0:00:00.333,0:00:02.335 - [Voiceover] In this video,[br]we're going to be talking about 0:00:02.335,0:00:06.072 using the small x approximation to solve 0:00:06.072,0:00:09.312 equilibrium problems when Kc is large. 0:00:09.312,0:00:14.047 And by large, I mean[br]that Kc is greater than 0:00:14.047,0:00:17.244 or approximately equal[br]to 10 to the fourth. 0:00:18.752,0:00:22.155 So we have another video[br]before this that talks about 0:00:22.155,0:00:26.392 using the small x approximation[br]for when Kc is small, 0:00:26.392,0:00:28.561 and I would say that's probably[br]a simpler place to start 0:00:28.561,0:00:31.164 if this is the first[br]time you're seeing it. 0:00:31.164,0:00:33.533 And in this video, we'll be talking about 0:00:33.533,0:00:35.535 the opposite situation. 0:00:35.535,0:00:37.504 As we talked about in the previous video, 0:00:37.504,0:00:41.174 the steps for solving this[br]kind of problem are fourfold. 0:00:42.064,0:00:44.624 So there are four steps. 0:00:44.624,0:00:49.582 The first step is to assume[br]that the reaction goes 100% 0:00:49.582,0:00:51.584 in the favored direction. 0:00:51.584,0:00:55.221 So the favored direction[br]when K is really large 0:00:55.221,0:00:58.882 is that means we're going[br]all the way to products. 0:01:00.393,0:01:04.214 We're assuming we have almost[br]no starting materials left. 0:01:05.298,0:01:08.201 The second step is to set up an ICE table. 0:01:08.201,0:01:10.403 And then to solve for x in our ICE table, 0:01:10.403,0:01:14.441 assuming that x is small,[br]hence small x approximation. 0:01:14.441,0:01:17.577 And then the last and possibly[br]the most important step 0:01:17.577,0:01:19.218 is to check our answer. 0:01:19.218,0:01:21.815 So make sure that the size[br]of x is actually small 0:01:21.815,0:01:24.250 compared to what we said[br]it was smaller than, 0:01:24.250,0:01:27.353 and also to make sure that[br]it gives us the right answer 0:01:27.353,0:01:30.123 when we plug it back in to calculate Kc. 0:01:30.123,0:01:32.759 In this video, we're going to[br]go through an example problem. 0:01:32.759,0:01:36.724 The example is the reaction of 0:01:37.611,0:01:42.434 NO gas reacting with 0:01:42.434,0:01:47.307 chlorine gas to make 0:01:47.307,0:01:52.307 2NOCl gas. 0:01:52.389,0:01:56.478 And this particular reaction has a K value 0:01:56.478,0:02:01.294 that is equal to 6.25 times 10 0:02:01.294,0:02:03.177 to the fourth. 0:02:03.177,0:02:06.326 So that is indeed on the[br]order of 10 to the fourth, 0:02:06.326,0:02:10.448 so we should be able to use[br]the small x approximation. 0:02:10.448,0:02:12.932 We're going to do it two ways. 0:02:12.932,0:02:15.195 So we're gonna set up our ICE table, 0:02:16.302,0:02:21.302 and we know the initial[br]concentrations are 2.0 molar 0:02:21.307,0:02:25.545 for NO, and 2.0 molar for Cl2. 0:02:25.545,0:02:29.048 And we have no NOCl at the beginning. 0:02:29.048,0:02:32.986 So if we are setting up an[br]ICE table as we normally did, 0:02:32.986,0:02:36.356 we would say, "Okay,[br]we'll, we're going to make 0:02:36.356,0:02:38.858 "some amount of NOCl." 0:02:38.858,0:02:42.695 So we'll just say minus 2x for our NO 0:02:42.695,0:02:47.365 since for every one Cl2 we use up, 0:02:47.365,0:02:51.894 we'll use up 2NO, which means[br]we'll have minus x for Cl2, 0:02:51.894,0:02:56.635 and we'll make 2x molar[br]concentration of our product. 0:02:57.610,0:02:59.913 And in the end, once we reach equilibrium, 0:02:59.913,0:03:02.949 we'll have to 2.0 molar minus 2x, 0:03:02.949,0:03:06.152 and we just get that by adding[br]the initial concentration 0:03:06.152,0:03:08.154 with the change. 0:03:08.154,0:03:13.154 In the same way, we get 2.0[br]molar minus x for chlorine gas. 0:03:13.893,0:03:17.830 And for a product, we just[br]get 2x at equilibrium. 0:03:17.830,0:03:21.334 Let's look back at our original[br]steps that we talked about 0:03:21.334,0:03:23.525 at the very beginning of this video. 0:03:24.671,0:03:28.608 We just made an ICE[br]table, so that's step two. 0:03:28.608,0:03:31.778 Did we assume that the[br]reaction goes to 100% 0:03:31.778,0:03:34.452 in our favored direction? 0:03:34.452,0:03:35.682 It turns out we didn't. 0:03:35.682,0:03:39.319 We actually skipped step[br]one to set up our ICE table 0:03:39.319,0:03:42.474 in what seems like a pretty natural way. 0:03:42.474,0:03:44.057 And it turns out that's not good. 0:03:44.057,0:03:47.299 So we'll see what happens[br]when we skip step one 0:03:47.299,0:03:48.834 and keep going. 0:03:49.696,0:03:53.299 So if we keep going, we can solve for x, 0:03:53.299,0:03:57.455 assuming x is small, and we'll get for 0:03:58.471,0:04:00.506 our equilibrium expression, 0:04:00.506,0:04:05.478 we get Kc is equal to 2x squared 0:04:05.478,0:04:10.012 divided by 2.0 molar 0:04:10.012,0:04:13.646 minus x, which our Cl2 concentration, 0:04:14.654,0:04:18.591 and multiply that by 2.0 molar minus 2x, 0:04:18.591,0:04:21.861 and that's all squared since[br]we have that stoichiometric 0:04:21.861,0:04:24.097 coefficient of two in front. 0:04:24.097,0:04:26.099 Just to point out, we already started 0:04:26.099,0:04:28.101 with a balanced reaction,[br]that's pretty important, 0:04:28.101,0:04:31.137 or else our Kc expression would be wrong. 0:04:31.137,0:04:34.398 So make sure everything is[br]balanced before you get started. 0:04:35.274,0:04:38.344 So we've written our expression for Kc, 0:04:38.344,0:04:42.081 and now we are going[br]to erroneously assume, 0:04:42.081,0:04:43.680 we're just going to be[br]like, "Oh, I don't care." 0:04:43.680,0:04:45.049 (chuckles)[br]"I'm just going to assume 0:04:45.049,0:04:46.152 "x is small. 0:04:46.152,0:04:47.963 "I don't care if that's a[br]good assumption at all." 0:04:47.963,0:04:51.090 So if x is small, what[br]we're really seeing is x 0:04:51.090,0:04:54.279 is a lot smaller than[br]two molar, here and here. 0:04:54.279,0:04:57.463 So we're really seeing 2x is[br]a lot smaller than two molar. 0:04:57.463,0:05:00.833 So we're seeing this is[br]approximately equal to 0:05:00.833,0:05:04.671 2x squared divided by 0:05:04.671,0:05:06.194 2.0 molar. 0:05:07.140,0:05:09.142 Since x is a lot smaller than two molar, 0:05:09.142,0:05:11.210 we'll just ignore it entirely, 0:05:11.210,0:05:13.813 and x is a lot smaller than 2x, 0:05:13.813,0:05:16.783 or sorry, x is a lot[br]smaller than 2.0 molar 0:05:16.783,0:05:18.785 even if you multiply x by two, 0:05:18.785,0:05:22.076 so this is going to be 2.0 molar squared. 0:05:24.223,0:05:28.928 And so if we multiply this out,[br]we get that this is equal to 0:05:28.928,0:05:32.532 x squared over 0:05:32.532,0:05:37.370 this two squared is[br]canceled out by this two, 0:05:37.370,0:05:41.040 so we actually get x[br]squared over two is equal to 0:05:41.040,0:05:44.310 our Kc, which is 6.25 0:05:44.310,0:05:46.795 times 10 to the fourth. 0:05:47.647,0:05:50.083 So that's what we get for x squared. 0:05:50.083,0:05:53.524 So if we multiply both sides by two, 0:05:53.524,0:05:56.547 we get that x squared is equal to 0:05:56.547,0:06:00.893 1.25 times 10 to the fifth, 0:06:00.893,0:06:05.298 which means x is equal to[br]the square root of that, 0:06:05.298,0:06:07.075 which is 354. 0:06:08.186,0:06:09.936 So this is where... 0:06:09.936,0:06:10.903 Well, okay. 0:06:10.903,0:06:13.206 So we've gone through steps two and three, 0:06:13.206,0:06:15.174 and now we're going to check our answer. 0:06:15.174,0:06:18.157 We're going to see if[br]our answer makes sense. 0:06:18.157,0:06:21.983 So we're saying that the[br]change in concentration here 0:06:22.882,0:06:24.884 is 354. 0:06:24.884,0:06:28.988 That clearly doesn't make sense[br]because it actually gives us 0:06:28.988,0:06:31.290 a negative concentration here. 0:06:31.290,0:06:35.294 So we're getting negative[br]concentrations for NO 0:06:35.294,0:06:37.463 and Cl2, so that's bad. 0:06:37.463,0:06:38.564 So that doesn't make sense. 0:06:38.564,0:06:42.802 The other thing is that[br]this already tells us 0:06:42.802,0:06:44.804 that our assumption was really bad. 0:06:44.804,0:06:47.573 We assumed to solve our equation for Kc. 0:06:47.573,0:06:50.235 We assumed that K is a[br]lot smaller than two, 0:06:51.244,0:06:53.513 or sorry, we assumed that x[br]is a lot smaller than two, 0:06:53.513,0:06:57.450 but then we got that x is 350,[br]which is clearly not smaller 0:06:57.450,0:06:58.451 than two. 0:06:58.451,0:07:02.655 So it looks like skipping[br]step one was bad. 0:07:02.655,0:07:04.891 So let's give this another try. 0:07:04.891,0:07:06.588 Try number two. 0:07:06.588,0:07:08.461 What was step one? 0:07:08.461,0:07:12.064 Step one says assume[br]the reaction goes 100% 0:07:12.064,0:07:14.467 in the favored direction. 0:07:14.467,0:07:18.704 And since Kc is really large[br]for this particular problem, 0:07:18.704,0:07:23.176 we're saying it's going[br]all the way to products. 0:07:23.176,0:07:26.179 What this basically means[br]is that we're gonna set up 0:07:26.179,0:07:28.447 our ICE table again, 0:07:28.447,0:07:31.150 except that this time,[br]our initial concentrations 0:07:31.150,0:07:33.920 are going to be assuming that 0:07:33.920,0:07:37.115 our reaction already went[br]all the way to products. 0:07:40.059,0:07:41.527 And we can figure that out, 0:07:41.527,0:07:43.596 we can figure out those[br]initial concentrations 0:07:43.596,0:07:45.565 using stoichiometry. 0:07:45.565,0:07:49.692 So in the beginning, we[br]started out with two molar NO 0:07:49.692,0:07:52.238 and two molar Cl2, 0:07:52.238,0:07:56.375 and we know that the NO and[br]Cl2 react in a 2:1 ratio. 0:07:56.375,0:07:59.145 So since we have the same[br]amount of both of these, 0:07:59.145,0:08:00.963 our limiting reactant 0:08:02.759,0:08:04.739 will be the NO. 0:08:06.252,0:08:08.254 And so that will get used up completely 0:08:08.254,0:08:10.256 when we get to equilibrium. 0:08:10.256,0:08:13.593 So that means if we assume[br]that it goes all the way to 0:08:13.593,0:08:18.064 products, we'll have zero molar NO left, 0:08:18.064,0:08:22.034 and will make 2.0 molar of our product. 0:08:22.034,0:08:26.005 And since we had our Cl2 in excess, 0:08:26.005,0:08:29.508 we'll use one molar of it[br]to react with two molar 0:08:29.508,0:08:32.359 of our NO, and we'll[br]have one molar left over. 0:08:33.980,0:08:37.450 So this is the special step[br]where we actually followed 0:08:37.450,0:08:41.267 step one and assumed 0:08:42.922,0:08:47.649 we got 100% of product. 0:08:49.161,0:08:51.697 And the reason why we made this assumption 0:08:51.697,0:08:54.333 was because we know[br]that K is really large. 0:08:54.333,0:08:58.004 So we know that at equilibrium,[br]we should have all product 0:08:58.004,0:08:59.639 or mostly product. 0:08:59.639,0:09:01.641 So now let's go through the other steps. 0:09:01.641,0:09:04.043 We assume that it's going to all product, 0:09:04.043,0:09:06.112 but we're not quite at equilibrium. 0:09:06.112,0:09:10.483 So if we're going to[br]equilibrium, that means we assume 0:09:10.483,0:09:14.453 that we'll see a little[br]bit of NO at equilibrium, 0:09:14.453,0:09:17.283 so that'll actually be plus 2x. 0:09:18.190,0:09:21.894 And we'll expect to see[br]a little bit more Cl2 0:09:21.894,0:09:22.895 at equilibrium. 0:09:22.895,0:09:25.807 So that'll be plus x because[br]of the stoichiometry, 0:09:26.732,0:09:31.732 and that would give us[br]minus 2x for the NOCl. 0:09:32.004,0:09:34.707 We expect a little of[br]this to get used up to go 0:09:34.707,0:09:36.709 in the reverse reaction. 0:09:36.709,0:09:39.045 So then if we add everything[br]together from the initial 0:09:39.045,0:09:42.836 and change, then we get 2x 0:09:42.836,0:09:45.684 for our NO concentration, 0:09:45.684,0:09:48.087 we get x for our Cl2 concentration, 0:09:48.087,0:09:52.141 and we get 2.0 minus 2x 0:09:52.141,0:09:54.882 for our product concentration. 0:09:57.863,0:09:59.952 So far, so good. 0:10:01.792,0:10:05.004 So now we're going to set up 0:10:05.004,0:10:08.007 our Kc expression just like we did before. 0:10:08.007,0:10:11.811 So Kc is still equal to 6.25 0:10:11.811,0:10:14.280 times 10 to the fourth, 0:10:14.280,0:10:18.772 and this is equal to the[br]concentration of NOCl squared. 0:10:19.618,0:10:24.123 So that is 2.0 minus 2x squared 0:10:24.123,0:10:27.159 divided by 2x squared, 0:10:27.159,0:10:30.596 which is our concentration[br]of NO squared times x. 0:10:30.596,0:10:32.631 Oh, oops, I made a mistake. 0:10:32.631,0:10:36.802 This is actually 1.0 plus x, sorry. 0:10:36.802,0:10:40.006 So this should be x plus 1.0 molar. 0:10:41.574,0:10:44.443 So that is our full expression for Kc 0:10:44.443,0:10:49.091 using our equilibrium concentrations, 0:10:49.091,0:10:52.469 and we have made zero approximations of R. 0:10:53.586,0:10:56.022 But now we're gonna[br]assume that we can assume, 0:10:56.022,0:10:57.120 (chuckles)[br]now we're gonna assume 0:10:57.120,0:10:57.952 that x is small. 0:10:57.952,0:10:59.688 We're gonna assume that assume. 0:10:59.688,0:11:02.017 So if x is small, what[br]we really mean is that x 0:11:02.017,0:11:05.031 is a lot smaller than one molar. 0:11:05.031,0:11:08.501 And we're also saying it's a[br]lot smaller than two molar. 0:11:08.501,0:11:12.438 So then we're saying that our[br]numerator is approximately 0:11:12.438,0:11:15.408 equal to 2.0 molar squared, 0:11:15.408,0:11:17.410 because we're saying that this is small. 0:11:17.410,0:11:20.179 I'm gonna make this an[br]approximately equal sign. 0:11:20.179,0:11:22.882 The 2x squared stays the same. 0:11:22.882,0:11:25.918 Since we're assuming x is[br]a lot more than 1.0 molar, 0:11:25.918,0:11:27.820 this just becomes 1.0 molar. 0:11:27.820,0:11:32.681 So now if we multiply this[br]all out, we get that four 0:11:33.759,0:11:38.130 divided by 4x squared 0:11:38.130,0:11:40.099 is equal to Kc, 0:11:40.099,0:11:42.168 and the fours cancel out. 0:11:42.168,0:11:46.005 So x squared is equal to one over Kc, 0:11:46.005,0:11:49.762 or 6.25 times 10 to the fourth. 0:11:51.143,0:11:53.145 And then if we take the square[br]root of both sides here, 0:11:53.145,0:11:56.415 we get that x is equal to, let's see, 0:11:56.415,0:11:59.685 we get x is equal to 4.0 times 10 0:11:59.685,0:12:02.588 to the minus three molar. 0:12:02.588,0:12:05.925 So this is where common sense is needed 0:12:05.925,0:12:09.328 to make sure that, well, 0:12:09.328,0:12:11.330 everything worked this time around. 0:12:11.330,0:12:13.332 So first of all, we can ask ourselves, 0:12:13.332,0:12:15.935 "Okay, is this x actually a[br]lot smaller than the numbers 0:12:15.935,0:12:17.903 "we said it was smaller than?" 0:12:17.903,0:12:21.207 So now we're comparing x to one, 0:12:21.207,0:12:23.776 which, it's about three[br]orders of magnitude smaller. 0:12:23.776,0:12:24.743 So that's good. 0:12:24.743,0:12:27.980 And we can also compare 2x to two. 0:12:27.980,0:12:30.216 And again, it's about three[br]orders of magnitude smaller. 0:12:30.216,0:12:31.684 So, so far, so good. 0:12:31.684,0:12:34.920 But our final test will[br]be to plug it back in 0:12:34.920,0:12:36.580 our Kc expression. 0:12:37.423,0:12:41.127 So if we plug in our[br]value of x, we get that Kc 0:12:41.127,0:12:42.962 is equal to 2.0 molar 0:12:42.962,0:12:47.166 minus two times 4.0 0:12:47.166,0:12:51.670 times 10 to the minus[br]three molar, all squared, 0:12:51.670,0:12:56.308 divided by two times 4.0 0:12:56.308,0:13:00.127 times 10 to the minus three molar, 0:13:00.127,0:13:02.314 and that's also squared, four, 0:13:02.314,0:13:06.452 and that's the NO[br]concentration at equilibrium. 0:13:06.452,0:13:10.222 And then the last part[br]is our Cl2 concentration, 0:13:10.222,0:13:15.222 which is 4.0 times 10[br]to the minus three molar 0:13:16.295,0:13:18.017 plus 1.0 molar. 0:13:19.498,0:13:23.636 And so if we multiply that all[br]out, what you get is that Kc 0:13:23.636,0:13:28.199 is equal to 6.23 times 10 to the fourth. 0:13:29.308,0:13:31.844 And so we can compare[br]that to the value of Kc 0:13:31.844,0:13:33.846 we started out our problem with, 0:13:33.846,0:13:34.847 which is, let's see. 0:13:34.847,0:13:37.616 It is 6.25 times 10 to the fourth. 0:13:37.616,0:13:39.718 And this is actually pretty good. 0:13:39.718,0:13:42.288 We did make an approximation,[br]so our answer isn't 0:13:42.288,0:13:43.289 exactly right. 0:13:43.289,0:13:44.356 But it's pretty close. 0:13:44.356,0:13:46.358 And so if we wanted to get it even closer, 0:13:46.358,0:13:48.761 there are other methods we could use. 0:13:48.761,0:13:52.131 But for most purposes, this is actually, 0:13:52.131,0:13:55.034 this tells us that our[br]approximation was good. 0:13:55.034,0:13:58.971 So we can see that when[br]Kc is really large, 0:13:58.971,0:14:01.941 what we need to do is assume that we have 0:14:01.941,0:14:04.577 100% product when we're[br]setting up our ICE table, 0:14:04.577,0:14:07.613 and that'll help us safely[br]assume that x is small.