WEBVTT chemistry/general-chemistry/ow 00:00:00.000 --> 00:00:03.000 Hi, welcome back to Educator.com. 00:00:03.000 --> 00:00:09.700 Today's lecture from general chemistry is on the principles of chemical equilibrium. 00:00:09.700 --> 00:00:12.300 Let's go ahead and take a look at the lesson overview. 00:00:12.300 --> 00:00:16.700 We will first start off with a brief introduction and then get in right to 00:00:16.700 --> 00:00:21.300 the core of everything which is basically what we call the equilibrium constant. 00:00:21.300 --> 00:00:26.300 We are going to first define this equilibrium constant and then go into different 00:00:26.300 --> 00:00:33.200 types of equilibria including what we call homogeneous equilibria and heterogeneous equilibria. 00:00:33.200 --> 00:00:38.900 We will then go and see how we can change the value of Keq. 00:00:38.900 --> 00:00:42.400 We will also introduce something called the reaction quotient. 00:00:42.400 --> 00:00:46.500 We then jump into a very fundamental principle in all of general chemistry. 00:00:46.500 --> 00:00:50.200 That is called Le Chatelier's principle. 00:00:50.200 --> 00:00:56.500 After that we will get into the quantitative part of the chapter which involves what we call ICE tables. 00:00:56.500 --> 00:01:04.900 We will wrap everything up with a very brief summary followed by a pair of sample problems. 00:01:04.900 --> 00:01:07.800 Chemical equilibrium, exactly what this is. 00:01:07.800 --> 00:01:14.800 This refers to the simultaneous occurrence of a forward and a reverse reaction at the same rate. 00:01:14.800 --> 00:01:18.900 We see that equilibrium is a dynamic process, not a static process. 00:01:18.900 --> 00:01:26.000 For example, let's say you had two beakers here; these beakers are closed. 00:01:26.000 --> 00:01:31.000 Let's say that this one beaker had H2O in it. 00:01:31.000 --> 00:01:33.800 Let's just say this is time zero. 00:01:33.800 --> 00:01:36.900 At time zero, I have nothing but pure liquid water. 00:01:36.900 --> 00:01:40.800 But we know this from everyday experience inside a water bottle. 00:01:40.800 --> 00:01:48.000 If you let some time progress, we notice that the water level is going to drop a little. 00:01:48.000 --> 00:01:56.100 That is because some of the liquid water has entered the gas phase. 00:01:56.100 --> 00:02:05.200 But there is also going to be a point in time where this water level is not going to drop forever. 00:02:05.200 --> 00:02:08.800 It is going to reach a minimum. 00:02:08.800 --> 00:02:22.100 It reaches a minimum because as soon as the vaporization process occurs, the condensation process also occurs. 00:02:22.100 --> 00:02:38.500 We say for this case that the rate of evaporation is equal to the rate of condensation. 00:02:38.500 --> 00:02:44.300 If we were to write this out in a chemical reaction, this would be H2O liquid. 00:02:44.300 --> 00:02:49.400 Now we introduce a new type of reaction arrow which is this. 00:02:49.400 --> 00:03:00.400 That goes to H2O gas; this is our equilibrium arrow. 00:03:00.400 --> 00:03:09.800 It basically shows that the forward direction is happening simultaneously with the reverse direction. 00:03:09.800 --> 00:03:18.000 Let's now get into how we can represent equilibrium numerically. 00:03:18.000 --> 00:03:19.400 Consider the following reaction. 00:03:19.400 --> 00:03:27.500 Small a big A plus small b big B, equilibrium sign and then small c big C small d big D. 00:03:27.500 --> 00:03:42.100 In this reaction, let the lowercase letters represent the stoichiometric coefficients. 00:03:42.100 --> 00:03:45.800 You know the moles after we balance the chemical equation. 00:03:45.800 --> 00:03:54.900 It turns out that after a chemical reaction has reached equilibrium, it is experimentally determined that the ratio of 00:03:54.900 --> 00:04:02.800 product to reactant concentration raised to the stoichiometric coefficients is actually constant at any given temperature. 00:04:02.800 --> 00:04:08.800 Basically the concentration of C raised to the c power times the 00:04:08.800 --> 00:04:16.400 concentration of D raised to the d power over the concentration of A 00:04:16.400 --> 00:04:21.700 raised to the a power times the concentration of B raised to the b power. 00:04:21.700 --> 00:04:31.800 This whole ratio of products to reactants raised to the coefficients is equal to some constant that we call K. 00:04:31.800 --> 00:04:35.000 Sometimes you are going to see this called Keq. 00:04:35.000 --> 00:04:45.300 This is formally what we call the equilibrium constant. 00:04:45.300 --> 00:04:49.900 The only thing that can change the value of Keq is temperature. 00:04:49.900 --> 00:05:02.200 The equilibrium constant is temperature dependent. 00:05:02.200 --> 00:05:07.400 We can typically represent K in two different ways. 00:05:07.400 --> 00:05:25.800 Kc is when molarities are used; Kp is when partial pressures are used. 00:05:25.800 --> 00:05:28.000 Every textbook is a little different. 00:05:28.000 --> 00:05:38.600 But for the partial pressures, the typical units are going to be atmosphere or bar. 00:05:38.600 --> 00:05:43.300 Once again Kc and Kp are just the equilibrium expressions when 00:05:43.300 --> 00:05:53.100 molarities and partial pressures are used for Kc and Kp respectively. 00:05:53.100 --> 00:06:00.700 For solutions, the equilibrium constant can be expressed in units of molarity just like we discussed. 00:06:00.700 --> 00:06:04.100 However it turns out that the equilibrium constant is unitless. 00:06:04.100 --> 00:06:08.300 Kc, how do we go ahead and do that? 00:06:08.300 --> 00:06:18.200 Mole over liter raised to some power divided by mole over liter raised to some power. 00:06:18.200 --> 00:06:25.200 It turns out that the equilibrium expression is always referenced to 1 molarity. 00:06:25.200 --> 00:06:34.000 This is really moles over liter per 1 molar. 00:06:34.000 --> 00:06:42.000 This is moles over liter per one molar raised to the y power and raised to the x power. 00:06:42.000 --> 00:06:51.000 As you can see that after this is done, we see that all units cancel. 00:06:51.000 --> 00:07:00.500 Keq is actually one of the few unitless values in all of your general chemistry studies. 00:07:00.500 --> 00:07:02.200 That is Kc. 00:07:02.200 --> 00:07:09.200 It turns out that if we choose to do our problem with Kp, maybe this is going to be in atmospheres. 00:07:09.200 --> 00:07:18.500 This is also relative to 1 atm, raised to some power divided by atm raised to some power. 00:07:18.500 --> 00:07:31.500 We see it again that the units cancel; Keq again is unitless. 00:07:31.500 --> 00:07:35.400 What is the relationship between Kc and Kp? 00:07:35.400 --> 00:07:41.400 Kc and Kp are directly proportional to each other. 00:07:41.400 --> 00:07:45.100 Depending on the reaction, sometimes they are nearly identical. 00:07:45.100 --> 00:07:49.300 But the exact relationship between the two is the following where 00:07:49.300 --> 00:07:58.900 Kp is equal to Kc times RT over Δn where Δn is 00:07:58.900 --> 00:08:09.100 equal to the moles of gaseous product minus the moles of gaseous reactant. 00:08:09.100 --> 00:08:15.100 Once again Kp and Kc can be very similar. 00:08:15.100 --> 00:08:18.700 However they are not quite the same thing. 00:08:18.700 --> 00:08:27.100 You should really check with your instructor to see what he or she prefers. 00:08:27.100 --> 00:08:32.400 For chemical equilibria, we are usually dealing with very small concentrations. 00:08:32.400 --> 00:08:40.200 Because of this, we assume that pure solids and pure liquids are going to remain relatively unchanged. 00:08:40.200 --> 00:08:52.400 If you have the following reaction, A solid plus B aqueous goes on to form C aqueous plus D solid, 00:08:52.400 --> 00:09:06.000 it is only the aqueous species that are going to affect the equilibrium, affect Keq. 00:09:06.000 --> 00:09:14.700 Hence pure solids and pure liquids are going to be assigned an arbitrary value of 1 when incorporated into the expression for K. 00:09:14.700 --> 00:09:16.900 That is they have no effect. 00:09:16.900 --> 00:09:26.400 Kc for this reaction here would be simply the concentration of C times 1 divided by the 00:09:26.400 --> 00:09:33.000 concentration of B times 1 which is just equal to the concentration of C over concentration of B. 00:09:33.000 --> 00:09:52.900 Once again pure liquids and pure solids do not appear in the expression for K, in the Keq expression. 00:09:52.900 --> 00:09:59.600 Again this is going to be specifically for heterogeneous equilibria. 00:09:59.600 --> 00:10:02.200 What are some ways where we can manipulate K? 00:10:02.200 --> 00:10:08.300 The first way of manipulating K is by multiplying an entire chemical equation by a factor. 00:10:08.300 --> 00:10:20.700 For example, let's take A aqueous plus 2B aqueous goes on to form 3C aqueous. 00:10:20.700 --> 00:10:26.900 In this case, K as you see is equal to the concentration of C cubed 00:10:26.900 --> 00:10:33.400 divided by the concentration of A times the concentration of B squared. 00:10:33.400 --> 00:10:38.000 Let's go ahead and take this chemical reaction and multiply everything through by 2. 00:10:38.000 --> 00:10:47.100 2A aqueous plus 4B aqueous goes on to form 6C aqueous. 00:10:47.100 --> 00:10:54.100 It turns out therefore that this Knew is going to be equal to concentration of C to the sixth 00:10:54.100 --> 00:11:01.300 divided by the concentration of A squared times the concentration of B raised to the fourth power. 00:11:01.300 --> 00:11:10.300 We see very nicely that Knew is simply equal to the original Kc squared. 00:11:10.300 --> 00:11:12.500 The rule of thumb is the following. 00:11:12.500 --> 00:11:34.900 That when multiplying a reaction by a factor, K is going to be raised to that factor. 00:11:34.900 --> 00:11:38.800 K is raised to this factor. 00:11:38.800 --> 00:11:42.100 That is the first way of algebraically manipulating K. 00:11:42.100 --> 00:11:48.700 Once again this is by multiplying through a chemical reaction by a factor. 00:11:48.700 --> 00:11:51.900 The second way is by taking the reverse reaction. 00:11:51.900 --> 00:11:56.900 For example, let's take not A plus 2B going to 3C 00:11:56.900 --> 00:12:06.500 but 3C aqueous goes on to form A aqueous plus 2B aqueous. 00:12:06.500 --> 00:12:18.800 In this case, K is equal to concentration of A times the concentration of B squared divided by the concentration of C cubed. 00:12:18.800 --> 00:12:24.900 We see that this is going to be 1 over kforward. 00:12:24.900 --> 00:12:33.000 kreverse is simply the reciprocal of kforward. 00:12:33.000 --> 00:12:40.100 The third way of changing K is by adding a series of individual chemical reactions together to form a net balanced chemical equation. 00:12:40.100 --> 00:12:45.100 Let's go ahead and see. 00:12:45.100 --> 00:12:57.600 For example, A aqueous plus B aqueous goes on to form C aqueous. 00:12:57.600 --> 00:13:08.200 C aqueous plus D aqueous goes on to form E aqueous. 00:13:08.200 --> 00:13:10.400 Let's go ahead and add these two together. 00:13:10.400 --> 00:13:23.400 When we add these two together, we are going to get A aqueous plus B aqueous plus D aqueous goes on to form E aqueous. 00:13:23.400 --> 00:13:35.500 You see that C is going to be cancelled out because it is going to be formed and consumed simultaneously. 00:13:35.500 --> 00:13:39.100 What we want to look at now are the expressions for k for each of these. 00:13:39.100 --> 00:13:43.200 Here k, I will call this k1 for reaction one, is equal to the 00:13:43.200 --> 00:13:49.800 concentration of C divided by the concentration of A times the concentration of B. 00:13:49.800 --> 00:14:01.200 Here k2 is equal to the concentration of E divided by the concentration of C times the concentration of D. 00:14:01.200 --> 00:14:05.600 Here I will call this third one knet. 00:14:05.600 --> 00:14:15.600 That is equal to the concentration of E divided by the concentration of A times the concentration of B times the concentration of D. 00:14:15.600 --> 00:14:24.300 We see that when we add individual reactions together to get a net reaction, the equilibrium constant 00:14:24.300 --> 00:14:34.000 of the net reaction is simply equal to the product of each individual equilibrium expression constant. 00:14:34.000 --> 00:14:43.700 That is the third way of algebraically manipulating Keq. 00:14:43.700 --> 00:14:47.200 Let's now go on to another aspect of the equilibrium constant. 00:14:47.200 --> 00:14:49.700 It is what we call the reaction quotient. 00:14:49.700 --> 00:14:54.500 The equilibrium constant is good when we actually have the values at equilibrium. 00:14:54.500 --> 00:14:58.900 But what happens if we use values not at equilibrium? 00:14:58.900 --> 00:15:06.000 When concentrations or pressures are inserted into the expression for K that are not at equilibrium, 00:15:06.000 --> 00:15:10.800 the ratio of products to reactants is now what we call the reaction quotient symbolized Q. 00:15:10.800 --> 00:15:16.600 Q is going to be equal to products raised to some power divided by reactants raised to 00:15:16.600 --> 00:15:42.300 some power except that these molarities and partial pressures are not at equilibrium. 00:15:42.300 --> 00:15:44.500 The significance is the following. 00:15:44.500 --> 00:16:16.200 Q can be used to predict which way a reaction will shift to reobtain a state of equilibrium. 00:16:16.200 --> 00:16:20.200 Basically we are going to have the following general rules. 00:16:20.200 --> 00:16:27.300 If Q is greater than K, that means we have too much product, not enough reactant. 00:16:27.300 --> 00:16:32.200 We are going to shift left. 00:16:32.200 --> 00:16:41.000 If Q is less than K, we have too much reactant relative to product. 00:16:41.000 --> 00:16:44.700 We are going to shift right. 00:16:44.700 --> 00:16:55.300 Finally if Q is identical to K, we are at a equilibrium state. 00:16:55.300 --> 00:17:07.700 Shifting left is the same as making more reactant. 00:17:07.700 --> 00:17:16.400 Shifting right is the same as making more product. 00:17:16.400 --> 00:17:24.400 At equilibrium, there is no net change; neither direction is favored over the other. 00:17:24.400 --> 00:17:34.900 Once again Q can be very useful for determining what direction a reaction will shift if any. 00:17:34.900 --> 00:17:40.600 Now we have come into one of the most fundamental principles from all of general chemistry. 00:17:40.600 --> 00:17:43.000 This is called Le Chatelier's principle. 00:17:43.000 --> 00:17:48.100 Le Chatelier's principle tells us that when a system at equilibrium is disturbed, 00:17:48.100 --> 00:17:57.700 it will react to counteract the disturbance in an attempt to restore equilibrium. 00:17:57.700 --> 00:18:01.700 One of the best examples we can think of is from us. 00:18:01.700 --> 00:18:10.900 When we bleed, when we lose blood, what is the natural thing that our body does in order to counteract this? 00:18:10.900 --> 00:18:12.600 Your body is going to try to make more blood. 00:18:12.600 --> 00:18:32.600 When we lose blood, our bodies try to make more in order to counteract the lost. 00:18:32.600 --> 00:18:34.600 This is a nice example of Le Chatelier's principle. 00:18:34.600 --> 00:18:39.600 But now we are going to apply this to chemical reactions. 00:18:39.600 --> 00:18:43.800 In Le Chatelier's principle, you notice that there is the word disturb. 00:18:43.800 --> 00:18:47.600 There are several ways of disturbing a chemical system that is at equilibrium already. 00:18:47.600 --> 00:18:51.700 Number one is a change in reactant or product concentration. 00:18:51.700 --> 00:18:57.600 Number two if we are dealing with gases, it is going to be a change in reactant or product partial pressure. 00:18:57.600 --> 00:19:00.600 The third one is the change in reaction volume. 00:19:00.600 --> 00:19:03.700 The fourth one is going to be a change in temperature. 00:19:03.700 --> 00:19:10.000 Let's now study each of these. 00:19:10.000 --> 00:19:22.700 Basically if the concentration of reactant goes up, then we are going to shift away from it. 00:19:22.700 --> 00:19:25.300 We are going to shift right. 00:19:25.300 --> 00:19:30.600 If the concentration of product goes down, we don't have enough of it. 00:19:30.600 --> 00:19:38.600 We are going to shift right; that is one situation. 00:19:38.600 --> 00:19:48.900 If the partial pressure of a reactant goes up, partial pressure is just the same as concentration. 00:19:48.900 --> 00:19:54.300 We know from ideal gas law that pressure is proportional to amount. 00:19:54.300 --> 00:20:03.800 If partial pressure of the reactant goes up, we are going to shift right. 00:20:03.800 --> 00:20:15.300 If the partial pressure of the product goes down, we are going to shift right. 00:20:15.300 --> 00:20:28.200 For temperature, I am going to do this on the last slide because I am running out of room right now. 00:20:28.200 --> 00:20:34.900 I will come back to changes in temperature and also to change in vessel volume. 00:20:34.900 --> 00:20:38.500 But now let's get into some problem solving with some ICE tables. 00:20:38.500 --> 00:20:41.600 We will now approach chemical equilibrium from a quantitative view. 00:20:41.600 --> 00:20:46.600 An ICE table allows for one to study a chemical system at three points in time. 00:20:46.600 --> 00:20:50.200 Basically what are all reactant and product initial amounts? 00:20:50.200 --> 00:20:51.700 That is what the I stands for. 00:20:51.700 --> 00:20:56.500 During the reaction on the way to equilibrium, what are their changes in concentration/pressure? 00:20:56.500 --> 00:20:57.900 That is what the C stands for. 00:20:57.900 --> 00:21:02.300 Finally what are their final volumes once equilibrium has been achieved? 00:21:02.300 --> 00:21:03.500 That is what the E stands for. 00:21:03.500 --> 00:21:06.200 We are going to look at this right now. 00:21:06.200 --> 00:21:11.100 Consider 2 water gas goes to 2H2 gas plus O2 gas. 00:21:11.100 --> 00:21:14.700 K is equal to 2.4 times 10^-5 at some temperature. 00:21:14.700 --> 00:21:19.200 At equilibrium, the concentration of H2O gas is 0.11 molar. 00:21:19.200 --> 00:21:25.400 The concentration of H2 gas is 0.019 molar; calculate O2 at equilibrium. 00:21:25.400 --> 00:21:29.200 The very first thing we want to do is set up our ICE table. 00:21:29.200 --> 00:21:36.000 2H2O gas goes on to form 2H2 gas plus O2 gas. 00:21:36.000 --> 00:21:40.100 What I like to do, I just like to set up the letters I-C-E right underneath it. 00:21:40.100 --> 00:21:43.200 We just fill in this table right now. 00:21:43.200 --> 00:21:53.200 You are told that at equilibrium, the water is 0.11 molar and that the H2 is 0.019 molar. 00:21:53.200 --> 00:21:54.700 We don't know what O2 is. 00:21:54.700 --> 00:22:02.400 We can just call that the concentration of O2 at equilibrium. 00:22:02.400 --> 00:22:08.000 We know the expression for K is equal to 2.4 times 10^-5. 00:22:08.000 --> 00:22:12.200 That is going to be equal to the concentration of H2 squared times 00:22:12.200 --> 00:22:17.800 the concentration of O2 divided by the concentration of H2O squared. 00:22:17.800 --> 00:22:28.900 That is going to be equal to 0.019 squared times the concentration of O2 at equilibrium divided by 0.11 squared. 00:22:28.900 --> 00:22:39.000 When all is said and done, the concentration of O2 at equilibrium is going to be 8.0 times 10^-4 molar. 00:22:39.000 --> 00:22:42.900 Don't forget the units. 00:22:42.900 --> 00:22:45.700 You notice that we didn't have to fill in the rest of the table 00:22:45.700 --> 00:22:50.800 because we were already at halfway there to the equilibrium values. 00:22:50.800 --> 00:22:54.500 This is a nice usage of the ICE table. 00:22:54.500 --> 00:22:57.200 Let's go ahead and do another example though. 00:22:57.200 --> 00:23:02.200 A 1 liter flask was filled with so much of SO2 and so much NO2 at some temperature. 00:23:02.200 --> 00:23:05.300 After equilibrium was reached, 1.3 moles of NO was present. 00:23:05.300 --> 00:23:10.100 The reaction is the following; calculate the value of Kc at this temperature. 00:23:10.100 --> 00:23:16.600 The problem wants us to calculate Kc at this temperature. 00:23:16.600 --> 00:23:26.000 We are going to go ahead and set up the ICE table--I, C, and E. 00:23:26.000 --> 00:23:31.600 SO2 here is going to be 2.00 molar initially. 00:23:31.600 --> 00:23:35.400 O2 is going to be 2.00 molar also. 00:23:35.400 --> 00:23:39.300 After equilibrium was reached, 1.30 moles of NO gas was present. 00:23:39.300 --> 00:23:45.800 That is what goes right here, 1.30 molar. 00:23:45.800 --> 00:23:47.100 Let's fill in the rest of the table. 00:23:47.100 --> 00:23:53.700 If no initial values are mentioned of SO3 and NO, it is safe to assume that they are zero. 00:23:53.700 --> 00:23:57.700 Let's go ahead and do the change. 00:23:57.700 --> 00:24:00.800 The SO2 is going to go down by some amount x. 00:24:00.800 --> 00:24:03.500 O2 is going to go down by some amount x. 00:24:03.500 --> 00:24:12.300 SO3 goes up by some amount x. 00:24:12.300 --> 00:24:16.700 O is going to go up by some amount x. 00:24:16.700 --> 00:24:25.200 We notice because these are all 1:1 mole ratio. 00:24:25.200 --> 00:24:32.000 At equilibrium, it is just going to be the addition of the two rows added together; sum. 00:24:32.000 --> 00:24:39.000 That is going to be 2.00 minus x, 2.00 minus x, x, and x. 00:24:39.000 --> 00:24:45.000 But guess what? x is 1.30 molar because we were told that in the beginning; that is very nice. 00:24:45.000 --> 00:24:54.200 Therefore this is 1.30 molar; this is 0.70 molar; this is 0.70 molar. 00:24:54.200 --> 00:24:58.100 We know enough to calculate the value of Kc. 00:24:58.100 --> 00:25:04.200 Kc is going to be 1.30 squared divided by 0.70 squared. 00:25:04.200 --> 00:25:17.100 That is going to give us an answer of 3.5 for our answer. 00:25:17.100 --> 00:25:26.200 This is how we use the ICE tables to help us come up with the equilibrium constant. 00:25:26.200 --> 00:25:30.800 I want to quickly summarize this and then jump into sample problems. 00:25:30.800 --> 00:25:36.300 Chemical equilibrium is a dynamic process where the rate of forward and reverse reactions is equal. 00:25:36.300 --> 00:25:42.700 Equilibrium constant K quantifies how reactant or product favored a reaction is once it has reached equilibrium. 00:25:42.700 --> 00:25:52.400 If K is between 0 and 1, we say that it is reactant favored. 00:25:52.400 --> 00:25:59.000 If K is greater than 1, we say that it is product favored. 00:25:59.000 --> 00:26:05.600 Le Chatelier's principle states that a system at equilibrium when disturbed will react to counteract the disturbance. 00:26:05.600 --> 00:26:12.800 We already saw the concentration and partial pressures. 00:26:12.800 --> 00:26:19.500 But let's go ahead and see what I left out before; this is now temperature. 00:26:19.500 --> 00:26:33.000 For an exothermic reaction, if the temperature increases, it turns out we are going to shift left. 00:26:33.000 --> 00:26:39.200 If the temperature decreases, we are going to shift right. 00:26:39.200 --> 00:26:49.200 Basically you treat the word heat as a molecule. 00:26:49.200 --> 00:26:56.600 Again if you treat the word heat as a molecule on the product side, it becomes 00:26:56.600 --> 00:27:02.300 a lot more intuitive as you will see what happens when we result in the change. 00:27:02.300 --> 00:27:09.300 Another thing we can disturb an equilibrium with is with the volume. 00:27:09.300 --> 00:27:16.300 If the volume of the reaction vessel goes up, that means the pressure is lower. 00:27:16.300 --> 00:27:19.800 That means we are going to shift to make more gases. 00:27:19.800 --> 00:27:34.600 Shift toward side of reaction with more gas molecules. 00:27:34.600 --> 00:27:41.800 Once again if the volume goes up, we are going to shift toward the side of the reaction with more gas molecules. 00:27:41.800 --> 00:27:48.000 Those are the ways we can disrupt a system that is at equilibrium and see how it is going to counteract the stress. 00:27:48.000 --> 00:27:55.800 Finally ICE tables allow for determination of the reactant and product concentrations/pressures given the equilibrium constant K. 00:27:55.800 --> 00:27:57.800 That is our summary. 00:27:57.800 --> 00:28:01.000 Now let's go ahead and jump into a pair of sample problems. 00:28:01.000 --> 00:28:05.100 Consider 2SO3 gas goes to 2SO2 plus O2. 00:28:05.100 --> 00:28:09.400 Initially 12 moles of SO3 is placed into a 3 liter flask at some temperature. 00:28:09.400 --> 00:28:12.200 At equilibrium, 3 moles of SO2 is present. 00:28:12.200 --> 00:28:16.100 Calculate the value of Kc at this temperature. 00:28:16.100 --> 00:28:18.100 Let's go ahead and rewrite this. 00:28:18.100 --> 00:28:26.700 2SO3 gas goes on to form 2SO2 gas plus O2 gas. 00:28:26.700 --> 00:28:32.100 Let's go ahead and see what we can fill in, I-C-E. 00:28:32.100 --> 00:28:36.300 Initially 12 moles is placed in a 3 liter flask of SO3. 00:28:36.300 --> 00:28:40.900 That is going to be 4.0 molar. 00:28:40.900 --> 00:28:48.400 At equilibrium, SO2 is present, 3 moles; that is going to be 1.0 molar. 00:28:48.400 --> 00:28:53.200 Usually you can fill in two-thirds of the table with just one or two sentences. 00:28:53.200 --> 00:28:55.600 SO2 initial is not mentioned; that is zero. 00:28:55.600 --> 00:29:00.100 O2 initial is not mentioned; that is zero also. 00:29:00.100 --> 00:29:02.200 We have to pay attention to the coefficents. 00:29:02.200 --> 00:29:07.500 The coefficients tells us the relative amount of reactant and product that is going to be involved. 00:29:07.500 --> 00:29:12.400 Really this is going to be -2x; SO2 is going to be +2x. 00:29:12.400 --> 00:29:15.800 O2 is just going to be +x. 00:29:15.800 --> 00:29:23.400 At equilibrium, we get 4.0 minus 2x, 1 molar, and x. 00:29:23.400 --> 00:29:36.100 The nice thing about this is that because we know the 2x and the 1 molar, we can conclude that x is 0.5 molar. 00:29:36.100 --> 00:29:40.400 When we do that, we are going to get here 0.5 molar. 00:29:40.400 --> 00:29:46.600 Here for SO3, we are going to get 3.0 molar. 00:29:46.600 --> 00:29:51.200 Great, we now have all of our molarities at equilibrium. 00:29:51.200 --> 00:29:57.300 Kc is equal to the concentration of SO2 squared times the concentration 00:29:57.300 --> 00:30:05.000 of O2 divided by concentration of SO3 squared, all equilibrium values. 00:30:05.000 --> 00:30:14.800 This is going to be 0.5 squared times 0.5 divided by 3.0. 00:30:14.800 --> 00:30:27.600 When all is said and done, this is going to give us a Kc of 0.056 for this reaction. 00:30:27.600 --> 00:30:31.100 That is another usage of the ICE table. 00:30:31.100 --> 00:30:34.000 Finally the last sample problem is sample problem number two. 00:30:34.000 --> 00:30:37.300 At a particular temperature, K is equal to 3.75. 00:30:37.300 --> 00:30:43.600 If all four gases had initial concentrations of 0.800, calculate their equilibrium concentrations. 00:30:43.600 --> 00:30:48.200 Let's go ahead and set it up--I, C, and E. 00:30:48.200 --> 00:30:58.200 0.800 molar initial, 0.800 molar, 0.800 molar, and 0.800 molar. 00:30:58.200 --> 00:31:08.400 Because we have nonzero amounts of all reactants and products, we don't know which way we are going to shift. 00:31:08.400 --> 00:31:12.500 This is a twist on things; now we have to solve for Q. 00:31:12.500 --> 00:31:15.000 Q is going to tell us which way we are going to shift. 00:31:15.000 --> 00:31:22.300 This is going to be equal to concentration of SO3 initial times the concentration of 00:31:22.300 --> 00:31:31.400 NO initial divided by concentration of SO2 initial times the concentration of NO2 initial. 00:31:31.400 --> 00:31:40.300 That is just going to be equal to 1, 1.00 which is going to be less than K. 00:31:40.300 --> 00:31:47.100 Because Q is less than K, that means we are going to shift to the right. 00:31:47.100 --> 00:31:51.100 Therefore the sign that goes here is going to be ?x. 00:31:51.100 --> 00:31:56.600 This is ?x; this is +x; this is +x. 00:31:56.600 --> 00:32:03.600 We are going to get 0.800 minus x; 0.800 minus x. 00:32:03.600 --> 00:32:06.300 This is going to be 0.800 plus x. 00:32:06.300 --> 00:32:14.200 This is going to be 0.800 plus x; K is equal to 3.75. 00:32:14.200 --> 00:32:25.700 That is equal to 0.800 plus x squared divided by 0.800 minus x squared. 00:32:25.700 --> 00:32:30.700 We can use what we call a perfect square to solve for this. 00:32:30.700 --> 00:32:35.900 The square root of both sides is what we are going to do next. 00:32:35.900 --> 00:32:46.500 When we go ahead and do that, we are going to get 1.93 is equal to 0.800 plus x divided by 0.800 minus x. 00:32:46.500 --> 00:32:52.200 When all is said and done, x is going to be equal to 0.25 molar. 00:32:52.200 --> 00:32:58.000 All we do next is simply plug it back into all of the expressions at equilibrium. 00:32:58.000 --> 00:33:05.800 The concentration of SO3 at equilibrium is equal to the concentration of NO at equilibrium. 00:33:05.800 --> 00:33:12.600 That is going to be 0.800 plus 0.25. 00:33:12.600 --> 00:33:23.800 I will let you guys do the sum; that is going to be 1.050 molar. 00:33:23.800 --> 00:33:32.200 The concentration of SO2 at equilibrium is equal to the concentration of NO2 at equilibrium. 00:33:32.200 --> 00:33:40.600 That is going to be equal to 0.800 minus 0.25. 00:33:40.600 --> 00:33:46.200 We are going to get 0.55 molar for these guys. 00:33:46.200 --> 00:33:53.900 That is use of the ICE table when we have all four initial values present. 00:33:53.900 --> 00:33:56.500 Once again you have to find which way it is going to shift. 00:33:56.500 --> 00:33:59.500 You do that by evaluating Q. 00:33:59.500 --> 00:34:05.000 That is our lecture from general chemistry on the principles of chemical equilibria. 00:34:05.000 --> 00:34:06.300 I want to thank you for your time. 00:34:06.300 --> 00:34:09.000 I will see you next time on Educator.com.