":"00:37.920 ","End":"00:43.665","Text":"to Delta G0 of reaction divided by RT and there\u0027s a minus sign."},{"Start":"00:43.665 ","End":"00:49.280","Text":"Ln K is equal to minus Delta G reaction 0,"},{"Start":"00:49.280 ","End":"00:55.275","Text":"or Delta G0 of reaction divided by RT."},{"Start":"00:55.275 ","End":"00:58.525","Text":"We\u0027re going to call that Equation 1."},{"Start":"00:58.525 ","End":"01:03.545","Text":"Now, let\u0027s recall what we learned in thermodynamics."},{"Start":"01:03.545 ","End":"01:05.810","Text":"The relation between Gibbs free energy,"},{"Start":"01:05.810 ","End":"01:08.050","Text":"the enthalpy and the entropy."},{"Start":"01:08.050 ","End":"01:11.850","Text":"We learned that Delta G0 of reaction is equal to"},{"Start":"01:11.850 ","End":"01:16.280","Text":"Delta H0 of reaction minus T Delta S0 of reaction."},{"Start":"01:16.280 ","End":"01:24.100","Text":"Delta G is always equal to Delta H minus T Delta S. We\u0027ll call that Equation 2."},{"Start":"01:24.620 ","End":"01:31.670","Text":"Now we\u0027re going to combine 1 and 2 and get the relation between the equilibrium constant,"},{"Start":"01:31.670 ","End":"01:34.355","Text":"the enthalpy and entropy."},{"Start":"01:34.355 ","End":"01:37.920","Text":"We\u0027re combining 1 and 2."},{"Start":"01:38.360 ","End":"01:46.025","Text":"Ln K is equal to minus Delta G reaction divided by RT."},{"Start":"01:46.025 ","End":"01:48.890","Text":"Instead of Delta G reaction,"},{"Start":"01:48.890 ","End":"01:52.260","Text":"we\u0027re going to insert this equation here."},{"Start":"01:52.260 ","End":"01:59.430","Text":"Now we get Ln K is equal to minus Delta G0 of reaction divided by RT is equal to"},{"Start":"01:59.430 ","End":"02:07.590","Text":"minus Delta H0 of reaction divided by RT plus those have minus,"},{"Start":"02:07.590 ","End":"02:14.870","Text":"minus T Delta S reaction divided by RT."},{"Start":"02:14.870 ","End":"02:17.889","Text":"We had T divided by RT."},{"Start":"02:17.889 ","End":"02:24.710","Text":"We\u0027re left just with the R. Now we have Ln K is equal to minus Delta H0"},{"Start":"02:24.710 ","End":"02:34.055","Text":"of reaction divided by RT plus Delta S0 of reaction divided by R. Now,"},{"Start":"02:34.055 ","End":"02:37.135","Text":"if we take the exponential of each side,"},{"Start":"02:37.135 ","End":"02:42.185","Text":"that e to the power Ln K is just K."},{"Start":"02:42.185 ","End":"02:49.345","Text":"We get each of the power Ln K is K and then the exponential of all this,"},{"Start":"02:49.345 ","End":"02:53.530","Text":"in the brackets minus Delta H0 reaction divided by RT plus"},{"Start":"02:53.530 ","End":"02:57.460","Text":"Delta S0 of reaction divided by R. Now"},{"Start":"02:57.460 ","End":"03:06.659","Text":"if we have the exponential of a sum x plus y,"},{"Start":"03:06.659 ","End":"03:13.260","Text":"that\u0027s equal to the exponential of x times exponential of y."},{"Start":"03:13.260 ","End":"03:19.680","Text":"Now we have exponential of minus Delta H0 of reaction divided"},{"Start":"03:19.680 ","End":"03:25.700","Text":"by RT times exp Delta S0 of reaction divided by R,"},{"Start":"03:25.700 ","End":"03:29.785","Text":"so we divided it into 2 separate components."},{"Start":"03:29.785 ","End":"03:35.010","Text":"Here\u0027s the equation K is equal to this."},{"Start":"03:35.010 ","End":"03:39.395","Text":"Now supposing we have a strongly exothermic reaction."},{"Start":"03:39.395 ","End":"03:46.775","Text":"That means that Delta H0 of reaction divided by RT will be very negative."},{"Start":"03:46.775 ","End":"03:48.950","Text":"If it\u0027s very negative,"},{"Start":"03:48.950 ","End":"03:54.170","Text":"that means the exponential minus Delta H0 of reaction"},{"Start":"03:54.170 ","End":"04:00.120","Text":"divided by RT will be positive and large,"},{"Start":"04:00.120 ","End":"04:03.480","Text":"so the equilibrium constant K will be very large,"},{"Start":"04:03.480 ","End":"04:05.475","Text":"it\u0027d be much greater than 1."},{"Start":"04:05.475 ","End":"04:07.875","Text":"If it\u0027s much greater than 1,"},{"Start":"04:07.875 ","End":"04:18.340","Text":"recall that K has the products in the numerator and the reactants in the denominator."},{"Start":"04:19.900 ","End":"04:23.105","Text":"That means if K is very large,"},{"Start":"04:23.105 ","End":"04:27.500","Text":"there are many more products than there are reactions."},{"Start":"04:27.500 ","End":"04:32.100","Text":"That means that reaction goes to completion."},{"Start":"04:33.040 ","End":"04:36.070","Text":"If it\u0027s an endothermic reaction,"},{"Start":"04:36.070 ","End":"04:39.245","Text":"Delta H0 of reaction is positive,"},{"Start":"04:39.245 ","End":"04:44.180","Text":"the equilibrium constant will be positive K is less than 1,"},{"Start":"04:44.180 ","End":"04:46.729","Text":"and not a lot of products will be formed,"},{"Start":"04:46.729 ","End":"04:49.105","Text":"so K will be small."},{"Start":"04:49.105 ","End":"04:58.310","Text":"Now K can be greater than 1 for an endothermic reaction if we have a very high entropy,"},{"Start":"04:58.310 ","End":"05:02.830","Text":"Delta S0 of reaction divided by R is very large."},{"Start":"05:02.830 ","End":"05:08.795","Text":"Endothermic reactions can occur spontaneously if there\u0027s a large increase"},{"Start":"05:08.795 ","End":"05:15.845","Text":"in entropy and that\u0027s the conclusion that we reached when we studied thermodynamics."},{"Start":"05:15.845 ","End":"05:19.849","Text":"In this video, we learned about the connection between the equilibrium"},{"Start":"05:19.849 ","End":"05:25.020","Text":"constant and the standard enthalpy and entropy of reaction."}],"ID":30458},{"Watched":false,"Name":"Dynamic Equilibrium","Duration":"2m 19s","ChapterTopicVideoID":26271,"CourseChapterTopicPlaylistID":253355,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.640","Text":"In previous video,"},{"Start":"00:01.640 ","End":"00:05.280","Text":"we talked about dynamic equilibrium in physical processes."},{"Start":"00:05.280 ","End":"00:10.560","Text":"In this video, we\u0027ll learn about dynamic equilibrium in chemical reactions."},{"Start":"00:10.560 ","End":"00:14.940","Text":"We\u0027re talking about dynamic equilibrium in chemical reactions."},{"Start":"00:14.940 ","End":"00:18.240","Text":"Now we all know that some chemical reaction seem to go to"},{"Start":"00:18.240 ","End":"00:22.695","Text":"completion with all the reactants converted into products."},{"Start":"00:22.695 ","End":"00:25.785","Text":"But some reactions stop at a certain stage"},{"Start":"00:25.785 ","End":"00:28.920","Text":"so that reactants and products are both present."},{"Start":"00:28.920 ","End":"00:32.415","Text":"We say that the reaction has reached equilibrium."},{"Start":"00:32.415 ","End":"00:38.175","Text":"Now at equilibrium, the forward reaction A to B and the reverse reaction B to"},{"Start":"00:38.175 ","End":"00:44.240","Text":"A take place at the same rate so that the amounts of A and B are constant."},{"Start":"00:44.240 ","End":"00:46.820","Text":"Of course this takes a certain amount of time."},{"Start":"00:46.820 ","End":"00:53.150","Text":"We write the dynamic equilibrium is A and this double arrow B."},{"Start":"00:53.150 ","End":"00:55.730","Text":"Here\u0027s an example,"},{"Start":"00:55.730 ","End":"01:00.410","Text":"nitrogen reacting with hydrogen to form ammonia."},{"Start":"01:00.410 ","End":"01:02.315","Text":"They\u0027re all in the gas phase,"},{"Start":"01:02.315 ","End":"01:06.180","Text":"and this happens in the presence of a catalyst."},{"Start":"01:06.290 ","End":"01:16.720","Text":"Now, supposing, we start off with hydrogen and nitrogen, but no ammonia."},{"Start":"01:16.720 ","End":"01:19.280","Text":"After certain amount of time,"},{"Start":"01:19.280 ","End":"01:21.410","Text":"the hydrogen will have decreased,"},{"Start":"01:21.410 ","End":"01:24.005","Text":"the nitrogen will have decreased,"},{"Start":"01:24.005 ","End":"01:26.630","Text":"and the ammonia will be formed."},{"Start":"01:26.630 ","End":"01:31.395","Text":"They seemed to reach a plateau where nothing is happening,"},{"Start":"01:31.395 ","End":"01:34.295","Text":"they\u0027ve reached a certain constant."},{"Start":"01:34.295 ","End":"01:35.750","Text":"On the other hand,"},{"Start":"01:35.750 ","End":"01:38.230","Text":"if we start off with ammonia,"},{"Start":"01:38.230 ","End":"01:41.685","Text":"the amount of ammonia decreases,"},{"Start":"01:41.685 ","End":"01:47.415","Text":"nitrogen is formed, and hydrogen is formed,"},{"Start":"01:47.415 ","End":"01:51.565","Text":"and once again, we reach a plateau."},{"Start":"01:51.565 ","End":"01:53.600","Text":"This is summarized here."},{"Start":"01:53.600 ","End":"01:55.055","Text":"On the left-hand side,"},{"Start":"01:55.055 ","End":"01:56.780","Text":"the reactants decrease,"},{"Start":"01:56.780 ","End":"02:01.505","Text":"products are formed and the reaction reaches equilibrium."},{"Start":"02:01.505 ","End":"02:03.785","Text":"Then the right-hand side,"},{"Start":"02:03.785 ","End":"02:06.215","Text":"the products decrease,"},{"Start":"02:06.215 ","End":"02:12.665","Text":"reactants are formed, and the reaction again reaches equilibrium."},{"Start":"02:12.665 ","End":"02:19.710","Text":"In this video, we talked about dynamic equilibrium in chemical reactions."}],"ID":30459},{"Watched":false,"Name":"Gibbs Free Energy and Reaction Composition 1","Duration":"9m 1s","ChapterTopicVideoID":26275,"CourseChapterTopicPlaylistID":253355,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.620","Text":"In previous videos,"},{"Start":"00:01.620 ","End":"00:03.690","Text":"we defined the equilibrium constant."},{"Start":"00:03.690 ","End":"00:06.390","Text":"In this video, we\u0027ll talk about the relation between"},{"Start":"00:06.390 ","End":"00:11.175","Text":"the Gibbs free energy of reactions and the reaction components."},{"Start":"00:11.175 ","End":"00:16.890","Text":"Let\u0027s recall the general expression we had for the equilibrium constant."},{"Start":"00:16.890 ","End":"00:20.700","Text":"For the reaction a moles of A plus b moles of B to give"},{"Start":"00:20.700 ","End":"00:24.465","Text":"c moles of C and d moles of D at equilibrium."},{"Start":"00:24.465 ","End":"00:28.530","Text":"Equilibrium is often given by this Omega."},{"Start":"00:28.530 ","End":"00:35.025","Text":"K the equilibrium constant is equal to the activity of C^c,"},{"Start":"00:35.025 ","End":"00:37.005","Text":"that\u0027s this little c here,"},{"Start":"00:37.005 ","End":"00:40.775","Text":"times the activity of D^d,"},{"Start":"00:40.775 ","End":"00:48.215","Text":"divided by the activity of A^a and the activity of B^b,"},{"Start":"00:48.215 ","End":"00:50.015","Text":"that\u0027s a and b."},{"Start":"00:50.015 ","End":"00:59.620","Text":"On the top we have the products and on the bottom we have the reactants."},{"Start":"01:02.620 ","End":"01:09.620","Text":"Now, we\u0027re going to talk about the reaction quotient in terms of activities."},{"Start":"01:09.620 ","End":"01:13.250","Text":"It\u0027s very similar to the equilibrium constant,"},{"Start":"01:13.250 ","End":"01:20.075","Text":"except that it can happen at any point in the reaction not just at equilibrium."},{"Start":"01:20.075 ","End":"01:24.590","Text":"For the same reaction at any stage of reaction,"},{"Start":"01:24.590 ","End":"01:29.060","Text":"Q is equal to the same expression as we had before,"},{"Start":"01:29.060 ","End":"01:33.155","Text":"but we are referring to the activities"},{"Start":"01:33.155 ","End":"01:37.955","Text":"or the concentrations or pressures at any point in the reaction."},{"Start":"01:37.955 ","End":"01:41.705","Text":"Obviously, when we have an equilibrium,"},{"Start":"01:41.705 ","End":"01:47.240","Text":"Q will be equal to K. At equilibrium Q"},{"Start":"01:47.240 ","End":"01:53.765","Text":"is equal to K. Now let\u0027s look at the Gibbs free energy of reaction."},{"Start":"01:53.765 ","End":"02:00.280","Text":"The Gibbs free energy of reaction changes as the reaction proceeds."},{"Start":"02:00.280 ","End":"02:03.840","Text":"We use a picture of G of the reaction,"},{"Start":"02:03.840 ","End":"02:09.070","Text":"the Gibbs free energy of reaction as a function of the progress of the reaction."},{"Start":"02:09.070 ","End":"02:17.350","Text":"Here we have only reactants and at the end we have the products."},{"Start":"02:17.350 ","End":"02:22.330","Text":"We see that the Gibbs free energy decreases"},{"Start":"02:22.330 ","End":"02:27.805","Text":"until it gets to a minimum and then increases."},{"Start":"02:27.805 ","End":"02:32.350","Text":"Now, if we look at the slope of the graph,"},{"Start":"02:32.350 ","End":"02:36.160","Text":"before we get to the minimum the slope is"},{"Start":"02:36.160 ","End":"02:43.235","Text":"negative and after the minimum the slope is positive."},{"Start":"02:43.235 ","End":"02:48.350","Text":"Delta G reaction is the slope of the graph."},{"Start":"02:48.350 ","End":"02:51.215","Text":"Negative before equilibrium, negative here,"},{"Start":"02:51.215 ","End":"02:56.820","Text":"0 to equilibrium and then positive after the minimum."},{"Start":"02:58.150 ","End":"03:03.350","Text":"Now the standard Gibbs free energy of reaction is a constant."},{"Start":"03:03.350 ","End":"03:08.030","Text":"That\u0027s the difference between the products and the reactants."},{"Start":"03:08.030 ","End":"03:09.740","Text":"This far we\u0027re here,"},{"Start":"03:09.740 ","End":"03:14.430","Text":"this is Delta G_rxn with a"},{"Start":"03:14.430 ","End":"03:20.135","Text":"0 on the top indicating that it\u0027s a standard Gibbs free energy."},{"Start":"03:20.135 ","End":"03:23.090","Text":"We can write an expression for it,"},{"Start":"03:23.090 ","End":"03:27.740","Text":"Delta G of reaction is the sum over the number"},{"Start":"03:27.740 ","End":"03:32.975","Text":"of moles of each product times its Delta G of formation"},{"Start":"03:32.975 ","End":"03:39.380","Text":"minus a sum over all the reactants of the number"},{"Start":"03:39.380 ","End":"03:45.350","Text":"of moles of each reactant times the Delta G^0 formation of each reaction,"},{"Start":"03:45.350 ","End":"03:47.880","Text":"the f is for formation."},{"Start":"03:49.840 ","End":"03:52.550","Text":"Now, what\u0027s the relation between"},{"Start":"03:52.550 ","End":"03:56.140","Text":"the Gibbs free energy of reaction and the reaction quotient?"},{"Start":"03:56.140 ","End":"04:00.140","Text":"I\u0027m going to give the proof in the next video."},{"Start":"04:00.140 ","End":"04:07.250","Text":"Delta G of reaction at any point in the reaction is equal to Delta G^0 of reaction,"},{"Start":"04:07.250 ","End":"04:12.865","Text":"that\u0027s the standard Gibbs free energy of reaction,"},{"Start":"04:12.865 ","End":"04:15.889","Text":"plus RT ln of Q,"},{"Start":"04:15.889 ","End":"04:19.470","Text":"where Q is the reaction quotient."},{"Start":"04:22.000 ","End":"04:33.645","Text":"Now at equilibrium, Delta G of reaction is equal to Delta G^0 of reaction plus RT ln Q."},{"Start":"04:33.645 ","End":"04:37.259","Text":"That equilibrium, this is 0."},{"Start":"04:37.360 ","End":"04:40.645","Text":"If this is 0,"},{"Start":"04:40.645 ","End":"04:45.645","Text":"we can write that Delta G^0 of reaction,"},{"Start":"04:45.645 ","End":"04:49.355","Text":"that\u0027s the Gibbs free energy at equilibrium,"},{"Start":"04:49.355 ","End":"04:54.350","Text":"is equal to minus RT In Q,"},{"Start":"04:54.350 ","End":"05:01.425","Text":"but Q is equal to K. We have Delta G^0 reaction"},{"Start":"05:01.425 ","End":"05:09.925","Text":"is equal to minus RT ln K. This is a very important equation."},{"Start":"05:09.925 ","End":"05:12.980","Text":"It\u0027s important equation because it links"},{"Start":"05:12.980 ","End":"05:20.220","Text":"the thermal dynamic data or Delta G^0 and the equilibrium constant."},{"Start":"05:20.440 ","End":"05:28.925","Text":"Now supposing Delta G^0 of reaction is negative."},{"Start":"05:28.925 ","End":"05:33.880","Text":"If Delta G^0 of reaction is negative,"},{"Start":"05:33.880 ","End":"05:39.110","Text":"In K will be positive because this is negative sign here."},{"Start":"05:39.110 ","End":"05:41.240","Text":"If In K is positive,"},{"Start":"05:41.240 ","End":"05:47.929","Text":"it means that K must be greater than 1 and the products are favored at equilibrium,"},{"Start":"05:47.929 ","End":"05:50.300","Text":"remembering K and Q,"},{"Start":"05:50.300 ","End":"05:59.460","Text":"the products are in the numerator and the reactants are in the denominator."},{"Start":"05:59.500 ","End":"06:01.580","Text":"If K is large,"},{"Start":"06:01.580 ","End":"06:04.985","Text":"it means there are many more products than reactants."},{"Start":"06:04.985 ","End":"06:08.530","Text":"The products are favored at equilibrium."},{"Start":"06:08.530 ","End":"06:13.490","Text":"If on the other hand Delta G^0 of reaction is positive,"},{"Start":"06:13.490 ","End":"06:18.305","Text":"then In K must be negative and K must be less than 1."},{"Start":"06:18.305 ","End":"06:21.920","Text":"That means that the reactants are favored at equilibrium,"},{"Start":"06:21.920 ","End":"06:27.140","Text":"the reactants will be more important than the products."},{"Start":"06:27.140 ","End":"06:31.555","Text":"Now if we combine equations 1 and 2,"},{"Start":"06:31.555 ","End":"06:33.480","Text":"here\u0027s equation 1,"},{"Start":"06:33.480 ","End":"06:41.925","Text":"Delta G reaction is equal to Delta G^0 of reaction plus RT ln Q and here\u0027s equation 2,"},{"Start":"06:41.925 ","End":"06:48.150","Text":"Delta G^0 of reaction is minus RT ln K. If we combine them,"},{"Start":"06:48.150 ","End":"06:55.325","Text":"we get Delta G of reaction is equal to Delta G^0 of reaction plus RT ln Q,"},{"Start":"06:55.325 ","End":"07:01.925","Text":"that gives us Delta G of reaction is equal to minus RT ln K,"},{"Start":"07:01.925 ","End":"07:09.680","Text":"because Delta G^0 is minus RT ln K. Then we have plus RT ln Q."},{"Start":"07:09.680 ","End":"07:16.065","Text":"Remember, if we subtract 2 logarithms,"},{"Start":"07:16.065 ","End":"07:20.490","Text":"then we get the ratio between them."},{"Start":"07:20.490 ","End":"07:24.810","Text":"This becomes RT ln Q,"},{"Start":"07:24.810 ","End":"07:27.980","Text":"that\u0027s on top, divided by K,"},{"Start":"07:27.980 ","End":"07:31.590","Text":"where the minus sign gives us the denominator."},{"Start":"07:32.450 ","End":"07:42.735","Text":"We have Delta G of reaction is equal to RT ln Q over K. Let\u0027s look at this equation."},{"Start":"07:42.735 ","End":"07:45.000","Text":"If Q is equal to K,"},{"Start":"07:45.000 ","End":"07:47.250","Text":"that means we\u0027re at equilibrium,"},{"Start":"07:47.250 ","End":"07:54.980","Text":"then the ln of 1 is 0 and Delta G of reaction must be equal to 0."},{"Start":"07:54.980 ","End":"07:58.610","Text":"The reaction is at equilibrium."},{"Start":"07:58.610 ","End":"08:02.620","Text":"If Q is smaller than K,"},{"Start":"08:02.620 ","End":"08:08.900","Text":"Delta G reaction must be negative because the In of a fraction is negative."},{"Start":"08:08.900 ","End":"08:13.610","Text":"If Delta G is negative,"},{"Start":"08:13.610 ","End":"08:17.360","Text":"it means the forward reaction is spontaneous."},{"Start":"08:17.360 ","End":"08:23.359","Text":"If we have A to B forward and B to A the backward reaction,"},{"Start":"08:23.359 ","End":"08:25.025","Text":"that A to B,"},{"Start":"08:25.025 ","End":"08:29.865","Text":"the Delta G of reaction always refers to forward direction."},{"Start":"08:29.865 ","End":"08:37.790","Text":"That means that the forward reaction is spontaneous but if Q is greater than K,"},{"Start":"08:37.790 ","End":"08:44.870","Text":"then Delta G of reaction will be positive and the reverse reaction is spontaneous."},{"Start":"08:44.870 ","End":"08:48.409","Text":"A to B will not be spontaneous,"},{"Start":"08:48.409 ","End":"08:51.360","Text":"but B to A will be."},{"Start":"08:51.850 ","End":"08:56.300","Text":"In this video, we showed the connection between Gibbs free energy of the reaction"},{"Start":"08:56.300 ","End":"09:00.810","Text":"and the composition of the reactants and products."}],"ID":30460},{"Watched":false,"Name":"Gibbs Free Energy and Reaction Composition 2","Duration":"6m 15s","ChapterTopicVideoID":26276,"CourseChapterTopicPlaylistID":253355,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.740","Text":"In the previous video,"},{"Start":"00:01.740 ","End":"00:04.470","Text":"we talked about the relation between Gibbs free energy of"},{"Start":"00:04.470 ","End":"00:08.235","Text":"reactions and the composition of the reaction components."},{"Start":"00:08.235 ","End":"00:12.450","Text":"In this video, we\u0027ll prove the relation that we quoted there."},{"Start":"00:12.450 ","End":"00:15.120","Text":"We\u0027re going to prove"},{"Start":"00:15.120 ","End":"00:20.355","Text":"the relation between Gibbs free energy of reaction and the reaction quotient."},{"Start":"00:20.355 ","End":"00:22.500","Text":"This is what we want to prove."},{"Start":"00:22.500 ","End":"00:29.260","Text":"Delta G of reaction is equal to Delta G^0 of reaction plus RTlnQ."},{"Start":"00:29.540 ","End":"00:32.225","Text":"Now how are we going to do that?"},{"Start":"00:32.225 ","End":"00:34.340","Text":"At a particular stage of the action,"},{"Start":"00:34.340 ","End":"00:39.185","Text":"we can write Delta G of reaction is equal to the sum over all the products,"},{"Start":"00:39.185 ","End":"00:44.610","Text":"where we multiply the number of moles of each product times G_m,"},{"Start":"00:44.610 ","End":"00:48.515","Text":"the Gibbs free energy of that particular substance."},{"Start":"00:48.515 ","End":"00:52.370","Text":"Then we subtract from that the sum over all the reactants of"},{"Start":"00:52.370 ","End":"00:56.710","Text":"the number of moles of each reactant times G_m for each reactant."},{"Start":"00:56.710 ","End":"01:00.680","Text":"Now this G_m is the molar Gibbs free energy"},{"Start":"01:00.680 ","End":"01:04.600","Text":"of the reactants or products at a particular stage."},{"Start":"01:04.600 ","End":"01:07.414","Text":"Now it\u0027s not something you can actually measure,"},{"Start":"01:07.414 ","End":"01:14.920","Text":"we can only really measure Delta G. But it\u0027s useful for this proof."},{"Start":"01:15.640 ","End":"01:20.735","Text":"Delta G reaction changes throughout the reaction,"},{"Start":"01:20.735 ","End":"01:27.580","Text":"whereas Delta G^0 reaction is a constant."},{"Start":"01:27.580 ","End":"01:35.110","Text":"Words about units G_m Delta G and RT all have the units kilo joules per mole."},{"Start":"01:35.110 ","End":"01:41.425","Text":"The coefficients n in this particular equation are unit-less."},{"Start":"01:41.425 ","End":"01:43.945","Text":"Just take the numbers."},{"Start":"01:43.945 ","End":"01:46.130","Text":"Now we\u0027re going to use an equation,"},{"Start":"01:46.130 ","End":"01:49.550","Text":"I\u0027m not going to prove is the relation between the molar"},{"Start":"01:49.550 ","End":"01:54.000","Text":"Gibbs free energy over a substance A and its activity."},{"Start":"01:54.260 ","End":"01:59.033","Text":"G_m is equal to G_m^0 plus RTlna,"},{"Start":"01:59.033 ","End":"02:01.755","Text":"the activity of A."},{"Start":"02:01.755 ","End":"02:05.290","Text":"Of course this is G_m of A."},{"Start":"02:06.890 ","End":"02:09.296","Text":"Now for an ideal gas,"},{"Start":"02:09.296 ","End":"02:16.095","Text":"G_m is G_m^0 plus RTln of P_A divide by P^0."},{"Start":"02:16.095 ","End":"02:17.585","Text":"P_A, if you remember,"},{"Start":"02:17.585 ","End":"02:25.240","Text":"is the partial pressure and P^0 is just 1 bar."},{"Start":"02:27.100 ","End":"02:29.930","Text":"For a solute in an ideal solution,"},{"Start":"02:29.930 ","End":"02:38.129","Text":"G_m is equal to G_m^0 plus RTln of the concentration of A divided by C^0."},{"Start":"02:38.129 ","End":"02:41.220","Text":"C^0 is 1 mole per liter."},{"Start":"02:42.130 ","End":"02:47.515","Text":"For a pure solid or liquid a=1,"},{"Start":"02:47.515 ","End":"02:53.880","Text":"so the ln of 1 is 0 and G_m is equal to G_m^0."},{"Start":"02:53.880 ","End":"02:58.100","Text":"Now we\u0027re going to prove equation 1."},{"Start":"02:58.100 ","End":"03:04.325","Text":"I\u0027m going to do it for the reaction of a moles of A plus b moles of B,"},{"Start":"03:04.325 ","End":"03:11.100","Text":"giving c moles of C and d moles of D. We\u0027re going to use equations 2 and 3."},{"Start":"03:11.540 ","End":"03:16.569","Text":"Here\u0027s equation 2,"},{"Start":"03:16.569 ","End":"03:19.475","Text":"written out specifically for this reaction."},{"Start":"03:19.475 ","End":"03:24.050","Text":"We have Delta G reaction is equal to c moles"},{"Start":"03:24.050 ","End":"03:30.965","Text":"times G_m of C plus d moles of G_m of D. It\u0027s not really moles,"},{"Start":"03:30.965 ","End":"03:37.905","Text":"it\u0027s just the number of moles minus a times G_m of A plus b, G_m of B."},{"Start":"03:37.905 ","End":"03:48.280","Text":"This is first of all for the products minus the reactants."},{"Start":"03:50.030 ","End":"03:58.020","Text":"Now I\u0027m going to use equation 3 in this equation here."},{"Start":"03:58.020 ","End":"04:01.380","Text":"Instead of G_m of C,"},{"Start":"04:01.380 ","End":"04:07.635","Text":"we have G_m^0 of C plus RTclna_c."},{"Start":"04:07.635 ","End":"04:09.360","Text":"We do that for each of them."},{"Start":"04:09.360 ","End":"04:10.930","Text":"We get 1 line,"},{"Start":"04:10.930 ","End":"04:15.250","Text":"which is the same as the first line except for the 0s."},{"Start":"04:15.250 ","End":"04:20.135","Text":"That line is equal to Delta G^0 of reaction."},{"Start":"04:20.135 ","End":"04:28.050","Text":"Then we have the RTlnRT times clna_C plus dlna_D minus"},{"Start":"04:28.050 ","End":"04:37.090","Text":"aIna_A plus bln_B and we have to combine this together to get this expression."},{"Start":"04:37.090 ","End":"04:44.330","Text":"We know that if we have In of x^y,"},{"Start":"04:44.330 ","End":"04:48.760","Text":"that is equal to yInx."},{"Start":"04:48.760 ","End":"04:56.790","Text":"Here clna_C is a_C^c."},{"Start":"04:56.790 ","End":"04:58.490","Text":"We can do that for each of these,"},{"Start":"04:58.490 ","End":"05:00.590","Text":"and then we have to combine them together."},{"Start":"05:00.590 ","End":"05:11.385","Text":"We know that ln of x plus ln of y is equal to ln of x times y."},{"Start":"05:11.385 ","End":"05:20.650","Text":"The ln of x minus the ln of y is equal to the In x divided by y."},{"Start":"05:20.650 ","End":"05:23.510","Text":"We take that into consideration."},{"Start":"05:23.510 ","End":"05:29.755","Text":"Then we get the ln outside of a_C^c times a_D^d,"},{"Start":"05:29.755 ","End":"05:32.490","Text":"that\u0027s a positive sign,"},{"Start":"05:32.490 ","End":"05:34.760","Text":"minus in the denominator,"},{"Start":"05:34.760 ","End":"05:39.500","Text":"we have a_A^a times a_B^b."},{"Start":"05:39.500 ","End":"05:44.030","Text":"Recall that this together is equal to Q."},{"Start":"05:44.030 ","End":"05:54.275","Text":"What we finally get is Delta G^0 of reaction here plus RTlnQ."},{"Start":"05:54.275 ","End":"05:58.280","Text":"What we proved was that Delta G of reaction is equal to"},{"Start":"05:58.280 ","End":"06:03.095","Text":"Delta G^0 of reaction plus RTlnQ,"},{"Start":"06:03.095 ","End":"06:06.989","Text":"and that\u0027s precisely the equation we wanted to prove."},{"Start":"06:07.460 ","End":"06:11.630","Text":"In this video, we gave a proof of the equation relating"},{"Start":"06:11.630 ","End":"06:16.380","Text":"Gibbs free energy of reaction to the reaction quotient."}],"ID":30461},{"Watched":false,"Name":"Direction of Reaction","Duration":"4m 20s","ChapterTopicVideoID":26277,"CourseChapterTopicPlaylistID":253355,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.055","Text":"In the previous video,"},{"Start":"00:02.055 ","End":"00:07.335","Text":"we talked about whether reactants or products are favored at equilibrium."},{"Start":"00:07.335 ","End":"00:13.440","Text":"In this video, we\u0027ll talk about the situation when the reaction is not at equilibrium."},{"Start":"00:13.440 ","End":"00:16.950","Text":"Let\u0027s recall the expression that relates"},{"Start":"00:16.950 ","End":"00:20.210","Text":"the Gibbs free energy of reaction and the reaction quotient,"},{"Start":"00:20.210 ","End":"00:28.620","Text":"that\u0027s Q. Delta G of reaction is equal to RTln(Q divided by K),"},{"Start":"00:28.620 ","End":"00:32.070","Text":"Q can be smaller than K,"},{"Start":"00:32.070 ","End":"00:34.410","Text":"or equal to K,"},{"Start":"00:34.410 ","End":"00:42.110","Text":"or Q can be larger than K. Now supposing we have a reaction A to"},{"Start":"00:42.110 ","End":"00:50.815","Text":"B and we should recall that Delta G of reaction relates to the forward reaction A to B."},{"Start":"00:50.815 ","End":"00:54.800","Text":"Now what happens is the reaction will proceed towards"},{"Start":"00:54.800 ","End":"00:59.730","Text":"the minimum value of the Gibbs free energy of reaction."},{"Start":"01:00.010 ","End":"01:02.885","Text":"We\u0027ve seen this picture before."},{"Start":"01:02.885 ","End":"01:05.239","Text":"This is the extent of the reaction."},{"Start":"01:05.239 ","End":"01:08.760","Text":"Here we start off with reactants."},{"Start":"01:09.590 ","End":"01:13.180","Text":"We end up with a products."},{"Start":"01:13.880 ","End":"01:17.645","Text":"Now the Gibbs free energy decreases."},{"Start":"01:17.645 ","End":"01:20.870","Text":"The slope here is negative."},{"Start":"01:20.870 ","End":"01:26.910","Text":"Here it\u0027s 0, and here is positive."},{"Start":"01:28.300 ","End":"01:32.090","Text":"Now if Q is smaller than K,"},{"Start":"01:32.090 ","End":"01:38.030","Text":"then Delta G reaction will be negative and the forward reaction will be spontaneous."},{"Start":"01:38.030 ","End":"01:43.475","Text":"The reaction will proceed in the direction of the products until equilibrium is achieved."},{"Start":"01:43.475 ","End":"01:51.365","Text":"We\u0027re starting this side and proceeding down to Delta G equals 0, down to the minimum."},{"Start":"01:51.365 ","End":"01:55.640","Text":"Now, if Q is greater than K,"},{"Start":"01:55.640 ","End":"01:59.390","Text":"Delta G of reaction will be greater than 0."},{"Start":"01:59.390 ","End":"02:03.040","Text":"The reverse reaction will be spontaneous."},{"Start":"02:03.040 ","End":"02:08.060","Text":"Here, Delta G for the forward reaction will be positive."},{"Start":"02:08.060 ","End":"02:09.830","Text":"It will not be spontaneous."},{"Start":"02:09.830 ","End":"02:12.065","Text":"The forward reaction will not be spontaneous,"},{"Start":"02:12.065 ","End":"02:14.660","Text":"but the reverse reaction will be."},{"Start":"02:14.660 ","End":"02:18.095","Text":"We\u0027ll start here and go down to the minimum."},{"Start":"02:18.095 ","End":"02:21.770","Text":"The reaction will proceed in the direction of the reactants,"},{"Start":"02:21.770 ","End":"02:26.245","Text":"this direction until equilibrium is achieved."},{"Start":"02:26.245 ","End":"02:31.040","Text":"Of course, if we start off at the minimum, nothing will happen."},{"Start":"02:31.040 ","End":"02:32.660","Text":"Q will be equal to K,"},{"Start":"02:32.660 ","End":"02:37.020","Text":"Delta G reaction will be equal to 0. Nothing will happen."},{"Start":"02:37.120 ","End":"02:45.175","Text":"Now what happens if the minimum is near the products or reactants?"},{"Start":"02:45.175 ","End":"02:47.180","Text":"On the left-hand side,"},{"Start":"02:47.180 ","End":"02:50.570","Text":"the minimum is near the products and the right-hand side,"},{"Start":"02:50.570 ","End":"02:53.740","Text":"the minimum is near the reactants."},{"Start":"02:53.740 ","End":"02:57.400","Text":"This is Delta G^0."},{"Start":"02:57.890 ","End":"03:02.500","Text":"The products minus reactants."},{"Start":"03:08.450 ","End":"03:12.530","Text":"If the minimum is near the pure products,"},{"Start":"03:12.530 ","End":"03:15.560","Text":"that means that the numerator of K,"},{"Start":"03:15.560 ","End":"03:21.960","Text":"remember K is products over reactants,"},{"Start":"03:23.870 ","End":"03:28.760","Text":"K will be much greater than 1 for near the products,"},{"Start":"03:28.760 ","End":"03:32.600","Text":"Delta G^0 reaction will be negative."},{"Start":"03:32.600 ","End":"03:37.680","Text":"We\u0027re going from this minus this number, so that\u0027s negative."},{"Start":"03:37.930 ","End":"03:43.085","Text":"The reaction will go to completion or almost to completion."},{"Start":"03:43.085 ","End":"03:46.010","Text":"Now, the minimum is near the pure reactants,"},{"Start":"03:46.010 ","End":"03:49.220","Text":"it\u0027s here, K will be much"},{"Start":"03:49.220 ","End":"03:53.315","Text":"less than 1 because the reactants were much greater, the products."},{"Start":"03:53.315 ","End":"03:56.990","Text":"Delta G^0 of reaction will be positive."},{"Start":"03:56.990 ","End":"04:01.760","Text":"We\u0027re going from a higher point to a lower point here."},{"Start":"04:01.760 ","End":"04:06.780","Text":"This is Delta G^0 will be positive."},{"Start":"04:06.780 ","End":"04:10.040","Text":"The reaction will hardly occur."},{"Start":"04:10.040 ","End":"04:14.615","Text":"Just saying here, near the pure reactants."},{"Start":"04:14.615 ","End":"04:20.520","Text":"In this video, we talked about the direction in which the reaction will proceed."}],"ID":30462}],"Thumbnail":null,"ID":253355},{"Name":"Tools For Equilibrium Calculations","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Equilibrium Constant Using Gas Concentrations","Duration":"7m 53s","ChapterTopicVideoID":26285,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:06.705","Text":"In previous videos, we defined the equilibrium constant for gases using pressures."},{"Start":"00:06.705 ","End":"00:13.035","Text":"In this video we\u0027ll talk about the equilibrium constant for gases using concentrations."},{"Start":"00:13.035 ","End":"00:19.110","Text":"Let\u0027s recall the equilibrium constant for reactions involving gases."},{"Start":"00:19.110 ","End":"00:23.850","Text":"Supposing we have a moles of A plus b of B in"},{"Start":"00:23.850 ","End":"00:27.870","Text":"equilibrium c moles of C and d moles of D and A,"},{"Start":"00:27.870 ","End":"00:30.850","Text":"B, C and D are all gases."},{"Start":"00:30.850 ","End":"00:36.110","Text":"Then K, which is defined in terms of the activities."},{"Start":"00:36.110 ","End":"00:41.270","Text":"K is equal to the activity of C^c times activity of D^d"},{"Start":"00:41.270 ","End":"00:46.361","Text":"divide by activity of A^a and activity of B^b,"},{"Start":"00:46.361 ","End":"00:50.570","Text":"when the gases, we can write this is the pressure of C divided"},{"Start":"00:50.570 ","End":"00:57.980","Text":"by P^0^c times the pressure of D divided by P^0^d divided by the pressure of A,"},{"Start":"00:57.980 ","End":"01:04.475","Text":"divided by p^0^a times the pressure of B divided by P^0^b."},{"Start":"01:04.475 ","End":"01:09.090","Text":"P^0 you will recall is 1 bar."},{"Start":"01:09.880 ","End":"01:15.420","Text":"Now sometimes, this is called K_p rather than just"},{"Start":"01:15.420 ","End":"01:20.810","Text":"K. That\u0027s to remind us that it\u0027s in terms of pressures."},{"Start":"01:20.810 ","End":"01:24.230","Text":"We want to write the equilibrium constant using"},{"Start":"01:24.230 ","End":"01:31.560","Text":"gas concentrations and we\u0027re going to assume that the gas is ideal."},{"Start":"01:31.870 ","End":"01:34.655","Text":"The pressure of gas a,"},{"Start":"01:34.655 ","End":"01:38.090","Text":"we can write as n_A the number of moles of a times"},{"Start":"01:38.090 ","End":"01:43.250","Text":"RT gas constant times the temperature in Kelvin divided by V,"},{"Start":"01:43.250 ","End":"01:45.460","Text":"the volume in liters."},{"Start":"01:45.460 ","End":"01:50.970","Text":"That\u0027s just from PV equals nRT, the gas law."},{"Start":"01:50.970 ","End":"01:54.480","Text":"Now n_A divided by V,"},{"Start":"01:54.480 ","End":"02:00.025","Text":"the number of moles of a divided by the volume is just the concentration of A."},{"Start":"02:00.025 ","End":"02:03.990","Text":"We can write this now as RT times"},{"Start":"02:03.990 ","End":"02:09.200","Text":"the concentration of A. P_A is RT times the concentration."},{"Start":"02:09.200 ","End":"02:15.050","Text":"Now, in order to use concentrations in dimensionless units,"},{"Start":"02:15.050 ","End":"02:18.710","Text":"we can divide the concentration by C^0."},{"Start":"02:18.710 ","End":"02:22.130","Text":"If you remember, C^0 is 1 mole per"},{"Start":"02:22.130 ","End":"02:30.760","Text":"liter and if we divide by C^0 then we have to multiply by C^0 so that we get 1."},{"Start":"02:30.890 ","End":"02:38.580","Text":"Let\u0027s rewrite the K. Instead of P_C divided by P^0^0."},{"Start":"02:38.580 ","End":"02:44.110","Text":"We have C, the concentration of C divided by C^0"},{"Start":"02:44.110 ","End":"02:50.005","Text":"and then outside we have C^0 RT divided by P^0,"},{"Start":"02:50.005 ","End":"02:54.725","Text":"and that\u0027s to the power of c. We can do that for each of the expressions."},{"Start":"02:54.725 ","End":"03:01.410","Text":"P_D divided by p^0^d is the concentration of D divided by"},{"Start":"03:01.410 ","End":"03:09.290","Text":"C^0^d multiplied by C^0 RT divided by P^0^d, and so on."},{"Start":"03:09.290 ","End":"03:11.870","Text":"We can do that also for the denominator."},{"Start":"03:11.870 ","End":"03:19.070","Text":"In the end, we get this expression involving concentrations multiplied"},{"Start":"03:19.070 ","End":"03:28.010","Text":"by C^0 RT divided by P^0^c plus d minus a plus b."},{"Start":"03:28.010 ","End":"03:35.810","Text":"Now we\u0027re going to call that Delta n. That\u0027s the number of moles of the products"},{"Start":"03:35.810 ","End":"03:44.540","Text":"minus the number of moles of the reactants and that\u0027s Delta n. Now,"},{"Start":"03:44.540 ","End":"03:48.995","Text":"if we defined this expression here as K_c,"},{"Start":"03:48.995 ","End":"03:52.910","Text":"that\u0027s the equilibrium constant terms of concentrations,"},{"Start":"03:52.910 ","End":"03:59.805","Text":"then we see that K in pressures is equal to K_c"},{"Start":"03:59.805 ","End":"04:05.250","Text":"multiplied by C^0 RT divided"},{"Start":"04:05.250 ","End":"04:11.030","Text":"by P^0^Delta n. Instead of C plus T minus a plus b,"},{"Start":"04:11.030 ","End":"04:17.555","Text":"we\u0027re writing Delta n. Now we have the connection between K and K_c."},{"Start":"04:17.555 ","End":"04:21.170","Text":"Here\u0027s our K_c and here\u0027s our Delta n."},{"Start":"04:21.170 ","End":"04:25.630","Text":"The number of moles of products minus the number of moles of reactants."},{"Start":"04:25.630 ","End":"04:28.560","Text":"Now, which value of R should we use?"},{"Start":"04:28.560 ","End":"04:32.985","Text":"We know that P^0 is 1 bar C^0 is 1 mole per liter."},{"Start":"04:32.985 ","End":"04:35.130","Text":"We use R in these units."},{"Start":"04:35.130 ","End":"04:43.745","Text":"R is 8.3145 times 10^minus 2 liters times bar per Kelvin per mole."},{"Start":"04:43.745 ","End":"04:46.640","Text":"Let\u0027s look at an example."},{"Start":"04:46.640 ","End":"04:51.290","Text":"Here\u0027s our reaction 2SO_2 in gas phase plus"},{"Start":"04:51.290 ","End":"04:58.490","Text":"O_2 gas phase in equilibrium with 2SO_3 in the gas phase and the K,"},{"Start":"04:58.490 ","End":"05:00.545","Text":"which we defined before."},{"Start":"05:00.545 ","End":"05:02.270","Text":"We can do it again."},{"Start":"05:02.270 ","End":"05:07.805","Text":"It\u0027s P_SO_3 divided by P^0^2 divide by"},{"Start":"05:07.805 ","End":"05:14.695","Text":"P_SO_2 divided by P^0^2 times P of oxygen divide by P^0."},{"Start":"05:14.695 ","End":"05:25.210","Text":"The 2 here comes from this 2 and the 2 here from this 2 and here\u0027s just to the power 1."},{"Start":"05:25.630 ","End":"05:34.445","Text":"This K is equal to 3.0 times 10^4 at 700 Kelvin."},{"Start":"05:34.445 ","End":"05:38.285","Text":"Now, we\u0027re asked what is the value of K_c."},{"Start":"05:38.285 ","End":"05:40.655","Text":"We know K or K_p,"},{"Start":"05:40.655 ","End":"05:45.660","Text":"whatever you want to call it and now we want to find K_c."},{"Start":"05:45.660 ","End":"05:49.950","Text":"Now we know that Delta n is equal to minus 1 because we have"},{"Start":"05:49.950 ","End":"05:55.200","Text":"2 moles of SO_3 and we have 3 moles of reactants,"},{"Start":"05:55.200 ","End":"05:57.840","Text":"2 of SO_2 and another 1 of O_2."},{"Start":"05:57.840 ","End":"06:00.920","Text":"Here we have 3 moles and here we have 2."},{"Start":"06:00.920 ","End":"06:05.085","Text":"So 2 minus 3 is equal to minus 1."},{"Start":"06:05.085 ","End":"06:11.960","Text":"Now, we saw above that K is equal to K_c times this ratio to the power"},{"Start":"06:11.960 ","End":"06:18.140","Text":"Delta n. K_c is equal to K times the same ratio,"},{"Start":"06:18.140 ","End":"06:23.540","Text":"but to the power minus Delta n. Here it\u0027s written K_c is equal to"},{"Start":"06:23.540 ","End":"06:32.460","Text":"K times C^0 RT divided by P^0^minus Delta n and Delta n is minus 1."},{"Start":"06:32.460 ","End":"06:35.610","Text":"This is going to be the power 1."},{"Start":"06:35.610 ","End":"06:43.330","Text":"We have K_c is equal to K times C^0 RT divided by P^0."},{"Start":"06:43.330 ","End":"06:48.265","Text":"Let\u0027s substitute the values we have above."},{"Start":"06:48.265 ","End":"06:53.070","Text":"K is 3.0 times 10^4."},{"Start":"06:53.070 ","End":"06:55.410","Text":"C^0 is 1 mole per liter."},{"Start":"06:55.410 ","End":"07:03.525","Text":"R is 8.3145 times 10^minus 2 liter times bar K^minus 1,"},{"Start":"07:03.525 ","End":"07:11.910","Text":"mole to the power minus 1 and T is 700 K and P^0 is 1 bar."},{"Start":"07:11.910 ","End":"07:13.860","Text":"We can look at the units,"},{"Start":"07:13.860 ","End":"07:16.650","Text":"liter to the power minus 1 goes with liter,"},{"Start":"07:16.650 ","End":"07:20.400","Text":"mole goes with mole to the power minus 1,"},{"Start":"07:20.400 ","End":"07:22.470","Text":"bar goes with bar,"},{"Start":"07:22.470 ","End":"07:26.065","Text":"K goes with K^minus 1."},{"Start":"07:26.065 ","End":"07:31.520","Text":"As we expect, we\u0027re going to get a result which is dimensionless."},{"Start":"07:31.520 ","End":"07:33.215","Text":"We multiply all this out,"},{"Start":"07:33.215 ","End":"07:37.865","Text":"we get 1.7 times 10^6."},{"Start":"07:37.865 ","End":"07:44.065","Text":"K_c is a larger volume than K_p."},{"Start":"07:44.065 ","End":"07:48.200","Text":"In this video, we discussed the equilibrium constant for"},{"Start":"07:48.200 ","End":"07:53.910","Text":"gases when the concentrations are used rather than pressures."}],"ID":30463},{"Watched":false,"Name":"Exercise 1","Duration":"3m 21s","ChapterTopicVideoID":28915,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.820","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.820 ","End":"00:07.050","Text":"Calculate the equilibrium constant K_p for the following reaction."},{"Start":"00:07.050 ","End":"00:10.030","Text":"Carbon dioxide and hydrogen,"},{"Start":"00:10.030 ","End":"00:11.160","Text":"we have on our reactant side,"},{"Start":"00:11.160 ","End":"00:14.740","Text":"and on our product side, we have carbon monoxide and water."},{"Start":"00:14.930 ","End":"00:19.560","Text":"We also know K_c equals 1.4 at"},{"Start":"00:19.560 ","End":"00:24.330","Text":"1,200 K. Now since we\u0027re asked to calculate K_p and we know K_c,"},{"Start":"00:24.330 ","End":"00:29.490","Text":"we\u0027re going to use the equation K_p equals"},{"Start":"00:29.490 ","End":"00:37.530","Text":"K_c times the standard"},{"Start":"00:37.530 ","End":"00:41.125","Text":"concentration times RT,"},{"Start":"00:41.125 ","End":"00:44.300","Text":"R is the gas constant and T is the temperature,"},{"Start":"00:44.300 ","End":"00:49.360","Text":"and this is divided by the standard pressure."},{"Start":"00:49.360 ","End":"00:56.110","Text":"Now the standard concentration c equals 1 mole per liter,"},{"Start":"00:56.300 ","End":"01:02.100","Text":"and the standard pressure equals 1 bar."},{"Start":"01:02.100 ","End":"01:06.120","Text":"c times RT divided by the standard pressure,"},{"Start":"01:06.120 ","End":"01:12.680","Text":"is raised to the power of delta n. It\u0027s raised to the power of delta n gas."},{"Start":"01:12.680 ","End":"01:14.839","Text":"First of all, we take into consideration"},{"Start":"01:14.839 ","End":"01:19.025","Text":"only the gaseous products and the gaseous reactants."},{"Start":"01:19.025 ","End":"01:25.100","Text":"Delta n means that we take the number of moles or the sum of the coefficients of"},{"Start":"01:25.100 ","End":"01:31.824","Text":"the products minus the number of moles or the sum of the coefficients of the reactants."},{"Start":"01:31.824 ","End":"01:34.500","Text":"We\u0027re going to do that in a minute."},{"Start":"01:34.500 ","End":"01:39.690","Text":"First of all, K_p equals K_c and we know K_c is"},{"Start":"01:39.690 ","End":"01:46.155","Text":"1.4 at 1,200 K. That\u0027s going to be times,"},{"Start":"01:46.155 ","End":"01:49.430","Text":"here we can put 1 mole per liter and here 1 bar."},{"Start":"01:49.430 ","End":"01:50.945","Text":"But before we do that,"},{"Start":"01:50.945 ","End":"01:55.190","Text":"I want to calculate delta n just to show you something."},{"Start":"01:55.190 ","End":"01:59.209","Text":"If we take delta n of the gaseous products and reactants,"},{"Start":"01:59.209 ","End":"02:02.075","Text":"we can see that we have the coefficient of"},{"Start":"02:02.075 ","End":"02:06.185","Text":"carbon monoxide as 1 and the coefficient of water is also 1."},{"Start":"02:06.185 ","End":"02:07.874","Text":"That\u0027s going to be 1 plus 1,"},{"Start":"02:07.874 ","End":"02:11.570","Text":"you can call that the number of moles of the product,"},{"Start":"02:11.570 ","End":"02:16.565","Text":"minus the number of moles of the sum of the coefficients of the reactants."},{"Start":"02:16.565 ","End":"02:18.553","Text":"In our case, the carbon dioxide also has 1,"},{"Start":"02:18.553 ","End":"02:23.070","Text":"so that\u0027s 1, plus the hydrogen also has 1,"},{"Start":"02:23.070 ","End":"02:24.795","Text":"so that\u0027s 1 plus 1."},{"Start":"02:24.795 ","End":"02:27.390","Text":"We have 1 plus 1 minus 1 plus 1,"},{"Start":"02:27.390 ","End":"02:30.530","Text":"meaning 2 minus 2. That\u0027s going to give us a 0."},{"Start":"02:30.530 ","End":"02:37.040","Text":"Meaning everything that\u0027s in parentheses is going to be raised to the power of 0."},{"Start":"02:37.040 ","End":"02:40.670","Text":"Now, anything raised to the power of 0 equals 1."},{"Start":"02:40.670 ","End":"02:51.085","Text":"That\u0027s just going to give us 1.4 times 1, which equals 1.4."},{"Start":"02:51.085 ","End":"02:58.475","Text":"There\u0027s no reason to solve inside the parentheses since our delta n equals 0,"},{"Start":"02:58.475 ","End":"03:03.840","Text":"again, and therefore we get 1 times 1.4, which equals 1.4."},{"Start":"03:03.840 ","End":"03:07.350","Text":"Again, since our delta n equals 0,"},{"Start":"03:07.350 ","End":"03:12.210","Text":"K_p actually equals K_c because it equals K_c times 1."},{"Start":"03:12.210 ","End":"03:17.780","Text":"So K_p also equals 1.4 in this case."},{"Start":"03:17.780 ","End":"03:19.280","Text":"That is our final answer."},{"Start":"03:19.280 ","End":"03:21.810","Text":"Thank you very much for watching."}],"ID":30464},{"Watched":false,"Name":"Exercise 2","Duration":"4m 13s","ChapterTopicVideoID":28916,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.519","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.650 ","End":"00:06.240","Text":"Calculate the equilibrium constant,"},{"Start":"00:06.240 ","End":"00:08.535","Text":"K_p, for the following reaction."},{"Start":"00:08.535 ","End":"00:11.190","Text":"We have dinitrogen tetroxide,"},{"Start":"00:11.190 ","End":"00:13.619","Text":"and on the product side,"},{"Start":"00:13.619 ","End":"00:17.010","Text":"we have 2 nitrogen dioxide."},{"Start":"00:17.010 ","End":"00:22.515","Text":"We know K_c equals 4.61 times 10 to the negative 3 at 298K."},{"Start":"00:22.515 ","End":"00:25.905","Text":"We\u0027re asked to calculate K_p."},{"Start":"00:25.905 ","End":"00:29.205","Text":"We know K_c, and we also have the reaction."},{"Start":"00:29.205 ","End":"00:32.970","Text":"Therefore, we\u0027re going to use the equation K_p equals"},{"Start":"00:32.970 ","End":"00:38.760","Text":"K_c times the standard concentration times RT,"},{"Start":"00:38.760 ","End":"00:40.470","Text":"R is the gas constant,"},{"Start":"00:40.470 ","End":"00:42.030","Text":"and T is the temperature,"},{"Start":"00:42.030 ","End":"00:46.500","Text":"divided by the standard pressure."},{"Start":"00:46.500 ","End":"00:51.780","Text":"This is all raised to the power of Delta n gas."},{"Start":"00:51.780 ","End":"00:55.065","Text":"Delta n, we\u0027ll calculate this in a minute,"},{"Start":"00:55.065 ","End":"00:59.165","Text":"is the difference in the number of moles between the products and the reactants,"},{"Start":"00:59.165 ","End":"01:05.640","Text":"or the sum of the coefficients of the gaseous products and reactants."},{"Start":"01:05.640 ","End":"01:07.235","Text":"We\u0027re going to see that in a minute."},{"Start":"01:07.235 ","End":"01:11.150","Text":"Now, I just want to mention that for simplicity\u0027s sake, many times,"},{"Start":"01:11.150 ","End":"01:17.340","Text":"we will just see K_p equals K_c times"},{"Start":"01:17.340 ","End":"01:19.500","Text":"RT to the power of"},{"Start":"01:19.500 ","End":"01:26.720","Text":"Delta n. They just don\u0027t write the standard pressure and the standard concentration."},{"Start":"01:26.720 ","End":"01:30.330","Text":"They just avoid the units altogether."},{"Start":"01:30.890 ","End":"01:34.925","Text":"K_p, which we want to calculate, equals K_c."},{"Start":"01:34.925 ","End":"01:38.070","Text":"We know K_c equals 4.61."},{"Start":"01:38.680 ","End":"01:44.410","Text":"We know that K_c equals 4.61 times 10 to the negative 3."},{"Start":"01:44.410 ","End":"01:46.820","Text":"Now, let\u0027s look in the parentheses."},{"Start":"01:46.820 ","End":"01:49.550","Text":"We have the standard concentration,"},{"Start":"01:49.550 ","End":"01:52.535","Text":"which we know equals 1 mole per liter,"},{"Start":"01:52.535 ","End":"01:55.355","Text":"times the gas constant."},{"Start":"01:55.355 ","End":"01:59.395","Text":"Since our standard pressure is 1 bar,"},{"Start":"01:59.395 ","End":"02:02.600","Text":"we\u0027re going to use the value of the gas constant,"},{"Start":"02:02.600 ","End":"02:08.300","Text":"which equals 8.3145 times 10 to"},{"Start":"02:08.300 ","End":"02:16.880","Text":"the minus 2 liters times bar divided by mole times kelvin."},{"Start":"02:16.880 ","End":"02:21.985","Text":"Now, we can see that the mole already cancel out and so do the liters."},{"Start":"02:21.985 ","End":"02:25.740","Text":"The bar is also going to cancel out."},{"Start":"02:25.740 ","End":"02:27.150","Text":"This is times the temperature."},{"Start":"02:27.150 ","End":"02:31.430","Text":"Our temperature is 298K."},{"Start":"02:31.430 ","End":"02:34.850","Text":"Also, our kelvin cancels out."},{"Start":"02:34.850 ","End":"02:37.325","Text":"Now, we\u0027re going to close the parentheses."},{"Start":"02:37.325 ","End":"02:42.960","Text":"Now, we need Delta n. Let\u0027s calculate Delta n. Again,"},{"Start":"02:42.960 ","End":"02:47.845","Text":"Delta n gas is the number of moles of"},{"Start":"02:47.845 ","End":"02:50.500","Text":"products or the coefficient of the products"},{"Start":"02:50.500 ","End":"02:53.860","Text":"minus the number of mole or the coefficient of the reactants."},{"Start":"02:53.860 ","End":"02:57.370","Text":"In our case, we have 2 as the coefficient or as the number of moles,"},{"Start":"02:57.370 ","End":"03:00.190","Text":"so it\u0027s going to be 2 in our product side because that\u0027s"},{"Start":"03:00.190 ","End":"03:07.595","Text":"2 nitrogen dioxide minus what we have on our reactant side, which is 1."},{"Start":"03:07.595 ","End":"03:10.980","Text":"This equals 1. Our Delta n,"},{"Start":"03:10.980 ","End":"03:13.085","Text":"again, it\u0027s gaseous products."},{"Start":"03:13.085 ","End":"03:15.760","Text":"Here in the products and reactants,"},{"Start":"03:15.760 ","End":"03:18.035","Text":"we only have gases."},{"Start":"03:18.035 ","End":"03:21.420","Text":"Delta n is gaseous products and reactants."},{"Start":"03:21.420 ","End":"03:25.250","Text":"This equals 1, so this is all going to be raised to the power of 1."},{"Start":"03:25.820 ","End":"03:28.499","Text":"After multiplying and dividing,"},{"Start":"03:28.499 ","End":"03:37.660","Text":"K_p equals 0.114."},{"Start":"03:37.660 ","End":"03:40.865","Text":"We found that K_p equals 0.114."},{"Start":"03:40.865 ","End":"03:43.070","Text":"Now, I just want to mention,"},{"Start":"03:43.070 ","End":"03:49.050","Text":"we use the standard pressure as 1 bar."},{"Start":"03:49.050 ","End":"03:52.500","Text":"We took the gas constant which fits the bar,"},{"Start":"03:52.500 ","End":"03:54.090","Text":"so the bar will be canceled."},{"Start":"03:54.090 ","End":"03:56.450","Text":"There are places which you will see that they"},{"Start":"03:56.450 ","End":"03:59.270","Text":"take the standard temperature as 1 atmosphere and then they"},{"Start":"03:59.270 ","End":"04:01.055","Text":"take the gas constant which fits"},{"Start":"04:01.055 ","End":"04:04.450","Text":"1 atmosphere since 1 atmosphere and 1 bar are very close."},{"Start":"04:04.450 ","End":"04:06.720","Text":"You might just see that in other places."},{"Start":"04:06.720 ","End":"04:10.040","Text":"Again, K_p equals 0.114."},{"Start":"04:10.040 ","End":"04:11.660","Text":"That is our final answer."},{"Start":"04:11.660 ","End":"04:14.310","Text":"Thank you very much for watching."}],"ID":30465},{"Watched":false,"Name":"Exercise 3","Duration":"4m 1s","ChapterTopicVideoID":28917,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.320","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.320 ","End":"00:03.880","Text":"Calculate the equilibrium constant,"},{"Start":"00:03.880 ","End":"00:06.140","Text":"K_c, for the following reaction."},{"Start":"00:06.140 ","End":"00:12.100","Text":"We have 4 potassium dioxide plus 2 carbon dioxide on the reactant side."},{"Start":"00:12.100 ","End":"00:17.555","Text":"On the product side, we have 2 potassium carbonate and 3 oxygen."},{"Start":"00:17.555 ","End":"00:20.490","Text":"We want to calculate K_c and we know K_p."},{"Start":"00:20.490 ","End":"00:22.410","Text":"We\u0027re going to use the equation,"},{"Start":"00:22.410 ","End":"00:31.690","Text":"K_p equals K_c times the standard concentration times RT."},{"Start":"00:31.690 ","End":"00:37.995","Text":"R is the gas constant and T is the temperature divided by the standard pressure."},{"Start":"00:37.995 ","End":"00:41.810","Text":"This is raised to the power of Delta n gas,"},{"Start":"00:41.810 ","End":"00:47.800","Text":"meaning the difference in the number of moles between the products and"},{"Start":"00:47.800 ","End":"00:54.670","Text":"the reactants of the gaseous products, or reactants."},{"Start":"00:54.670 ","End":"00:58.165","Text":"Meaning, if we look at our reaction here,"},{"Start":"00:58.165 ","End":"01:03.402","Text":"the potassium dioxide is solid and so is the potassium carbonate, it\u0027s solid."},{"Start":"01:03.402 ","End":"01:07.900","Text":"Therefore, we\u0027re only going to look at the carbon dioxide and"},{"Start":"01:07.900 ","End":"01:12.820","Text":"the oxygen when we calculate the Delta n. Now first of all,"},{"Start":"01:12.820 ","End":"01:15.940","Text":"we want to calculate K_c and therefore we\u0027re going to divide"},{"Start":"01:15.940 ","End":"01:23.650","Text":"both sides by the parentheses raised to the power of Delta n. K_c"},{"Start":"01:23.650 ","End":"01:28.730","Text":"equals K_p divided by"},{"Start":"01:29.500 ","End":"01:36.260","Text":"the standard concentration times RT divided by the standard pressure,"},{"Start":"01:36.260 ","End":"01:43.665","Text":"raised to the power of Delta n gas."},{"Start":"01:43.665 ","End":"01:49.105","Text":"First of all, let\u0027s just calculate Delta n right here."},{"Start":"01:49.105 ","End":"01:52.790","Text":"Again, of the gaseous products and reactants,"},{"Start":"01:52.790 ","End":"01:56.135","Text":"we\u0027re only looking at the gaseous products and reactants equals."},{"Start":"01:56.135 ","End":"01:58.085","Text":"First of all, we\u0027re going to look at our products."},{"Start":"01:58.085 ","End":"02:02.690","Text":"As we can see, our only gas is our oxygen and we have 3 moles,"},{"Start":"02:02.690 ","End":"02:03.830","Text":"or the coefficient is 3,"},{"Start":"02:03.830 ","End":"02:07.255","Text":"so that\u0027s 3 minus."},{"Start":"02:07.255 ","End":"02:09.455","Text":"Now we\u0027re going to look at our reactants,"},{"Start":"02:09.455 ","End":"02:13.355","Text":"and the only gaseous reactant is the carbon dioxide."},{"Start":"02:13.355 ","End":"02:15.435","Text":"We have 2 moles,"},{"Start":"02:15.435 ","End":"02:16.950","Text":"or again the coefficient is 2,"},{"Start":"02:16.950 ","End":"02:18.720","Text":"so that\u0027s 3 minus 2,"},{"Start":"02:18.720 ","End":"02:20.715","Text":"and that equals 1,"},{"Start":"02:20.715 ","End":"02:24.760","Text":"so our Delta n equals 1."},{"Start":"02:24.860 ","End":"02:27.420","Text":"Let\u0027s see. First of all,"},{"Start":"02:27.420 ","End":"02:29.790","Text":"K_p is given 28.5 at 298k,"},{"Start":"02:29.790 ","End":"02:35.149","Text":"so that\u0027s 28.5 divided"},{"Start":"02:35.149 ","End":"02:41.620","Text":"by our standard concentration is 1 mole per liter."},{"Start":"02:41.620 ","End":"02:49.460","Text":"The gas constant that we use in this case is 8.3145 times 10 to"},{"Start":"02:49.460 ","End":"02:56.190","Text":"the negative 2 liters times bar divided by"},{"Start":"02:56.190 ","End":"03:05.060","Text":"mole times k. This is multiplied by the temperature which equals 298k,"},{"Start":"03:05.060 ","End":"03:08.330","Text":"and this is all divided by the standard pressure,"},{"Start":"03:08.330 ","End":"03:10.440","Text":"which equals 1 bar."},{"Start":"03:10.440 ","End":"03:14.075","Text":"The moles per liter cancel out with liters per mole."},{"Start":"03:14.075 ","End":"03:16.115","Text":"The bar cancels out with bar,"},{"Start":"03:16.115 ","End":"03:19.025","Text":"and the Kelvin cancels out with Kelvin."},{"Start":"03:19.025 ","End":"03:21.020","Text":"This is all raised, remember,"},{"Start":"03:21.020 ","End":"03:23.655","Text":"to the power of 1."},{"Start":"03:23.655 ","End":"03:27.125","Text":"After multiplying and dividing everything,"},{"Start":"03:27.125 ","End":"03:35.330","Text":"K_c comes out to equal 1.15."},{"Start":"03:35.330 ","End":"03:37.745","Text":"Now, as I said in a previous video,"},{"Start":"03:37.745 ","End":"03:42.380","Text":"you might see cases where the standard pressure is taken as 1 atmosphere"},{"Start":"03:42.380 ","End":"03:46.850","Text":"and therefore the gas constant is taken with atmosphere also."},{"Start":"03:46.850 ","End":"03:49.685","Text":"Since 1 bar and 1 atmosphere are very close,"},{"Start":"03:49.685 ","End":"03:53.250","Text":"K_c will also come out very close."},{"Start":"03:53.270 ","End":"03:57.900","Text":"K_c equals 1.15 at 298k."},{"Start":"03:57.900 ","End":"03:59.075","Text":"That is our final answer."},{"Start":"03:59.075 ","End":"04:01.560","Text":"Thank you very much for watching."}],"ID":30466},{"Watched":false,"Name":"Equilibrium Calculations 1","Duration":"7m 13s","ChapterTopicVideoID":26287,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.120 ","End":"00:05.200","Text":"In previous videos, we talked about equilibrium constants."},{"Start":"00:05.200 ","End":"00:07.810","Text":"In this video and the next 3,"},{"Start":"00:07.810 ","End":"00:13.315","Text":"we\u0027ll give examples of calculations involving these constants."},{"Start":"00:13.315 ","End":"00:22.700","Text":"In equilibrium calculations, we often use a table called an ICE table, ICE."},{"Start":"00:23.450 ","End":"00:29.600","Text":"I refers to the initial concentrations given in the problem,"},{"Start":"00:29.670 ","End":"00:35.884","Text":"and C refers to the change in the concentrations at equilibrium,"},{"Start":"00:35.884 ","End":"00:40.800","Text":"and E refers to the equilibrium concentrations,"},{"Start":"00:40.800 ","End":"00:43.870","Text":"so we have ICE."},{"Start":"00:44.090 ","End":"00:49.799","Text":"Now there are different types of problems you may be asked to solve."},{"Start":"00:50.200 ","End":"00:56.015","Text":"The first thing to note is that you always need a balanced chemical equation."},{"Start":"00:56.015 ","End":"00:59.045","Text":"It doesn\u0027t matter what sort of problem you want to solve,"},{"Start":"00:59.045 ","End":"01:02.930","Text":"the equation should always be balanced."},{"Start":"01:02.930 ","End":"01:07.130","Text":"You might be asked to calculate the equilibrium constant given"},{"Start":"01:07.130 ","End":"01:14.330","Text":"the equilibrium concentrations and we had given examples of this in a previous video,"},{"Start":"01:14.330 ","End":"01:18.800","Text":"or you might be asked to calculate the equilibrium constant"},{"Start":"01:18.800 ","End":"01:24.180","Text":"given 1 initial and 1 equilibrium concentration,"},{"Start":"01:24.260 ","End":"01:28.790","Text":"or to calculate the equilibrium concentrations given"},{"Start":"01:28.790 ","End":"01:33.750","Text":"the equilibrium constant and 1 initial concentration."},{"Start":"01:33.820 ","End":"01:39.860","Text":"We will solve a problem of this type in this video,"},{"Start":"01:39.860 ","End":"01:45.390","Text":"and 1 of this type in the next video."},{"Start":"01:46.150 ","End":"01:50.959","Text":"In this video, we\u0027re going to calculate the equilibrium"},{"Start":"01:50.959 ","End":"01:57.150","Text":"constant given 1 initial and 1 equilibrium concentration."},{"Start":"01:57.610 ","End":"02:07.490","Text":"Here\u0027s the problem. Consider the reaction 2 NOCl in equilibrium with 2 NO and Cl_2,"},{"Start":"02:07.490 ","End":"02:09.430","Text":"and these are all gases."},{"Start":"02:09.430 ","End":"02:15.940","Text":"Initially, we have 1 mole of NOCl in a vessel of volume 2 liters,"},{"Start":"02:15.940 ","End":"02:19.145","Text":"and the temperature is 400 Kelvin."},{"Start":"02:19.145 ","End":"02:25.655","Text":"At equilibrium, there were 0.056 moles of CO_2."},{"Start":"02:25.655 ","End":"02:29.220","Text":"We\u0027re asked to calculate K_c."},{"Start":"02:30.760 ","End":"02:35.900","Text":"Here\u0027s our ICE table, I, C,"},{"Start":"02:35.900 ","End":"02:43.580","Text":"E, and we have 3 columns for the 3 ingredients in the reaction."},{"Start":"02:43.580 ","End":"02:45.740","Text":"NOCl which is reactant,"},{"Start":"02:45.740 ","End":"02:50.260","Text":"NO and Cl_2, which are products."},{"Start":"02:50.260 ","End":"02:56.055","Text":"Now we\u0027re told that initially we have 1 mole of NOCl,"},{"Start":"02:56.055 ","End":"02:58.515","Text":"0 moles of NO,"},{"Start":"02:58.515 ","End":"03:01.480","Text":"and 0 moles of CO_2."},{"Start":"03:01.820 ","End":"03:05.175","Text":"Then we\u0027re told that at equilibrium,"},{"Start":"03:05.175 ","End":"03:10.780","Text":"we have 0.056 moles of CO_2."},{"Start":"03:11.590 ","End":"03:17.315","Text":"If we have 0.056 moles of CO_2 at equilibrium,"},{"Start":"03:17.315 ","End":"03:24.960","Text":"the change from 0 moles is +0.056 moles."},{"Start":"03:25.130 ","End":"03:29.550","Text":"Now, we know the change in the Cl_ 2,"},{"Start":"03:29.550 ","End":"03:33.555","Text":"we can see what the change will be in NO and NOCl."},{"Start":"03:33.555 ","End":"03:40.005","Text":"Now, NO is 2 moles for every 1 mole of Cl_2,"},{"Start":"03:40.005 ","End":"03:44.800","Text":"so the change will be twice 0.056 moles."},{"Start":"03:44.800 ","End":"03:49.670","Text":"Here it is, +0.112 moles,"},{"Start":"03:49.670 ","End":"03:55.605","Text":"and NOCl also has 2 here,"},{"Start":"03:55.605 ","End":"04:00.000","Text":"2 moles, so it will be the same as NO,"},{"Start":"04:00.000 ","End":"04:01.905","Text":"but with the opposite sign,"},{"Start":"04:01.905 ","End":"04:04.870","Text":"because NOCl is reacting,"},{"Start":"04:04.870 ","End":"04:08.690","Text":"the number of moles will decrease,"},{"Start":"04:08.690 ","End":"04:14.980","Text":"so we have -0.112 moles for NOCl,"},{"Start":"04:14.980 ","End":"04:18.435","Text":"and +0.112 moles for NO,"},{"Start":"04:18.435 ","End":"04:23.090","Text":"and +0.056 moles for Cl_2."},{"Start":"04:23.090 ","End":"04:27.415","Text":"Now we can calculate what happens at equilibrium."},{"Start":"04:27.415 ","End":"04:33.445","Text":"We will have 1 mole minus 0.112 moles for NOCl."},{"Start":"04:33.445 ","End":"04:38.783","Text":"That\u0027s 0.888 moles."},{"Start":"04:38.783 ","End":"04:40.845","Text":"For NO,"},{"Start":"04:40.845 ","End":"04:45.210","Text":"we\u0027ll have 0 plus 0.112 moles."},{"Start":"04:45.210 ","End":"04:51.370","Text":"That gives us, at equilibrium, 0.112 moles."},{"Start":"04:52.040 ","End":"04:55.255","Text":"Now, we need the concentrations,"},{"Start":"04:55.255 ","End":"04:58.220","Text":"and we know that the volume is 2 liters,"},{"Start":"04:58.220 ","End":"05:04.865","Text":"so we can calculate the concentration at equilibrium to be"},{"Start":"05:04.865 ","End":"05:11.990","Text":"0.888 moles divided by 2 giving us 0.444 M,"},{"Start":"05:11.990 ","End":"05:14.650","Text":"that\u0027s the molar concentration."},{"Start":"05:14.650 ","End":"05:21.281","Text":"For NO, 0.112 moles divided by the volume,"},{"Start":"05:21.281 ","End":"05:26.515","Text":"giving us the molar concentration of 0.056."},{"Start":"05:26.515 ","End":"05:33.800","Text":"For Cl_2, we have 0.056 moles divided by 2 liters,"},{"Start":"05:33.800 ","End":"05:36.215","Text":"to give us the molar concentration"},{"Start":"05:36.215 ","End":"05:46.220","Text":"0.028 M. Now we have the 3 concentrations at equilibrium."},{"Start":"05:46.220 ","End":"05:49.700","Text":"Now K_c is equal to NO,"},{"Start":"05:49.700 ","End":"05:54.050","Text":"the concentration of NO^2 because of the 2 here,"},{"Start":"05:54.050 ","End":"05:57.215","Text":"times the concentration of Cl_2,"},{"Start":"05:57.215 ","End":"06:00.680","Text":"divided by the concentration of NOCL,"},{"Start":"06:00.680 ","End":"06:04.100","Text":"and this is squared because of the 2 here."},{"Start":"06:04.100 ","End":"06:12.070","Text":"Now we can substitute the values we found in the final row of this table."},{"Start":"06:12.070 ","End":"06:20.280","Text":"For NO, we have 0.056 and that\u0027s squared."},{"Start":"06:20.280 ","End":"06:27.525","Text":"For Cl_2, we have 0.028,"},{"Start":"06:27.525 ","End":"06:29.800","Text":"and for NOCl,"},{"Start":"06:29.800 ","End":"06:34.894","Text":"we have 0.444 and that\u0027s also squared."},{"Start":"06:34.894 ","End":"06:37.325","Text":"If we work that out,"},{"Start":"06:37.325 ","End":"06:40.345","Text":"and we forget about the units,"},{"Start":"06:40.345 ","End":"06:44.435","Text":"because K_c is without any dimensions,"},{"Start":"06:44.435 ","End":"06:53.130","Text":"we get 4.5x10^-4."},{"Start":"06:53.130 ","End":"07:03.110","Text":"We\u0027ve calculated Kc to be 4.5x10^-4."},{"Start":"07:03.110 ","End":"07:07.630","Text":"In this video, we solved an equilibrium problem exactly."},{"Start":"07:07.630 ","End":"07:10.430","Text":"In the next 3 videos,"},{"Start":"07:10.430 ","End":"07:13.530","Text":"we\u0027ll solve some more problems."}],"ID":30467},{"Watched":false,"Name":"Equilibrium Calculations 2","Duration":"6m 56s","ChapterTopicVideoID":26288,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.560","Text":"In the previous video,"},{"Start":"00:01.560 ","End":"00:04.725","Text":"we solved an equilibrium problem exactly,"},{"Start":"00:04.725 ","End":"00:09.555","Text":"and we\u0027ll solve another one exactly in this video."},{"Start":"00:09.555 ","End":"00:14.190","Text":"The problem we want to solve is to calculate"},{"Start":"00:14.190 ","End":"00:16.830","Text":"the equilibrium concentrations given"},{"Start":"00:16.830 ","End":"00:21.285","Text":"the initial concentrations and the equilibrium constant."},{"Start":"00:21.285 ","End":"00:30.090","Text":"Here\u0027s a problem. Consider the reaction H_2 plus CO_2 in equilibrium with H_2O and CO,"},{"Start":"00:30.090 ","End":"00:32.385","Text":"and they\u0027re all in the gas phase."},{"Start":"00:32.385 ","End":"00:34.920","Text":"Now, what\u0027s information we\u0027re given?"},{"Start":"00:34.920 ","End":"00:37.205","Text":"We\u0027re given the information that initially,"},{"Start":"00:37.205 ","End":"00:47.925","Text":"the concentrations of CO_2 and H_2 are 0.640 M and K_c is 0.106."},{"Start":"00:47.925 ","End":"00:51.625","Text":"This happens at 700 Kelvin."},{"Start":"00:51.625 ","End":"00:57.710","Text":"We\u0027re asked to calculate all the concentrations at equilibrium."},{"Start":"00:57.710 ","End":"01:02.695","Text":"Once again, we set up our ICE table."},{"Start":"01:02.695 ","End":"01:05.370","Text":"We have 4 columns this time,"},{"Start":"01:05.370 ","End":"01:06.705","Text":"one for H_2,"},{"Start":"01:06.705 ","End":"01:08.080","Text":"one for CO_2,"},{"Start":"01:08.080 ","End":"01:09.315","Text":"one for H_2O,"},{"Start":"01:09.315 ","End":"01:12.284","Text":"and one for CO. What are we told?"},{"Start":"01:12.284 ","End":"01:18.600","Text":"We\u0027re told that H_2 is initially 0.640 M, and CO_2,"},{"Start":"01:18.600 ","End":"01:22.980","Text":"0.640 M. There\u0027s no H_2O,"},{"Start":"01:22.980 ","End":"01:26.680","Text":"so that\u0027s 0 and no CO, so that\u0027s 0."},{"Start":"01:26.680 ","End":"01:30.505","Text":"Now, when we go to equilibrium,"},{"Start":"01:30.505 ","End":"01:34.920","Text":"the concentration of H_2 will decrease,"},{"Start":"01:34.920 ","End":"01:38.180","Text":"so the H_2O and CO can be formed."},{"Start":"01:38.180 ","End":"01:42.500","Text":"Similarly, concentration of CO_2 will decrease."},{"Start":"01:42.500 ","End":"01:47.078","Text":"Here, we write minus x because we have to find out what x is,"},{"Start":"01:47.078 ","End":"01:50.190","Text":"we don\u0027t know by how much it decreased."},{"Start":"01:50.360 ","End":"01:53.505","Text":"Now, if these two decrease,"},{"Start":"01:53.505 ","End":"01:59.890","Text":"and H_2O and CO will increase also by plus x."},{"Start":"02:00.140 ","End":"02:02.370","Text":"Here, we have minus x,"},{"Start":"02:02.370 ","End":"02:06.440","Text":"minus x, plus x, and plus x."},{"Start":"02:06.440 ","End":"02:09.580","Text":"Now, at equilibrium,"},{"Start":"02:09.580 ","End":"02:15.195","Text":"the concentration of H_2 is 0.640 minus x,"},{"Start":"02:15.195 ","End":"02:18.750","Text":"0.640 minus x, and of course,"},{"Start":"02:18.750 ","End":"02:23.115","Text":"this is capital M, it\u0027s a concentration."},{"Start":"02:23.115 ","End":"02:25.890","Text":"Then we go to CO_2, and again,"},{"Start":"02:25.890 ","End":"02:35.040","Text":"0.640 minus x. H_2O is 0 plus x,"},{"Start":"02:35.040 ","End":"02:37.720","Text":"so it\u0027s just plus x."},{"Start":"02:38.240 ","End":"02:43.900","Text":"The same is true for CO, also plus x."},{"Start":"02:45.470 ","End":"02:48.469","Text":"Now, we\u0027re going to have to use"},{"Start":"02:48.469 ","End":"02:52.180","Text":"the solution of a quadratic equation where we work out K_c,"},{"Start":"02:52.180 ","End":"02:56.645","Text":"we\u0027ll see that it involves a quadratic equation."},{"Start":"02:56.645 ","End":"03:00.970","Text":"Just to remind you how to solve these because I\u0027m sure you all know."},{"Start":"03:00.970 ","End":"03:03.605","Text":"If we have a quadratic equation,"},{"Start":"03:03.605 ","End":"03:07.280","Text":"ax^2 plus bx plus c equal to 0,"},{"Start":"03:07.280 ","End":"03:11.720","Text":"then the roots of this equation are x equal to minus"},{"Start":"03:11.720 ","End":"03:16.535","Text":"b plus or minus the square root of b^2 minus 4ac,"},{"Start":"03:16.535 ","End":"03:18.740","Text":"all divided by 2a."},{"Start":"03:18.740 ","End":"03:23.180","Text":"Now, you may be lucky enough that your university allows"},{"Start":"03:23.180 ","End":"03:27.140","Text":"you to use a calculator to solve quadratic equations."},{"Start":"03:27.140 ","End":"03:32.570","Text":"If not, I\u0027m afraid you\u0027ll just have to use this long way of doing it."},{"Start":"03:32.570 ","End":"03:36.785","Text":"Now, we can write the expression for K_c."},{"Start":"03:36.785 ","End":"03:43.220","Text":"K_c is equal to the concentration of H_2O times the concentration of CO,"},{"Start":"03:43.220 ","End":"03:49.580","Text":"divided by the concentration of H_2 multiplied by the concentration of CO_2."},{"Start":"03:49.580 ","End":"03:53.589","Text":"Now, we can substitute these values."},{"Start":"03:53.589 ","End":"03:57.380","Text":"We have H_2O which is x, CO is x,"},{"Start":"03:57.380 ","End":"04:05.900","Text":"we have x^2 in the numerator and H_2 and CO_2 are both 0.640 minus x."},{"Start":"04:05.900 ","End":"04:10.375","Text":"That\u0027s 0.640 minus x^2."},{"Start":"04:10.375 ","End":"04:12.390","Text":"We know the value of K_c,"},{"Start":"04:12.390 ","End":"04:17.165","Text":"that\u0027s given in the problem is 0.106."},{"Start":"04:17.165 ","End":"04:24.520","Text":"Now, we can solve this equation here."},{"Start":"04:25.250 ","End":"04:29.480","Text":"Now, we have to write it out as a quadratic equation."},{"Start":"04:29.480 ","End":"04:38.365","Text":"First of all, we multiply 0.106 by 0.640 minus x^2."},{"Start":"04:38.365 ","End":"04:42.715","Text":"That\u0027s here, and that\u0027s equal to x^2."},{"Start":"04:42.715 ","End":"04:49.635","Text":"When we work out the square and do a little bit of algebra,"},{"Start":"04:49.635 ","End":"04:52.835","Text":"we find this quadratic equation."},{"Start":"04:52.835 ","End":"04:57.230","Text":"Now, when we solve the quadratic equation, first,"},{"Start":"04:57.230 ","End":"05:04.515","Text":"we have to note that a is 0.894 and b is 0.136,"},{"Start":"05:04.515 ","End":"05:09.015","Text":"and c is minus 0.0435,"},{"Start":"05:09.015 ","End":"05:19.020","Text":"and the roots are then x equal to minus 0.309 and 0.157."},{"Start":"05:19.020 ","End":"05:22.119","Text":"Now, when we look here,"},{"Start":"05:22.119 ","End":"05:29.315","Text":"we see that the concentration of H_2 has to decrease,"},{"Start":"05:29.315 ","End":"05:32.425","Text":"so x must be positive."},{"Start":"05:32.425 ","End":"05:34.929","Text":"If x has to be positive,"},{"Start":"05:34.929 ","End":"05:37.765","Text":"means only 1 root is relevant,"},{"Start":"05:37.765 ","End":"05:41.050","Text":"x is equal to 0.157."},{"Start":"05:41.050 ","End":"05:47.920","Text":"That means that the concentration of H_2O and CO are equal to 0.157."},{"Start":"05:47.920 ","End":"05:52.120","Text":"We could either write M or not because in the end,"},{"Start":"05:52.120 ","End":"05:54.130","Text":"we\u0027re going to ignore the units."},{"Start":"05:54.130 ","End":"06:02.955","Text":"The concentration of H_2 and CO_2 is 0.640 minus 0.157 minus x,"},{"Start":"06:02.955 ","End":"06:06.940","Text":"and that\u0027s 0.483, and then again,"},{"Start":"06:06.940 ","End":"06:11.710","Text":"that\u0027s M. I suppose if we\u0027re really being asked concentrations,"},{"Start":"06:11.710 ","End":"06:17.720","Text":"then we need to include the M. If we are going to re-establish this K_c,"},{"Start":"06:17.720 ","End":"06:20.930","Text":"if we\u0027re going to use these values to calculate K_c,"},{"Start":"06:20.930 ","End":"06:23.510","Text":"then we don\u0027t need the Ms. Now,"},{"Start":"06:23.510 ","End":"06:30.145","Text":"we can check the result simply by substituting these values,"},{"Start":"06:30.145 ","End":"06:40.705","Text":"and we get 0.157^2 divided by 0.483^2 and we see that it\u0027s indeed 0.106."},{"Start":"06:40.705 ","End":"06:44.930","Text":"This is just a check that will solve the problem correctly."},{"Start":"06:44.930 ","End":"06:49.180","Text":"In this video, we solved an equilibrium problem exactly,"},{"Start":"06:49.180 ","End":"06:51.260","Text":"and in the next video, we\u0027re going to solve"},{"Start":"06:51.260 ","End":"06:55.800","Text":"an equilibrium problem using an approximation."}],"ID":30468},{"Watched":false,"Name":"Equilibrium Calculations with Approximation 1","Duration":"6m 20s","ChapterTopicVideoID":26294,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.440 ","End":"00:03.260","Text":"In previous 2 videos,"},{"Start":"00:03.260 ","End":"00:06.970","Text":"we solve problems involving equilibrium constants."},{"Start":"00:06.970 ","End":"00:09.655","Text":"In this video and the next,"},{"Start":"00:09.655 ","End":"00:15.190","Text":"we\u0027ll talk about approximations that can simplify these calculations."},{"Start":"00:15.190 ","End":"00:19.870","Text":"We\u0027re going to talk about approximations that can be used when"},{"Start":"00:19.870 ","End":"00:25.045","Text":"the equilibrium constant is either very small or very large."},{"Start":"00:25.045 ","End":"00:26.920","Text":"Just to remind us,"},{"Start":"00:26.920 ","End":"00:29.020","Text":"if K is very small,"},{"Start":"00:29.020 ","End":"00:32.920","Text":"there will be much more reactants than products at equilibrium."},{"Start":"00:32.920 ","End":"00:35.575","Text":"On the other hand, if K is very large,"},{"Start":"00:35.575 ","End":"00:38.980","Text":"there will be much more products than reactants at equilibrium."},{"Start":"00:38.980 ","End":"00:43.830","Text":"We\u0027re going to solve a problem of the first sort in this video,"},{"Start":"00:43.830 ","End":"00:49.240","Text":"and a problem of the second sort in the next video."},{"Start":"00:49.240 ","End":"00:53.240","Text":"We\u0027re going to solve an example in which we have to calculate"},{"Start":"00:53.240 ","End":"00:56.270","Text":"the equilibrium concentrations given"},{"Start":"00:56.270 ","End":"01:02.030","Text":"1 initial concentration and given a very small equilibrium constant."},{"Start":"01:02.030 ","End":"01:03.860","Text":"Here\u0027s a sample problem."},{"Start":"01:03.860 ","End":"01:11.350","Text":"Consider the reaction 2 H_2S in equilibrium with 2 H_2 and S_2,"},{"Start":"01:11.350 ","End":"01:13.280","Text":"and they\u0027re all gases."},{"Start":"01:13.280 ","End":"01:18.050","Text":"Initially, the concentration of H_2S is 0.600"},{"Start":"01:18.050 ","End":"01:24.960","Text":"M. K_c is 4.20 times 10^-6."},{"Start":"01:24.960 ","End":"01:30.170","Text":"This is at a temperature of 1,103 Kelvin."},{"Start":"01:30.170 ","End":"01:35.695","Text":"We\u0027re asked to calculate all the concentrations at equilibrium."},{"Start":"01:35.695 ","End":"01:39.660","Text":"Here\u0027s our table. We have I,"},{"Start":"01:39.660 ","End":"01:43.170","Text":"C and E. We have H_2S,"},{"Start":"01:43.170 ","End":"01:46.715","Text":"reactant, H_2 and S_2, the products."},{"Start":"01:46.715 ","End":"01:54.335","Text":"Now initially, we\u0027re given that the concentration of H_2S is 0.600 molar."},{"Start":"01:54.335 ","End":"01:56.960","Text":"There\u0027s no H_2 or S_2,"},{"Start":"01:56.960 ","End":"01:59.455","Text":"so the concentration is 0."},{"Start":"01:59.455 ","End":"02:04.250","Text":"Now, we need to change and we\u0027re going to write to change"},{"Start":"02:04.250 ","End":"02:09.095","Text":"the H_2S as -2x because the concentration,"},{"Start":"02:09.095 ","End":"02:13.790","Text":"which 2S is going to decrease as we approach equilibrium."},{"Start":"02:13.790 ","End":"02:17.490","Text":"The 2 is because of the 2 moles here."},{"Start":"02:17.490 ","End":"02:20.400","Text":"We have -2x,"},{"Start":"02:20.400 ","End":"02:24.990","Text":"an H_2 will increase by +2x,"},{"Start":"02:24.990 ","End":"02:27.910","Text":"an S_2 just by x."},{"Start":"02:28.160 ","End":"02:34.350","Text":"The equilibrium when I have 0.600 -2x,"},{"Start":"02:34.350 ","End":"02:40.090","Text":"we have 2x for H_2 and x for S_2."},{"Start":"02:40.580 ","End":"02:43.560","Text":"Now we can write the K_c,"},{"Start":"02:43.560 ","End":"02:49.350","Text":"which is equal to the concentration of H_2^2 times S_2 divide by"},{"Start":"02:49.350 ","End":"02:56.295","Text":"the concentration of H_2S^2 is equal to 2x, all squared."},{"Start":"02:56.295 ","End":"03:01.680","Text":"Here we have 2x times x with S_2 divided by"},{"Start":"03:01.680 ","End":"03:09.340","Text":"0.600 minus 2x all squared for the H_2S^2."},{"Start":"03:09.340 ","End":"03:16.325","Text":"We\u0027re told that that\u0027s equal to 4.20 times 10^-6."},{"Start":"03:16.325 ","End":"03:18.590","Text":"Now if we multiply this out,"},{"Start":"03:18.590 ","End":"03:22.680","Text":"we see this as 4x^3."},{"Start":"03:23.510 ","End":"03:27.610","Text":"That means we have a cubic equation."},{"Start":"03:31.310 ","End":"03:37.055","Text":"Now assuming that 2x is much smaller than 0.600,"},{"Start":"03:37.055 ","End":"03:41.860","Text":"that\u0027s equivalent to say it\u0027s only a small amount of reactants reacted."},{"Start":"03:41.860 ","End":"03:51.135","Text":"Then we can write that K_c is equal to 4x^3 divided by 0.600^2."},{"Start":"03:51.135 ","End":"03:56.000","Text":"We know that K_c is 4.20 times 10^-6."},{"Start":"03:56.000 ","End":"04:04.375","Text":"So from this, we can conclude the x^3 is equal to 0.378 times 10^-6,"},{"Start":"04:04.375 ","End":"04:08.005","Text":"then x will be the cube root of this,"},{"Start":"04:08.005 ","End":"04:12.705","Text":"which is 0.723 times 10^-2."},{"Start":"04:12.705 ","End":"04:16.865","Text":"Now that\u0027s approximately 0.007,"},{"Start":"04:16.865 ","End":"04:22.290","Text":"which is certainly a good bit smaller than 0.600."},{"Start":"04:22.660 ","End":"04:27.050","Text":"Now we can write the concentrations at equilibrium."},{"Start":"04:27.050 ","End":"04:33.335","Text":"We have H_2S will be 0.600 minus 2x."},{"Start":"04:33.335 ","End":"04:36.125","Text":"That\u0027s 0.586."},{"Start":"04:36.125 ","End":"04:40.235","Text":"The concentration of H_2 will be 2x."},{"Start":"04:40.235 ","End":"04:43.400","Text":"That\u0027s twice 0.007,"},{"Start":"04:43.400 ","End":"04:52.980","Text":"that 0.014, the concentration of S_2 will be x, which is 0.007."},{"Start":"04:52.980 ","End":"04:56.105","Text":"Of course, if we\u0027re asked for concentration,"},{"Start":"04:56.105 ","End":"05:01.610","Text":"we should write the capital M. Now we can check whether this result is"},{"Start":"05:01.610 ","End":"05:08.790","Text":"accurate by calculating K_c substituting the value of x."},{"Start":"05:08.790 ","End":"05:18.810","Text":"We have 4x^3 equal to 4 times 0.723 times"},{"Start":"05:18.810 ","End":"05:23.700","Text":"10^-2 all cubed divided by"},{"Start":"05:23.700 ","End":"05:30.935","Text":"0.600 minus 2x^2, and that\u0027s 0.586^2."},{"Start":"05:30.935 ","End":"05:36.815","Text":"Work that out. We get 4.4 times 10^-6."},{"Start":"05:36.815 ","End":"05:40.205","Text":"Now, is this a reasonable approximation?"},{"Start":"05:40.205 ","End":"05:45.835","Text":"We can compare the value we got for K_c with the value given in the problem."},{"Start":"05:45.835 ","End":"05:52.685","Text":"We have 4.40, that\u0027s the value we got just now, minus 4.20."},{"Start":"05:52.685 ","End":"05:57.050","Text":"The value we got to the problem, divide by 4.20."},{"Start":"05:57.050 ","End":"06:01.210","Text":"Of course, 10^-6 cancels top and bottom."},{"Start":"06:01.210 ","End":"06:05.990","Text":"That\u0027s multiplied by a 100 percent to get the percentage error."},{"Start":"06:05.990 ","End":"06:08.960","Text":"That works out as 4.8 percent,"},{"Start":"06:08.960 ","End":"06:11.540","Text":"so means it\u0027s not a wonderful approximation."},{"Start":"06:11.540 ","End":"06:14.585","Text":"But for many purposes it\u0027s sufficient."},{"Start":"06:14.585 ","End":"06:20.460","Text":"In this video, we solved an equilibrium problem using an approximation."}],"ID":30469},{"Watched":false,"Name":"Equilibrium Calculations with Approximations 2","Duration":"5m 20s","ChapterTopicVideoID":26284,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.270 ","End":"00:02.710","Text":"In the previous video,"},{"Start":"00:02.710 ","End":"00:07.150","Text":"we solved an equilibrium problem using an approximation."},{"Start":"00:07.150 ","End":"00:13.550","Text":"This video we\u0027ll solve another problem using an approximation."},{"Start":"00:13.860 ","End":"00:19.285","Text":"This time we\u0027re going to calculate equilibrium concentrations given"},{"Start":"00:19.285 ","End":"00:25.075","Text":"the initial concentration and a very large equilibrium constant."},{"Start":"00:25.075 ","End":"00:29.170","Text":"The strategy we\u0027re going to use is to assume first that"},{"Start":"00:29.170 ","End":"00:34.390","Text":"the reaction goes to completion and after that to equilibrium."},{"Start":"00:34.390 ","End":"00:36.355","Text":"Here\u0027s an example."},{"Start":"00:36.355 ","End":"00:44.500","Text":"Consider the reaction 2NO plus CL-2 in equilibrium with 2NOCL, all in the gas phase."},{"Start":"00:44.500 ","End":"00:51.050","Text":"We\u0027re given this initially the concentration of NO 0.100M,"},{"Start":"00:51.050 ","End":"00:55.070","Text":"and that\u0027s of CL-2 is 0.050M,"},{"Start":"00:55.070 ","End":"01:04.850","Text":"and K_c is 3.70 times 10^8 at 298 Kelvin."},{"Start":"01:04.850 ","End":"01:11.120","Text":"We\u0027re asked to calculate all the concentrations at equilibrium."},{"Start":"01:11.120 ","End":"01:13.940","Text":"Here is our table,"},{"Start":"01:13.940 ","End":"01:20.574","Text":"our ICE table, but we\u0027ve got an extra line, called completion."},{"Start":"01:20.574 ","End":"01:27.085","Text":"Initially, the concentration NO 0.100,"},{"Start":"01:27.085 ","End":"01:30.845","Text":"and of CL-2 0.050,"},{"Start":"01:30.845 ","End":"01:34.180","Text":"and of NO-CL 0."},{"Start":"01:34.180 ","End":"01:37.140","Text":"Now if it goes to completion,"},{"Start":"01:37.140 ","End":"01:39.880","Text":"all the NO will react,"},{"Start":"01:39.880 ","End":"01:46.505","Text":"it\u0027s exactly twice the number of moles as CL-2,"},{"Start":"01:46.505 ","End":"01:50.430","Text":"so they\u0027re Stoichiometric proportions."},{"Start":"01:55.910 ","End":"01:58.230","Text":"All the NO will react,"},{"Start":"01:58.230 ","End":"02:00.675","Text":"all the CL-2 will react,"},{"Start":"02:00.675 ","End":"02:04.930","Text":"so we\u0027ll be left with 00."},{"Start":"02:05.300 ","End":"02:10.470","Text":"Just as NO starts with 0.100 M,"},{"Start":"02:10.470 ","End":"02:18.645","Text":"NO-CL at completion will be 0.100M because here we have 2 moles and here we have 2 moles."},{"Start":"02:18.645 ","End":"02:23.550","Text":"Now we\u0027re going to step back from completion."},{"Start":"02:23.550 ","End":"02:28.140","Text":"NO-CL will be reduced by -2x."},{"Start":"02:28.140 ","End":"02:30.915","Text":"So the change is -2x."},{"Start":"02:30.915 ","End":"02:38.400","Text":"CL-2 will increase by x and NO will increase by 2x."},{"Start":"02:38.740 ","End":"02:42.410","Text":"Then at equilibrium we have plus"},{"Start":"02:42.410 ","End":"02:48.980","Text":"2x plus x"},{"Start":"02:48.980 ","End":"02:54.540","Text":"and 0.100 minus 2x."},{"Start":"02:56.060 ","End":"03:06.330","Text":"Now if we assume that x is much smaller than 0.050, in other words,"},{"Start":"03:06.330 ","End":"03:14.200","Text":"that 2x is much smaller than 0.100,"},{"Start":"03:16.340 ","End":"03:21.510","Text":"then instead of writing 0.100 minus"},{"Start":"03:21.510 ","End":"03:29.330","Text":"2x^2 for NOCl^2 and of course we just have 4x^3."},{"Start":"03:29.330 ","End":"03:31.625","Text":"There\u0027s no simplification here."},{"Start":"03:31.625 ","End":"03:42.415","Text":"For NO is 2x=4x^2 and CL-2 is x=4x^2 and x."},{"Start":"03:42.415 ","End":"03:47.355","Text":"Instead of writing 0.100 minus 2x,"},{"Start":"03:47.355 ","End":"03:52.390","Text":"we simplify that and write 0.100^2."},{"Start":"03:52.550 ","End":"04:00.345","Text":"We ignore the 2x and we know that K-c is 3.70 times 10^8,"},{"Start":"04:00.345 ","End":"04:08.070","Text":"then we can work out the x^3 is equal to 6.76 times 10^-12,"},{"Start":"04:08.070 ","End":"04:13.145","Text":"and then x is the cube root of that."},{"Start":"04:13.145 ","End":"04:18.000","Text":"That\u0027s 1.89 times 10^-4."},{"Start":"04:19.010 ","End":"04:24.610","Text":"That\u0027s certainly much smaller than 0.050."},{"Start":"04:25.960 ","End":"04:30.250","Text":"Let\u0027s look at the equilibrium concentrations."},{"Start":"04:30.250 ","End":"04:34.125","Text":"NO at equilibrium is supposed to be 2x."},{"Start":"04:34.125 ","End":"04:39.815","Text":"That\u0027s 2 times 1.89 times 10^-4."},{"Start":"04:39.815 ","End":"04:46.110","Text":"That\u0027s 3.78 times 10^-4."},{"Start":"04:46.110 ","End":"04:49.065","Text":"CL-2 has a concentration x,"},{"Start":"04:49.065 ","End":"04:54.240","Text":"so that\u0027s 1.89 times 10^-4."},{"Start":"04:55.360 ","End":"05:02.865","Text":"The concentration of NO-CL is 0.100 minus 3.78 times"},{"Start":"05:02.865 ","End":"05:11.795","Text":"10^-4 and that\u0027s just 0.100 to that same number of significant figures."},{"Start":"05:11.795 ","End":"05:15.290","Text":"In this video, we solved an equilibrium problem"},{"Start":"05:15.290 ","End":"05:20.850","Text":"approximately for the case of a large equilibrium constant."}],"ID":30470},{"Watched":false,"Name":"Exercise 4","Duration":"3m 46s","ChapterTopicVideoID":28918,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.280","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.280 ","End":"00:03.945","Text":"Given the following reaction,"},{"Start":"00:03.945 ","End":"00:08.550","Text":"and we can see that on the reactant side we have nitrogen plus 3 hydrogen gas,"},{"Start":"00:08.550 ","End":"00:11.250","Text":"and on our product side we have 2 ammonia,"},{"Start":"00:11.250 ","End":"00:17.010","Text":"and we\u0027re given the equilibrium constant K_p equals 4.34 times 10 to the minus 3 at"},{"Start":"00:17.010 ","End":"00:23.460","Text":"300 K. What is the value of the equilibrium constant K_p for the following reaction?"},{"Start":"00:23.460 ","End":"00:25.980","Text":"If we look at the reaction given,"},{"Start":"00:25.980 ","End":"00:29.310","Text":"we can see that it\u0027s actually the reverse of the first reaction."},{"Start":"00:29.310 ","End":"00:31.869","Text":"We have on our reactant side 2 ammonia"},{"Start":"00:31.869 ","End":"00:35.070","Text":"and then on our product side we have nitrogen and 3 hydrogen."},{"Start":"00:35.070 ","End":"00:40.685","Text":"When we reverse our reaction and we want to calculate our new equilibrium constant,"},{"Start":"00:40.685 ","End":"00:43.115","Text":"we\u0027ll just call that K_p1."},{"Start":"00:43.115 ","End":"00:49.265","Text":"All we do is we take our previous equilibrium constant K_p."},{"Start":"00:49.265 ","End":"00:55.620","Text":"We start with 1 divided by K_p,"},{"Start":"00:55.620 ","End":"00:59.620","Text":"our previous equilibrium constant."},{"Start":"00:59.780 ","End":"01:06.780","Text":"In this case, this equals 1 divided by 4.34"},{"Start":"01:06.780 ","End":"01:17.330","Text":"times 10 to the minus 3, which equals 230.41."},{"Start":"01:17.330 ","End":"01:20.090","Text":"Now, to explain why the new equilibrium constant equals"},{"Start":"01:20.090 ","End":"01:23.650","Text":"1 divided by our previous equilibrium constant,"},{"Start":"01:23.650 ","End":"01:25.680","Text":"let\u0027s take a look at our reaction."},{"Start":"01:25.680 ","End":"01:29.025","Text":"Again, we have ammonia and 3 hydrogen on our reactant side."},{"Start":"01:29.025 ","End":"01:32.900","Text":"Again, we have nitrogen and 3 hydrogen on our reactant side,"},{"Start":"01:32.900 ","End":"01:35.300","Text":"and we have 2 ammonia on our product side."},{"Start":"01:35.300 ","End":"01:39.540","Text":"Our K_p for this reaction is going to"},{"Start":"01:39.540 ","End":"01:45.210","Text":"be the pressure of ammonia raised to the second power,"},{"Start":"01:45.210 ","End":"01:47.035","Text":"since the coefficient is 2."},{"Start":"01:47.035 ","End":"01:50.840","Text":"Remember, we begin with our product side divided by our reactant side,"},{"Start":"01:50.840 ","End":"01:52.310","Text":"which is nitrogen,"},{"Start":"01:52.310 ","End":"01:54.650","Text":"so that\u0027s the pressure of nitrogen."},{"Start":"01:54.650 ","End":"01:59.750","Text":"The coefficient is 1 times the pressure of hydrogen raised to the third power,"},{"Start":"01:59.750 ","End":"02:01.610","Text":"since our coefficient is 3."},{"Start":"02:01.610 ","End":"02:04.400","Text":"That\u0027s our K_p for our first reaction."},{"Start":"02:04.400 ","End":"02:05.975","Text":"Now, as I said,"},{"Start":"02:05.975 ","End":"02:07.569","Text":"our secondary action,"},{"Start":"02:07.569 ","End":"02:09.170","Text":"we\u0027re going to call the equilibrium constant"},{"Start":"02:09.170 ","End":"02:13.610","Text":"K_p1 and this time on our product side we have nitrogen and hydrogen."},{"Start":"02:13.610 ","End":"02:17.120","Text":"That\u0027s going to be the pressure of nitrogen raised to"},{"Start":"02:17.120 ","End":"02:23.350","Text":"the first power times the pressure of hydrogen raised to the third power,"},{"Start":"02:23.350 ","End":"02:25.750","Text":"divided by our reactant side,"},{"Start":"02:25.750 ","End":"02:30.280","Text":"which is the pressure of the ammonia raised to the second power,"},{"Start":"02:30.280 ","End":"02:32.545","Text":"since our coefficient is 2."},{"Start":"02:32.545 ","End":"02:35.770","Text":"If you take a look at our first equilibrium constant K_p and our"},{"Start":"02:35.770 ","End":"02:39.220","Text":"second equilibrium constant for the reverse reaction K_p1,"},{"Start":"02:39.220 ","End":"02:43.240","Text":"we can see that what we had in our numerator actually is now in"},{"Start":"02:43.240 ","End":"02:47.484","Text":"our denominator and what we had in our denominator is now in our numerator."},{"Start":"02:47.484 ","End":"02:51.565","Text":"All we need to do is to divide K_p by 1,"},{"Start":"02:51.565 ","End":"02:56.395","Text":"and we will reach our new equilibrium constant K_p1."},{"Start":"02:56.395 ","End":"03:04.160","Text":"All we need to do is take 1 and divide it by K_p in order to reach K_p1."},{"Start":"03:04.160 ","End":"03:08.240","Text":"All we need to do is take our first equilibrium constant, K_p."},{"Start":"03:08.240 ","End":"03:12.140","Text":"We begin with 1 and we divide it by our first equilibrium constant,"},{"Start":"03:12.140 ","End":"03:16.050","Text":"K_p, and this equals the value of K_p1,"},{"Start":"03:16.050 ","End":"03:20.180","Text":"because in order to switch our denominator and numerator,"},{"Start":"03:20.180 ","End":"03:24.065","Text":"all we have to do is take 1 and divide it by K_p."},{"Start":"03:24.065 ","End":"03:28.700","Text":"That\u0027s the reason, again, that in order to get our new equilibrium constant K_p1,"},{"Start":"03:28.700 ","End":"03:33.760","Text":"all we have to do is take 1 divided by our equilibrium constant, K_p."},{"Start":"03:33.760 ","End":"03:36.860","Text":"K_p1, equilibrium constant for"},{"Start":"03:36.860 ","End":"03:43.685","Text":"the reverse reaction equals 1 divided by K_p, which equals 230.41."},{"Start":"03:43.685 ","End":"03:45.080","Text":"That is our final answer."},{"Start":"03:45.080 ","End":"03:47.460","Text":"Thank you very much for watching."}],"ID":30471},{"Watched":false,"Name":"Exercise 5","Duration":"4m 42s","ChapterTopicVideoID":28919,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.700","Text":"Hi. We\u0027re going to solve the following exercise given the following reaction."},{"Start":"00:03.700 ","End":"00:07.808","Text":"On the reactant side we have nitrogen plus 3 hydrogen gas,"},{"Start":"00:07.808 ","End":"00:10.148","Text":"and on our product side we have 2 ammonia,"},{"Start":"00:10.148 ","End":"00:11.954","Text":"and we have our equilibrium constant,"},{"Start":"00:11.954 ","End":"00:15.810","Text":"K_p, which equals 4.34 times 10 to the minus 3 at 300"},{"Start":"00:15.810 ","End":"00:20.865","Text":"K. What is the value of the equilibrium constant K_p for the following reactions?"},{"Start":"00:20.865 ","End":"00:22.740","Text":"Let\u0027s start with 1."},{"Start":"00:22.740 ","End":"00:29.790","Text":"If we take a look at our reaction in 1 and compare it to our first reaction,"},{"Start":"00:29.790 ","End":"00:33.470","Text":"we can see that the coefficients are all multiplied by a common factor."},{"Start":"00:33.470 ","End":"00:36.890","Text":"They\u0027re all multiplied by 2 instead of 1 in front of the nitrogen,"},{"Start":"00:36.890 ","End":"00:39.680","Text":"we have 2, in front of the hydrogen we have 6,"},{"Start":"00:39.680 ","End":"00:42.800","Text":"and in front of 2, in front of the ammonia we have 4."},{"Start":"00:42.800 ","End":"00:47.225","Text":"Again, we multiply the coefficients by a common factor 2."},{"Start":"00:47.225 ","End":"00:52.655","Text":"When this is done, in order to obtain the value of our new equilibrium constant,"},{"Start":"00:52.655 ","End":"00:55.920","Text":"we can call this K_p1."},{"Start":"00:56.050 ","End":"01:02.960","Text":"All we need to do is take our previous equilibrium constant, which, in this case,"},{"Start":"01:02.960 ","End":"01:05.360","Text":"is K_p in arrays,"},{"Start":"01:05.360 ","End":"01:08.960","Text":"the equilibrium constant to the power,"},{"Start":"01:08.960 ","End":"01:10.700","Text":"in this case, of 2."},{"Start":"01:10.700 ","End":"01:16.400","Text":"Again, what we did was we multiplied all our coefficients by a common factor,"},{"Start":"01:16.400 ","End":"01:18.850","Text":"which, in this case, equals 2."},{"Start":"01:18.850 ","End":"01:21.980","Text":"Then we need to take the equilibrium constant,"},{"Start":"01:21.980 ","End":"01:23.570","Text":"our first equilibrium constant,"},{"Start":"01:23.570 ","End":"01:29.210","Text":"and raise it to the power of the number that we multiplied our coefficients by."},{"Start":"01:29.210 ","End":"01:31.760","Text":"In this case, we multiplied our coefficients by 2."},{"Start":"01:31.760 ","End":"01:37.070","Text":"Therefore, our first equilibrium constant will be raised to the power of 2."},{"Start":"01:37.070 ","End":"01:40.640","Text":"If we were to multiply our coefficients by 4,"},{"Start":"01:40.640 ","End":"01:45.000","Text":"we would raise the equilibrium constant to the power of 4."},{"Start":"01:45.340 ","End":"01:53.940","Text":"In this case, we have 4.34 times 10 to the negative 3 and this is squared."},{"Start":"01:53.940 ","End":"02:03.120","Text":"This equals 1.89 times 10 to the minus 5."},{"Start":"02:03.120 ","End":"02:08.370","Text":"K_p1 equals 1.89 times 10 to the minus 5."},{"Start":"02:08.370 ","End":"02:10.685","Text":"Again, this happens when we"},{"Start":"02:10.685 ","End":"02:14.510","Text":"multiply our coefficients by a common factor, a common number,"},{"Start":"02:14.510 ","End":"02:18.950","Text":"and then we take our first equilibrium constant,"},{"Start":"02:18.950 ","End":"02:21.335","Text":"and raise it to the power of that number."},{"Start":"02:21.335 ","End":"02:25.045","Text":"Now, let\u0027s take a look at the second reaction."},{"Start":"02:25.045 ","End":"02:28.000","Text":"In our second reaction,"},{"Start":"02:28.000 ","End":"02:30.920","Text":"if we take a look at our new coefficients,"},{"Start":"02:30.920 ","End":"02:32.880","Text":"we have 2 ways of looking at this."},{"Start":"02:32.880 ","End":"02:40.885","Text":"Either we took our previous coefficients and multiply them all by 1/3,"},{"Start":"02:40.885 ","End":"02:44.370","Text":"our nitrogen had a coefficient of 1,"},{"Start":"02:44.370 ","End":"02:45.735","Text":"so it equals 1/3,"},{"Start":"02:45.735 ","End":"02:48.675","Text":"our hydrogen had a coefficient of 3, so it equals 1."},{"Start":"02:48.675 ","End":"02:50.790","Text":"Since 3 times 1/3 equals 1,"},{"Start":"02:50.790 ","End":"02:57.050","Text":"and our ammonia had a coefficient of 2 and it equals now 2/3 since we multiply it by 1/3."},{"Start":"02:57.050 ","End":"03:02.455","Text":"Again, we can take a look at this as multiplying our coefficients all by 1/3."},{"Start":"03:02.455 ","End":"03:09.140","Text":"In this case, so we multiply all our coefficients by a common factor which equals 1/3,"},{"Start":"03:09.140 ","End":"03:12.365","Text":"we can do the same thing that we did in 1."},{"Start":"03:12.365 ","End":"03:17.425","Text":"K_p2 our new equilibrium constant is going to equal K_p,"},{"Start":"03:17.425 ","End":"03:19.385","Text":"our former equilibrium constant,"},{"Start":"03:19.385 ","End":"03:27.510","Text":"raised to the power of 1/3 and this equals the 3rd root of K_p."},{"Start":"03:27.580 ","End":"03:30.980","Text":"Now, another similar way to look at this"},{"Start":"03:30.980 ","End":"03:34.490","Text":"is instead of multiplying all our coefficients by 1/3,"},{"Start":"03:34.490 ","End":"03:38.080","Text":"we actually took all of our coefficients and divided them by 3."},{"Start":"03:38.080 ","End":"03:40.530","Text":"Now, you can see 1 became 1/3,"},{"Start":"03:40.530 ","End":"03:42.908","Text":"3 became 1 because 3 divided by 3 equals 1,"},{"Start":"03:42.908 ","End":"03:45.435","Text":"and 2 became 2/3."},{"Start":"03:45.435 ","End":"03:47.820","Text":"Then our new equilibrium constant,"},{"Start":"03:47.820 ","End":"03:55.990","Text":"K_p2 equals the 3rd root of K_p."},{"Start":"03:55.990 ","End":"04:00.665","Text":"Again, since we divided our coefficients by a common factor 3,"},{"Start":"04:00.665 ","End":"04:03.980","Text":"the new equilibrium constant equals the 3rd root of K_p."},{"Start":"04:03.980 ","End":"04:07.490","Text":"For example, if we were to divide our coefficients by 2,"},{"Start":"04:07.490 ","End":"04:10.580","Text":"we would have the square root of K_p and so on."},{"Start":"04:10.580 ","End":"04:11.660","Text":"As you can see,"},{"Start":"04:11.660 ","End":"04:14.330","Text":"this is essentially the same and the value,"},{"Start":"04:14.330 ","End":"04:15.680","Text":"obviously, it\u0027s going to come out the same,"},{"Start":"04:15.680 ","End":"04:18.200","Text":"but it\u0027s a different way to look at it."},{"Start":"04:18.200 ","End":"04:21.980","Text":"In both cases, K_p2 equals the 3rd root of K_p,"},{"Start":"04:21.980 ","End":"04:25.145","Text":"so that equals the 3rd root of"},{"Start":"04:25.145 ","End":"04:33.920","Text":"4.34 times 10 to the minus 3 and this equals 0.163,"},{"Start":"04:33.920 ","End":"04:39.065","Text":"so our equilibrium constant K_p2 equals 0.163."},{"Start":"04:39.065 ","End":"04:40.280","Text":"That is our final answer."},{"Start":"04:40.280 ","End":"04:42.720","Text":"Thank you very much for watching."}],"ID":30472},{"Watched":false,"Name":"Exercise 6","Duration":"7m 50s","ChapterTopicVideoID":28920,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.299","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.299 ","End":"00:08.250","Text":"What is the value of the equilibrium constant K_c for the following reaction?"},{"Start":"00:08.250 ","End":"00:09.990","Text":"On our reactant side, we can see that we have"},{"Start":"00:09.990 ","End":"00:13.829","Text":"2 hydrophilic acid and we also have the oxalate ion and on"},{"Start":"00:13.829 ","End":"00:20.745","Text":"our product side we have 2 times the fluoride ion and we also have oxalic acid."},{"Start":"00:20.745 ","End":"00:23.355","Text":"We\u0027re also given 2 different reactions."},{"Start":"00:23.355 ","End":"00:24.450","Text":"We have 1 and 2,"},{"Start":"00:24.450 ","End":"00:27.855","Text":"which we know our equilibrium constants for."},{"Start":"00:27.855 ","End":"00:30.210","Text":"First, what we\u0027re going to do is we\u0027re going to take"},{"Start":"00:30.210 ","End":"00:32.070","Text":"these 2 reactions and we\u0027re going to see what we need"},{"Start":"00:32.070 ","End":"00:35.690","Text":"to do to them in order to achieve our reaction."},{"Start":"00:35.690 ","End":"00:37.350","Text":"If we look at our first reaction,"},{"Start":"00:37.350 ","End":"00:42.705","Text":"we can see that we have a hydrofluoric acid also as a reactant."},{"Start":"00:42.705 ","End":"00:45.410","Text":"If you can see, the reaction that we\u0027re trying to achieve,"},{"Start":"00:45.410 ","End":"00:47.870","Text":"we have 2 times the hydrofluoric acid."},{"Start":"00:47.870 ","End":"00:49.160","Text":"What we can do is first,"},{"Start":"00:49.160 ","End":"00:53.400","Text":"we\u0027re going to try to multiply our first reaction by 2."},{"Start":"00:53.540 ","End":"00:57.965","Text":"We have our first reaction, we\u0027re going to multiply everything by 2."},{"Start":"00:57.965 ","End":"01:00.535","Text":"That\u0027s 2 HF."},{"Start":"01:00.535 ","End":"01:09.715","Text":"Then our product side, we have 2 protons plus 2 times our fluoride ion."},{"Start":"01:09.715 ","End":"01:14.045","Text":"Now, since we multiplied all our coefficients by 2,"},{"Start":"01:14.045 ","End":"01:16.070","Text":"our new equilibrium constant,"},{"Start":"01:16.070 ","End":"01:18.020","Text":"we can call it K_c1,"},{"Start":"01:18.020 ","End":"01:21.990","Text":"is going to be equal to our equilibrium constant K_c."},{"Start":"01:22.610 ","End":"01:26.480","Text":"Now, since we multiplied all our coefficients by 2,"},{"Start":"01:26.480 ","End":"01:28.130","Text":"our new equilibrium constant,"},{"Start":"01:28.130 ","End":"01:35.780","Text":"let\u0027s just call it K_c1* is going to equal our equilibrium constant for this reaction,"},{"Start":"01:35.780 ","End":"01:39.785","Text":"which we\u0027re going to call K_c1 so that we have a difference between"},{"Start":"01:39.785 ","End":"01:44.745","Text":"the first equilibrium constant and the equilibrium constant for the second reaction."},{"Start":"01:44.745 ","End":"01:46.620","Text":"We\u0027re going to call it K_c1."},{"Start":"01:46.620 ","End":"01:50.960","Text":"Remember that since we multiplied all our coefficients by a common factor by 2,"},{"Start":"01:50.960 ","End":"01:53.480","Text":"we need to raise K_c1 by a power of 2,"},{"Start":"01:53.480 ","End":"01:56.365","Text":"meaning this is going to be K_c1^2."},{"Start":"01:56.365 ","End":"01:58.790","Text":"Again, our new equilibrium constant,"},{"Start":"01:58.790 ","End":"02:02.180","Text":"we\u0027re just calling it K_c1* equals K_c1^2."},{"Start":"02:02.180 ","End":"02:06.935","Text":"Now, let\u0027s take a look at our second reaction."},{"Start":"02:06.935 ","End":"02:10.775","Text":"We can see that in our second reaction we have our oxalic acid"},{"Start":"02:10.775 ","End":"02:14.540","Text":"as our reactant and not as our product,"},{"Start":"02:14.540 ","End":"02:16.380","Text":"as in the final reaction."},{"Start":"02:16.380 ","End":"02:22.220","Text":"We can also see that we have oxalate ion in our product rather than in our reactant,"},{"Start":"02:22.220 ","End":"02:25.145","Text":"as we need to have in our final reaction."},{"Start":"02:25.145 ","End":"02:29.670","Text":"Therefore, what we can do is reverse this reaction. Let\u0027s reverse it."},{"Start":"02:29.670 ","End":"02:38.400","Text":"We have 2 times proton plus the oxalate ion."},{"Start":"02:38.400 ","End":"02:40.265","Text":"On our product side,"},{"Start":"02:40.265 ","End":"02:42.995","Text":"we\u0027re going to have the oxalic acid."},{"Start":"02:42.995 ","End":"02:44.705","Text":"Since we reversed the reaction,"},{"Start":"02:44.705 ","End":"02:46.955","Text":"our new equilibrium constant,"},{"Start":"02:46.955 ","End":"02:55.440","Text":"which we\u0027re just going to call K_c2* is going to equal 1 divided by K_c2,"},{"Start":"02:55.440 ","End":"02:58.780","Text":"is 1 divided by the equilibrium constant."},{"Start":"02:58.780 ","End":"03:05.075","Text":"Now, after multiplying our first reaction by 2 and reversing our second reaction,"},{"Start":"03:05.075 ","End":"03:10.110","Text":"we can add 1 to the other in order to achieve our final reaction."},{"Start":"03:10.130 ","End":"03:12.495","Text":"We\u0027re just going to add 1 to another."},{"Start":"03:12.495 ","End":"03:18.885","Text":"We have 2 HF aqueous"},{"Start":"03:18.885 ","End":"03:26.960","Text":"plus 2 protons plus oxalate ion and on our product side,"},{"Start":"03:26.960 ","End":"03:29.960","Text":"we\u0027re going to have 2 H plus,"},{"Start":"03:29.960 ","End":"03:36.570","Text":"plus our 2 fluoride ion plus our oxalic acid."},{"Start":"03:36.650 ","End":"03:43.010","Text":"Right away we can see that our protons cancel out on both sides."},{"Start":"03:43.010 ","End":"03:46.525","Text":"After canceling out our protons,"},{"Start":"03:46.525 ","End":"03:49.770","Text":"we\u0027ll write that reaction again."},{"Start":"03:49.770 ","End":"03:57.650","Text":"What we\u0027re left with is 2 HF plus our oxalate ion and on our product side,"},{"Start":"03:57.650 ","End":"04:07.275","Text":"we have our fluoride anion times 2 plus our oxalic acid."},{"Start":"04:07.275 ","End":"04:10.190","Text":"This is the reaction that we wanted to achieve."},{"Start":"04:10.190 ","End":"04:13.208","Text":"Now, what we did in order to achieve this reaction again,"},{"Start":"04:13.208 ","End":"04:16.540","Text":"first of all, we multiply their first reaction by 2."},{"Start":"04:16.540 ","End":"04:18.770","Text":"This is a reaction that we wanted to achieve."},{"Start":"04:18.770 ","End":"04:20.180","Text":"Again, to achieve this reaction,"},{"Start":"04:20.180 ","End":"04:23.405","Text":"we took our first reaction and multiplied it by 2,"},{"Start":"04:23.405 ","End":"04:25.115","Text":"that common factor of 2."},{"Start":"04:25.115 ","End":"04:29.585","Text":"Therefore, our new equilibrium constant for the first reaction equals"},{"Start":"04:29.585 ","End":"04:35.450","Text":"our equilibrium constant for our first reaction squared or to the power of 2."},{"Start":"04:35.450 ","End":"04:39.220","Text":"Second, we took the second reaction and we reversed it."},{"Start":"04:39.220 ","End":"04:41.120","Text":"Since we\u0027ve reversed the second reaction,"},{"Start":"04:41.120 ","End":"04:45.080","Text":"our equilibrium constant for our second reaction equals 1 divided"},{"Start":"04:45.080 ","End":"04:49.650","Text":"by the equilibrium constant. This is 1 and 2."},{"Start":"04:49.650 ","End":"04:52.025","Text":"The third and after"},{"Start":"04:52.025 ","End":"04:55.063","Text":"multiplying the first reaction by 2 and reversing the second reaction,"},{"Start":"04:55.063 ","End":"05:01.610","Text":"what we did is we added reaction 1 to reaction 2 to achieve our final reaction."},{"Start":"05:01.610 ","End":"05:04.700","Text":"Now, when reactions are added,"},{"Start":"05:04.700 ","End":"05:07.970","Text":"the equilibrium constants are multiplied to obtain"},{"Start":"05:07.970 ","End":"05:12.235","Text":"the new equilibrium constant or the equilibrium constant of the overall reaction."},{"Start":"05:12.235 ","End":"05:15.140","Text":"Now that we added these 2 reactions,"},{"Start":"05:15.140 ","End":"05:17.675","Text":"what we need to do is to multiply"},{"Start":"05:17.675 ","End":"05:20.840","Text":"our new equilibrium constant to"},{"Start":"05:20.840 ","End":"05:24.155","Text":"achieve the equilibrium constant for the overall reaction."},{"Start":"05:24.155 ","End":"05:28.565","Text":"Meaning the equilibrium constant for our overall reaction is going to"},{"Start":"05:28.565 ","End":"05:36.920","Text":"equal K_c1* times K_c2*."},{"Start":"05:36.920 ","End":"05:43.400","Text":"Remember, K_c1 star equals K_c1^2 and K_c1 is the equilibrium constant given for"},{"Start":"05:43.400 ","End":"05:51.255","Text":"reaction 1 and this is multiplied by K_c2* and K_c2* equals 1 divided by K_c2."},{"Start":"05:51.255 ","End":"05:59.040","Text":"Remember K_c2 is the equilibrium constant which is given for our second reaction."},{"Start":"05:59.410 ","End":"06:02.090","Text":"Our first reaction, as we said,"},{"Start":"06:02.090 ","End":"06:06.555","Text":"we multiplied our coefficients by 2."},{"Start":"06:06.555 ","End":"06:09.965","Text":"Therefore, our equilibrium constant,"},{"Start":"06:09.965 ","End":"06:14.585","Text":"which we called K_c1* equals K_c1,"},{"Start":"06:14.585 ","End":"06:16.730","Text":"which is the equilibrium constant for"},{"Start":"06:16.730 ","End":"06:21.785","Text":"the first reaction squared or raised to the second power."},{"Start":"06:21.785 ","End":"06:24.638","Text":"Again, since we multiplied all our coefficients by 2,"},{"Start":"06:24.638 ","End":"06:29.585","Text":"and therefore we need to take our equilibrium constant and raise it to the power of 2."},{"Start":"06:29.585 ","End":"06:31.070","Text":"For the second reaction,"},{"Start":"06:31.070 ","End":"06:35.210","Text":"what we did is reversed the reaction and in reversing the reaction,"},{"Start":"06:35.210 ","End":"06:36.470","Text":"the new equilibrium constant,"},{"Start":"06:36.470 ","End":"06:41.680","Text":"which we called K_c2* equals 1 divided by K_c2."},{"Start":"06:41.680 ","End":"06:44.990","Text":"K_c2 is the equilibrium constant given for"},{"Start":"06:44.990 ","End":"06:50.174","Text":"the second reaction and our new equilibrium constant is 1 divided by this constant,"},{"Start":"06:50.174 ","End":"06:53.615","Text":"again, since we\u0027ve reversed our reaction."},{"Start":"06:53.615 ","End":"06:57.665","Text":"Now, after multiplying our first reaction by 2 and reversing our second reaction,"},{"Start":"06:57.665 ","End":"07:03.250","Text":"I\u0027m reminding you that we added 1 to the other to achieve our overall reaction."},{"Start":"07:03.250 ","End":"07:06.270","Text":"Since we added 1 reaction to the other reaction,"},{"Start":"07:06.270 ","End":"07:09.079","Text":"our equilibrium constant is actually"},{"Start":"07:09.079 ","End":"07:13.825","Text":"the first equilibrium constant multiplied by second equilibrium constant."},{"Start":"07:13.825 ","End":"07:15.705","Text":"This equals, we have K_c1^2,"},{"Start":"07:15.705 ","End":"07:22.560","Text":"K_c1 is 6.8 times 10 to the minus 4."},{"Start":"07:22.560 ","End":"07:27.440","Text":"This is squared, and this is all multiplied by 1 divided by K_c2."},{"Start":"07:27.440 ","End":"07:33.885","Text":"That\u0027s times 1 divided by K_c2 is 3.8 times 10 to the minus 6,"},{"Start":"07:33.885 ","End":"07:39.570","Text":"and this equals 0.121."},{"Start":"07:39.570 ","End":"07:46.415","Text":"Our equilibrium constant for our overall reaction equals 0.121."},{"Start":"07:46.415 ","End":"07:47.930","Text":"That is our final answer."},{"Start":"07:47.930 ","End":"07:50.640","Text":"Thank you very much for watching."}],"ID":30473},{"Watched":false,"Name":"Combining Equilibrium Constants","Duration":"8m 5s","ChapterTopicVideoID":26286,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.910","Text":"In previous videos we defined"},{"Start":"00:02.910 ","End":"00:06.990","Text":"the equilibrium constant for a particular chemical reaction."},{"Start":"00:06.990 ","End":"00:10.050","Text":"For this video, we will learn how the equilibrium"},{"Start":"00:10.050 ","End":"00:14.310","Text":"constant changes when we modify the chemical equation."},{"Start":"00:14.310 ","End":"00:20.140","Text":"The first thing we\u0027re going to do is to change the stoichiometric coefficients."},{"Start":"00:20.140 ","End":"00:27.500","Text":"Now this is the reaction as we\u0027ve written it many times in moles of A and b moles of B in"},{"Start":"00:27.500 ","End":"00:35.160","Text":"equilibrium with c moles of C and d moles of D. Here\u0027s expression we\u0027ll seen many times."},{"Start":"00:35.160 ","End":"00:44.180","Text":"Activity of C^c activity of D^t divided by the activity of A^a times the activity of B^b."},{"Start":"00:44.180 ","End":"00:46.730","Text":"We\u0027re going to call this K_1."},{"Start":"00:46.730 ","End":"00:52.190","Text":"Now, we\u0027re going to multiply the stoichiometric coefficients by N. N"},{"Start":"00:52.190 ","End":"00:58.105","Text":"can be whole number or a fraction it could be positive or negative."},{"Start":"00:58.105 ","End":"01:06.125","Text":"Negative and it can be a fraction like a half or so whatever."},{"Start":"01:06.125 ","End":"01:10.590","Text":"Now we\u0027re looking at the reaction na moles of"},{"Start":"01:10.590 ","End":"01:14.865","Text":"A and nb moles of B and nc moles of C and nd"},{"Start":"01:14.865 ","End":"01:18.430","Text":"moles of D. We are going to write"},{"Start":"01:18.430 ","End":"01:23.750","Text":"the equilibrium constant for this we\u0027re going to call it K_2,"},{"Start":"01:23.750 ","End":"01:25.330","Text":"so again we have a_C,"},{"Start":"01:25.330 ","End":"01:27.370","Text":"but this time it\u0027s the power nc,"},{"Start":"01:27.370 ","End":"01:33.460","Text":"a_D^nd, a_A^na, a_B^nb,"},{"Start":"01:33.460 ","End":"01:39.290","Text":"so it\u0027s like K_1^n"},{"Start":"01:39.290 ","End":"01:45.160","Text":"that means when we multiply the stoichiometric coefficients by a number n,"},{"Start":"01:45.160 ","End":"01:50.690","Text":"we raise the equilibrium constant to"},{"Start":"01:50.690 ","End":"01:57.690","Text":"the power n, two is K_1^n."},{"Start":"01:57.690 ","End":"02:01.940","Text":"Now suppose we reverse the chemical equation."},{"Start":"02:01.940 ","End":"02:07.925","Text":"Instead of having aA and bB as reactions."},{"Start":"02:07.925 ","End":"02:12.530","Text":"Now we have c moles of C and d moles of D as the reactions,"},{"Start":"02:12.530 ","End":"02:19.970","Text":"and a moles of A and b moles of B as the products, so reverse reaction."},{"Start":"02:19.970 ","End":"02:27.935","Text":"Now, K_3 will call it is aA that snow A is now product."},{"Start":"02:27.935 ","End":"02:35.000","Text":"a_A^a times a_B^b divided by a_C^c times a_D^d,"},{"Start":"02:35.000 ","End":"02:39.755","Text":"and that\u0027s like K_1 inverted."},{"Start":"02:39.755 ","End":"02:43.265","Text":"K_3 is 1 over K_1."},{"Start":"02:43.265 ","End":"02:48.110","Text":"That means that when we reverse the chemical equation,"},{"Start":"02:48.110 ","End":"02:55.445","Text":"then we are taking the inverse of the chemical equilibrium constant."},{"Start":"02:55.445 ","End":"02:59.495","Text":"K_3 is equal to 1 over K_1."},{"Start":"02:59.495 ","End":"03:04.460","Text":"Now we\u0027re going to combine the equilibrium constants."},{"Start":"03:04.460 ","End":"03:08.135","Text":"Here we\u0027ll do shortly example."},{"Start":"03:08.135 ","End":"03:13.805","Text":"Combine the equilibrium constants to the following chemical reactions."},{"Start":"03:13.805 ","End":"03:17.390","Text":"We have solid sulfur reacted with oxygen to give"},{"Start":"03:17.390 ","End":"03:21.450","Text":"SO_2 in the gas phase and 2 moles of vessels,"},{"Start":"03:21.450 ","End":"03:27.485","Text":"3 in the gas phase to give 2 moles of SO_2 in the gas phase plus oxygen in the gas phase."},{"Start":"03:27.485 ","End":"03:30.180","Text":"Called his K_1, K_2."},{"Start":"03:30.680 ","End":"03:35.540","Text":"Going to combine these 2 to give the equilibrium constant for"},{"Start":"03:35.540 ","End":"03:39.380","Text":"the reaction 2 S and the solid phase"},{"Start":"03:39.380 ","End":"03:43.790","Text":"plus 3 O_2 in the gas phase giving 2 moles of SO_3 in the gas phase."},{"Start":"03:43.790 ","End":"03:46.470","Text":"I am going to call that K_3."},{"Start":"03:46.640 ","End":"03:51.710","Text":"We want to know how to combine these 2 equations to give this one."},{"Start":"03:51.710 ","End":"03:58.100","Text":"Then, how will we combine K_1 and K_2 to give k _3?"},{"Start":"03:58.100 ","End":"04:02.570","Text":"Now supposing we multiply the first equation by 2,"},{"Start":"04:02.570 ","End":"04:06.815","Text":"and reverse the second equation and then sum them,"},{"Start":"04:06.815 ","End":"04:08.540","Text":"then we\u0027ll get the third equation."},{"Start":"04:08.540 ","End":"04:10.580","Text":"Let\u0027s see if that\u0027s true."},{"Start":"04:10.580 ","End":"04:14.270","Text":"We\u0027re multiplying the first equation by 2,"},{"Start":"04:14.270 ","End":"04:22.270","Text":"so instead of S plus O_2 gives me SO_2 is 2S plus 2O_2 to give 2SO_2,"},{"Start":"04:22.270 ","End":"04:26.050","Text":"so if we call the equilibrium constant K_4,"},{"Start":"04:26.050 ","End":"04:28.705","Text":"that will be K_1 squared."},{"Start":"04:28.705 ","End":"04:31.455","Text":"We\u0027ve multiplied this equation by 2,"},{"Start":"04:31.455 ","End":"04:36.360","Text":"so we square the equilibrium constant."},{"Start":"04:36.820 ","End":"04:41.800","Text":"Now if we invert the second equation,"},{"Start":"04:41.800 ","End":"04:47.340","Text":"now we have 2SO_2 plus O_2 giving 2SO_3,"},{"Start":"04:47.340 ","End":"04:50.690","Text":"then K_5 equilibrium constant for"},{"Start":"04:50.690 ","End":"04:57.305","Text":"this equation will be equal to the equilibrium constant we had before,"},{"Start":"04:57.305 ","End":"04:59.675","Text":"to the power minus 1."},{"Start":"04:59.675 ","End":"05:04.055","Text":"Now let\u0027s sum up these 2 equations."},{"Start":"05:04.055 ","End":"05:08.000","Text":"We see we have to S,"},{"Start":"05:08.000 ","End":"05:13.850","Text":"not here 2 O_2 and another O_2 that gives us 3O_2,"},{"Start":"05:13.850 ","End":"05:18.005","Text":"2SO_2 cancels with 2SO_2,"},{"Start":"05:18.005 ","End":"05:22.165","Text":"and on the right-hand side we\u0027re left with 2SO_3."},{"Start":"05:22.165 ","End":"05:24.995","Text":"When we add these 2 equations,"},{"Start":"05:24.995 ","End":"05:28.920","Text":"we get the equation we were looking for."},{"Start":"05:28.920 ","End":"05:36.915","Text":"The question is how can we express this K_3 in terms of K_4 and K_5?"},{"Start":"05:36.915 ","End":"05:40.065","Text":"Let\u0027s write out K_4."},{"Start":"05:40.065 ","End":"05:44.628","Text":"Just need to look at this equation It\u0027s the pressure of SO_2^2"},{"Start":"05:44.628 ","End":"05:48.980","Text":"divided by the pressure of O_2^2,"},{"Start":"05:48.980 ","End":"05:54.450","Text":"and S, sulfur is pure solid,"},{"Start":"05:54.450 ","End":"05:56.580","Text":"so that isn\u0027t included."},{"Start":"05:56.580 ","End":"05:58.695","Text":"We look at K_5,"},{"Start":"05:58.695 ","End":"06:02.660","Text":"that\u0027s the pressure of SO_3^2 divided by the pressure,"},{"Start":"06:02.660 ","End":"06:07.060","Text":"SO_2^2 times the pressure of oxygen."},{"Start":"06:07.060 ","End":"06:10.380","Text":"Now we have K_4 and K_5."},{"Start":"06:10.380 ","End":"06:12.930","Text":"Let\u0027s look at K_3."},{"Start":"06:12.930 ","End":"06:17.900","Text":"K_3 is the pressure of SO_3^2 divided by"},{"Start":"06:17.900 ","End":"06:23.730","Text":"the pressure of oxygen cubed to the power 3,"},{"Start":"06:23.730 ","End":"06:27.450","Text":"and again the sulfur isn\u0027t included."},{"Start":"06:27.450 ","End":"06:29.820","Text":"Let\u0027s compare K_4,"},{"Start":"06:29.820 ","End":"06:31.605","Text":"K_5, and K_3."},{"Start":"06:31.605 ","End":"06:37.545","Text":"We can see that K_3 is the product of K_4 and K_5."},{"Start":"06:37.545 ","End":"06:41.120","Text":"SO_2^2 goes with this."},{"Start":"06:41.120 ","End":"06:47.750","Text":"We\u0027re left with P SO_3^2 divided by pressure of oxygen squared times"},{"Start":"06:47.750 ","End":"06:50.840","Text":"the pressure of oxygen which is the pressure of the oxygen to"},{"Start":"06:50.840 ","End":"06:54.195","Text":"the power 3, so here\u0027s K_3."},{"Start":"06:54.195 ","End":"06:57.990","Text":"K_3 is equal to the product of K_4 and K_5."},{"Start":"06:57.990 ","End":"07:00.585","Text":"If we look what we did before,"},{"Start":"07:00.585 ","End":"07:07.740","Text":"we saw K_4 was K_1^2 and K_5 was K_2^ minus 1,"},{"Start":"07:07.740 ","End":"07:14.700","Text":"so now we can see that K_3 is equal to K_4 times K_5,"},{"Start":"07:14.700 ","End":"07:21.180","Text":"and that\u0027s equal to K_1^2times K_2^ minus 1."},{"Start":"07:21.180 ","End":"07:27.545","Text":"That tells us that when we sum two equations to get the third equation"},{"Start":"07:27.545 ","End":"07:35.000","Text":"the equilibrium constant for the third 1 is the product of the two individual equations,"},{"Start":"07:35.000 ","End":"07:38.855","Text":"so here\u0027s the rule for combining equilibrium constant."},{"Start":"07:38.855 ","End":"07:42.950","Text":"The equilibrium constant for the overall reaction is the product"},{"Start":"07:42.950 ","End":"07:47.255","Text":"of the equilibrium constants of the individual reactions."},{"Start":"07:47.255 ","End":"07:48.960","Text":"We sum these two,"},{"Start":"07:48.960 ","End":"07:56.955","Text":"so the equilibrium constant for the sum K_3 is equal to the product K_4 times K_5,"},{"Start":"07:56.955 ","End":"08:00.920","Text":"so in this video we showed how the equilibrium constant"},{"Start":"08:00.920 ","End":"08:05.760","Text":"change when the chemical equations are modified."}],"ID":30474},{"Watched":false,"Name":"Exercise 7","Duration":"4m 29s","ChapterTopicVideoID":28921,"CourseChapterTopicPlaylistID":253356,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.995","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.995 ","End":"00:09.895","Text":"Calculate the value of K_c for the following reaction if 0.19 moles of carbon dioxide,"},{"Start":"00:09.895 ","End":"00:11.875","Text":"0.1 mole of hydrogen,"},{"Start":"00:11.875 ","End":"00:14.695","Text":"0.0096 moles of carbon monoxide,"},{"Start":"00:14.695 ","End":"00:18.910","Text":"and 0.0096 moles of water are present in"},{"Start":"00:18.910 ","End":"00:23.670","Text":"a 2 liter reaction vessel at equilibrium and we have a reaction."},{"Start":"00:23.670 ","End":"00:27.430","Text":"In our reactants we have carbon dioxide and hydrogen gas;"},{"Start":"00:27.430 ","End":"00:28.600","Text":"and in our products,"},{"Start":"00:28.600 ","End":"00:31.765","Text":"we have carbon monoxide and water."},{"Start":"00:31.765 ","End":"00:35.380","Text":"You have to calculate the equilibrium constant, K_c."},{"Start":"00:35.380 ","End":"00:39.480","Text":"K_c equals, first we\u0027re going to look at our products."},{"Start":"00:39.480 ","End":"00:46.390","Text":"It\u0027s the concentration of the carbon monoxide times the concentration of the water."},{"Start":"00:46.390 ","End":"00:50.095","Text":"Since the coefficient for carbon monoxide and water equals 1,"},{"Start":"00:50.095 ","End":"00:52.220","Text":"it\u0027s raised to the power of 1."},{"Start":"00:52.220 ","End":"00:54.000","Text":"This is divided by,"},{"Start":"00:54.000 ","End":"00:55.830","Text":"and now we\u0027re going to look at our reactants."},{"Start":"00:55.830 ","End":"01:00.100","Text":"Again, we have a coefficient of 1 also for carbon dioxide and also for the hydrogen."},{"Start":"01:00.100 ","End":"01:02.350","Text":"There\u0027s just going to be the concentration of"},{"Start":"01:02.350 ","End":"01:08.360","Text":"the carbon dioxide times the concentration of the hydrogen."},{"Start":"01:08.360 ","End":"01:13.405","Text":"All of these concentrations have to be the concentrations at equilibrium."},{"Start":"01:13.405 ","End":"01:15.730","Text":"We\u0027re given the number of moles at equilibrium,"},{"Start":"01:15.730 ","End":"01:19.210","Text":"of the reactants and of the products and we also know"},{"Start":"01:19.210 ","End":"01:22.810","Text":"that they\u0027re in a 2 liter reaction vessel."},{"Start":"01:22.810 ","End":"01:27.745","Text":"We\u0027re going to calculate the concentrations of our products and reactants."},{"Start":"01:27.745 ","End":"01:30.140","Text":"We are talking about the molar concentrations."},{"Start":"01:30.140 ","End":"01:33.425","Text":"Remember that molarity M equals N,"},{"Start":"01:33.425 ","End":"01:35.420","Text":"which is the number of moles divided by"},{"Start":"01:35.420 ","End":"01:38.330","Text":"V. The number of moles of the solute divided by V,"},{"Start":"01:38.330 ","End":"01:42.005","Text":"which is the volume of the solution."},{"Start":"01:42.005 ","End":"01:44.485","Text":"Now, here we\u0027re dealing with gases,"},{"Start":"01:44.485 ","End":"01:51.005","Text":"and therefore the concentration of carbon monoxide at equilibrium,"},{"Start":"01:51.005 ","End":"01:53.630","Text":"is going to be equal to the number of moles,"},{"Start":"01:53.630 ","End":"02:02.090","Text":"which equals 0.0096 moles divided by the volume of the vessel."},{"Start":"02:02.090 ","End":"02:05.285","Text":"The volume of the vessel, in our case, is 2 liters."},{"Start":"02:05.285 ","End":"02:15.580","Text":"This equals 0.0048 mole per liter, which equals molar."},{"Start":"02:15.620 ","End":"02:17.930","Text":"That\u0027s the carbon monoxide."},{"Start":"02:17.930 ","End":"02:22.190","Text":"If we look at our water,"},{"Start":"02:22.190 ","End":"02:26.750","Text":"the number of moles of the water also 0.0096."},{"Start":"02:26.750 ","End":"02:30.975","Text":"This is going to give us the same molarity."},{"Start":"02:30.975 ","End":"02:34.080","Text":"That\u0027s 0.0096 mole divided by 2 liters,"},{"Start":"02:34.080 ","End":"02:39.535","Text":"and this also comes out to 0.0048 molar."},{"Start":"02:39.535 ","End":"02:43.400","Text":"Now, our carbon dioxide at equilibrium,"},{"Start":"02:43.400 ","End":"02:46.025","Text":"we have 0.19 moles."},{"Start":"02:46.025 ","End":"02:50.940","Text":"In this case, we also have a 2 liter vessel."},{"Start":"02:51.280 ","End":"03:00.020","Text":"That equals 0.095 and for our hydrogen concentration,"},{"Start":"03:00.020 ","End":"03:06.090","Text":"the number of moles equals 0.1, and again,"},{"Start":"03:06.090 ","End":"03:13.050","Text":"that\u0027s divided by 2 liters and that equals 0.05M."},{"Start":"03:13.050 ","End":"03:17.090","Text":"Now, we can continue with the equilibrium constant K_c."},{"Start":"03:17.090 ","End":"03:23.165","Text":"Now, we can continue with calculating the equilibrium constant,"},{"Start":"03:23.165 ","End":"03:25.040","Text":"now that we know our concentration."},{"Start":"03:25.040 ","End":"03:28.760","Text":"K_c equals the concentration of the carbon monoxide,"},{"Start":"03:28.760 ","End":"03:34.540","Text":"which equals 0.0048 times the concentration of the water,"},{"Start":"03:34.540 ","End":"03:42.400","Text":"which equals 0.0048, divided by the concentration of the carbon dioxide,"},{"Start":"03:42.400 ","End":"03:48.190","Text":"which equals 0.095 times the concentration of the hydrogen,"},{"Start":"03:48.190 ","End":"03:51.183","Text":"which equals 0.05,"},{"Start":"03:51.183 ","End":"03:57.325","Text":"and this equals 0.0049."},{"Start":"03:57.325 ","End":"04:02.080","Text":"Now, I just want to remind you that this equation is actually a simplified version of"},{"Start":"04:02.080 ","End":"04:04.450","Text":"a regular equation that we need to take"},{"Start":"04:04.450 ","End":"04:07.780","Text":"each concentration and divide it by the standard concentration,"},{"Start":"04:07.780 ","End":"04:08.980","Text":"which is 1 mole per liter."},{"Start":"04:08.980 ","End":"04:12.775","Text":"In this case, I just didn\u0027t write the units."},{"Start":"04:12.775 ","End":"04:14.875","Text":"Now, we just multiply and divide."},{"Start":"04:14.875 ","End":"04:20.690","Text":"Our equilibrium constant that we found equals 0.0049."},{"Start":"04:20.690 ","End":"04:26.210","Text":"Remember, the concentrations need to be the concentrations in equilibrium."},{"Start":"04:26.210 ","End":"04:27.500","Text":"That is our final answers."},{"Start":"04:27.500 ","End":"04:29.940","Text":"Thank you very much for watching."}],"ID":30475}],"Thumbnail":null,"ID":253356},{"Name":"Equilibria and Changes in Conditions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Le Chateliers Principle - Adding or Removing Products","Duration":"7m 46s","ChapterTopicVideoID":26279,"CourseChapterTopicPlaylistID":253357,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"In previous videos,"},{"Start":"00:01.890 ","End":"00:04.305","Text":"we learned about chemical equilibrium."},{"Start":"00:04.305 ","End":"00:05.940","Text":"The next few videos,"},{"Start":"00:05.940 ","End":"00:11.700","Text":"we\u0027ll study what happens to the equilibrium when we change the reaction conditions."},{"Start":"00:11.700 ","End":"00:17.115","Text":"We\u0027re interested in what happens when we change the reaction conditions."},{"Start":"00:17.115 ","End":"00:22.935","Text":"Now, when a reaction is in dynamic equilibrium and the conditions have changed,"},{"Start":"00:22.935 ","End":"00:28.140","Text":"the reaction tends to return to a new equilibrium state."},{"Start":"00:28.140 ","End":"00:32.055","Text":"It\u0027s the same volume of the equilibrium constant,"},{"Start":"00:32.055 ","End":"00:36.105","Text":"but new values of the concentrations,"},{"Start":"00:36.105 ","End":"00:39.710","Text":"that\u0027s called a new equilibrium state."},{"Start":"00:39.710 ","End":"00:43.055","Text":"Now, how can we change the conditions?"},{"Start":"00:43.055 ","End":"00:47.675","Text":"The conditions can be changed by adding or removing a reactant or product,"},{"Start":"00:47.675 ","End":"00:51.335","Text":"by changing the volume of the reaction vessel or its pressure,"},{"Start":"00:51.335 ","End":"00:53.495","Text":"or by changing the temperature."},{"Start":"00:53.495 ","End":"00:57.514","Text":"We\u0027ll talk about all these in the next few videos."},{"Start":"00:57.514 ","End":"01:01.744","Text":"Now, the person responsible for"},{"Start":"01:01.744 ","End":"01:06.835","Text":"our understanding of these changes is called Le Chatelier,"},{"Start":"01:06.835 ","End":"01:10.220","Text":"and he lived from 1850-1936."},{"Start":"01:10.220 ","End":"01:14.270","Text":"He formulated important qualitative principle"},{"Start":"01:14.270 ","End":"01:18.775","Text":"that helps us predict the response of the reaction to the change."},{"Start":"01:18.775 ","End":"01:22.070","Text":"This principle, this qualitative principle is"},{"Start":"01:22.070 ","End":"01:25.340","Text":"consistent with thermodynamics as we\u0027ll see."},{"Start":"01:25.340 ","End":"01:29.015","Text":"Now, what is Le Chatelier\u0027s principle?"},{"Start":"01:29.015 ","End":"01:35.270","Text":"It says that when a stress is applied to a system in dynamic equilibrium,"},{"Start":"01:35.270 ","End":"01:41.290","Text":"the equilibrium tends to adjust to minimize the effects of the stress."},{"Start":"01:41.290 ","End":"01:44.330","Text":"We\u0027ll see several examples of this."},{"Start":"01:44.330 ","End":"01:50.660","Text":"Let\u0027s start by considering what happens when we add or remove reactants or products."},{"Start":"01:50.660 ","End":"01:54.410","Text":"Now let\u0027s consider the reaction a moles of A plus b of"},{"Start":"01:54.410 ","End":"01:59.280","Text":"B in equilibrium with c moles of C and d moles of D. Now,"},{"Start":"01:59.280 ","End":"02:03.680","Text":"the stress is going to be adding A or B."},{"Start":"02:03.680 ","End":"02:05.300","Text":"We\u0027re going to add A,"},{"Start":"02:05.300 ","End":"02:07.040","Text":"or B, or both."},{"Start":"02:07.040 ","End":"02:13.069","Text":"The result will be that the reaction goes from left to right,"},{"Start":"02:13.069 ","End":"02:15.570","Text":"reducing A and B,"},{"Start":"02:15.570 ","End":"02:17.870","Text":"so A and B will be reduced,"},{"Start":"02:17.870 ","End":"02:24.150","Text":"and increasing C and D. A and B will go down,"},{"Start":"02:25.990 ","End":"02:29.280","Text":"C and D will go up."},{"Start":"02:31.340 ","End":"02:34.095","Text":"Let\u0027s put it more clearly."},{"Start":"02:34.095 ","End":"02:35.580","Text":"A and B will go down,"},{"Start":"02:35.580 ","End":"02:40.480","Text":"and C and D will go up."},{"Start":"02:42.830 ","End":"02:47.305","Text":"The reaction is going from left to right."},{"Start":"02:47.305 ","End":"02:52.340","Text":"Now, supposing the stress as we add C or D,"},{"Start":"02:52.340 ","End":"02:56.630","Text":"then the result will be the reaction goes from right to left,"},{"Start":"02:56.630 ","End":"03:01.495","Text":"reducing C and D and increasing A and B."},{"Start":"03:01.495 ","End":"03:03.615","Text":"In this case,"},{"Start":"03:03.615 ","End":"03:07.800","Text":"the reaction\u0027s going to go from right to left,"},{"Start":"03:08.440 ","End":"03:12.120","Text":"reducing C and D,"},{"Start":"03:13.690 ","End":"03:17.640","Text":"and increasing A and B."},{"Start":"03:18.850 ","End":"03:21.860","Text":"Now we have to note, as I\u0027ve said before,"},{"Start":"03:21.860 ","End":"03:25.790","Text":"the new equilibrium concentrations are produced."},{"Start":"03:25.790 ","End":"03:29.150","Text":"The previous ones are not restored."},{"Start":"03:29.150 ","End":"03:33.359","Text":"This is called a new equilibrium condition;"},{"Start":"03:33.359 ","End":"03:36.780","Text":"a new position of equilibrium."},{"Start":"03:40.760 ","End":"03:46.420","Text":"Now, how does this compare to thermodynamics?"},{"Start":"03:46.700 ","End":"03:52.780","Text":"Let\u0027s write the Gibbs free energy of reaction and the reaction quotient."},{"Start":"03:52.780 ","End":"03:58.310","Text":"This is the Gibbs free energy of reaction at any point in the reaction."},{"Start":"04:00.630 ","End":"04:08.050","Text":"We have Delta G of reaction is equal to RT ln Q/K."},{"Start":"04:08.050 ","End":"04:11.935","Text":"This is for any stage of the reaction."},{"Start":"04:11.935 ","End":"04:17.620","Text":"We can write Q as the activity of C to the power of"},{"Start":"04:17.620 ","End":"04:23.530","Text":"c times the derivative of D to the power d divided by the activity of A to the power a,"},{"Start":"04:23.530 ","End":"04:26.530","Text":"times the activity of B to the power b."},{"Start":"04:26.530 ","End":"04:29.300","Text":"We\u0027ve seen this many times before."},{"Start":"04:29.300 ","End":"04:34.460","Text":"Recall that equilibrium Q=K."},{"Start":"04:34.460 ","End":"04:36.166","Text":"We have Q/K,"},{"Start":"04:36.166 ","End":"04:38.780","Text":"they are equal, that\u0027s just 1."},{"Start":"04:38.780 ","End":"04:41.135","Text":"Ln of 1 is 0."},{"Start":"04:41.135 ","End":"04:46.650","Text":"So Delta G reaction will be equal to 0 at equilibrium."},{"Start":"04:48.340 ","End":"04:51.260","Text":"Now let\u0027s consider the stress."},{"Start":"04:51.260 ","End":"04:53.480","Text":"We\u0027re going to add A or B."},{"Start":"04:53.480 ","End":"04:58.310","Text":"If we add A or B, the denominator increases."},{"Start":"04:58.310 ","End":"05:05.120","Text":"Q will now be smaller than K. To restore equilibrium,"},{"Start":"05:05.120 ","End":"05:11.180","Text":"Q must increase until it\u0027s equal to K. How does it increase?"},{"Start":"05:11.180 ","End":"05:15.485","Text":"By reducing A and B and increasing C and D,"},{"Start":"05:15.485 ","End":"05:17.435","Text":"just as we saw before."},{"Start":"05:17.435 ","End":"05:20.660","Text":"So A must go down,"},{"Start":"05:20.660 ","End":"05:23.285","Text":"B must go down,"},{"Start":"05:23.285 ","End":"05:27.095","Text":"C will go up, D will go up."},{"Start":"05:27.095 ","End":"05:31.475","Text":"The reaction goes from left to right."},{"Start":"05:31.475 ","End":"05:35.930","Text":"When we decrease A and B and increase C and D,"},{"Start":"05:35.930 ","End":"05:39.390","Text":"the reaction is going from left to right."},{"Start":"05:40.370 ","End":"05:47.630","Text":"Now, supposing the stress as we add C or D at equilibrium,"},{"Start":"05:47.630 ","End":"05:49.430","Text":"we\u0027re adding C or D,"},{"Start":"05:49.430 ","End":"05:52.325","Text":"so Q will increase,"},{"Start":"05:52.325 ","End":"06:02.255","Text":"Q will now be greater than K. To restore the equilibrium, Q must decrease."},{"Start":"06:02.255 ","End":"06:04.175","Text":"How can we decrease?"},{"Start":"06:04.175 ","End":"06:10.550","Text":"By reducing the amount of C and D and increasing A and B."},{"Start":"06:10.550 ","End":"06:13.730","Text":"C is going to decrease,"},{"Start":"06:13.730 ","End":"06:18.920","Text":"D will decrease, A will increase, B will increase."},{"Start":"06:18.920 ","End":"06:23.020","Text":"That\u0027s exactly what Le Chatelier taught us."},{"Start":"06:23.020 ","End":"06:25.325","Text":"Here\u0027s an example."},{"Start":"06:25.325 ","End":"06:30.830","Text":"N_2 plus 3H_2 in equilibrium with 2 NH_3,"},{"Start":"06:30.830 ","End":"06:33.030","Text":"and they\u0027re all gases."},{"Start":"06:33.160 ","End":"06:36.155","Text":"If we add N_2,"},{"Start":"06:36.155 ","End":"06:39.074","Text":"so we\u0027re adding N_2 here,"},{"Start":"06:39.074 ","End":"06:43.170","Text":"the reaction will go from left to right."},{"Start":"06:44.000 ","End":"06:47.550","Text":"That means the N_2 will now be"},{"Start":"06:47.550 ","End":"06:55.130","Text":"reduced as will H_2 and NH_3 will increase."},{"Start":"06:55.130 ","End":"06:58.040","Text":"The reaction is going from left to right."},{"Start":"06:58.040 ","End":"07:00.245","Text":"If we add NH_3,"},{"Start":"07:00.245 ","End":"07:03.810","Text":"the reaction will go from right to left."},{"Start":"07:05.210 ","End":"07:11.020","Text":"This time, NH_3 will decrease,"},{"Start":"07:11.120 ","End":"07:16.020","Text":"N_2 will increase, and so will H_2."},{"Start":"07:16.020 ","End":"07:19.930","Text":"Now, if we keep removing NH_3,"},{"Start":"07:19.930 ","End":"07:26.300","Text":"the reaction will keep going from left to right until it goes to completion."},{"Start":"07:26.300 ","End":"07:34.530","Text":"This is a way of making a reaction go to completion by simply removing the product."},{"Start":"07:35.830 ","End":"07:40.760","Text":"In this video, we talked about Le Chatelier\u0027s principle and"},{"Start":"07:40.760 ","End":"07:45.990","Text":"applied it to adding or removing reactants or products."}],"ID":27193},{"Watched":false,"Name":"Adding or Removing Reagent - Example","Duration":"5m 6s","ChapterTopicVideoID":26281,"CourseChapterTopicPlaylistID":253357,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.055","Text":"In the previous video,"},{"Start":"00:02.055 ","End":"00:07.335","Text":"we talked about the effect of adding or removing reagents on a reaction at equilibrium."},{"Start":"00:07.335 ","End":"00:11.805","Text":"In this video, we\u0027ll show how to calculate the new equilibrium."},{"Start":"00:11.805 ","End":"00:18.360","Text":"We\u0027re going to solve a problem involving the adding or removing of a reagent,"},{"Start":"00:18.360 ","End":"00:20.265","Text":"and here\u0027s the problem."},{"Start":"00:20.265 ","End":"00:22.019","Text":"In the reaction FeO,"},{"Start":"00:22.019 ","End":"00:23.250","Text":"which is a solid,"},{"Start":"00:23.250 ","End":"00:25.110","Text":"plus CO, the gas,"},{"Start":"00:25.110 ","End":"00:34.605","Text":"in equilibrium with Fe as a solid plus CO_2 as a gas, K=0.403."},{"Start":"00:34.605 ","End":"00:36.320","Text":"Now, at equilibrium,"},{"Start":"00:36.320 ","End":"00:40.160","Text":"the partial pressure of CO is 4.24 bar,"},{"Start":"00:40.160 ","End":"00:45.690","Text":"and the partial pressure of CO_2 is 1.71 bar."},{"Start":"00:45.690 ","End":"00:51.215","Text":"Now, if we add 1 bar of CO_2 to the reaction at equilibrium,"},{"Start":"00:51.215 ","End":"00:55.850","Text":"what are the new equilibrium concentrations?"},{"Start":"00:55.850 ","End":"01:00.710","Text":"In other words, what is the new equilibrium?"},{"Start":"01:00.710 ","End":"01:07.915","Text":"Of course, we don\u0027t include the solids in the expression of the equilibrium constant."},{"Start":"01:07.915 ","End":"01:11.050","Text":"Here\u0027s our ICE table."},{"Start":"01:11.050 ","End":"01:14.580","Text":"We have CO and CO_2, the 2 gases."},{"Start":"01:14.580 ","End":"01:18.090","Text":"Now, the first equilibrium, E_1,"},{"Start":"01:18.090 ","End":"01:23.980","Text":"has 4.24 bars of CO and 1.71 bars of CO_2."},{"Start":"01:23.980 ","End":"01:28.590","Text":"Then we add 1 bar to CO and nothing to CO_2,"},{"Start":"01:28.590 ","End":"01:33.850","Text":"so the new initial state is 5.24 bars,"},{"Start":"01:33.850 ","End":"01:37.070","Text":"4.24 plus 1,"},{"Start":"01:38.990 ","End":"01:45.705","Text":"and the initial state of CO_2 stays 1.71 bars."},{"Start":"01:45.705 ","End":"01:52.455","Text":"Now, what\u0027s the change when the new equilibrium is obtained?"},{"Start":"01:52.455 ","End":"01:59.970","Text":"CO decreases by x minus x and CO_2 increases by x,"},{"Start":"01:59.970 ","End":"02:02.500","Text":"so it\u0027s plus x."},{"Start":"02:02.650 ","End":"02:11.420","Text":"Now, the new equilibrium is 5.24 minus x and 1.71 plus x,"},{"Start":"02:11.420 ","End":"02:17.985","Text":"5.24 minus x for CO and 1.71 plus x for CO_2."},{"Start":"02:17.985 ","End":"02:22.555","Text":"Now, we can write the expression for K. It\u0027s the pressure of"},{"Start":"02:22.555 ","End":"02:27.214","Text":"CO_2 divided by the pressure of CO. As I said before,"},{"Start":"02:27.214 ","End":"02:30.365","Text":"the solids don\u0027t play any part in this."},{"Start":"02:30.365 ","End":"02:39.570","Text":"We have 1.71 plus x for CO_2 and 5.24 minus x"},{"Start":"02:39.570 ","End":"02:49.195","Text":"for CO. We divided these 2 and we know the value of the equilibrium constant is 0.403,"},{"Start":"02:49.195 ","End":"02:56.720","Text":"so we can solve this equation and we get 1.403x is equal to 0.400,"},{"Start":"02:56.720 ","End":"03:03.930","Text":"so that x is equal to 0.400 divided by 1.403."},{"Start":"03:03.930 ","End":"03:08.410","Text":"That\u0027s equal to 0.285."},{"Start":"03:10.850 ","End":"03:15.370","Text":"We can work out that the partial pressure of CO_2 is"},{"Start":"03:15.370 ","End":"03:22.550","Text":"1.71 plus x and x is 0.285, that\u0027s x."},{"Start":"03:22.550 ","End":"03:27.630","Text":"We get, when we add these 2, 1.995,"},{"Start":"03:27.630 ","End":"03:36.610","Text":"and the partial pressure of CO is 5.24 minus x,"},{"Start":"03:36.610 ","End":"03:39.695","Text":"and x is 0.285."},{"Start":"03:39.695 ","End":"03:44.980","Text":"When we subtract these 2 numbers, we get 4.955."},{"Start":"03:46.130 ","End":"03:50.605","Text":"Now we can check whether these values are reasonable"},{"Start":"03:50.605 ","End":"03:55.320","Text":"by re-calculating K, the equilibrium constant."},{"Start":"03:55.320 ","End":"03:59.780","Text":"K is the pressure of CO_2 divided by the pressure of CO,"},{"Start":"03:59.780 ","End":"04:10.100","Text":"that\u0027s 1.995 divided by 4.995 and that gives us 0.403,"},{"Start":"04:10.100 ","End":"04:13.025","Text":"which is the volume we were given initially."},{"Start":"04:13.025 ","End":"04:20.745","Text":"We\u0027ve maintained the equilibrium constant but changed the equilibrium concentrations,"},{"Start":"04:20.745 ","End":"04:23.860","Text":"that\u0027s a new equilibrium."},{"Start":"04:26.090 ","End":"04:29.725","Text":"Now, I just want to make a comment or 2."},{"Start":"04:29.725 ","End":"04:33.855","Text":"This is a particularly easy example that we had here and"},{"Start":"04:33.855 ","End":"04:38.180","Text":"often they are much more complicated and 1 has to solve a quadratic,"},{"Start":"04:38.180 ","End":"04:41.000","Text":"cubic or even higher-order equation."},{"Start":"04:41.000 ","End":"04:45.455","Text":"However, even in complicated quantitative problems,"},{"Start":"04:45.455 ","End":"04:51.410","Text":"Le Chatelier\u0027s principle will give us a reliable qualitative answer."},{"Start":"04:51.410 ","End":"04:56.315","Text":"Even if you can\u0027t solve the problem quantitatively,"},{"Start":"04:56.315 ","End":"05:00.155","Text":"you can always get a qualitative answer."},{"Start":"05:00.155 ","End":"05:06.990","Text":"In this video, we solved a problem involving the adding of a reagent."}],"ID":27195},{"Watched":false,"Name":"Effect of Changing Volume or Pressure","Duration":"6m 50s","ChapterTopicVideoID":26278,"CourseChapterTopicPlaylistID":253357,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.250","Text":"In the previous video,"},{"Start":"00:02.250 ","End":"00:06.195","Text":"we learned about the effect of adding or removing reagents."},{"Start":"00:06.195 ","End":"00:10.050","Text":"In this video, we\u0027ll learn about the effect of changing the volume,"},{"Start":"00:10.050 ","End":"00:11.895","Text":"or pressure of the system."},{"Start":"00:11.895 ","End":"00:16.755","Text":"Now the reactions we\u0027re going to consider involve gases."},{"Start":"00:16.755 ","End":"00:20.070","Text":"That\u0027s because changing the volume of the system has"},{"Start":"00:20.070 ","End":"00:23.445","Text":"little effect on solids, or liquids."},{"Start":"00:23.445 ","End":"00:28.025","Text":"We\u0027re going to consider a reaction involving gases in equilibrium,"},{"Start":"00:28.025 ","End":"00:30.665","Text":"and at constant temperature."},{"Start":"00:30.665 ","End":"00:33.583","Text":"Now how can we change the volume of the system?"},{"Start":"00:33.583 ","End":"00:38.360","Text":"Of course, we can always move the reaction to a large or a smaller vessel."},{"Start":"00:38.360 ","End":"00:40.100","Text":"That\u0027s pretty obvious."},{"Start":"00:40.100 ","End":"00:44.479","Text":"Another way is to add an inert gas at constant pressure."},{"Start":"00:44.479 ","End":"00:47.165","Text":"We\u0027ll talk about that in the next video."},{"Start":"00:47.165 ","End":"00:50.210","Text":"Another thing we can do is to reduce"},{"Start":"00:50.210 ","End":"00:55.565","Text":"the total volume of the system by increasing the external pressure."},{"Start":"00:55.565 ","End":"01:00.950","Text":"For example, pressing down harder on this piston."},{"Start":"01:00.950 ","End":"01:02.465","Text":"The gas is in here,"},{"Start":"01:02.465 ","End":"01:05.695","Text":"and this is a piston that we can move."},{"Start":"01:05.695 ","End":"01:11.690","Text":"Now, let\u0027s first consider a quantitative treatment of this problem."},{"Start":"01:11.690 ","End":"01:14.545","Text":"Consider the general equation,"},{"Start":"01:14.545 ","End":"01:16.255","Text":"a moles of A,"},{"Start":"01:16.255 ","End":"01:21.890","Text":"b moles of B in equilibrium with c moles of C and d moles of D,"},{"Start":"01:21.890 ","End":"01:23.525","Text":"and they\u0027re all gases."},{"Start":"01:23.525 ","End":"01:25.570","Text":"Let\u0027s write K_c,"},{"Start":"01:25.570 ","End":"01:30.440","Text":"the equilibrium constant in terms of concentrations."},{"Start":"01:30.440 ","End":"01:34.730","Text":"I will just take everything to be dimensionless."},{"Start":"01:34.730 ","End":"01:40.400","Text":"We have the concentration of C^c times the concentration of D^d,"},{"Start":"01:40.400 ","End":"01:45.985","Text":"divided by the concentration of A^a and the concentration of B^b."},{"Start":"01:45.985 ","End":"01:50.030","Text":"Concentration is number of moles divided by the volume."},{"Start":"01:50.030 ","End":"01:53.935","Text":"Again, we\u0027re just assuming everything\u0027s without units."},{"Start":"01:53.935 ","End":"01:59.880","Text":"We have n_C divided by V^c times n_D divided by V^d."},{"Start":"01:59.880 ","End":"02:04.395","Text":"All that divided by n_A divided by V^a,"},{"Start":"02:04.395 ","End":"02:08.235","Text":"and n_B divided by V^b."},{"Start":"02:08.235 ","End":"02:12.965","Text":"Let\u0027s separate the number of moles from the volume."},{"Start":"02:12.965 ","End":"02:20.695","Text":"We have (n_C)^c times (n_D)^d divided by (n_A)a times (n_B)^b."},{"Start":"02:20.695 ","End":"02:23.115","Text":"Let\u0027s look at the volume."},{"Start":"02:23.115 ","End":"02:26.560","Text":"We have 1 over volume,"},{"Start":"02:27.410 ","End":"02:31.110","Text":"c plus d from here,"},{"Start":"02:31.110 ","End":"02:35.200","Text":"minus a plus b."},{"Start":"02:35.390 ","End":"02:41.280","Text":"So c plus d minus a plus b is Delta n. We"},{"Start":"02:41.280 ","End":"02:48.750","Text":"have v^minus Delta n. 1/v is v^minus 1."},{"Start":"02:48.750 ","End":"02:55.800","Text":"We have v^minus Delta n. Now,"},{"Start":"02:55.800 ","End":"03:00.870","Text":"if we increase v^minus Delta n,"},{"Start":"03:00.870 ","End":"03:10.390","Text":"then (n_C)^c times (n_D)^d divided by (n_A)^a times (n_B)^b,"},{"Start":"03:10.390 ","End":"03:14.825","Text":"this must decrease to maintain the volume of K_c."},{"Start":"03:14.825 ","End":"03:17.285","Text":"K_c is a constant."},{"Start":"03:17.285 ","End":"03:20.045","Text":"Once again, if this goes up,"},{"Start":"03:20.045 ","End":"03:23.165","Text":"this must go down or vice versa."},{"Start":"03:23.165 ","End":"03:24.695","Text":"If this goes down,"},{"Start":"03:24.695 ","End":"03:26.615","Text":"this must go up."},{"Start":"03:26.615 ","End":"03:29.450","Text":"Let\u0027s consider an example."},{"Start":"03:29.450 ","End":"03:36.955","Text":"Consider the reaction to 2H_2S in equilibrium with 2H_2 and S_2."},{"Start":"03:36.955 ","End":"03:39.420","Text":"They\u0027re all gases."},{"Start":"03:39.420 ","End":"03:43.680","Text":"Delta n here is 2 plus 1,"},{"Start":"03:43.680 ","End":"03:47.010","Text":"that\u0027s 3 minus 2."},{"Start":"03:47.010 ","End":"03:50.685","Text":"That\u0027s equal to 1."},{"Start":"03:50.685 ","End":"03:52.665","Text":"Delta n is 1,"},{"Start":"03:52.665 ","End":"03:59.350","Text":"and K_c is (n_H_2)^2."},{"Start":"03:59.350 ","End":"04:06.120","Text":"(n_H_2)^2 times n_S_2 divided by (n_H_2S)^2."},{"Start":"04:06.120 ","End":"04:08.765","Text":"That\u0027s because of the 2s in this equation."},{"Start":"04:08.765 ","End":"04:10.320","Text":"Here we have 2 moles,"},{"Start":"04:10.320 ","End":"04:12.030","Text":"2 moles in 1 mole."},{"Start":"04:12.030 ","End":"04:17.955","Text":"Now, that is multiplied by v^minus 1."},{"Start":"04:17.955 ","End":"04:22.050","Text":"Supposing the volume decreases,"},{"Start":"04:22.050 ","End":"04:27.865","Text":"then 1/V will increase, this will increase."},{"Start":"04:27.865 ","End":"04:36.440","Text":"The ratio of the numbers of moles must decrease."},{"Start":"04:36.440 ","End":"04:40.340","Text":"The reaction will proceed from right to left."},{"Start":"04:40.340 ","End":"04:42.500","Text":"Because if this decreases,"},{"Start":"04:42.500 ","End":"04:44.475","Text":"it means H_2 goes down,"},{"Start":"04:44.475 ","End":"04:48.140","Text":"S_2 goes down, H_2S goes up."},{"Start":"04:48.140 ","End":"04:52.730","Text":"That means the reaction is going from right to left."},{"Start":"04:52.730 ","End":"04:56.300","Text":"Let\u0027s make a general statement."},{"Start":"04:56.300 ","End":"05:01.265","Text":"When the volume of an equilibrium mixture of gases decreases,"},{"Start":"05:01.265 ","End":"05:07.240","Text":"the reaction proceeds in the direction that produces fewer moles of gas."},{"Start":"05:07.240 ","End":"05:11.390","Text":"When the volume of an equilibrium mixture of gases increases,"},{"Start":"05:11.390 ","End":"05:18.350","Text":"the reaction proceeds in the direction that produces more moles of gas."},{"Start":"05:18.350 ","End":"05:22.490","Text":"Now, this is an agreement with Le Chatelier\u0027s principle."},{"Start":"05:22.490 ","End":"05:27.860","Text":"Let\u0027s see how. When the volume decreases, the pressure increases."},{"Start":"05:27.860 ","End":"05:29.660","Text":"The pressure increases,"},{"Start":"05:29.660 ","End":"05:32.365","Text":"that\u0027s going to be our stress."},{"Start":"05:32.365 ","End":"05:34.590","Text":"To minimize the effect,"},{"Start":"05:34.590 ","End":"05:38.600","Text":"the reaction moves in the direction where there are fewer moles"},{"Start":"05:38.600 ","End":"05:43.740","Text":"of gases and less pressure."},{"Start":"05:47.780 ","End":"05:50.635","Text":"When the volume increases,"},{"Start":"05:50.635 ","End":"05:53.020","Text":"that\u0027s equivalent to the pressure decreasing,"},{"Start":"05:53.020 ","End":"05:56.080","Text":"so the stress is the pressure decreasing."},{"Start":"05:56.080 ","End":"05:58.329","Text":"To minimize the effect,"},{"Start":"05:58.329 ","End":"06:04.105","Text":"the reaction moves in the direction where there are more moles of gas."},{"Start":"06:04.105 ","End":"06:06.480","Text":"That\u0027s to increase the pressure."},{"Start":"06:06.480 ","End":"06:10.410","Text":"The stress is the pressure decreasing and we increase"},{"Start":"06:10.410 ","End":"06:16.345","Text":"the pressure by moving the reaction in the direction where there are more moles of gas."},{"Start":"06:16.345 ","End":"06:22.825","Text":"Now, supposing we want to change the total pressure rather than the volume."},{"Start":"06:22.825 ","End":"06:27.280","Text":"If you need to know what happens when the pressure is increased or decreased,"},{"Start":"06:27.280 ","End":"06:32.315","Text":"translate the problem into one of decreasing or increasing volume,"},{"Start":"06:32.315 ","End":"06:34.565","Text":"and then proceed as before."},{"Start":"06:34.565 ","End":"06:37.760","Text":"We\u0027re always going to work in terms of volume,"},{"Start":"06:37.760 ","End":"06:41.660","Text":"which is easier than in terms of pressure."},{"Start":"06:41.660 ","End":"06:45.230","Text":"In this video, we talked about how changing"},{"Start":"06:45.230 ","End":"06:50.430","Text":"the volume affects the equilibrium concentrations."}],"ID":27192},{"Watched":false,"Name":"Effect of Adding Inert Gas and Effect of Catalyst","Duration":"2m 54s","ChapterTopicVideoID":26282,"CourseChapterTopicPlaylistID":253357,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.965","Text":"In the previous video,"},{"Start":"00:01.965 ","End":"00:04.830","Text":"we talked about the effect of changing the volume."},{"Start":"00:04.830 ","End":"00:07.890","Text":"In this video, we\u0027ll talk about the effect of adding"},{"Start":"00:07.890 ","End":"00:13.695","Text":"an inert gas and also the effect of a catalyst on equilibrium."},{"Start":"00:13.695 ","End":"00:18.735","Text":"What happens when we add an inert gas at constant pressure?"},{"Start":"00:18.735 ","End":"00:24.630","Text":"When we add an inert gas to a reaction in equilibrium at constant pressure,"},{"Start":"00:24.630 ","End":"00:28.440","Text":"we increase the volume that the gases take up."},{"Start":"00:28.440 ","End":"00:31.230","Text":"We\u0027ve got another additional gas."},{"Start":"00:31.230 ","End":"00:34.575","Text":"As before, when the volume increases,"},{"Start":"00:34.575 ","End":"00:40.275","Text":"the reaction proceeds in the direction that produces more moles of gas."},{"Start":"00:40.275 ","End":"00:46.890","Text":"It\u0027s just the same as any increase in the volume for any reason."},{"Start":"00:46.890 ","End":"00:50.235","Text":"Now what happens when we add an inert gas at"},{"Start":"00:50.235 ","End":"00:53.895","Text":"constant volume rather than the constant pressure?"},{"Start":"00:53.895 ","End":"00:59.055","Text":"Now when we add an inert gas to a reaction in equilibrium at constant volume,"},{"Start":"00:59.055 ","End":"01:02.580","Text":"it doesn\u0027t change the concentrations which are n"},{"Start":"01:02.580 ","End":"01:06.900","Text":"divided by v and v is constant of the gases."},{"Start":"01:06.900 ","End":"01:08.970","Text":"Nothing happens at all."},{"Start":"01:08.970 ","End":"01:14.553","Text":"Something happens when we add the inert gas at constant pressure,"},{"Start":"01:14.553 ","End":"01:20.190","Text":"but nothing happens when we add an inert gas at constant volume."},{"Start":"01:20.190 ","End":"01:24.345","Text":"Now, what about the effect of adding a catalyst?"},{"Start":"01:24.345 ","End":"01:28.709","Text":"Now, the definition of a catalyst is that a catalyst"},{"Start":"01:28.709 ","End":"01:34.875","Text":"increases the rate of a chemical reaction without being consumed itself."},{"Start":"01:34.875 ","End":"01:37.380","Text":"As it isn\u0027t consumed,"},{"Start":"01:37.380 ","End":"01:41.835","Text":"it doesn\u0027t appear in the overall chemical equation for the reaction."},{"Start":"01:41.835 ","End":"01:48.150","Text":"What it does is to speed up the forward and the reverse reactions to the same extent,"},{"Start":"01:48.150 ","End":"01:52.530","Text":"so that it has no effect on the dynamic equilibrium."},{"Start":"01:52.530 ","End":"01:54.930","Text":"Later, when we study kinetics,"},{"Start":"01:54.930 ","End":"01:59.415","Text":"we\u0027ll see that the catalyst works by altering the reaction path."},{"Start":"01:59.415 ","End":"02:02.700","Text":"Now, a brief thermodynamic explanation of why"},{"Start":"02:02.700 ","End":"02:06.420","Text":"catalysts doesn\u0027t affect the equilibrium constant."},{"Start":"02:06.420 ","End":"02:11.025","Text":"Now a catalyst has no effect on Delta G0 reaction,"},{"Start":"02:11.025 ","End":"02:12.945","Text":"which is a state function,"},{"Start":"02:12.945 ","End":"02:18.640","Text":"and depends only on the products and reactants and not on the path."},{"Start":"02:22.060 ","End":"02:25.910","Text":"Ln K, where K is the equilibrium constant,"},{"Start":"02:25.910 ","End":"02:31.585","Text":"is equal to minus Delta G0 of reaction divided by RT."},{"Start":"02:31.585 ","End":"02:36.120","Text":"The catalyst has no effect on the equilibrium constant,"},{"Start":"02:36.120 ","End":"02:39.360","Text":"has no effect on delta G0, therefore,"},{"Start":"02:39.360 ","End":"02:43.950","Text":"no effect on K, the equilibrium constant."},{"Start":"02:43.950 ","End":"02:46.680","Text":"In this video, we talked about the effect of"},{"Start":"02:46.680 ","End":"02:49.620","Text":"adding an inert gas to the reaction vessel,"},{"Start":"02:49.620 ","End":"02:53.890","Text":"and also on the effect of adding a catalyst."}],"ID":27196},{"Watched":false,"Name":"Effect of Changing Temperature","Duration":"8m 38s","ChapterTopicVideoID":26283,"CourseChapterTopicPlaylistID":253357,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:02.520","Text":"In the previous videos,"},{"Start":"00:02.520 ","End":"00:05.760","Text":"we considered the effect of adding or removing regions"},{"Start":"00:05.760 ","End":"00:10.200","Text":"or changing the volume or pressure or adding an inert gas."},{"Start":"00:10.200 ","End":"00:15.300","Text":"In this video, we consider the effect of changing the temperature."},{"Start":"00:15.300 ","End":"00:21.915","Text":"We want to know what the effect of temperature is on the equilibrium constant."},{"Start":"00:21.915 ","End":"00:25.155","Text":"What happens when we change the temperature?"},{"Start":"00:25.155 ","End":"00:29.790","Text":"Now we know that the equilibrium constant is temperature-dependent."},{"Start":"00:29.790 ","End":"00:32.010","Text":"We\u0027ve seen this in previous videos."},{"Start":"00:32.010 ","End":"00:38.195","Text":"InK is equal to minus Delta G^0 reaction divided by RT."},{"Start":"00:38.195 ","End":"00:42.530","Text":"Delta G can be written as Delta H minus T Delta S,"},{"Start":"00:42.530 ","End":"00:47.555","Text":"which gives us minus Delta H^0 reaction divided by RT"},{"Start":"00:47.555 ","End":"00:53.180","Text":"plus Delta S reaction divided by R. Delta H,"},{"Start":"00:53.180 ","End":"00:54.649","Text":"of course, is enthalpy,"},{"Start":"00:54.649 ","End":"00:56.630","Text":"Delta S is entropy,"},{"Start":"00:56.630 ","End":"01:00.305","Text":"and Delta G is Gibbs free energy."},{"Start":"01:00.305 ","End":"01:06.110","Text":"Now, we can write this equation for two different temperatures."},{"Start":"01:06.110 ","End":"01:07.880","Text":"We have InK_1,"},{"Start":"01:07.880 ","End":"01:09.545","Text":"that\u0027s one temperature,"},{"Start":"01:09.545 ","End":"01:13.565","Text":"equal to minus Delta H divided by RT_1,"},{"Start":"01:13.565 ","End":"01:15.770","Text":"that\u0027s the temperature we\u0027re talking about,"},{"Start":"01:15.770 ","End":"01:22.085","Text":"that\u0027s the value of K_1 at T_1 plus Delta S divided by R,"},{"Start":"01:22.085 ","End":"01:25.670","Text":"and that\u0027s independent of temperature."},{"Start":"01:25.670 ","End":"01:29.165","Text":"We can write the same thing at another temperature,"},{"Start":"01:29.165 ","End":"01:38.325","Text":"InK_2 is equal to minus Delta H reaction divided by RT_2 so K_2 goes with T_2,"},{"Start":"01:38.325 ","End":"01:40.523","Text":"whereas K_1 goes to T_1,"},{"Start":"01:40.523 ","End":"01:44.765","Text":"and again we have plus Delta S divided by R,"},{"Start":"01:44.765 ","End":"01:46.910","Text":"which is temperature independent."},{"Start":"01:46.910 ","End":"01:53.630","Text":"If we want to get rid of this Delta S over R,"},{"Start":"01:53.630 ","End":"01:56.390","Text":"we can simply subtract the two equations."},{"Start":"01:56.390 ","End":"02:01.100","Text":"Then we get InK_2 minus InK_1 is equal to"},{"Start":"02:01.100 ","End":"02:10.440","Text":"minus Delta H divided by RT_2 minus Delta H divided by RT_1,"},{"Start":"02:10.440 ","End":"02:14.770","Text":"and the Delta S disappears."},{"Start":"02:15.140 ","End":"02:25.165","Text":"Now we know the InK_2 minus InK_1 is equal to ln of the ratio of K_2 and K_1,"},{"Start":"02:25.165 ","End":"02:28.125","Text":"that\u0027s InK_2 over K_1."},{"Start":"02:28.125 ","End":"02:32.060","Text":"Here, we can take out of the brackets"},{"Start":"02:32.060 ","End":"02:38.720","Text":"minus Delta H reaction divided by R. We\u0027ve taken it out."},{"Start":"02:38.720 ","End":"02:40.845","Text":"We\u0027ve also changed the sign."},{"Start":"02:40.845 ","End":"02:43.795","Text":"Instead of having minus 1 over T_2,"},{"Start":"02:43.795 ","End":"02:45.850","Text":"minus 1 over T_1,"},{"Start":"02:45.850 ","End":"02:51.240","Text":"we have 1 over T_1 minus 1 over T_2."},{"Start":"02:51.240 ","End":"02:53.520","Text":"Here\u0027s our final equation."},{"Start":"02:53.520 ","End":"02:59.060","Text":"Now, this equation is called the Van\u0027t Hoff equation."},{"Start":"02:59.060 ","End":"03:04.190","Text":"Because of course, Van\u0027t Hoff was the first to write it down."},{"Start":"03:04.190 ","End":"03:06.845","Text":"Now, there\u0027s an assumption here."},{"Start":"03:06.845 ","End":"03:10.550","Text":"It assumes that the enthalpy doesn\u0027t change with temperature."},{"Start":"03:10.550 ","End":"03:14.165","Text":"Now the enthalpy does change slightly with temperature,"},{"Start":"03:14.165 ","End":"03:16.850","Text":"but not to any large degree."},{"Start":"03:16.850 ","End":"03:21.500","Text":"It\u0027s a very useful first approximation."},{"Start":"03:21.500 ","End":"03:25.220","Text":"Let\u0027s look at what this equation says."},{"Start":"03:25.220 ","End":"03:29.780","Text":"Well, first take the example of an endothermic reaction."},{"Start":"03:29.780 ","End":"03:33.560","Text":"We know that for an endothermic reaction,"},{"Start":"03:33.560 ","End":"03:37.460","Text":"Delta H of reaction is positive."},{"Start":"03:37.460 ","End":"03:40.547","Text":"Now, if we look at the expression in the brackets,"},{"Start":"03:40.547 ","End":"03:43.376","Text":"1 over T_1 minus 1 over T_2,"},{"Start":"03:43.376 ","End":"03:46.340","Text":"if T_2 is greater than T_1,"},{"Start":"03:46.340 ","End":"03:49.590","Text":"that means we\u0027re increasing the temperature."},{"Start":"03:53.540 ","End":"03:58.590","Text":"Then 1 over T_1 minus 1 over T_2 will be positive"},{"Start":"03:58.590 ","End":"04:03.095","Text":"because 1 over T_2 will be smaller than 1 over T_1."},{"Start":"04:03.095 ","End":"04:04.960","Text":"That\u0027s positive."},{"Start":"04:04.960 ","End":"04:07.480","Text":"Delta H is positive."},{"Start":"04:07.480 ","End":"04:11.290","Text":"That means that InK_2 over K_1 will also be"},{"Start":"04:11.290 ","End":"04:17.065","Text":"positive because it\u0027s a multiplication of Delta H over R,"},{"Start":"04:17.065 ","End":"04:21.154","Text":"and this expression brackets involving temperature."},{"Start":"04:21.154 ","End":"04:24.430","Text":"If both of them are positive,"},{"Start":"04:24.430 ","End":"04:29.935","Text":"then InK_2 over K_1 will also be positive."},{"Start":"04:29.935 ","End":"04:36.345","Text":"We know that if InK_2 over K_1 is positive,"},{"Start":"04:36.345 ","End":"04:42.640","Text":"it means that K_2 divided by K_1 is greater than 1."},{"Start":"04:44.530 ","End":"04:47.165","Text":"If that\u0027s greater than 1,"},{"Start":"04:47.165 ","End":"04:49.890","Text":"then K_2 is greater than K_1."},{"Start":"04:51.130 ","End":"04:58.115","Text":"That tells us that increasing the temperature increases K,"},{"Start":"04:58.115 ","End":"05:00.410","Text":"it leads to an increasing K,"},{"Start":"05:00.410 ","End":"05:05.375","Text":"and an increase in K favors products."},{"Start":"05:05.375 ","End":"05:12.180","Text":"Reaction\u0027s going from left to right when we increase the temperature."},{"Start":"05:15.650 ","End":"05:21.035","Text":"Raising the temperature favors the forward reaction,"},{"Start":"05:21.035 ","End":"05:25.100","Text":"which we know is endothermic because that\u0027s what we\u0027re talking about."},{"Start":"05:25.100 ","End":"05:30.170","Text":"Raising the temperature favors the forward endothermic reaction."},{"Start":"05:30.170 ","End":"05:36.050","Text":"Now, what happens when the forward reaction is exothermic?"},{"Start":"05:36.050 ","End":"05:40.770","Text":"That\u0027s the forward reaction is exothermic."},{"Start":"05:41.360 ","End":"05:45.715","Text":"Now, Delta H is negative."},{"Start":"05:45.715 ","End":"05:48.440","Text":"Again, we\u0027re raising the temperature,"},{"Start":"05:48.440 ","End":"05:52.585","Text":"so 1 over T_1 minus 1 over T_2 is positive."},{"Start":"05:52.585 ","End":"05:58.040","Text":"That means that In of K_2 over K_1 is going to be negative because it\u0027s"},{"Start":"05:58.040 ","End":"06:03.335","Text":"the multiplication of a negative number and a positive number so it\u0027s negative."},{"Start":"06:03.335 ","End":"06:08.385","Text":"We know that that means that K_2 over K_1"},{"Start":"06:08.385 ","End":"06:16.630","Text":"is less than 1 and that means that K_2 is less than K_1."},{"Start":"06:17.380 ","End":"06:22.970","Text":"Now, what has happened when we raise the temperature is"},{"Start":"06:22.970 ","End":"06:28.205","Text":"that the value of K has decreased."},{"Start":"06:28.205 ","End":"06:30.980","Text":"K has decreased."},{"Start":"06:30.980 ","End":"06:36.605","Text":"That means that the reaction is going to go from right to left."},{"Start":"06:36.605 ","End":"06:39.020","Text":"It will favor the reactants."},{"Start":"06:39.020 ","End":"06:42.655","Text":"Decreasing K favors reactants."},{"Start":"06:42.655 ","End":"06:48.229","Text":"We can say that raising the temperature favors the backward reaction."},{"Start":"06:48.229 ","End":"06:51.875","Text":"If the forward reaction is exothermic,"},{"Start":"06:51.875 ","End":"06:57.710","Text":"then the back reaction will be endothermic."},{"Start":"06:57.710 ","End":"07:01.970","Text":"Raising the temperature when the forward reaction is"},{"Start":"07:01.970 ","End":"07:08.245","Text":"exothermic favors the backward reaction, which is endothermic."},{"Start":"07:08.245 ","End":"07:12.600","Text":"Does this agree with Le Chatelier\u0027s principle?"},{"Start":"07:12.600 ","End":"07:16.339","Text":"Let\u0027s see. An increase in temperature."},{"Start":"07:16.339 ","End":"07:18.230","Text":"That\u0027s going to be our stress."},{"Start":"07:18.230 ","End":"07:24.020","Text":"The increase in temperatures as stress is minimized by shifting the equilibrium in"},{"Start":"07:24.020 ","End":"07:30.455","Text":"the direction of the endothermic reaction because the endothermic reaction absorbs heat."},{"Start":"07:30.455 ","End":"07:32.300","Text":"We\u0027ve raised the heat,"},{"Start":"07:32.300 ","End":"07:33.800","Text":"we\u0027ve raised the temperature,"},{"Start":"07:33.800 ","End":"07:38.840","Text":"and now the endothermic reaction will respond to this by absorbing"},{"Start":"07:38.840 ","End":"07:44.360","Text":"heat so the endothermic reaction will be favored."},{"Start":"07:44.360 ","End":"07:48.140","Text":"Now, if we decrease the temperature,"},{"Start":"07:48.140 ","End":"07:50.960","Text":"then that\u0027s going to be our stress this time,"},{"Start":"07:50.960 ","End":"07:53.138","Text":"a decrease in temperature."},{"Start":"07:53.138 ","End":"07:57.230","Text":"This will be minimized by shifting the equilibrium in the direction of"},{"Start":"07:57.230 ","End":"08:02.524","Text":"the exothermic reaction because the exothermic reaction releases heat."},{"Start":"08:02.524 ","End":"08:06.161","Text":"Reducing the temperature, reducing the amount of heat,"},{"Start":"08:06.161 ","End":"08:09.140","Text":"and the exothermic reaction will"},{"Start":"08:09.140 ","End":"08:13.895","Text":"release heat to compensate for the decrease in temperature."},{"Start":"08:13.895 ","End":"08:16.970","Text":"What we found before is that raising"},{"Start":"08:16.970 ","End":"08:20.180","Text":"the temperature favors the forward reaction and raising"},{"Start":"08:20.180 ","End":"08:23.300","Text":"the temperature favors the backward reaction where we"},{"Start":"08:23.300 ","End":"08:26.825","Text":"have endothermic or exothermic reactions."},{"Start":"08:26.825 ","End":"08:30.739","Text":"This is an agreement with Le Chatelier\u0027s principle."},{"Start":"08:30.739 ","End":"08:33.860","Text":"In this video, we talked about the effects of changing"},{"Start":"08:33.860 ","End":"08:38.040","Text":"the temperature on a reaction at equilibrium."}],"ID":27197},{"Watched":false,"Name":"Van\u0027t Hoff Equation-Example","Duration":"7m 37s","ChapterTopicVideoID":26280,"CourseChapterTopicPlaylistID":253357,"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":"00:04.365","Text":"we developed the Van\u0027t Hoff equation."},{"Start":"00:04.365 ","End":"00:07.845","Text":"In this video, we\u0027ll show how to use it."},{"Start":"00:07.845 ","End":"00:10.575","Text":"Here\u0027s the Van\u0027t Hoff equation."},{"Start":"00:10.575 ","End":"00:15.210","Text":"The ln of K_2 divided by K_1 is equal to Delta H^0 of"},{"Start":"00:15.210 ","End":"00:20.760","Text":"reaction divided by R times 1 over T_1 minus 1 over T_2,"},{"Start":"00:20.760 ","End":"00:26.340","Text":"where K_1 is the equilibrium constant at T_1,"},{"Start":"00:26.340 ","End":"00:30.150","Text":"and K_2 is equilibrium constant at T_2."},{"Start":"00:30.150 ","End":"00:38.970","Text":"Here\u0027s example. For the reaction N_2O_4 in equilibrium with 2NO_2, both gases,"},{"Start":"00:38.970 ","End":"00:48.220","Text":"Delta H^0 reaction is equal to 57.2 kilojoules per mole and K is equal to 0.113,"},{"Start":"00:48.220 ","End":"00:53.490","Text":"and all this is at 298 Kelvin."},{"Start":"00:53.490 ","End":"00:56.580","Text":"Now, the 2 parts to the question."},{"Start":"00:56.580 ","End":"01:02.425","Text":"Part a, calculate K at 350 K,"},{"Start":"01:02.425 ","End":"01:04.745","Text":"and part b,"},{"Start":"01:04.745 ","End":"01:10.230","Text":"at what temperature will K be 0.0113?"},{"Start":"01:10.240 ","End":"01:14.720","Text":"The first thing to note is the reaction is endothermic,"},{"Start":"01:14.720 ","End":"01:16.400","Text":"Delta H is positive,"},{"Start":"01:16.400 ","End":"01:23.735","Text":"so we expect K to increase when the temperature is raised."},{"Start":"01:23.735 ","End":"01:30.440","Text":"Another important point is we should only round up numbers at the end."},{"Start":"01:30.440 ","End":"01:33.350","Text":"The reason for this is often we get"},{"Start":"01:33.350 ","End":"01:38.650","Text":"very small numbers for the difference between 1 over T_1 and 1 over T_2."},{"Start":"01:38.650 ","End":"01:42.425","Text":"If we don\u0027t take enough significant figures,"},{"Start":"01:42.425 ","End":"01:45.690","Text":"we end up with an incorrect answer."},{"Start":"01:45.790 ","End":"01:47.990","Text":"Here\u0027s the first part."},{"Start":"01:47.990 ","End":"01:55.294","Text":"We\u0027re given K_1 and that\u0027s 0.113 at T_1 298 Kelvin."},{"Start":"01:55.294 ","End":"01:57.380","Text":"We\u0027re not given K_2,"},{"Start":"01:57.380 ","End":"02:03.760","Text":"we\u0027re expected to find K_2 at T_2, 350 Kelvin."},{"Start":"02:03.860 ","End":"02:06.345","Text":"Let\u0027s write the equation,"},{"Start":"02:06.345 ","End":"02:13.240","Text":"ln of K_2 over K_1 is equal to ln of K_2 over K_1, which is 0.113."},{"Start":"02:13.250 ","End":"02:17.955","Text":"That\u0027s equal to 57.2,"},{"Start":"02:17.955 ","End":"02:23.908","Text":"that\u0027s the value of Delta H^0 but it\u0027s given in kilojoules,"},{"Start":"02:23.908 ","End":"02:27.860","Text":"we need to convert it to joules because R,"},{"Start":"02:27.860 ","End":"02:30.410","Text":"the gas constant is given in joules."},{"Start":"02:30.410 ","End":"02:36.320","Text":"It\u0027s 57.2 times 1,000 joules per mole divide by R,"},{"Start":"02:36.320 ","End":"02:41.440","Text":"which is 8.3145 joules per mole per Kelvin."},{"Start":"02:41.440 ","End":"02:44.520","Text":"Then we have 1 over T_1,"},{"Start":"02:44.520 ","End":"02:48.240","Text":"which is 1 over 298 minus 1 over T_2,"},{"Start":"02:48.240 ","End":"02:51.970","Text":"which is 1 over 350."},{"Start":"02:51.970 ","End":"02:56.375","Text":"The units of that are Kelvin to the power of minus 1."},{"Start":"02:56.375 ","End":"02:58.955","Text":"Let\u0027s look at the units first."},{"Start":"02:58.955 ","End":"03:06.270","Text":"Joules cancels, moles with power minus 1 cancels, K^-1 cancels."},{"Start":"03:06.270 ","End":"03:08.270","Text":"As we need to have,"},{"Start":"03:08.270 ","End":"03:12.427","Text":"this is dimensionless because the ln is always dimensionless,"},{"Start":"03:12.427 ","End":"03:15.720","Text":"and this is dimensionless."},{"Start":"03:17.230 ","End":"03:20.795","Text":"Now, if we work out the right-hand side here,"},{"Start":"03:20.795 ","End":"03:23.770","Text":"we get for this first factor,"},{"Start":"03:23.770 ","End":"03:25.965","Text":"Delta H over R,"},{"Start":"03:25.965 ","End":"03:30.630","Text":"6.8795 times 10^3,"},{"Start":"03:30.630 ","End":"03:33.035","Text":"and for the difference in the temperatures,"},{"Start":"03:33.035 ","End":"03:36.410","Text":"4.986 times 10^-4,"},{"Start":"03:36.410 ","End":"03:38.650","Text":"you see it\u0027s a small number."},{"Start":"03:38.650 ","End":"03:43.020","Text":"When we multiply this out, we get 3.430."},{"Start":"03:43.020 ","End":"03:52.870","Text":"The conclusion is ln K_2 over 0.113 is equal to 3.430."},{"Start":"03:53.240 ","End":"03:57.715","Text":"Now we want to find K_2,"},{"Start":"03:57.715 ","End":"04:00.800","Text":"so we take the exponential of both sides of the equation."},{"Start":"04:00.800 ","End":"04:10.185","Text":"Exponential of ln K_2 over 0.113 is just K_2 over 0.113."},{"Start":"04:10.185 ","End":"04:12.746","Text":"Then we have that\u0027s equal to the e^3.430."},{"Start":"04:12.746 ","End":"04:14.790","Text":"The e^3.430"},{"Start":"04:14.790 ","End":"04:24.870","Text":"is 30.87."},{"Start":"04:24.870 ","End":"04:30.660","Text":"We have K_2 over 0.113 is equal to 30.87."},{"Start":"04:30.660 ","End":"04:39.510","Text":"We can see that K_2 is equal to 0.113 times 30.87, and that\u0027s 3.49."},{"Start":"04:39.510 ","End":"04:44.535","Text":"We see that K_2 is larger than K_1."},{"Start":"04:44.535 ","End":"04:48.220","Text":"Here\u0027s K_2, 3.49,"},{"Start":"04:49.130 ","End":"04:53.735","Text":"it\u0027s larger than K_1 as we expected."},{"Start":"04:53.735 ","End":"04:56.330","Text":"Let\u0027s look at the second part of the problem."},{"Start":"04:56.330 ","End":"05:01.399","Text":"We\u0027re given K_1, 0.113 at T_1, 298 Kelvin,"},{"Start":"05:01.399 ","End":"05:03.230","Text":"we\u0027re given K_2,"},{"Start":"05:03.230 ","End":"05:08.020","Text":"0.0113, and we have to find T_2."},{"Start":"05:08.020 ","End":"05:14.340","Text":"So ln K_2 over K_1 is equal to ln 0.0113,"},{"Start":"05:14.340 ","End":"05:18.480","Text":"that\u0027s K_2 divided by 0.113 K_1,"},{"Start":"05:18.480 ","End":"05:25.265","Text":"and that\u0027s the ln of 0.1 and that\u0027s minus 2.30259."},{"Start":"05:25.265 ","End":"05:28.265","Text":"That\u0027s the left-hand side of the equation."},{"Start":"05:28.265 ","End":"05:31.970","Text":"We look at the right-hand side of the Van\u0027t Hoff equation."},{"Start":"05:31.970 ","End":"05:35.195","Text":"First of all, we have Delta H over R,"},{"Start":"05:35.195 ","End":"05:38.060","Text":"and that we found out before."},{"Start":"05:38.060 ","End":"05:48.535","Text":"Here it is, so we can put it in here, 6.8795 times 10^3."},{"Start":"05:48.535 ","End":"05:53.400","Text":"That\u0027s multiplied by 1 over 298 minus 1 over T_2,"},{"Start":"05:53.400 ","End":"05:56.545","Text":"and this is the T_2 we need to find."},{"Start":"05:56.545 ","End":"06:00.695","Text":"Let\u0027s isolate this factor in the brackets."},{"Start":"06:00.695 ","End":"06:05.160","Text":"1 over 298 minus 1 over T_2 is equal to minus"},{"Start":"06:05.160 ","End":"06:12.495","Text":"2.30259 divided by 6.8795 times 10^3."},{"Start":"06:12.495 ","End":"06:14.055","Text":"We divide these 2,"},{"Start":"06:14.055 ","End":"06:20.860","Text":"we get minus 3.3470 times 10^-4."},{"Start":"06:22.160 ","End":"06:25.950","Text":"So 1 over T_2 is equal to"},{"Start":"06:25.950 ","End":"06:33.330","Text":"3.3570 times 10^-3,"},{"Start":"06:33.330 ","End":"06:40.560","Text":"that\u0027s 1 over 298 plus 3.3470 times 10^-4,"},{"Start":"06:40.560 ","End":"06:44.295","Text":"that\u0027s this number here."},{"Start":"06:44.295 ","End":"06:53.830","Text":"Now we worked that out, we get 1 over T_2 is equal to 3.6904 times 10^-3."},{"Start":"06:54.350 ","End":"06:59.760","Text":"Now, we\u0027re in a position to find out the value of T_2."},{"Start":"06:59.760 ","End":"07:05.402","Text":"T_2 is just equal to the inverse of this number here,"},{"Start":"07:05.402 ","End":"07:13.530","Text":"so T_2 is 270.97 and we can round that up to 271."},{"Start":"07:13.700 ","End":"07:17.910","Text":"We can write that here, 271."},{"Start":"07:17.910 ","End":"07:26.475","Text":"You can see that T_1 is larger than T_2,"},{"Start":"07:26.475 ","End":"07:32.605","Text":"and that fits in with K_1 being larger than K_2."},{"Start":"07:32.605 ","End":"07:37.409","Text":"In this video, we showed how to use the Van\u0027t Hoff equation."}],"ID":27194}],"Thumbnail":null,"ID":253357}]

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