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Gibbs Free Energy - Equilibrium Constant, Enthalpy & Entropy - Equations & Practice Problems
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00:00:00.000 in this video we're going to focus on 00:00:02.94900:00:02.959 Delta s entropy enthalpy Delta H and 00:00:07.03900:00:07.049 Delta G gives free energy so let's start 00:00:10.49000:00:10.500 with Delta s Delta s is associated with 00:00:14.62900:00:14.639 entropy and it's basically a measure of 00:00:17.51000:00:17.520 disorder so the more disorder a system 00:00:21.23000:00:21.240 has the higher the entropy so we need to 00:00:24.95000:00:24.960 know is that solids solids have low 00:00:30.74000:00:30.750 entropy values because the atoms in the 00:00:34.45900:00:34.469 solid are more organized than the atoms 00:00:37.01000:00:37.020 in the gas gas molecules they have a 00:00:41.33000:00:41.340 very high entropy value so let's say if 00:00:48.08000:00:48.090 you have a reaction let's say carbon 00:00:56.74000:00:56.750 reacts with oxygen gas to produce two 00:01:06.64900:01:06.659 carbon dioxide molecules now this 00:01:12.38000:01:12.390 reaction would you say the change in 00:01:15.17000:01:15.180 entropy is positive or is it negative 00:01:19.45000:01:19.460 how would you predict the sign of Delta 00:01:21.85900:01:21.869 s so notice that we have two gas 00:01:25.21900:01:25.229 molecules on the right side but only one 00:01:27.56000:01:27.570 gas model can left and solid so we said 00:01:30.53000:01:30.540 have a very low entropy value so we 00:01:32.51000:01:32.520 don't have to worry about it so because 00:01:34.52000:01:34.530 we have more gas molecules on the right 00:01:36.76000:01:36.770 the reaction went from a low entropy 00:01:40.24900:01:40.259 State to a high entropy State so 00:01:42.14000:01:42.150 therefore we can predict that the sign 00:01:44.12000:01:44.130 of Delta s will be positive we can say 00:01:46.34000:01:46.350 that the entropy increases as the 00:01:48.77000:01:48.780 reaction goes from left to right now 00:01:54.05000:01:54.060 what about 00:01:57.03000:01:57.040 the condensation of water as water goes 00:02:00.69000:02:00.700 from a gas to a liquid predict the sign 00:02:03.81000:02:03.820 of Delta s for this reaction so as we 00:02:08.07000:02:08.080 mentioned before gases they have high 00:02:10.05000:02:10.060 entropy values and liquids relative to a 00:02:13.95000:02:13.960 gas they have low entropy values so if 00:02:16.62000:02:16.630 you're going from high to low the change 00:02:18.54000:02:18.550 in entropy is negative the entropy of 00:02:21.00000:02:21.010 the system is decreasing now what about 00:02:27.98000:02:27.990 this example let's say if we have two 00:02:32.25000:02:32.260 sulfur dioxide molecules and reacts with 00:02:36.45000:02:36.460 a single oxygen molecule to produce two 00:02:43.02000:02:43.030 sulphur trioxide molecules is the entry 00:02:48.21000:02:48.220 change of the system is it positive or 00:02:49.92000:02:49.930 is it negative what would you say so the 00:02:54.18000:02:54.190 phases are all the same so let's look at 00:02:55.89000:02:55.900 the number of gas molecules we have 00:02:57.18000:02:57.190 three gas molecules on the left two on 00:02:59.16000:02:59.170 the right so we're going from a system 00:03:01.71000:03:01.720 of high entropy to a system of low 00:03:04.35000:03:04.360 entropy so since the change in entropy 00:03:06.93000:03:06.940 is negative Delta s or the entropy is 00:03:10.97900:03:10.989 decreasing so Delta s is negative 00:03:16.00900:03:16.019 consider this example let's say if we 00:03:18.12000:03:18.130 have magnesium oxide which is a solid 00:03:21.33000:03:21.340 and it reacts with gaseous carbon 00:03:25.17000:03:25.180 dioxide to produce solid magnesium 00:03:30.44900:03:30.459 carbonate predict a sign of Delta s for 00:03:35.34000:03:35.350 this reaction so we have a solid and a 00:03:39.03000:03:39.040 gas on the left side so because we have 00:03:41.25000:03:41.260 the gas we have a system of high entropy 00:03:43.86000:03:43.870 on the left side we only have a solid on 00:03:46.08000:03:46.090 the right so the entropy on the right is 00:03:49.50000:03:49.510 low so as we go from high to low Delta s 00:03:52.32000:03:52.330 is negative since the entropy is 00:03:54.66000:03:54.670 decreasing as you go from left to right 00:04:00.05000:04:00.060 now let's say if you have two gases 00:04:04.47000:04:04.480 let's say carbon dioxide but one of them 00:04:09.42000:04:09.430 is at a temperature of 300 Kelvin and 00:04:11.28000:04:11.290 the other sample is at 400 Kelvin which 00:04:15.80000:04:15.810 system has a higher entropy value is it 00:04:21.42000:04:21.430 the one on the left or the one on the 00:04:22.74000:04:22.750 right now if you have the same substance 00:04:27.12000:04:27.130 at the same phase it's going to be the 00:04:30.06000:04:30.070 one with the higher temperature this is 00:04:32.43000:04:32.440 going to have a higher entropy system 00:04:33.96000:04:33.970 because at a higher temperature the gas 00:04:36.71900:04:36.729 molecules move faster in more chaotic 00:04:38.79000:04:38.800 and so you have less organization and 00:04:42.81000:04:42.820 keep in mind as you increase the 00:04:43.86000:04:43.870 temperature of a solid the solid will 00:04:46.46900:04:46.479 eventually melt into a liquid and if you 00:04:48.57000:04:48.580 continue to heat it it will become a gas 00:04:50.31000:04:50.320 so raising the temperature causes the 00:04:52.52900:04:52.539 system to be to have a higher entropy 00:04:55.05000:04:55.060 value to be more chaotic and so the one 00:04:59.01000:04:59.020 with the higher temperature has the 00:05:00.39000:05:00.400 higher entry value and the one of the 00:05:02.10000:05:02.110 lower temperature has the lower entropy 00:05:04.17000:05:04.180 value now let's say if you have three 00:05:09.87000:05:09.880 compounds calcium carbonate actually 00:05:14.25000:05:14.260 let's say calcium oxide 00:05:21.17000:05:21.180 liquid water 00:05:27.62000:05:27.630 selenium dioxide gas and selenium 00:05:33.59000:05:33.600 trioxide gas if you have to rank these 00:05:39.26000:05:39.270 in order of increase in entropy how 00:05:42.59000:05:42.600 would you do it let's say number one has 00:05:44.00000:05:44.010 the lowest entropy and number four has 00:05:46.07000:05:46.080 the highest well we know solids have 00:05:49.64000:05:49.650 very low entropy so that's going to be 00:05:51.20000:05:51.210 the lowest then it's the liquid and then 00:05:53.15000:05:53.160 it's the gas now between these two gas 00:05:55.19000:05:55.200 molecules which one has more entropy 00:05:58.24000:05:58.250 when comparing two molecules at the same 00:06:00.56000:06:00.570 temperature the one that has a higher 00:06:02.87000:06:02.880 entropy is the one that has more atoms 00:06:04.25000:06:04.260 two one it's more complicated in 00:06:05.63000:06:05.640 structure so this is going to be number 00:06:08.03000:06:08.040 three and because this has four atoms 00:06:09.74000:06:09.750 instead of three this is number four 00:06:11.84000:06:11.850 this one has the highest entropy now 00:06:16.49000:06:16.500 consider this reaction 00:06:32.29000:06:32.300 let's say if you're given the entropy 00:06:36.40000:06:36.410 values for each of these gases so let's 00:06:43.54000:06:43.550 say this reaction occurs at a 00:06:44.83000:06:44.840 temperature where water is in a gaseous 00:06:47.14000:06:47.150 state as opposed to liquid state and 00:06:49.54000:06:49.550 let's say the entry value for hydrogen 00:06:51.85000:06:51.860 is about 131 402 it's about 205 and for 00:06:57.70000:06:57.710 water let's say it's about 189 how can 00:07:01.27000:07:01.280 you use this data to calculate the 00:07:02.74000:07:02.750 change in entropy for the reaction to 00:07:06.03000:07:06.040 find the change in entropy for the 00:07:08.08000:07:08.090 reaction you would you can use the same 00:07:09.85000:07:09.860 method as you would to find the change 00:07:12.10000:07:12.110 in enthalpy Delta H for the reaction 00:07:13.89000:07:13.900 which is basically the sum of the 00:07:16.00000:07:16.010 products minus the sum of the reactants 00:07:21.48000:07:21.490 so for the products we have just two 00:07:25.27000:07:25.280 water molecules and for the reactants 00:07:29.98000:07:29.990 which is on the Left we have two 00:07:31.96000:07:31.970 hydrogen molecules one oxygen molecule 00:07:33.82000:07:33.830 and so we just got to plug in the 00:07:35.50000:07:35.510 information so it's going to be two 00:07:38.02000:07:38.030 times 189 minus 2 times 1 31 00:07:46.29000:07:46.300 Plus 205 for o2 00:07:56.31000:07:56.320 so two times 189 that's about 378 and 00:08:03.81000:08:03.820 don't forget to distribute the negative 00:08:05.62000:08:05.630 sign 2 times 131 that's negative 262 00:08:08.92000:08:08.930 we're going to have negative 205 so if 00:08:12.61000:08:12.620 you combine these numbers you should get 00:08:15.51000:08:15.520 an entropy value of negative 89 now if 00:08:20.47000:08:20.480 you're using a standard thermodynamic 00:08:21.58000:08:21.590 table it's going to have the unit's 00:08:23.86000:08:23.870 joules per mole per Kelvin and so that's 00:08:27.43000:08:27.440 how you can calculate the entropy of a 00:08:29.38000:08:29.390 reaction now there are some other 00:08:33.33900:08:33.349 equations that you need to know 00:08:34.71900:08:34.729 regarding entropy Delta s can be 00:08:38.32000:08:38.330 calculated by taking Q the amount of 00:08:41.35000:08:41.360 heat that's absorbed or released in the 00:08:43.33000:08:43.340 system divided by the temperature and 00:08:46.74000:08:46.750 this must be a reversible process by the 00:08:49.42000:08:49.430 way so keep in mind Delta s has unit 00:08:54.37000:08:54.380 joules per mole per Kelvin which means Q 00:08:58.81000:08:58.820 has to have the unit's joules per mole 00:09:01.06000:09:01.070 and the temperature of course is going 00:09:03.43000:09:03.440 to be in Kelvin now there are some other 00:09:06.55000:09:06.560 things need to know about entropy 00:09:09.87000:09:09.880 especially as it relates to the second 00:09:12.28000:09:12.290 law of thermodynamics for a spontaneous 00:09:17.83000:09:17.840 reaction the entropy of the universe 00:09:21.00000:09:21.010 which is the sum of the change in 00:09:23.89000:09:23.900 entropy of the system plus the change in 00:09:28.84000:09:28.850 entropy of the surroundings is always 00:09:33.43000:09:33.440 greater than zero for process that is 00:09:36.13000:09:36.140 spontaneous would basically the second 00:09:38.44000:09:38.450 law of thermodynamics states that the 00:09:40.00000:09:40.010 entropy of the universe increases for a 00:09:42.52000:09:42.530 spontaneous process now for a process 00:09:47.47000:09:47.480 that is reversible where the system is 00:09:51.34000:09:51.350 at equilibrium than the entropy of the 00:09:53.86000:09:53.870 universe which is the sum of the entropy 00:09:59.98000:09:59.990 of the system plus the change in entropy 00:10:02.80000:10:02.810 of the surroundings that's going to be 00:10:07.18000:10:07.190 equal to zero 00:10:09.18000:10:09.190 if it's a reversible process at 00:10:13.09000:10:13.100 equilibrium 00:10:21.07900:10:21.089 now if it's spontaneous then it's going 00:10:25.69900:10:25.709 to be greater than zero the entropy of 00:10:27.13900:10:27.149 the universe increases for a spontaneous 00:10:29.05900:10:29.069 process but if it's reversible then the 00:10:33.49900:10:33.509 entropy of the universe can equal zero 00:10:35.47900:10:35.489 if it's at a system of equilibrium 00:10:39.18900:10:39.199 now another equation that you need to 00:10:41.41900:10:41.429 know is this one Delta G which 00:10:44.80900:10:44.819 represents the change in free energy 00:10:47.34900:10:47.359 also known as Gibbs free energy is equal 00:10:49.84900:10:49.859 to Delta H to change the enthalpy change 00:10:52.35900:10:52.369 minus T times Delta s now Delta G 00:10:57.87900:10:57.889 represents the amount of free energy 00:11:01.12900:11:01.139 that's available to do useful work and 00:11:04.77900:11:04.789 it usually has the unit's it could be in 00:11:07.66900:11:07.679 joules per mole or kilojoules per mole 00:11:11.61900:11:11.629 Delta H is usually provided in 00:11:13.84900:11:13.859 kilojoules per mole the temperature is 00:11:17.41900:11:17.429 in Kelvin now Delta s is usually 00:11:20.11900:11:20.129 provided in joules per mole but you need 00:11:21.70900:11:21.719 to convert it to kilojoules per mole to 00:11:25.12900:11:25.139 make this equation work now Delta H is 00:11:30.16900:11:30.179 usually the amount of heat energy that's 00:11:32.38900:11:32.399 absorbed or release in a system at a 00:11:34.63900:11:34.649 constant pressure Delta G is the amount 00:11:37.66900:11:37.679 of free energy that's available to do 00:11:39.25900:11:39.269 useful work Delta S which is entropy we 00:11:43.72900:11:43.739 described it as a measure of disorder of 00:11:45.55900:11:45.569 a system but it can also be described as 00:11:47.86900:11:47.879 the energy that's unavailable to do 00:11:51.58900:11:51.599 useful work so typically when you 00:11:54.43900:11:54.449 whenever you have a process let's say 00:11:55.75900:11:55.769 like a combustible heat engine and if 00:12:00.27900:12:00.289 you can generate heat let's say if you 00:12:04.09900:12:04.109 have a system at high temperature and 00:12:07.72900:12:07.739 let's say the surroundings is at low 00:12:09.25900:12:09.269 temperature heat flows from hot to cold 00:12:14.10900:12:14.119 now some of that heat energy can be used 00:12:16.57900:12:16.589 to do useful work however some of the 00:12:20.86900:12:20.879 heat energy will also radiate out into 00:12:23.74900:12:23.759 the surroundings and that energy will be 00:12:25.57900:12:25.589 lost and so you can think of entropy as 00:12:30.37900:12:30.389 the dispersal of energy in a way 00:12:33.62000:12:33.630 as that heat energy leaves into the 00:12:35.44900:12:35.459 surrounding to the temperature of the 00:12:36.74000:12:36.750 surrounding to go up and whenever the 00:12:38.30000:12:38.310 temperature goes up the entropy of that 00:12:40.46000:12:40.470 of the surroundings will go up as well 00:12:42.62000:12:42.630 and so that heat energy loss it's hard 00:12:46.97000:12:46.980 to get it back it's hard to use it to do 00:12:48.86000:12:48.870 what useful work because it's gone so 00:12:53.30000:12:53.310 another way to describe entropy is the 00:12:55.73000:12:55.740 dispersal of energy or the energy that's 00:12:58.06900:12:58.079 unavailable to do useful work now there 00:13:02.26900:13:02.279 are some other things you need to know 00:13:03.61000:13:03.620 when Delta G is negative the reaction is 00:13:09.37900:13:09.389 spontaneous in the forward direction so 00:13:13.10000:13:13.110 if it's spontaneous in the forward 00:13:14.30000:13:14.310 direction that means it's a non 00:13:16.43000:13:16.440 spontaneous in the reverse direction 00:13:18.97000:13:18.980 when Delta G is zero the system is at 00:13:23.15000:13:23.160 equilibrium and so the rate of the 00:13:27.01900:13:27.029 forward reaction equals the rate of the 00:13:28.75900:13:28.769 reverse reaction so at that point the 00:13:31.73000:13:31.740 reaction is reversible and when Delta G 00:13:34.93900:13:34.949 is positive the reaction is non 00:13:38.68900:13:38.699 spontaneous in the four direction which 00:13:42.43900:13:42.449 means that it's spontaneous in the 00:13:44.30000:13:44.310 reverse direction so those are some 00:13:47.36000:13:47.370 things that you need to know regarding 00:13:48.67900:13:48.689 the sign of Delta G now when Delta H is 00:13:55.63900:13:55.649 negative that means that the reaction is 00:13:58.24900:13:58.259 exothermic that means that heat energy 00:14:01.12000:14:01.130 is being released from the system into 00:14:04.51900:14:04.529 the ceramics and when Delta H is 00:14:07.16000:14:07.170 positive the reaction is endothermic 00:14:09.74000:14:09.750 that means heat energy is being absorbed 00:14:12.53000:14:12.540 by the system from the surroundings now 00:14:20.92900:14:20.939 here's the question for you let's say if 00:14:22.62900:14:22.639 you have the enthalpy change of the 00:14:25.46000:14:25.470 reaction let's say it's negative 100 00:14:31.60000:14:31.610 kilojoules per mole and you also have 00:14:35.62900:14:35.639 the entropy change of the reaction let's 00:14:38.96000:14:38.970 say this is positive 200 joules per mole 00:14:45.04900:14:45.059 per Kelvin 00:14:48.85000:14:48.860 S is always positive but the change in s 00:14:52.18900:14:52.199 can be positive or negative by the way 00:14:55.68900:14:55.699 now let's say the temperature is 300 00:14:58.42900:14:58.439 Kelvin using this information how can 00:15:01.18900:15:01.199 you calculate Delta G of this reaction 00:15:04.11900:15:04.129 how can you find the amount of free 00:15:07.34000:15:07.350 energy that is available to do useful 00:15:09.47000:15:09.480 work so let's use the equation Delta G 00:15:15.04900:15:15.059 is equal to Delta H minus T Delta s so 00:15:19.30900:15:19.319 Delta H is in kilojoules per mole minus 00:15:24.98000:15:24.990 the temperature now Delta s is in joules 00:15:27.55900:15:27.569 per mole per Kelvin so if you plug in 00:15:29.96000:15:29.970 200 you will not get the answer right 00:15:31.46000:15:31.470 you need to make sure the unit's match 00:15:32.98900:15:32.999 so we need to divide 200 by a thousand 00:15:35.98900:15:35.999 to convert joules in to kilojoules so 00:15:39.17000:15:39.180 200 joules is the same as point to 00:15:41.42000:15:41.430 kilojoules and now it can work so this 00:15:44.48000:15:44.490 is going to be negative 100 and 300 00:15:47.36000:15:47.370 times 0.2 well 3 times 300 times 2 is 00:15:52.24900:15:52.259 600 so 300 times 0.2 must be 60 so this 00:15:56.15000:15:56.160 is going to be negative 60 and so Delta 00:16:01.42900:16:01.439 G is equal to negative 116 kilojoules 00:16:04.73000:16:04.740 per mole for this particular example so 00:16:07.51900:16:07.529 whenever you use the equation make sure 00:16:10.00900:16:10.019 that these two units match if you want 00:16:12.88900:16:12.899 to get the right answer now let's say if 00:16:17.38900:16:17.399 you have Delta H for a certain process 00:16:21.74000:16:21.750 let's say it's 300 kilojoules per mole 00:16:25.61000:16:25.620 and you also have Delta S which is 00:16:35.56000:16:35.570 150 joules per mole per Kelvin now let's 00:16:41.96000:16:41.970 say these values correspond to the 00:16:46.19000:16:46.200 vaporization of a substance how can you 00:16:48.92000:16:48.930 calculate the boiling point of that 00:16:52.07000:16:52.080 substance used in Delta H and Delta s so 00:16:55.25000:16:55.260 let's say the substance is B we'll call 00:16:57.80000:16:57.810 it um some substance B at the boiling 00:17:03.50000:17:03.510 point the liquid and the gas phase they 00:17:07.76000:17:07.770 coexist so at that temperature these two 00:17:11.09000:17:11.100 they're reversible they can the liquid 00:17:13.34000:17:13.350 can go into the gas phase and the gas 00:17:14.99000:17:15.000 molecules can go back into the liquid 00:17:16.46000:17:16.470 phase so we have a reversible system so 00:17:19.64000:17:19.650 how can we estimate the temperature of 00:17:21.61000:17:21.620 the boiling point of the substance so we 00:17:27.98000:17:27.990 can use this equation Delta G is equal 00:17:29.93000:17:29.940 to Delta H minus T Delta s so at the 00:17:35.00000:17:35.010 boiling point or at the phase change of 00:17:38.00000:17:38.010 a substance in this case the liquid and 00:17:42.20000:17:42.210 gas coexist because we have a reversible 00:17:44.21000:17:44.220 reaction the system is at equilibrium 00:17:46.27000:17:46.280 whenever the system is at equilibrium 00:17:48.08000:17:48.090 Delta G is zero and so since we have the 00:17:51.83000:17:51.840 values of Delta H and Delta s we can 00:17:54.26000:17:54.270 solve for T let's add T Delta s to both 00:17:59.00000:17:59.010 sides 00:18:04.30000:18:04.310 you 00:18:10.97000:18:10.980 so this is what we now have now to solve 00:18:14.00000:18:14.010 for the temperature and when you set 00:18:15.23000:18:15.240 divide both sides by Delta s and so the 00:18:20.15000:18:20.160 temperature or the boiling point 00:18:23.09000:18:23.100 temperature is equal to the change in 00:18:25.58000:18:25.590 enthalpy divided by the change in 00:18:27.23000:18:27.240 entropy so that's how you can find the 00:18:29.48000:18:29.490 boiling point of a substance if you know 00:18:32.78000:18:32.790 Delta H Delta F so you could just divide 00:18:34.52000:18:34.530 the two so Delta H is 300 kilojoules 00:18:41.50000:18:41.510 well actually let's convert kilojoules 00:18:44.00000:18:44.010 inch into joules because these units 00:18:45.74000:18:45.750 they have to match so 300 kilojoules is 00:18:51.23000:18:51.240 the same as 300 thousand joules to 00:18:54.92000:18:54.930 convert kilojoules to joules you need to 00:18:57.83000:18:57.840 multiply by a thousand and to convert 00:19:00.32000:19:00.330 Joules to kilojoules he needs to divide 00:19:03.20000:19:03.210 by a thousand so since we're going from 00:19:05.90000:19:05.910 kilojoules to joules we'll multiply 00:19:07.16000:19:07.170 three hundred by thousand so we have 00:19:09.23000:19:09.240 three hundred thousand joules per mole 00:19:10.61000:19:10.620 and Delta s we can leave it as 150 00:19:13.25000:19:13.260 joules per mole per Kelvin 00:19:19.96000:19:19.970 so the units joules per mole will cancel 00:19:23.11000:19:23.120 given us the Kelvin temperature so 00:19:26.25900:19:26.269 what's three hundred thousand divided by 00:19:28.36000:19:28.370 150 well we know we can at least cancel 00:19:33.00900:19:33.019 a zero and so what we now have is 30,000 00:19:38.85000:19:38.860 divided by 15 30,000 is basically 30 00:19:43.02900:19:43.039 times a thousand and 30 is 15 times two 00:19:49.76900:19:49.779 so we can cancel the 15s so what we have 00:19:53.91900:19:53.929 left over is two times a thousand which 00:19:56.37900:19:56.389 is two thousand so for this particular 00:19:58.99000:19:59.000 substance based on the information that 00:20:00.66900:20:00.679 we have the boiling point is two 00:20:03.12900:20:03.139 thousand Kelvin now if we want to we can 00:20:09.73000:20:09.740 convert Kelvin into Celsius by 00:20:12.82000:20:12.830 subtracting it by 273 so 17 this is 00:20:18.82000:20:18.830 going to be 17 27 degrees Celsius make 00:20:24.22000:20:24.230 sure my math is correct yeah that's 00:20:27.58000:20:27.590 right so keep in mind the Kelvin 00:20:29.35000:20:29.360 temperature is the Celsius temperature 00:20:31.45000:20:31.460 plus 273 so if you need to find a 00:20:33.85000:20:33.860 Celsius temperature and subtract the 00:20:35.28900:20:35.299 Kelvin temperature by 273 now the next 00:20:42.78900:20:42.799 thing we need to talk about is how to 00:20:44.74000:20:44.750 tell if a reaction is going to be 00:20:47.25900:20:47.269 spontaneous at low temperature high 00:20:50.32000:20:50.330 temperature or if it's always 00:20:54.85000:20:54.860 spontaneous or never spontaneous so you 00:20:58.50900:20:58.519 can find this out based on the equation 00:21:00.46000:21:00.470 Delta G is equal to Delta H minus T 00:21:02.47000:21:02.480 Delta s but first let me give you the 00:21:04.45000:21:04.460 table they need to know 00:21:43.88000:21:43.890 l stands for low temperature H stands 00:21:46.66900:21:46.679 for high temperature 00:21:55.07000:21:55.080 so let's say if Delta H is positive and 00:21:59.04900:21:59.059 if Delta s is positive is it going to be 00:22:03.16900:22:03.179 spontaneous at low temperature or at 00:22:04.97000:22:04.980 high temperature now if both were 00:22:09.68000:22:09.690 positive it's going to be spontaneous at 00:22:13.02900:22:13.039 high temperature so that means that it's 00:22:18.13900:22:18.149 going to be spawned non spontaneous at 00:22:20.06000:22:20.070 low temperature so at little temperature 00:22:24.35000:22:24.360 the sine is positive which means that 00:22:26.29900:22:26.309 it's non spontaneous and at high 00:22:28.94000:22:28.950 temperature it's negative which means 00:22:32.29900:22:32.309 that it's spontaneous so I want you to 00:22:34.46000:22:34.470 understand the way I wrote it 00:22:36.22000:22:36.230 so I'm going to highlight at high 00:22:39.44000:22:39.450 temperature it's spontaneous so those 00:22:44.23900:22:44.249 two correspond to each other now what 00:22:45.88900:22:45.899 about if Delta H is positive and if 00:22:48.79900:22:48.809 Delta s is negative is the spontaneous 00:22:52.31000:22:52.320 non spontaneous how would you describe 00:22:55.19000:22:55.200 it now Delta H is positive for this term 00:22:58.54900:22:58.559 is positive and if Delta s is negative 00:23:00.87900:23:00.889 and while T Reddy has a negative sign 00:23:03.32000:23:03.330 plus this is negative which means it's 00:23:05.06000:23:05.070 positive so everything is positive so it 00:23:08.09000:23:08.100 doesn't matter what the temperature is 00:23:09.16900:23:09.179 it's always going to be positive so 00:23:12.97900:23:12.989 regardless if the system is at low 00:23:15.20000:23:15.210 temperature or at high temperature Delta 00:23:17.45000:23:17.460 G is always positive which means that 00:23:19.03900:23:19.049 it's non spontaneous for all temperature 00:23:21.49900:23:21.509 whenever Delta H is positive and Delta s 00:23:23.84000:23:23.850 is negative now what about the next one 00:23:26.57000:23:26.580 when Delta H is negative but Delta s is 00:23:28.70000:23:28.710 positive so if Delta s is positive what 00:23:34.46000:23:34.470 that means is that the first term is 00:23:36.56000:23:36.570 going to be negative and the second term 00:23:38.62900:23:38.639 negative and a positive is negative 00:23:39.91900:23:39.929 whenever you add two negative numbers is 00:23:42.01900:23:42.029 always going to be negative so at any 00:23:44.59900:23:44.609 temperature under these conditions is 00:23:47.06000:23:47.070 always Delta G will always be negative 00:23:48.56000:23:48.570 so it's always going to be spontaneous 00:23:50.52900:23:50.539 so I'm just going to put a negative sign 00:23:52.54900:23:52.559 which means that it's always spontaneous 00:23:54.28900:23:54.299 regardless of the temperature if it's a 00:23:56.59900:23:56.609 low or high now when Delta H and Delta s 00:24:00.44000:24:00.450 are both negative it's going to be 00:24:03.32000:24:03.330 spontaneous at low temperature which 00:24:06.59000:24:06.600 means that high temperature is not 00:24:08.02900:24:08.039 enos so I'm going to put a box for 00:24:10.43000:24:10.440 also-- at low temperature and 00:24:11.65900:24:11.669 spontaneous because sometimes you might 00:24:15.01900:24:15.029 have a question saying okay under what 00:24:18.44000:24:18.450 conditions of enthalpy and entropy is it 00:24:20.41900:24:20.429 always spontaneous that's the only time 00:24:24.04900:24:24.059 it's always spontaneous if you have an 00:24:26.41900:24:26.429 exothermic reaction and if Delta s for 00:24:28.87900:24:28.889 the reaction is positive then it's 00:24:30.56000:24:30.570 always spontaneous so they might ask you 00:24:34.19000:24:34.200 okay if Delta H is positive and if Delta 00:24:37.07000:24:37.080 s is positive is the reaction 00:24:39.48900:24:39.499 spontaneous that low temperature or at 00:24:42.32000:24:42.330 high temperature you'd have to say it's 00:24:44.29900:24:44.309 spontaneous at high temperature so you 00:24:47.69000:24:47.700 need to be familiar with this table 00:24:49.03900:24:49.049 because it's going to help you to answer 00:24:50.29900:24:50.309 certain questions so let's say if you 00:24:59.02900:24:59.039 have the following reaction sulfur plus 00:25:03.85900:25:03.869 oxygen reacts to form actually think of 00:25:10.51900:25:10.529 a different reaction so hydrogen gas 00:25:15.64900:25:15.659 reacts with nitrogen gas to produce 00:25:18.04900:25:18.059 ammonia and if we balance it it's going 00:25:20.71900:25:20.729 to be to a 1 and a 3 now all of these 00:25:24.25900:25:24.269 are gases and let's say that this 00:25:26.53900:25:26.549 reaction is an exothermic reaction so 00:25:31.81000:25:31.820 here's a question for you will this 00:25:34.70000:25:34.710 reaction be spontaneous at low 00:25:36.73900:25:36.749 temperature at high temperature or is it 00:25:39.16900:25:39.179 always spontaneous or never spontaneous 00:25:41.18000:25:41.190 based on the way it's written in the 00:25:43.24900:25:43.259 forward direction so we know that Delta 00:25:47.06000:25:47.070 H is negative because it's an exothermic 00:25:49.27900:25:49.289 reaction and we can predict the sign of 00:25:52.39900:25:52.409 Delta s by looking at the coefficients 00:25:54.44000:25:54.450 we have for gas molecules on the left 00:25:57.49900:25:57.509 side two on the right side so going from 00:26:00.25900:26:00.269 a system of high entropy to a system of 00:26:02.65900:26:02.669 low entropy which means that Delta s is 00:26:04.87900:26:04.889 negative now when Delta H and Delta s 00:26:07.51900:26:07.529 are both negative is it spontaneous at 00:26:10.51900:26:10.529 high temperatures low temperatures or 00:26:12.91900:26:12.929 all temperatures or at no temperatures 00:26:15.54900:26:15.559 so that's when you need to use a table 00:26:17.93000:26:17.940 if you go back to the table you can see 00:26:20.74900:26:20.759 that it's pond 00:26:21.89000:26:21.900 only at low temperatures but now let's 00:26:26.09000:26:26.100 say if you don't have the table with you 00:26:27.20000:26:27.210 what can you do now the first thing I 00:26:31.58000:26:31.590 will do is start with equation Delta G 00:26:34.25000:26:34.260 is equal to Delta H minus T Delta s let 00:26:39.02000:26:39.030 me just clear away the page first 00:26:45.94000:26:45.950 now choose a value for Delta H and Delta 00:26:48.86000:26:48.870 s just to keep things simple let's make 00:26:51.65000:26:51.660 sure that they're the same what I would 00:26:53.78000:26:53.790 do is choose 100 now Delta H is negative 00:26:56.51000:26:56.520 so I'm going to pick a negative 100 and 00:26:59.44000:26:59.450 Delta s is also negative so I'm going to 00:27:03.20000:27:03.210 choose 100 let's not worry about the 00:27:05.27000:27:05.280 unit's here what we need to focus on is 00:27:07.90000:27:07.910 plugging in a low temperature value and 00:27:10.25000:27:10.260 a high temperature value and see what 00:27:11.75000:27:11.760 happens to the sign of Delta G so you 00:27:15.20000:27:15.210 want to plug in the number that's less 00:27:17.21000:27:17.220 than 140 and the number that's greater 00:27:18.95000:27:18.960 than 140 if Delta H Delta s are both the 00:27:22.04000:27:22.050 same so make sure you just plug in 100 00:27:23.81000:27:23.820 for H and 100 for s so let's plug in a 00:27:27.20000:27:27.210 low value for T let's say 0.1 so this is 00:27:31.49000:27:31.500 going to be negative 100 point one times 00:27:34.16000:27:34.170 100 is 10 so this is new positive 10 so 00:27:37.04000:27:37.050 this is going to be negative 90 so we 00:27:38.93000:27:38.940 can see that it's spontaneous when we 00:27:41.12000:27:41.130 plug in a low temperature a temperature 00:27:42.89000:27:42.900 less than one now what's going to happen 00:27:45.38000:27:45.390 if we plug in a high temperature so 00:27:49.88000:27:49.890 Delta H is going to be the same and 00:27:51.58000:27:51.590 Delta s is still going to be negative 00:27:53.60000:27:53.610 100 but instead of plugging point 1 00:27:55.91000:27:55.920 let's plug in a number that's greater 00:27:56.87000:27:56.880 than 1 like 10 so this is going to be 00:28:00.59000:28:00.600 negative 110 times 1 hundreds of 00:28:02.33000:28:02.340 thousand but it's going to be positive 00:28:03.68000:28:03.690 thousand so now negative 100 plus a 00:28:06.77000:28:06.780 thousand is positive 900 so as we can 00:28:10.16000:28:10.170 see when we plug in a relatively higher 00:28:12.08000:28:12.090 temperature the Delta H became more 00:28:15.77000:28:15.780 positive so it's not spontaneous at high 00:28:21.77000:28:21.780 temperatures but at low temperatures its 00:28:26.75000:28:26.760 spontaneous 00:28:32.48000:28:32.490 so as you can see under low temperature 00:28:36.26900:28:36.279 conditions when Delta H is negative and 00:28:38.34000:28:38.350 Delta s is negative the reaction will be 00:28:40.98000:28:40.990 spontaneous in the forward direction if 00:28:42.65900:28:42.669 you increase the temperature is going to 00:28:44.25000:28:44.260 be non spontaneous so that's how you can 00:28:46.32000:28:46.330 tell if you ever forget the table just 00:28:48.89900:28:48.909 make sure you plug in two values for H 00:28:50.73000:28:50.740 and s then plug in a high temperature 00:28:52.91900:28:52.929 value and the low temperature value and 00:28:54.63000:28:54.640 see what happens to the sign of Delta G 00:29:00.47000:29:00.480 now let's say if we have this reaction 00:29:24.53900:29:24.549 and let's say we're given the values for 00:29:27.95900:29:27.969 the standard change in Delta G for so2 00:29:32.33900:29:32.349 let's say it's about negative 304 h2s 00:29:36.97900:29:36.989 negative 33 and for h2o the gaseous form 00:29:41.43000:29:41.440 let's say negative 229 how can you use 00:29:45.38900:29:45.399 these values to calculate the Delta G of 00:29:50.54900:29:50.559 the reaction to calculate Delta G we can 00:29:54.61900:29:54.629 simply use the sum of the products minus 00:29:58.37900:29:58.389 the sum of the reactants if you're given 00:30:01.85900:30:01.869 the Delta G for each reactant and 00:30:03.77900:30:03.789 product of course so on a product side 00:30:07.25900:30:07.269 we have H 2 s plus 2 H 2 O and on the 00:30:11.43000:30:11.440 reactant side we have simply 3 h2 plus 00:30:15.56900:30:15.579 so2 now the value for H 2 s is negative 00:30:24.11900:30:24.129 33 and for H 2 O it's going to be 2 00:30:28.25900:30:28.269 times negative 2 29 now H 2 is the pure 00:30:35.12900:30:35.139 element so that's going to be 0 and for 00:30:38.65900:30:38.669 so2 it's a negative 300 so this is going 00:30:45.93000:30:45.940 to be negative 33 and 2 times 2 29 00:30:50.93000:30:50.940 that's 458 but it's going to be negative 00:30:53.63900:30:53.649 458 and we got to distribute this 00:30:56.09900:30:56.109 negative sign to this one so it's going 00:30:58.49900:30:58.509 to plus 300 so if we add these numbers 00:31:01.91900:31:01.929 negative 33 minus 458 plus 300 that is 00:31:06.69000:31:06.700 equal to negative 191 kilojoules per 00:31:10.19900:31:10.209 mole 00:31:16.70000:31:16.710 so now that we have the Delta G of the 00:31:20.43000:31:20.440 reaction which is so negative 91 91 00:31:24.84000:31:24.850 kilojoules per mole how can we calculate 00:31:27.96000:31:27.970 the equilibrium constant for the 00:31:30.99000:31:31.000 reaction how can we calculate K using 00:31:34.98000:31:34.990 this value now keep in mind Delta G is 00:31:37.77000:31:37.780 equal to negative RT natural log of K 00:31:41.79000:31:41.800 and this is of course the standard Delta 00:31:44.46000:31:44.470 G value so we got to isolate K the first 00:31:49.14000:31:49.150 thing I will do is divide both sides by 00:31:51.00000:31:51.010 negative RT so then Ln K is equal to 00:31:57.00000:31:57.010 Delta G divided by negative RT now the 00:32:01.50000:32:01.510 base of natural log is e and a property 00:32:06.03000:32:06.040 of logs allows you to convert it to its 00:32:08.91000:32:08.920 exponential form let's say if you have 00:32:10.20000:32:10.210 log base a of B is equal to C a raised 00:32:14.19000:32:14.200 to the C power is equal to B so you can 00:32:16.59000:32:16.600 convert it to this form if you want 00:32:19.66900:32:19.679 that's what we need to do in this 00:32:22.11000:32:22.120 particular example so e raised to 00:32:28.04900:32:28.059 everything on the right is equal to 00:32:30.29900:32:30.309 what's inside of Ln which is K so 00:32:34.56000:32:34.570 therefore we're going to have this 00:32:37.02000:32:37.030 equation K is equal to e raised to the 00:32:44.66900:32:44.679 negative Delta G divided by RT now you 00:32:48.45000:32:48.460 need to pay attention to the units of 00:32:49.98000:32:49.990 our our it's nine point zero eight 206 00:32:53.97000:32:53.980 in this particular equation its 8.3145 00:32:57.06000:32:57.070 and it has units joules per mole per 00:33:00.39000:33:00.400 Kelvin now because R is in joules Delta 00:33:03.36000:33:03.370 G has to be in joules so you can't plug 00:33:05.73000:33:05.740 in negative 191 because that's in 00:33:07.29000:33:07.300 kilojoules it's not going to work you 00:33:09.06000:33:09.070 won't get the right answer you have to 00:33:10.68000:33:10.690 convert kilojoules into joules so to 00:33:14.22000:33:14.230 convert kilojoules into tools we need to 00:33:16.04900:33:16.059 multiply by a thousand if you want to 00:33:18.41900:33:18.429 write out the conversion it's going to 00:33:20.22000:33:20.230 look something like this so we know that 00:33:23.25000:33:23.260 one kilojoule is equal to a thousand 00:33:25.98000:33:25.990 joules and so as we can see the unit's 00:33:28.91900:33:28.929 can 00:33:29.28000:33:29.290 so it's going to be 191 times a thousand 00:33:31.77000:33:31.780 which is 191,000 so that's what we need 00:33:37.71000:33:37.720 to plug in for Delta G so k is equal to 00:33:43.68000:33:43.690 e raised to the negative 191,000 divided 00:33:51.36000:33:51.370 by 8.3145 now let's say the temperature 00:33:55.11000:33:55.120 is at 300 Kelvin so let's multiply that 00:33:58.29000:33:58.300 by 300 so negative one hundred ninety 00:34:04.02000:34:04.030 one thousand divided by 8.3145 that's 00:34:08.85000:34:08.860 like negative twenty three thousand four 00:34:10.62000:34:10.630 hundred fifty and if you divide that by 00:34:12.41900:34:12.429 three hundred you're going to get 00:34:15.47000:34:15.480 negative 78 point one seven now I almost 00:34:23.07000:34:23.080 made a mistake because there's a 00:34:25.98000:34:25.990 negative sign here and Delta G itself is 00:34:29.13000:34:29.140 negative so this should have been like 00:34:30.71000:34:30.720 negative negative so that's two 00:34:34.23000:34:34.240 negatives which is going to be positive 00:34:36.12000:34:36.130 so this is really II to the positive 78 00:34:40.98000:34:40.990 point if I'm one my answer to be more 00:34:44.01000:34:44.020 accurate it's like one six six it's a 00:34:51.90000:34:51.910 lot of sixes 00:34:54.27000:34:54.280 so it's that number so it's e raise to 00:35:00.55000:35:00.560 the seventy eight point one six six 00:35:02.61000:35:02.620 which is about eight point eight five 00:35:07.30000:35:07.310 times ten to the thirty three so that's 00:35:11.26000:35:11.270 the equilibrium constant K for this 00:35:13.15000:35:13.160 reaction notice that K is very large 00:35:16.84000:35:16.850 when Delta G is negative so let's talk 00:35:19.99000:35:20.000 about that when Delta G is a very large 00:35:26.05000:35:26.060 negative value K is going to be 00:35:29.20000:35:29.210 significantly then one in the last 00:35:31.93000:35:31.940 example Delta G was negative 191 00:35:33.91000:35:33.920 kilojoules per mole and K was like 8 00:35:36.16000:35:36.170 times 10 to the 33 that's a very huge 00:35:38.32000:35:38.330 number when K significantly large the 00:35:41.14000:35:41.150 reaction is product favored and so 00:35:44.62000:35:44.630 whenever it's a product favored the 00:35:47.05000:35:47.060 reaction is also spontaneous now when 00:35:53.68000:35:53.690 Delta G is zero this is the standard 00:35:58.51000:35:58.520 Delta G of the reaction K is going to be 00:36:01.27000:36:01.280 approximately close to 1 and so the 00:36:06.40000:36:06.410 system is for the most part at 00:36:08.26000:36:08.270 equilibrium now keep in mind Delta G is 00:36:14.59000:36:14.600 negative RT Ln K so if K is 1 the 00:36:20.86000:36:20.870 natural log of one is equal to zero so 00:36:22.93000:36:22.940 therefore Delta G is zero now the last 00:36:25.99000:36:26.000 one when Delta G is positive K is going 00:36:28.93000:36:28.940 to be significantly less than one but K 00:36:31.15000:36:31.160 is never negative so K is going to be 00:36:32.77000:36:32.780 between zero and one so K is going to be 00:36:38.41000:36:38.420 a very small number like 5 times 10 to 00:36:40.87000:36:40.880 the negative 21 K significally small 00:36:44.20000:36:44.210 it's going to be reactant favored so the 00:36:51.28000:36:51.290 reaction is going to be non spontaneous 00:36:52.96000:36:52.970 in the forward direction which means 00:36:55.99000:36:56.000 it's spontaneous in the reverse 00:36:57.61000:36:57.620 direction or towards the reactant side 00:37:00.19000:37:00.200 cents into the active favored so if it's 00:37:02.92000:37:02.930 to go into the left it's going to be 00:37:04.15000:37:04.160 spontaneous to the right not spontaneous 00:37:07.09900:37:07.109 now for this one is spontaneous in the 00:37:09.31900:37:09.329 forward direction towards the product 00:37:10.84900:37:10.859 side because it's product favored so 00:37:15.38000:37:15.390 here's another problem for you let's say 00:37:16.57900:37:16.589 if K is equal to 2 times 10 to the 12 00:37:21.81900:37:21.829 calculate Delta G of the reaction to 00:37:25.46000:37:25.470 standard Delta G so let's use the 00:37:29.18000:37:29.190 equation Delta G is equal to negative RT 00:37:31.33900:37:31.349 Ln K so R that's going to be negative 00:37:37.63000:37:37.640 8.3145 and I keep again in the 00:37:43.43000:37:43.440 temperature let's say the temperature is 00:37:46.63000:37:46.640 300 Kelvin or 298 Kelvin let's go with 00:37:53.56900:37:53.579 that 00:37:58.28900:37:58.299 and then it's going to be Ln two times 00:38:01.14000:38:01.150 10 to the 12 so if you have K it's 00:38:04.43900:38:04.449 fairly easy to find Delta G you just got 00:38:06.15000:38:06.160 a plug name 298 of 8.3145 times 273 and 00:38:22.07900:38:22.089 79 but the units is in joules per mole 00:38:25.16900:38:25.179 because that sentence of art it has the 00:38:27.92900:38:27.939 units of ira joules per mole per Kelvin 00:38:29.30900:38:29.319 and it's going to cancel with the Kelvin 00:38:32.72900:38:32.739 temperature so R has unit joules per 00:38:35.54900:38:35.559 mole per Kelvin and the temperature is 00:38:38.48900:38:38.499 in Kelvin and usually K typically would 00:38:42.35900:38:42.369 have to worry about the units of K so 00:38:44.66900:38:44.679 Delta G is going to have the unit's 00:38:46.41000:38:46.420 joules per mole now sometimes in a 00:38:49.22900:38:49.239 multiple-choice exam they may want you 00:38:51.41900:38:51.429 to report the answer in kilojoules per 00:38:52.94900:38:52.959 mole so to convert jewels into 00:38:55.43900:38:55.449 kilojoules you need to divide it by a 00:38:57.02900:38:57.039 thousand so this is going to be negative 00:38:59.13000:38:59.140 70 point two kilojoules per mole and so 00:39:03.95900:39:03.969 now you know how to calculate Delta G of 00:39:05.91000:39:05.920 the reaction if you're given the 00:39:08.06900:39:08.079 equilibrium constant K now let's go back 00:39:14.13000:39:14.140 to the reaction that we had earlier and 00:39:18.41000:39:18.420 what we're going to do this time we're 00:39:21.05900:39:21.069 going to calculate Delta G under 00:39:23.59900:39:23.609 non-standard conditions so all of these 00:39:31.89000:39:31.900 are in the gas phase now let's say that 00:39:38.13000:39:38.140 the partial pressure for h2s let's say 00:39:42.68900:39:42.699 it's 0.1 and for h2o it's a point two 00:39:47.99900:39:48.009 and for h2 let's say the partial 00:39:51.83900:39:51.849 pressure is three and for so2 its to how 00:39:56.81900:39:56.829 can you calculate the standard I mean 00:39:59.64000:39:59.650 the non-standard 00:40:00.39000:40:00.400 Delta G for this reaction now we have 00:40:03.95900:40:03.969 the standard value the standard value is 00:40:07.70900:40:07.719 negative 191 kilojoules per mole what's 00:40:10.41000:40:10.420 the non-standard value 00:40:13.57900:40:13.589 here's the equation that you need to 00:40:15.69000:40:15.700 find the non-standard value Delta G is 00:40:19.25900:40:19.269 equal to Delta G naught plus RT Ln Q so 00:40:26.33900:40:26.349 the standard value you have this 00:40:29.78900:40:29.799 whenever the partial pressures of all 00:40:31.92000:40:31.930 reactants and products is equal to one 00:40:34.47000:40:34.480 ATM or if you're dealing with 00:40:37.82900:40:37.839 concentration has to be one that's the 00:40:40.17000:40:40.180 standard value if it's not one then 00:40:43.34900:40:43.359 Delta G is going to change so you got to 00:40:45.77900:40:45.789 calculate the non-standard Delta G value 00:40:49.78900:40:49.799 now because we have our R is in the 00:40:53.67000:40:53.680 units of joules per mole per Kelvin that 00:40:55.65000:40:55.660 means Delta G naught has to be in joules 00:40:59.03900:40:59.049 per mole so we got to convert it right 00:41:01.55900:41:01.569 now the only thing that's different here 00:41:03.18000:41:03.190 is Q we need to understand how to 00:41:05.91000:41:05.920 calculate the reaction quotient Q so I 00:41:09.21000:41:09.220 think we should do that first and then 00:41:10.38000:41:10.390 we'll get back to the equation Q is the 00:41:15.72000:41:15.730 reaction quotient and like K you can 00:41:18.21000:41:18.220 find it by dividing the products by the 00:41:21.08900:41:21.099 reactants so the products are h2s and 00:41:25.73000:41:25.740 h2o but there's a two in front of h2o so 00:41:30.34900:41:30.359 we're going to put a two on the exponent 00:41:33.50900:41:33.519 of h2 them and then divided by h2 x SL 00:41:39.12000:41:39.130 two so the coefficient for h2 is 3 so 00:41:42.99000:41:43.000 let's put a 3 here so the value for h2s 00:41:47.40000:41:47.410 is 0.1 and for h2o its point two squared 00:41:52.78900:41:52.799 divided by the value for H 2 which is 3 00:41:55.97000:41:55.980 cubed times that for so2 which is 2 so 00:42:01.97000:42:01.980 0.1 times 0.2 squared that's about point 00:42:06.69000:42:06.700 zero zero four three to the third power 00:42:10.41000:42:10.420 is 27 times 2 that's 54 so if you take 00:42:16.28900:42:16.299 point zero zero four divided by 54 you 00:42:19.65000:42:19.660 should get seven point four zero seven 00:42:22.82900:42:22.839 times 10 to negative 5 00:42:24.91000:42:24.920 so that's the value of the reaction 00:42:26.96000:42:26.970 quotient Q so with that we can now find 00:42:32.06000:42:32.07000:42:33.80000:42:33.810 Delta G value 00:42:43.53000:42:43.540 so let's plug in negative 191,000 joules 00:42:48.84000:42:48.850 for Delta G naught it has to be in 00:42:50.31000:42:50.320 joules per mole so R is going to be 00:42:53.54000:42:53.550 8.3145 00:42:54.81000:42:54.820 and let's say the temperature is 298 00:42:57.18000:42:57.190 Kelvin times the natural log of Q which 00:43:03.33000:43:03.340 was seven point four zero seven times 10 00:43:07.56000:43:07.570 to negative five the natural log of 00:43:11.25000:43:11.260 seven point four zero seven times ten to 00:43:14.16000:43:14.170 95 that's negative nine point five one 00:43:18.03000:43:18.040 if you multiply that by 298 and 8.3145 00:43:22.79000:43:22.800 you should get negative twenty three 00:43:30.15000:43:30.160 thousand five hundred sixty-four so 00:43:34.32000:43:34.330 notice that the contribution of RT Ln Q 00:43:39.24000:43:39.250 is negative that means that the reaction 00:43:41.19000:43:41.200 is becoming more spontaneous notice that 00:43:46.68000:43:46.690 we have a large amount of reactants and 00:43:49.17000:43:49.180 a small amount of products if you 00:43:52.08000:43:52.090 increase the reactants according to 00:43:55.74000:43:55.750 Lateisha Talia's principle the reaction 00:43:58.83000:43:58.840 is going to shift to the right and as it 00:44:03.45000:44:03.460 shifts to the right it becomes more 00:44:05.04000:44:05.050 spontaneous and so that's why the Delta 00:44:07.59000:44:07.600 G is going to be more negative so if we 00:44:10.11000:44:10.120 add negative one ninety one thousand and 00:44:13.44000:44:13.450 twenty three thousand five sixty four 00:44:16.58000:44:16.590 the Delta G the non standard Delta G is 00:44:19.83000:44:19.840 going to be negative two hundred 00:44:20.85000:44:20.860 fourteen thousand five hundred sixty 00:44:23.25000:44:23.260 four joules per mole and if you want to 00:44:26.28000:44:26.290 convert it to kilojoules per mole 00:44:27.45000:44:27.460 divided by thousand so it's going to be 00:44:29.19000:44:29.200 negative two fourteen point six 00:44:31.16000:44:31.170 kilojoules per mole so anytime you 00:44:33.96000:44:33.970 increase the reactants the reaction is 00:44:35.70000:44:35.710 going to shift to the right and Delta G 00:44:38.64000:44:38.650 will be more negative it's going to be 00:44:39.96000:44:39.970 more spontaneous whenever it shifts to 00:44:43.11000:44:43.120 the right 00:44:47.08900:44:47.099 now consider the following reversible 00:44:51.01900:44:51.029 reaction let's say we have a which is a 00:44:53.95900:44:53.969 solid plus B which is a gas and let's 00:44:57.38000:44:57.390 say we have two of B and this turns into 00:45:00.41000:45:00.420 C which is a liquid and D which is a gas 00:45:08.10900:45:08.119 so what's going to happen if we increase 00:45:11.01900:45:11.029 the concentration of B will the reaction 00:45:13.64000:45:13.650 shift to the right or to the left 00:45:15.14000:45:15.150 well according to Lateisha Talia's 00:45:17.32900:45:17.339 principle whenever you impose a change 00:45:19.78900:45:19.799 on a system the system is going to try 00:45:21.79900:45:21.809 to undo that change so if you increase 00:45:24.25900:45:24.269 the reactants the system is going to try 00:45:26.42000:45:26.430 to decrease the value of the reactants 00:45:28.06900:45:28.079 so it's going to shift to the right 00:45:31.09900:45:31.109 because when it shifts to the right the 00:45:33.17000:45:33.180 value of the products go up but the 00:45:35.53900:45:35.549 value of the reactants go down and 00:45:36.89000:45:36.900 that's how it bring this back down the 00:45:39.41000:45:39.420 value of the reactants so whenever you 00:45:42.22900:45:42.239 increase the concentration of reactants 00:45:44.39000:45:44.400 is going to shift to the right and any 00:45:45.76900:45:45.779 time the reaction shifts to the right 00:45:47.22900:45:47.239 Delta G will become more negative 00:45:50.15000:45:50.160 meaning that Delta G will decrease and 00:45:52.60900:45:52.619 so the Spontini of the reaction will 00:45:57.28900:45:57.299 increase the reaction will become more 00:45:58.93900:45:58.949 spontaneous now what about D if we 00:46:02.93000:46:02.940 increase the value of D if you increase 00:46:05.50900:46:05.519 the concentration or the partial 00:46:07.67000:46:07.680 pressure of a product the reaction is 00:46:09.89000:46:09.900 going to shift to left it does that to 00:46:13.54900:46:13.559 decrease the value of the products since 00:46:16.24900:46:16.259 you increased it it's going to try to 00:46:17.56900:46:17.579 undo the change that you impose to them 00:46:19.30900:46:19.319 anytime the reaction goes to the left 00:46:21.25900:46:21.269 the value of the products go down and 00:46:23.42000:46:23.430 the value of the reactants go up so as 00:46:27.28900:46:27.299 it shifts to the left 00:46:28.18900:46:28.199 Delta G that will increase in value it's 00:46:30.76900:46:30.779 going to become more more positive 00:46:32.38000:46:32.390 meaning that the reaction is going to be 00:46:36.01900:46:36.029 it's going to become more and not 00:46:37.84900:46:37.859 spontaneous in the four direction but 00:46:40.45900:46:40.46900:46:41.66000:46:41.670 because it's going to the left so in a 00:46:44.59900:46:44.609 forward direction it's going to be non 00:46:46.64000:46:46.65000:46:51.76900:46:51.779 now what happens if we increase the 00:46:54.50900:46:54.519 amount of reactant a now I need to keep 00:46:58.23000:46:58.240 in mind that solids and liquids have no 00:47:00.02900:47:00.039 effect on the position of equilibrium so 00:47:02.43000:47:02.440 if you increase the amount of a the 00:47:04.62000:47:04.630 reaction will not shift to the right or 00:47:06.15000:47:06.160 to the left so therefore there's going 00:47:08.91000:47:08.920 to be no change and if it doesn't shift 00:47:12.56900:47:12.579 to the right or to the left 00:47:13.81900:47:13.829 Delta G will remain constant now let's 00:47:20.97000:47:20.980 say if we increase the volume of the 00:47:23.40000:47:23.410 container that holds the reaction what's 00:47:25.01900:47:25.029 going to happen well whenever you 00:47:27.08900:47:27.099 increase the volume since we have gases 00:47:29.87000:47:29.880 the pressure will be reduced and so 00:47:34.95000:47:34.960 therefore whenever the pressure is 00:47:37.04900:47:37.059 reduced the system is going to try to 00:47:39.25900:47:39.269 undo the change that you impose to it 00:47:41.54900:47:41.559 it's going to try to increase the 00:47:42.53900:47:42.549 pressure in order for it to increase the 00:47:44.37000:47:44.380 pressure it has to shift to the side 00:47:46.47000:47:46.480 with more moles of gas now we have two 00:47:50.03900:47:50.049 gas molecules on the left only one on 00:47:52.31900:47:52.329 the right so it's going to shift to the 00:47:54.05900:47:54.069 left to increase the pressure to bring 00:47:56.78900:47:56.799 it back to where it was and so as it 00:47:59.73000:47:59.740 shifts to the left 00:48:00.42000:48:00.430 we know that its reactant favored so 00:48:04.20000:48:04.210 it's going to be non spontaneous in the 00:48:06.10900:48:06.119 four direction so Delta G is going to 00:48:09.96000:48:09.970 increase in value it's going to become 00:48:12.21000:48:12.220 more positive if whenever the reaction 00:48:14.37000:48:14.380 shifts to the left now let's say if we 00:48:17.64000:48:17.650 have an endothermic reaction where Delta 00:48:22.89000:48:22.900 H is positive what's going to happen if 00:48:26.13000:48:26.140 we increase the temperature if you raise 00:48:29.27900:48:29.289 the temperature then energy is going to 00:48:35.33900:48:35.349 go into the system whenever you raise 00:48:37.10900:48:37.119 the temperature particularly the 00:48:38.43000:48:38.440 surroundings so if you raise the 00:48:40.10900:48:40.119 temperature for an endothermic reaction 00:48:41.77900:48:41.789 the reaction is going to shift to the 00:48:44.00900:48:44.019 right a simple way that you can find out 00:48:46.58900:48:46.599 the direction if it's an O thermic put 00:48:48.53900:48:48.549 Delta H on the left side if its X with 00:48:50.84900:48:50.859 let me put it on the right side so 00:48:52.58900:48:52.599 whenever you raise the value of the 00:48:53.84900:48:53.859 reactants the reaction is going to shift 00:48:55.79900:48:55.809 to the right likewise if you increase 00:48:58.04900:48:58.059 the temperature for an endothermic 00:48:59.06900:48:59.079 reaction it's going to shift to the 00:49:00.72000:49:00.730 right so Delta G will decrease in value 00:49:03.18000:49:03.190 which means that 00:49:04.80000:49:04.810 cause delta g is becoming more negative 00:49:06.12000:49:06.130 it's going to be more spontaneous and 00:49:11.69000:49:11.700 also the equally makan stink a will be 00:49:14.52000:49:14.530 affected if you increase the 00:49:17.01000:49:17.020 concentration of let's say reaction b d 00:49:19.44000:49:19.450 or if you change the pressure or volume 00:49:21.12000:49:21.130 the equilibrium constant is going to 00:49:23.31000:49:23.320 stay the same it's not going to change 00:49:24.69000:49:24.700 however the equilibrium constant will 00:49:27.42000:49:27.430 change due to changes in temperature so 00:49:29.91000:49:29.920 if you raise the temperature and if the 00:49:31.95000:49:31.960 reaction shifts to the right the value 00:49:34.23000:49:34.240 of the products will go up and the value 00:49:36.09000:49:36.100 of the reactants will go down and k is 00:49:39.83000:49:39.840 equal to the ratio of the products 00:49:42.60000:49:42.610 divided by the reactants whenever you 00:49:44.91000:49:44.920 increase the numerator of a fraction in 00:49:46.83000:49:46.840 this case the products the value of the 00:49:48.78000:49:48.790 whole fraction goes up and if you 00:49:51.99000:49:52.000 increase the denominator the value goes 00:49:54.24000:49:54.250 down but if you decrease the denominator 00:49:56.61000:49:56.620 or decrease the value of the reactants 00:49:58.62000:49:58.630 the value of the whole fraction goes up 00:50:00.62000:50:00.630 so as the reaction shifts to the right 00:50:03.12000:50:03.130 due to a temperature change okay the 00:50:06.12000:50:06.130 equilibrium constant is going to 00:50:07.38000:50:07.390 increase and the reaction will be more 00:50:09.54000:50:09.55000:50:12.92000:50:12.930 so now let's understand how the reaction 00:50:18.00000:50:18.010 is going to shift to the right if it's 00:50:20.04000:50:20.050 endothermic so Delta H let's say it's 00:50:24.75000:50:24.760 positive 50 now I'm going to draw a box 00:50:29.07000:50:29.080 inside the box represents a system now 00:50:34.17000:50:34.180 keep in mind you need to know the 00:50:35.34000:50:35.350 difference between an open system a 00:50:37.80000:50:37.810 closed system and the isolated system an 00:50:39.72000:50:39.730 open system is a system where mass and 00:50:42.96000:50:42.970 energy can be exchanged with the 00:50:44.55000:50:44.560 surroundings in a closed system the mass 00:50:49.14000:50:49.150 cannot be exchanged 00:50:50.61000:50:50.620 however energy like heat energy can flow 00:50:53.25000:50:53.260 into or out of a closed system 00:50:55.71000:50:55.720 but gas molecules can't enter a closed 00:50:58.44000:50:58.450 system now if you have an isolated 00:51:00.75000:51:00.760 system heat cannot flow into or out of 00:51:03.39000:51:03.400 an isolated system nor can a gas 00:51:05.61000:51:05.620 molecule enter or leave an isolated 00:51:08.28000:51:08.290 system nothing 00:51:09.69000:51:09.700 enters or leaves an isolated system but 00:51:12.72000:51:12.730 now the example that we are going to 00:51:14.07000:51:14.080 have is a closed system where heat can 00:51:16.11000:51:16.120 flow into or out of it so 00:51:18.51000:51:18.520 let's say we have the surroundings on 00:51:20.61000:51:20.620 the outside and the system on the inside 00:51:23.48000:51:23.490 so let's say the temperature the system 00:51:25.47000:51:25.480 is 50 but let's say that of let's say if 00:51:28.38000:51:28.390 we raise the temperature of the 00:51:29.97000:51:29.980 surroundings to about 200 degrees 00:51:31.92000:51:31.930 Celsius heat is going to flow from hot 00:51:35.16000:51:35.170 to cold so 200 is hot 50 is relatively 00:51:39.36000:51:39.370 cold compared to 200 so heat is going to 00:51:42.54000:51:42.550 flow into the system as heat flows into 00:51:46.89000:51:46.900 the system it's going to be ml thermic 00:51:51.71000:51:51.720 because the system is absorbing heat and 00:51:54.53000:51:54.540 so as it absorbs heat the reaction is 00:51:58.98000:51:58.990 going to shift to the right because one 00:52:02.10000:52:02.110 of the action shifts to the right it's 00:52:05.19000:52:05.200 going to be endothermic if you reverse 00:52:07.68000:52:07.690 the reaction is going to be exothermic 00:52:11.27000:52:11.280 so the only way to reverse the reaction 00:52:14.01000:52:14.020 is to decrease the temperature of the 00:52:15.72000:52:15.730 surroundings so let's say if we decrease 00:52:18.54000:52:18.550 it to about zero degrees Celsius heat is 00:52:21.18000:52:21.190 going to flow out of the system and the 00:52:23.82000:52:23.830 only way to extract heat from the system 00:52:25.74000:52:25.750 is if the reaction reverses direction so 00:52:29.70000:52:29.710 it's going to shift to the left if we 00:52:31.77000:52:31.780 reverse the reaction and then the 00:52:35.52000:52:35.530 enthalpy has to be reversed it's going 00:52:38.16000:52:38.170 to be instead of being positive 50 it's 00:52:41.10000:52:41.110 going to be negative 50 and so it's 00:52:44.31000:52:44.320 going to be an exothermic reaction so by 00:52:47.25000:52:47.260 cooling the surroundings we can cause 00:52:48.84000:52:48.850 the reaction to reverse in direction we 00:52:51.39000:52:51.400 can cause it to go to the left and if we 00:52:54.30000:52:54.310 increase the temperature of the 00:52:55.50000:52:55.510 surroundings we can cause the reaction 00:52:57.51000:52:57.520 become endothermic and it's going to 00:52:59.40000:52:59.410 shift to the right so we can control the 00:53:02.07000:53:02.080 direction of the reaction by controlling 00:53:04.20000:53:04.210 the temperature of the surroundings so 00:53:06.60000:53:06.610 if we increase the temperature of the 00:53:09.24000:53:09.250 surroundings we can see that the 00:53:11.67000:53:11.680 reaction is going to shift to the right 00:53:13.32000:53:13.330 because it's going to be endothermic 00:53:14.76000:53:14.770 heat is going to flow into the system 00:53:17.39000:53:17.400 and when it shifts to the right Delta G 00:53:20.64000:53:20.650 will decrease and the reaction will 00:53:22.98000:53:22.990 become more spontaneous in the four 00:53:25.17000:53:25.180 direction now if we decrease the 00:53:27.99000:53:28.000 temperature the reaction is going to 00:53:30.03000:53:30.040 shift to the left and Delta G will 00:53:31.98000:53:31.990 this can be more positive which means 00:53:34.05000:53:34.060 that it's more non-spontaneous in the 00:53:36.03000:53:36.040 four direction but because it's going to 00:53:38.25000:53:38.260 the left its spontaneous in the reverse 00:53:40.17000:53:40.180 direction and so you can adjust the 00:53:44.13000:53:44.140 value of Delta G by controlling the 00:53:46.20000:53:46.210 temperature of the surroundings but that 00:53:48.48000:53:48.490 is it for this video that's all I got 00:53:50.67000:53:50.680 hopefully you found it to be educational 00:53:52.65000:53:52.660 thanks for watching and have a great day
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