WEBVTT chemistry/general-chemistry/ow 00:00:00.000 --> 00:00:02.900 Hi, welcome back to Educator.com. 00:00:02.900 --> 00:00:09.100 Today's lesson from general chemistry is on entropy and free energy. 00:00:09.100 --> 00:00:13.600 This is basically going to be our second and final lecture on thermodynamics. 00:00:13.600 --> 00:00:17.600 As usual we are going to start off with a brief introduction. 00:00:17.600 --> 00:00:23.200 We are going to get into something very important in thermodynamics which is what we call entropy. 00:00:23.200 --> 00:00:27.400 It is going to be related to spontaneous processes. 00:00:27.400 --> 00:00:38.000 After we get into entropy, we are going to go through the second and third laws of thermodynamics followed by problem solving. 00:00:38.000 --> 00:00:47.500 After entropy, we are going to then get into what is called Gibbs free energy followed by definition and problem solving. 00:00:47.500 --> 00:00:55.300 Then we are going to go ahead and conclude with a summary and a pair of sample problems. 00:00:55.300 --> 00:01:03.100 We have consistently said, you have heard me say over and over again that nature favors states of low energy. 00:01:03.100 --> 00:01:08.700 In other words, reactions occur to obtain a lower energy state in the end. 00:01:08.700 --> 00:01:11.800 This is also known as a downhill reaction. 00:01:11.800 --> 00:01:19.400 In other words, nature tends to favor reactions that start in a high energy state and go to a low energy state. 00:01:19.400 --> 00:01:21.400 That is going to be the natural tendency. 00:01:21.400 --> 00:01:28.800 In this lecture, we are going to complete our discussion from thermodynamics and quantify the above statement. 00:01:28.800 --> 00:01:31.600 In the end, we are going to see that nature not only favors 00:01:31.600 --> 00:01:39.300 states of low energy but also states of what we call high entropy. 00:01:39.300 --> 00:01:44.600 Consider the following closed reaction vessels that contain gas molecules. 00:01:44.600 --> 00:01:49.300 Let me go ahead and draw a closed reaction vessel here. 00:01:49.300 --> 00:01:55.500 We can make this one gas particle; the red one can be another gas particle. 00:01:55.500 --> 00:02:00.100 The black one can be another gas particle; we are now going to... 00:02:00.100 --> 00:02:08.100 Because these are gases, gases naturally expand; what can happen is the following. 00:02:08.100 --> 00:02:14.700 What are some possible different configurations for these three gas particles if they naturally expand? 00:02:14.700 --> 00:02:20.300 Situation number one is when all three gas particles are on the same side. 00:02:20.300 --> 00:02:28.400 We can have a second situation when only the blue and the black are together and the red is different. 00:02:28.400 --> 00:02:42.200 We can have another situation where the blue is by itself and the red and black are together. 00:02:42.200 --> 00:02:48.900 You notice that these two situations here are actually the same 00:02:48.900 --> 00:02:57.000 where we have two gas particles on one side versus one on the other. 00:02:57.000 --> 00:03:16.400 It turns out that configuration number two actually has more what we call microstates. 00:03:16.400 --> 00:03:24.200 Once again configuration number two is going to have what we call more microstates. 00:03:24.200 --> 00:03:28.900 We can even do one more to show you how this is possible. 00:03:28.900 --> 00:03:35.300 Now red and blue can be together with black by itself. 00:03:35.300 --> 00:03:39.900 This is also considered a configuration number two. 00:03:39.900 --> 00:03:47.000 Configuration number one, there is only one way to put three particles on the same side. 00:03:47.000 --> 00:03:56.700 Configuration number one has one microstate. 00:03:56.700 --> 00:04:05.700 Because gases expand, we are going to see that configuration number two is going to be the more probable configuration. 00:04:05.700 --> 00:04:09.700 Why?--because there are a higher number of ways to do it. 00:04:09.700 --> 00:04:16.900 There are a higher number of microstates to achieve, configuration number two, where one gas particle is on one side. 00:04:16.900 --> 00:04:20.000 Two gas particles are on the other side. 00:04:20.000 --> 00:04:24.600 We symbolize the term microstate with capital W. 00:04:24.600 --> 00:04:33.300 For configuration number one, W is 1; for configuration number two, W is simply 3. 00:04:33.300 --> 00:04:41.600 We can quantify that this is going to be more probable. 00:04:41.600 --> 00:04:46.400 In other words, this is more likely to occur. 00:04:46.400 --> 00:04:48.900 The equation we use to quantify this is following. 00:04:48.900 --> 00:04:57.600 S is equal to kB times the natural log of W where S is what we call entropy. 00:04:57.600 --> 00:05:08.600 I want you to think of entropy as basically disorder or chaos. 00:05:08.600 --> 00:05:13.600 We see that S is directly proportional to W. 00:05:13.600 --> 00:05:18.600 In other words, as W goes up, so does S. 00:05:18.600 --> 00:05:20.900 What does that mean though? 00:05:20.900 --> 00:05:25.900 We already said that a high value of W means it is going to be more likely to occur 00:05:25.900 --> 00:05:37.000 which says that very high entropy values are going to relate to more likely processes. 00:05:37.000 --> 00:05:51.300 If W large, S is large; process more likely to occur. 00:05:51.300 --> 00:05:55.800 In this equation, we see that there is a proportionality constant called kB. 00:05:55.800 --> 00:06:03.600 kB is what we call the Boltzmann constant. 00:06:03.600 --> 00:06:09.400 That is equal to 1.38 times 10^-23 joules per kelvin. 00:06:09.400 --> 00:06:17.200 Because natural log of W is going to be unitless, that means the units of entropy are also joules per kelvin. 00:06:17.200 --> 00:06:23.100 This entire equation is known as the Boltzmann equation. 00:06:23.100 --> 00:06:26.000 S is equal to kB natural log of W. 00:06:26.000 --> 00:06:33.500 The units of entropy are going to be joules per kelvin. 00:06:33.500 --> 00:06:37.100 Let's go ahead and relate to entropy to what we have already discussed 00:06:37.100 --> 00:06:42.300 in our first lecture from thermodynamics which involved heat flow and energy. 00:06:42.300 --> 00:06:50.200 If you recall from our first thermodynamic lecture, if heat leaves the system, Q is going to be a negative value. 00:06:50.200 --> 00:06:55.600 If heat enters the system, Q is going to be a positive value. 00:06:55.600 --> 00:07:06.600 Entropy is going to be a state function and can be related to Q where ΔS is equal to Q over T. 00:07:06.600 --> 00:07:12.700 Temperature, because you see capital T, has to be in kelvin here. 00:07:12.700 --> 00:07:16.800 Again because this is a state function, what we are interested in is 00:07:16.800 --> 00:07:28.600 a change in entropy just like we talked about for ΔE or ΔU. 00:07:28.600 --> 00:07:40.600 If heat enters the system, Q is going to be positive which means ΔS is going to be positive. 00:07:40.600 --> 00:07:42.200 Let's see if that makes sense. 00:07:42.200 --> 00:07:49.700 Remember if heat enters the system, we are supplying thermal energy for molecules to absorb. 00:07:49.700 --> 00:07:56.700 If they are going to absorb this thermal energy, that gives them more molecular motion such as vibrations. 00:07:56.700 --> 00:08:02.900 That increases the number of states an electron can occupy. 00:08:02.900 --> 00:08:05.400 Basically we are increasing the amount of disorder. 00:08:05.400 --> 00:08:07.800 We are increasing the amount of chaos. 00:08:07.800 --> 00:08:14.300 Yes, it does make sense that ΔS is a positive value. 00:08:14.300 --> 00:08:20.000 ΔS is also related to heat flow during phase changes. 00:08:20.000 --> 00:08:26.100 It is found that Q is equal to n times ΔH of the phase change. 00:08:26.100 --> 00:08:33.700 This is from our first thermodynamics; from thermo I. 00:08:33.700 --> 00:08:45.400 Therefore it follows that ΔS is equal to Q over T which equals to nΔH of the phase change over the temperature. 00:08:45.400 --> 00:08:46.700 What does that mean then? 00:08:46.700 --> 00:09:00.000 That means for endothermic processes such as melting and vaporization, endothermic means Q is going to be a positive value. 00:09:00.000 --> 00:09:02.500 ΔS is positive. 00:09:02.500 --> 00:09:13.100 This agrees with the previous slide where when we put energy in, everything gets a little more chaotic and a little more disorderly. 00:09:13.100 --> 00:09:23.600 For example, as you go for melting, you go from a very orderly state, a solid, to a less orderly state, to a liquid. 00:09:23.600 --> 00:09:27.600 When go from a liquid to a gas, a liquid is less orderly. 00:09:27.600 --> 00:09:32.600 A gas is going to be the most disordered state. 00:09:32.600 --> 00:09:36.100 Everything is a lot more chaotic there. 00:09:36.100 --> 00:09:40.200 ΔS is going to be positive when heat enters the system. 00:09:40.200 --> 00:09:46.300 Here we have the phase change backing up this point. 00:09:46.300 --> 00:09:50.500 In other words then, we can summarize this very nicely. 00:09:50.500 --> 00:09:56.100 If a process is likely to occur, then ΔS is going to be a positive value. 00:09:56.100 --> 00:10:05.500 We say that the process is expected to be spontaneous if ΔS is positive. 00:10:05.500 --> 00:10:09.400 This brings us into our second law of thermodynamics. 00:10:09.400 --> 00:10:16.300 The second law of thermodynamics states that a process will be spontaneous if ΔS total is greater 00:10:16.300 --> 00:10:22.800 than 0 where ΔS total is equal to ΔS of the system plus ΔS of the surroundings. 00:10:22.800 --> 00:10:31.600 In other words, nature is going to favor processes that contribute to the overall entropy of the entire universe. 00:10:31.600 --> 00:10:39.800 In other words, nature favors processes where the amount of entropy or disorder increases as a result of the process. 00:10:39.800 --> 00:10:43.600 We see this occurring in nature quite a bit. 00:10:43.600 --> 00:10:48.100 If you think of a virus, what does a virus naturally do? 00:10:48.100 --> 00:10:57.700 A virus is going to naturally expand and go from cell to cell consuming the resources there and moving on and on. 00:10:57.700 --> 00:11:02.200 If viruses expand, they get more disorderly, more chaotic. 00:11:02.200 --> 00:11:06.900 We see this in herd migration. 00:11:06.900 --> 00:11:17.700 Herd migration, flocks of animals tend to naturally expand from area to area. 00:11:17.700 --> 00:11:22.100 They tend not to stay just in a single location. 00:11:22.100 --> 00:11:29.600 Of course the universe itself, the universe theoretically started off as a single point. 00:11:29.600 --> 00:11:34.400 It is just constantly expanding and expanding. 00:11:34.400 --> 00:11:44.000 Again we see that entropy is seen in several naturally occurring examples. 00:11:44.000 --> 00:11:54.900 Then you may be wondering, if ΔS positive is going to be for a spontaneous process, what does it mean when entropy is zero? 00:11:54.900 --> 00:11:58.500 That brings us into the third law of thermodynamics. 00:11:58.500 --> 00:12:08.000 Zero entropy really does not exist because the conditions to obtain zero entropy do not exist experimentally. 00:12:08.000 --> 00:12:16.500 The third law of thermodynamics states that entropy is zero only for a perfect flawless crystal at absolute zero. 00:12:16.500 --> 00:12:20.800 But absolute zero has not been obtained experimentally. 00:12:20.800 --> 00:12:30.000 Only in the most extreme conditions where we have zero thermo motion, that is molecules cease to move, entropy is completely zero. 00:12:30.000 --> 00:12:38.200 In other words, entropy is always going to be a nonzero quantity in practice. 00:12:38.200 --> 00:12:41.900 Let's now focus on problem solving involving entropy. 00:12:41.900 --> 00:12:45.300 The following are typical types of questions involving entropy. 00:12:45.300 --> 00:12:50.500 Sometimes a question can ask you simply to predict the sign of ΔS given a chemical reaction. 00:12:50.500 --> 00:12:56.800 This is going to be an example of a qualitative process, of a qualitative problem. 00:12:56.800 --> 00:13:01.400 Basically when the process is a physical change, ΔS is going to be positive for 00:13:01.400 --> 00:13:08.500 any endothermic process just like we talked about, including melting, vaporization, and sublimation. 00:13:08.500 --> 00:13:12.700 When you go from a solid to a liquid, from a liquid to a gas, or from a 00:13:12.700 --> 00:13:20.700 solid to a gas, ΔS of these processes are going to be expected to be positive. 00:13:20.700 --> 00:13:27.000 They are opposite processes, are all exothermic which means ΔS is expected to be negative. 00:13:27.000 --> 00:13:32.300 Liquid to solid, gas to liquid, and gas to solid. 00:13:32.300 --> 00:13:42.300 Freezing, condensation, and deposition; again ΔS is expected to be negative. 00:13:42.300 --> 00:13:47.700 Let's go on to another type of problem. 00:13:47.700 --> 00:13:53.800 Another type of problem involves a quantitative problem. 00:13:53.800 --> 00:13:58.000 What you want to do here is pay close attention to the physical states. 00:13:58.000 --> 00:14:01.600 From a solid to a liquid to a gas, entropy increases. 00:14:01.600 --> 00:14:03.200 We have already seen that. 00:14:03.200 --> 00:14:10.500 What you want to keep track of is the change in physical states from reactant to product side. 00:14:10.500 --> 00:14:15.700 Let's go ahead and look at an example. 00:14:15.700 --> 00:14:24.500 Fe solid plus O2 gas goes on to form Fe2O3 solid. 00:14:24.500 --> 00:14:29.500 We are going to need two of these, three of these, and two of these. 00:14:29.500 --> 00:14:33.000 Excuse me... we will need four irons. 00:14:33.000 --> 00:14:38.400 You notice that we start off with a solid and a gas. 00:14:38.400 --> 00:14:46.900 But on the product side, we have no gases here; no gases on product side. 00:14:46.900 --> 00:14:54.200 If we are eliminating the highest disorderly state which is a gas, we are starting off with it. 00:14:54.200 --> 00:14:58.500 We are not winding with it; we are actually getting less disorderly. 00:14:58.500 --> 00:15:02.700 ΔS for this process expected to be a negative value. 00:15:02.700 --> 00:15:06.400 Let's go on to another type of problem. 00:15:06.400 --> 00:15:09.300 If the reaction involves all gases, what you want to keep track of 00:15:09.300 --> 00:15:13.200 is the net change in gas molecules from reactant to product side. 00:15:13.200 --> 00:15:23.600 For example, N2 gas plus 3H2 gas goes on to form 2NH3 gas. 00:15:23.600 --> 00:15:29.900 On the left side, we start off with four gas molecules. 00:15:29.900 --> 00:15:35.200 On the right side, we wind up with only two gas molecules. 00:15:35.200 --> 00:15:44.900 For all intents and purposes, gas molecule A is going to be the same as gas molecule B for entropy counting purposes. 00:15:44.900 --> 00:15:49.300 Because we start off with four gas molecules and we only wind up with two, 00:15:49.300 --> 00:15:53.100 we are actually decreasing the amount of disorder in this process. 00:15:53.100 --> 00:15:57.700 ΔS is expected to be negative for this example. 00:15:57.700 --> 00:15:59.800 Sometimes the problem is going to ask you to calculate ΔS 00:15:59.800 --> 00:16:02.800 for what is called the system, surroundings, and total where 00:16:02.800 --> 00:16:07.200 ΔS total is equal to ΔS system plus ΔS of the surroundings. 00:16:07.200 --> 00:16:12.000 We have talked about system and surroundings in thermodynamics I. 00:16:12.000 --> 00:16:18.000 If you recall, Q of the system equals to ?Q of the surroundings. 00:16:18.000 --> 00:16:19.300 This is going to come into play. 00:16:19.300 --> 00:16:28.600 If a refrigerator coolant absorbs heat from stored food items, the heat then vaporizes the coolant which boils at -27 degrees Celsius. 00:16:28.600 --> 00:16:33.200 ΔH of vaporization is equal to -22.0 kilojoules per mole. 00:16:33.200 --> 00:16:41.800 Calculate ΔS total when 1.471 moles of the coolant vaporizes exchanging heat with the food items that are stored at 4 degrees Celsius. 00:16:41.800 --> 00:16:44.900 What we have to do is the following. 00:16:44.900 --> 00:16:47.300 We know we are using this equation for sure. 00:16:47.300 --> 00:16:57.800 ΔS total is equal to ΔS of the system plus ΔS of the surroundings. 00:16:57.800 --> 00:17:03.800 Let's go ahead and define the system; it doesn't matter which one as the coolant. 00:17:03.800 --> 00:17:11.200 Let's go ahead and define the surroundings as the food items inside the refrigerator. 00:17:11.200 --> 00:17:15.300 This is then going to be equal to... 00:17:15.300 --> 00:17:23.400 We know that ΔS is equal to Q of the system over T. 00:17:23.400 --> 00:17:30.000 The ΔS of the surroundings is going to be equal to Q of the surroundings over T. 00:17:30.000 --> 00:17:40.100 But we also know that this relationship here where Q of surroundings is equal to ?Q of the system. 00:17:40.100 --> 00:17:42.600 We can then go ahead and plug everything in. 00:17:42.600 --> 00:17:52.500 That is going to be equal to nΔH of vaporization over the temperature of the coolant. 00:17:52.500 --> 00:18:00.000 That is going to be equal to ?nΔH of vaporization but this time over the temperature of the food. 00:18:00.000 --> 00:18:05.200 Really the only difference is the temperature that the system and surroundings are stored at 00:18:05.200 --> 00:18:12.600 if we assume complete transfer of heat from the system to the surroundings. 00:18:12.600 --> 00:18:18.400 When all is said and done, we are going to get -0.248... don't forget the units. 00:18:18.400 --> 00:18:28.000 Here this is going to be kilojoules per kelvin or -248 joules per kelvin. 00:18:28.000 --> 00:18:33.600 Again you must be careful on how you define your system and surroundings. 00:18:33.600 --> 00:18:37.700 Be consistent throughout the entire equation. 00:18:37.700 --> 00:18:43.500 Finally another type of problem involves standard molar entropies of formations which is the 00:18:43.500 --> 00:18:51.600 change in entropy when a compound is formed from its constituent elements under standard conditions. 00:18:51.600 --> 00:18:57.400 Recall that we covered standard molar enthalpies of formation and used the summation equation. 00:18:57.400 --> 00:19:07.700 Similarly ΔS of the reaction is going to be equal to summation nS of the products minus summation nS of the reactants. 00:19:07.700 --> 00:19:17.300 These values here again just like for the enthalpy, you are going to look up in the appendix. 00:19:17.300 --> 00:19:28.700 Don't forget that n is going to be the stoichiometric coefficient. 00:19:28.700 --> 00:19:35.400 Unlike standard molar enthalpies of formation, known compound is going to have a standard molar enthalpy of formation. 00:19:35.400 --> 00:19:42.600 Why?--because the zero is only obtained under the conditions of the third law of thermodynamics. 00:19:42.600 --> 00:19:50.800 Let's go ahead and examine our... excuse me... this is very important. 00:19:50.800 --> 00:20:00.100 Again you never can assume that the molar entropy of formation is a zero value ever. 00:20:00.100 --> 00:20:07.400 It is only true for ΔH and, we are going to learn, for one more--what we call ΔG later. 00:20:07.400 --> 00:20:10.300 Let's examine our final state function. 00:20:10.300 --> 00:20:15.000 This is an introduction to Gibbs free energy; free energy is ΔG. 00:20:15.000 --> 00:20:19.900 It is defined as the amount of energy available to do work. 00:20:19.900 --> 00:20:24.900 We said that mother nature favors high entropy already and low energy. 00:20:24.900 --> 00:20:27.000 It is that energy that we are really talking about. 00:20:27.000 --> 00:20:41.100 We will see that the spontaneous processes will all have ΔG less than zero. 00:20:41.100 --> 00:20:57.800 To go ahead and do this, we can do ΔG is equal to ΔH minus TΔS. 00:20:57.800 --> 00:21:05.800 This is the equation that relates all three thermodynamic properties together. 00:21:05.800 --> 00:21:19.000 We already said that a positive ΔS is going to be a spontaneous process. 00:21:19.000 --> 00:21:26.900 We want ΔG to be negative for a spontaneous process. 00:21:26.900 --> 00:21:30.100 What conditions give you ΔG negative? 00:21:30.100 --> 00:21:51.900 ΔG negative is going to be spontaneous under the following conditions--for very large ΔS values and at high temperatures. 00:21:51.900 --> 00:21:58.000 As you can see, if ΔS is large and temperature is large, 00:21:58.000 --> 00:22:02.000 this term is going to be larger than this term giving us 00:22:02.000 --> 00:22:09.400 an overall negative ΔG value; very large ΔS at high temperatures. 00:22:09.400 --> 00:22:24.900 Basically ΔH is going to be a relatively negative value; ΔH, small as possible. 00:22:24.900 --> 00:22:30.300 Again this is what we call Gibbs free energy. 00:22:30.300 --> 00:22:35.400 Just like our previous state functions, we have standard molar free energies of formation 00:22:35.400 --> 00:22:42.800 where ΔG of the reaction is equal to summation nΔG of the products minus summation nΔG of the reactants. 00:22:42.800 --> 00:22:49.200 Like enthalpy, the free energies of formation are zero for all compounds in the standard state. 00:22:49.200 --> 00:23:05.100 Something like O2 gas, H2 gas, N2 gas, Cl2 gas, Br2 liquid, and I2 solid just to name a couple of examples. 00:23:05.100 --> 00:23:17.200 These are all going to have a ΔG of formation which is equal to ΔH of formation equal to zero. 00:23:17.200 --> 00:23:29.400 But again S, the molar entropy of formation, is not going to be zero whatsoever. 00:23:29.400 --> 00:23:33.900 Let's now study the temperature dependence of ΔG. 00:23:33.900 --> 00:23:43.900 It is found that we already said that ΔG of the system is equal to ΔH of the system minus TΔS of the system. 00:23:43.900 --> 00:23:49.400 The reason why this is important is because of the following--this term right here. 00:23:49.400 --> 00:23:53.000 ΔS of the system is in this equation. 00:23:53.000 --> 00:24:05.700 But the second law tells us that ΔS of the universe has to be a positive value. 00:24:05.700 --> 00:24:13.900 If we go strictly by entropy, we need to know the ΔS of the total, not just of the system or the surroundings. 00:24:13.900 --> 00:24:26.300 However this equation here tells us that all I have to know is ΔG of the system; that is all. 00:24:26.300 --> 00:24:34.100 Let's go ahead and see what conditions are going to be right for this. 00:24:34.100 --> 00:24:41.400 We can go ahead and make a chart here; ΔH, T, and ΔS. 00:24:41.400 --> 00:24:48.300 Then this is going to be sign of ΔG. 00:24:48.300 --> 00:24:54.900 Let's go ahead and delete temperature here--sign of ΔG, and then ΔH, ΔS. 00:24:54.900 --> 00:25:05.300 Let's say ΔH is a negative value--if this is less than zero and ΔS is a negative value. 00:25:05.300 --> 00:25:15.900 If ΔH is negative and ΔS is negative, it will be very difficult to get ΔG to be negative. 00:25:15.900 --> 00:25:23.400 The only time ΔG can be negative is going to be at a low temperature here. 00:25:23.400 --> 00:25:31.300 What if ΔH is negative and ΔS is positive? 00:25:31.300 --> 00:25:46.800 If ΔH is negative and ΔS is positive, then this sign of ΔG is going to be negative at high temperature. 00:25:46.800 --> 00:25:51.000 What if ΔH is positive and ΔS is positive? 00:25:51.000 --> 00:26:00.400 If ΔH is positive and ΔS is positive, ΔG is going to be negative at high temperatures. 00:26:00.400 --> 00:26:05.400 Finally what if ΔH is positive and ΔS is negative? 00:26:05.400 --> 00:26:14.100 If ΔH is positive and ΔS is negative, ΔG will never be negative according to the equation at any temperature. 00:26:14.100 --> 00:26:19.100 This is going to be greater than zero at all temperatures. 00:26:19.100 --> 00:26:25.000 Again as you can see, ΔG of the system is greatly influenced by 00:26:25.000 --> 00:26:35.200 not only the signs of ΔH and ΔS but also the kelvin temperature. 00:26:35.200 --> 00:26:42.400 Finally let's go ahead and examine the effect of reactant and product concentration on the sign of free energy. 00:26:42.400 --> 00:26:49.300 It is found that ΔG of the reaction is equal to ΔG^0 plus RT natural log of Q. 00:26:49.300 --> 00:26:57.700 Remember what Q is?--Q is what we call the reaction quotient. 00:26:57.700 --> 00:27:02.700 This equation relates reactant and product concentration to the free energy not at 00:27:02.700 --> 00:27:09.800 standard conditions which means the concentrations are not going to be 1 molar. 00:27:09.800 --> 00:27:19.500 Or the pressures are not going to be 1 atm; or partial pressures not equal to 1 atm. 00:27:19.500 --> 00:27:27.400 Finally it can also be shown at equilibrium that ΔG not of the reaction is equal to ?RT natural log of K 00:27:27.400 --> 00:27:34.400 where this is our equilibrium constant which we have discussed so much in the previous sections. 00:27:34.400 --> 00:27:36.200 This is any equilibrium constant. 00:27:36.200 --> 00:27:47.400 This can be Ka, Kb, Kp, Ksp, Kf, etc; any of them. 00:27:47.400 --> 00:27:52.500 Basically this equation tells us that if K is large, the reaction is product favored. 00:27:52.500 --> 00:27:57.300 ΔG of the reaction is a negative value; that makes sense. 00:27:57.300 --> 00:28:01.700 If K is very large, that means the reaction is highly product favored. 00:28:01.700 --> 00:28:07.000 In other words, the reaction as written is going to be likely to occur. 00:28:07.000 --> 00:28:14.200 If the reaction is likely to occur, ΔG of that reaction should be negative. 00:28:14.200 --> 00:28:20.700 Let's go ahead and summarize our thermodynamics lecture before getting into our problems. 00:28:20.700 --> 00:28:27.000 We see that nature not only favors states of low energy but also now high entropy. 00:28:27.000 --> 00:28:30.100 Entropy is basically disorder or chaos. 00:28:30.100 --> 00:28:33.500 It is the focus of the second and third laws of thermodynamics. 00:28:33.500 --> 00:28:42.600 Basically what we have seen in this lecture is that there is a series of equations that govern entropy and free energy. 00:28:42.600 --> 00:28:49.600 They allow us to calculate both of them under a given set of conditions. 00:28:49.600 --> 00:28:52.300 Let's go ahead and do sample problem number one. 00:28:52.300 --> 00:28:58.200 Calculate ΔS of the reaction under standard conditions for the following. 00:28:58.200 --> 00:29:06.500 We have aluminum oxide reacting with 3H2 going on to form two aluminums plus three waters. 00:29:06.500 --> 00:29:11.200 Anytime you see a balanced chemical equation and it asks you simply to calculate 00:29:11.200 --> 00:29:15.300 ΔS not of the reaction, you are going to use the summation equation. 00:29:15.300 --> 00:29:32.400 Use ΔS of the reaction is equal to summation n S molality of all products minus summation n S molality of all reactants. 00:29:32.400 --> 00:29:34.800 I am just going to setup the problem for you. 00:29:34.800 --> 00:29:39.500 Again you are going to look up these values in the appendix of your textbook. 00:29:39.500 --> 00:29:41.600 Let's go ahead and do the products first. 00:29:41.600 --> 00:29:59.200 This is going to be 2 moles of aluminum times the standard molar entropy of formation of aluminum solid. 00:29:59.200 --> 00:30:03.100 Because it is summation, I am going to add... plus. 00:30:03.100 --> 00:30:04.300 Let me put that in parentheses. 00:30:04.300 --> 00:30:13.600 Plus 3 moles of the water vapor times its standard molar entropy. 00:30:13.600 --> 00:30:19.600 This is all in brackets; that is going to be subtracted from the reactant part. 00:30:19.600 --> 00:30:30.300 Here this is going to be 1 mole of Al2O3 solid times 00:30:30.300 --> 00:30:48.200 its molar entropy plus 3 moles of H2 gas times its molar entropy. 00:30:48.200 --> 00:30:56.200 Again you must pay attention to the physical states because the molar entropy of formation for say water vapor 00:30:56.200 --> 00:31:01.900 is going to be different than from liquid water which is going to be different than from solid water. 00:31:01.900 --> 00:31:04.800 Once again pay attention to the physical states. 00:31:04.800 --> 00:31:07.900 When this is all said and done, you are going to get your answer 00:31:07.900 --> 00:31:14.200 in of course units of joules per kelvin or kilojoules per kelvin. 00:31:14.200 --> 00:31:19.800 It depends what the question specifies of course. 00:31:19.800 --> 00:31:23.500 Let's go ahead and do sample problem number two. 00:31:23.500 --> 00:31:29.200 The reaction N2O4 going to 2NO2 is not spontaneous under standard conditions. 00:31:29.200 --> 00:31:32.800 Calculate the temperature at which the reaction becomes spontaneous. 00:31:32.800 --> 00:31:41.700 What that translates to is, in other words, at what temperature does ΔG become negative? 00:31:41.700 --> 00:31:48.600 Anytime you are asked to calculate a temperature at which something becomes spontaneous, 00:31:48.600 --> 00:31:56.400 pretty much you always use this equation: ΔG is equal to ΔH minus TΔS. 00:31:56.400 --> 00:32:00.600 They are asking you when does this change sign? 00:32:00.600 --> 00:32:06.300 In other words, when does ΔG go from positive to negative? 00:32:06.300 --> 00:32:11.100 What you simply is you set this whole thing equal to zero. 00:32:11.100 --> 00:32:16.400 Zero is equal to ΔH minus TΔS; you solve for temperature. 00:32:16.400 --> 00:32:20.700 Temperature is going to be equal to ΔH over ΔS. 00:32:20.700 --> 00:32:25.700 This is all going to be standard conditions of course. 00:32:25.700 --> 00:32:30.500 To get ΔH, you are going to use the summation equation. 00:32:30.500 --> 00:32:33.900 To get ΔS, you are going to use the summation equation 00:32:33.900 --> 00:32:38.700 which means you have to look up the formation values in the appendix. 00:32:38.700 --> 00:32:49.100 Look up formation values in appendix; appendices. 00:32:49.100 --> 00:32:55.700 When you do this, you are going to get a temperature value. 00:32:55.700 --> 00:33:07.400 At temperatures lower than T, ΔG is going to be positive. 00:33:07.400 --> 00:33:25.700 At temperatures higher than what you calculate for T, ΔG is going to be a negative value. 00:33:25.700 --> 00:33:34.700 Once again anytime you are asked to calculate the temperature at which something becomes spontaneous, 00:33:34.700 --> 00:33:41.100 almost always you are going to use this equation: ΔG is equal to ΔH minus TΔS. 00:33:41.100 --> 00:33:49.200 Set ΔG equal to zero and solve for the temperature. 00:33:49.200 --> 00:33:56.200 Finally our third sample problem, calculate ΔG of the reaction at 25 degrees Celsius 00:33:56.200 --> 00:34:07.200 for the formation of 1 mole of C2H5OH gas from C2H4 gas and H2O gas. 00:34:07.200 --> 00:34:14.300 The first thing we probably want to do is translate this into a balanced chemical equation. 00:34:14.300 --> 00:34:23.100 The reactants are C2H4 gas; this is going to react with H2O gas. 00:34:23.100 --> 00:34:29.500 That is going to form 1 mole of C2H5OH gas. 00:34:29.500 --> 00:34:31.500 Let's see if everything is balanced. 00:34:31.500 --> 00:34:38.500 Two carbons, six hydrogens, one oxygen; yes, this is nicely balanced. 00:34:38.500 --> 00:34:42.800 You are asked to calculate ΔG of the reaction at 25 degrees Celsius. 00:34:42.800 --> 00:34:54.500 Again ΔG of the reaction can simply be calculated from ΔH of the reaction minus TΔS of the reaction. 00:34:54.500 --> 00:34:58.700 Again to get the ΔH, you are going to use the summation equation. 00:34:58.700 --> 00:35:02.000 To get the ΔS, you are going to use the summation equation. 00:35:02.000 --> 00:35:08.000 Again the kelvin temperature is already specified for you right there. 00:35:08.000 --> 00:35:14.300 That gives you ΔG of the reaction; they then ask you to calculate Kp. 00:35:14.300 --> 00:35:25.700 The only equation in this lecture that involves the equilibrium constant is ΔG is equal to ?RT natural log of K. 00:35:25.700 --> 00:35:30.800 We can then insert this ΔG of a reaction in there. 00:35:30.800 --> 00:35:40.300 K therefore is equal to 10 raised to the -ΔG^0 over RT 00:35:40.300 --> 00:35:50.900 where of course R is equal to 8.314 joules per k mole. 00:35:50.900 --> 00:35:52.300 Always double check your work. 00:35:52.300 --> 00:35:58.200 Make sure that K is a positive value because again 00:35:58.200 --> 00:36:05.000 we saw in previous lectures that K can never ever be negative. 00:36:05.000 --> 00:36:08.700 That is our lecture from general chemistry on thermodynamics. 00:36:08.700 --> 00:36:11.200 I want to thank you for your time and attention. 00:36:11.200 --> 00:36:13.000 I will see you next time on Educator.com.