WEBVTT 00:00:00.410 --> 00:00:02.730 I have this reaction here where if I had a mole of 00:00:02.730 --> 00:00:05.970 methane, and I react that with two moles of oxygen, I'll 00:00:05.970 --> 00:00:08.045 produce a mole of carbon dioxide and 00:00:08.045 --> 00:00:09.680 two moles of water. 00:00:09.680 --> 00:00:12.380 And what we want to answer in this video is whether this 00:00:12.380 --> 00:00:14.760 reaction is spontaneous. 00:00:14.760 --> 00:00:17.290 And we learned in the last video that to answer that 00:00:17.290 --> 00:00:19.810 question, we have to turn to Gibbs free energy, or the 00:00:19.810 --> 00:00:21.620 change in Gibbs free energy. 00:00:21.620 --> 00:00:25.320 And the change in Gibbs free energy is equal to the 00:00:25.320 --> 00:00:29.300 enthalpy change for the reaction minus the temperature 00:00:29.300 --> 00:00:32.820 at which it is occurring, times the change in entropy. 00:00:32.820 --> 00:00:35.840 And if this is less than zero, then it's 00:00:35.840 --> 00:00:39.240 a spontaneous reaction. 00:00:39.240 --> 00:00:41.860 So I gave us a little bit of a head start. 00:00:41.860 --> 00:00:46.070 I just calculated the change in enthalpy for this reaction, 00:00:46.070 --> 00:00:47.530 and that's right here. 00:00:47.530 --> 00:00:48.460 And we know how to do that. 00:00:48.460 --> 00:00:50.230 We've done that several videos ago. 00:00:50.230 --> 00:00:52.900 You could just look up the heats of formation of each of 00:00:52.900 --> 00:00:54.880 these products. 00:00:54.880 --> 00:00:56.950 For water you'll multiply it by 2, since you 00:00:56.950 --> 00:00:58.180 have 2 moles of it. 00:00:58.180 --> 00:01:00.510 And so you have the heats of formation of all the products, 00:01:00.510 --> 00:01:02.500 and then you subtract out the heats of 00:01:02.500 --> 00:01:03.780 formation of all the reactants. 00:01:03.780 --> 00:01:07.270 And of course the heat of formation of O2 is O, so this 00:01:07.270 --> 00:01:11.040 won't even show up in it, and you'll get minus 890.3 00:01:11.040 --> 00:01:11.590 kilojoules. 00:01:11.590 --> 00:01:15.090 Well, this tells us that this is an exothermic reaction. 00:01:15.090 --> 00:01:18.490 That this side of the equation has less energy in it-- you 00:01:18.490 --> 00:01:20.450 could kind of think of it that way-- is that side. 00:01:20.450 --> 00:01:22.450 So some energy must have been released. 00:01:22.450 --> 00:01:25.400 We could even put here, you know, plus e for energy. 00:01:25.400 --> 00:01:27.980 Let me write, plus some energy is going to be released. 00:01:27.980 --> 00:01:29.750 So that's why it's exothermic. 00:01:29.750 --> 00:01:32.010 But our question is, is this spontaneous? 00:01:32.010 --> 00:01:34.100 So to figure out if it's spontaneous, we also have to 00:01:34.100 --> 00:01:36.140 figure out our delta s. 00:01:39.450 --> 00:01:42.690 And to help figure out the delta s I, ahead of time, 00:01:42.690 --> 00:01:46.090 looked up the standard molar entropies for 00:01:46.090 --> 00:01:48.130 each of these molecules. 00:01:48.130 --> 00:01:50.460 So for example, the standard-- I'll write it here in a 00:01:50.460 --> 00:01:51.710 different color. 00:02:00.510 --> 00:02:04.980 The standard-- you put a little naught symbol there-- 00:02:04.980 --> 00:02:08.389 the standard molar entropy-- so when we say standard, it's 00:02:08.389 --> 00:02:11.039 at 298 degrees Kelvin. 00:02:11.039 --> 00:02:12.730 Actually, I shouldn't say degrees Kelvin. 00:02:12.730 --> 00:02:15.680 It's at 298 Kelvin You don't use the word degrees, 00:02:15.680 --> 00:02:17.440 necessarily, when you talk about Kelvin. 00:02:17.440 --> 00:02:21.020 So it's at 298 Kelvin, which is 25 degrees Celsius, so it's 00:02:21.020 --> 00:02:21.880 at room temperature. 00:02:21.880 --> 00:02:24.690 So that's why it's considered standard temperature. 00:02:24.690 --> 00:02:29.700 So the standard entropy of methane at room temperature is 00:02:29.700 --> 00:02:30.980 equal to this number right here. 00:02:30.980 --> 00:02:37.880 186 joules per Kelvin mole. 00:02:37.880 --> 00:02:42.320 So if I have 1 mole of methane, I have 186 joules per 00:02:42.320 --> 00:02:43.930 Kelvin of entropy. 00:02:43.930 --> 00:02:46.050 If I have 2 moles, I multiply that by 2. 00:02:46.050 --> 00:02:48.780 If I have 3 moles, I multiply that by 3. 00:02:48.780 --> 00:02:54.870 So the total change in entropy of this reaction is the total 00:02:54.870 --> 00:02:58.950 standard entropies of the products minus the total 00:02:58.950 --> 00:03:00.610 standard entropies of the reactants. 00:03:00.610 --> 00:03:02.450 Just like what we did with enthalpy. 00:03:02.450 --> 00:03:11.650 So that's going to be equal to 213.6 plus-- I have 2 moles of 00:03:11.650 --> 00:03:12.190 water here. 00:03:12.190 --> 00:03:17.810 So it's plus 2 times-- let's just write 70 there. 00:03:17.810 --> 00:03:20.090 69.9, almost 70. 00:03:20.090 --> 00:03:24.720 Plus 2 times 70, and then I want to subtract out the 00:03:24.720 --> 00:03:28.920 entropy of the reactants, or this side of the reaction. 00:03:28.920 --> 00:03:42.900 So the entropy of 1 mole of CH4 is 186 plus 2 times 205. 00:03:42.900 --> 00:03:45.430 So just eyeballing it already, this number is close to this 00:03:45.430 --> 00:03:48.220 number, but this number is much larger than this number. 00:03:48.220 --> 00:03:50.620 Liquid water has a much lower-- this is 00:03:50.620 --> 00:03:51.650 liquid water's entropy. 00:03:51.650 --> 00:03:54.970 It has a much lower entropy than oxygen gas. 00:03:54.970 --> 00:03:55.730 And that makes sense. 00:03:55.730 --> 00:04:00.150 Because liquid formed, there's a lot fewer states. 00:04:00.150 --> 00:04:02.630 It all falls to the bottom of the container, as opposed to 00:04:02.630 --> 00:04:04.680 kind of taking the shape of the room and expanding. 00:04:04.680 --> 00:04:06.600 So a gas is naturally going to have much higher 00:04:06.600 --> 00:04:08.230 entropy than a liquid. 00:04:08.230 --> 00:04:11.830 So just eyeballing it, we can already see that our products 00:04:11.830 --> 00:04:13.930 are going to have a lower entropy than our reactants. 00:04:13.930 --> 00:04:15.400 So this is probably going to be a negative number. 00:04:15.400 --> 00:04:19.430 But let's confirm that. 00:04:19.430 --> 00:04:30.610 So I have 200, 213.6 plus-- well, plus 140, right? 00:04:30.610 --> 00:04:31.270 2 times 70. 00:04:31.270 --> 00:04:35.540 Plus 140 is equal to 353.6. 00:04:35.540 --> 00:04:39.930 So this is 353.6. 00:04:39.930 --> 00:04:47.580 And then from that, I'm going to subtract out-- so 186 plus 00:04:47.580 --> 00:04:54.790 2 times 205 is equal to 596. 00:04:54.790 --> 00:04:58.900 So minus 596, and what is that equal to? 00:04:58.900 --> 00:05:06.780 So we put the minus 596, and then plus the 353.6, and we 00:05:06.780 --> 00:05:10.910 have minus 242.4. 00:05:10.910 --> 00:05:18.220 So this is equal to minus 242.4 joules per Kelvin is our 00:05:18.220 --> 00:05:21.820 delta s minus. 00:05:21.820 --> 00:05:23.800 So we lose that much entropy. 00:05:23.800 --> 00:05:26.020 And those units might not make sense to you right now, and 00:05:26.020 --> 00:05:28.780 actually you know these are but of arbitrary units. 00:05:28.780 --> 00:05:30.750 But you can just say, hey, this is getting more ordered. 00:05:30.750 --> 00:05:33.710 And it makes sense, because we have a ton of gas. 00:05:33.710 --> 00:05:36.080 We have 3 separate molecules, 1 here and 2 00:05:36.080 --> 00:05:38.230 molecules of oxygen. 00:05:38.230 --> 00:05:40.840 And then we go to 3 molecules again, but the 00:05:40.840 --> 00:05:42.310 water is now liquid. 00:05:42.310 --> 00:05:45.770 So it makes sense to me that we lose entropy. 00:05:45.770 --> 00:05:47.330 There's fewer states that the liquid, 00:05:47.330 --> 00:05:48.790 especially, can take on. 00:05:48.790 --> 00:05:51.650 But let's figure out whether this reaction is spontaneous. 00:05:51.650 --> 00:05:57.440 So our delta g is equal to our delta h. 00:05:57.440 --> 00:06:01.140 We're releasing energy, so it's minus 890. 00:06:01.140 --> 00:06:02.510 I'll just get rid of the decimals. 00:06:02.510 --> 00:06:04.490 We don't have to be that precise. 00:06:04.490 --> 00:06:06.350 Minus our temperature. 00:06:06.350 --> 00:06:09.220 We're assuming that we're at room temperature, or 298 00:06:09.220 --> 00:06:10.340 degrees Kelvin. 00:06:10.340 --> 00:06:13.400 That's 28-- I should just say, 298 Kelvin. 00:06:13.400 --> 00:06:14.590 I should get in the habit of not saying 00:06:14.590 --> 00:06:15.960 degrees when I say Kelvin. 00:06:15.960 --> 00:06:22.750 Which is 25 degrees Celsius, times our change in entropy. 00:06:22.750 --> 00:06:25.050 Now, this is going to be a minus. 00:06:25.050 --> 00:06:27.760 Now you might say, OK, minus 242, you might want to put 00:06:27.760 --> 00:06:28.530 that there. 00:06:28.530 --> 00:06:30.280 But you have to be very, very, very careful. 00:06:30.280 --> 00:06:33.160 This right here is in kilojoules. 00:06:33.160 --> 00:06:35.190 This right here is in joules. 00:06:35.190 --> 00:06:37.540 So if we want to write everything in kilojoules, 00:06:37.540 --> 00:06:39.660 since we already wrote that down, let's write this in 00:06:39.660 --> 00:06:40.370 kilojoules. 00:06:40.370 --> 00:06:46.620 So it's 0.242 kilojoules per Kelvin. 00:06:52.370 --> 00:06:55.790 And so now our Gibbs free energy right here is going to 00:06:55.790 --> 00:07:00.880 be minus 890 kilojoules minus 290-- so the minus and the 00:07:00.880 --> 00:07:02.760 minus, you get a plus. 00:07:02.760 --> 00:07:05.770 And that makes sense, that the entropy term is going to make 00:07:05.770 --> 00:07:08.480 our Gibbs free energy more positive. 00:07:08.480 --> 00:07:11.480 Which, as we know, since we want to get this thing below 00:07:11.480 --> 00:07:13.980 0, this is going to fight the spontaneity. 00:07:13.980 --> 00:07:18.840 But let's see if it can overwhelm the actual enthalpy, 00:07:18.840 --> 00:07:20.470 the exothermic nature of it. 00:07:20.470 --> 00:07:22.480 And it seems like it will, because you multiply a 00:07:22.480 --> 00:07:24.010 fraction times this, it's going to be a 00:07:24.010 --> 00:07:25.090 smaller number than that. 00:07:25.090 --> 00:07:27.560 But let's just figure it out. 00:07:27.560 --> 00:07:31.830 So divided by 1, 2, 3. 00:07:31.830 --> 00:07:37.170 That's our change in entropy times 298, that's our 00:07:37.170 --> 00:07:40.410 temperature, is minus 72. 00:07:40.410 --> 00:07:43.710 So this term becomes-- and then we put a minus there-- so 00:07:43.710 --> 00:07:47.360 it's plus 72.2. 00:07:47.360 --> 00:07:50.300 So this is the entropy term at standard temperature. 00:07:50.300 --> 00:07:51.370 It turns into that. 00:07:51.370 --> 00:07:52.970 And this is our enthalpy term. 00:07:52.970 --> 00:07:55.900 So we can already see that the enthalpy is a much more 00:07:55.900 --> 00:07:58.320 negative number than our positive term from our 00:07:58.320 --> 00:08:00.480 temperature times our change in entropy. 00:08:00.480 --> 00:08:04.530 So this term is going to win out. 00:08:04.530 --> 00:08:08.330 Even though we lose entropy in this reaction, it releases so 00:08:08.330 --> 00:08:10.840 much energy that's going to be spontaneous. 00:08:10.840 --> 00:08:14.340 This is definitely less than 0, so this is going to be a 00:08:14.340 --> 00:08:17.360 spontaneous reaction. 00:08:17.360 --> 00:08:19.520 As you can see, these Gibbs free energy problems, they're 00:08:19.520 --> 00:08:20.630 really not too difficult. 00:08:20.630 --> 00:08:23.900 You just really need to find these values. 00:08:23.900 --> 00:08:27.230 And to find these values, it'll either be given, the 00:08:27.230 --> 00:08:29.670 delta h, but we know how to solve for the delta h. 00:08:29.670 --> 00:08:31.720 You just look up the heats of formations of all the 00:08:31.720 --> 00:08:35.210 products, subtract out the reactants, and of course you 00:08:35.210 --> 00:08:38.110 wait by the coefficients. 00:08:38.110 --> 00:08:40.490 And then, to figure out the change in entropy, you do the 00:08:40.490 --> 00:08:40.909 same thing. 00:08:40.909 --> 00:08:44.020 You have to look up the standard molar entropies of 00:08:44.020 --> 00:08:46.840 the products' weight by the coefficients, subtract out the 00:08:46.840 --> 00:08:50.110 reactants, and then just substitute in here, and then 00:08:50.110 --> 00:08:51.850 you essentially have the Gibbs free energy. 00:08:51.850 --> 00:08:54.640 And in this case, it was negative. 00:08:54.640 --> 00:08:56.570 Now, you could imagine a situation where we're at a 00:08:56.570 --> 00:08:57.720 much higher temperature. 00:08:57.720 --> 00:09:00.890 Like the surface of the sun or something, where all of a 00:09:00.890 --> 00:09:06.380 sudden, instead of a 298 here, if you had like a 2,000 or a 00:09:06.380 --> 00:09:08.140 4,000 there. 00:09:08.140 --> 00:09:10.700 Then all of a sudden, things become interesting. 00:09:10.700 --> 00:09:14.710 If you could imagine, if you had a 40,000 Kelvin 00:09:14.710 --> 00:09:17.530 temperature here, then all of a sudden the entropy term, the 00:09:17.530 --> 00:09:20.150 loss of entropy, is going to matter a lot more. 00:09:20.150 --> 00:09:22.480 And so this term, this positive term, is going to 00:09:22.480 --> 00:09:25.870 outweigh this, and maybe it wouldn't be spontaneous at a 00:09:25.870 --> 00:09:27.960 very, very, very, very high temperature. 00:09:27.960 --> 00:09:29.160 Another way to think about it. 00:09:29.160 --> 00:09:34.560 A reaction that generates heat that lets out heat-- the heat 00:09:34.560 --> 00:09:37.090 being released doesn't matter so much when there's already a 00:09:37.090 --> 00:09:39.980 lot of heat or kinetic energy in the environment. 00:09:39.980 --> 00:09:42.920 If the temperature was high enough, this reaction would 00:09:42.920 --> 00:09:46.050 not be spontaneous, because maybe then the entropy term 00:09:46.050 --> 00:09:46.980 would win out. 00:09:46.980 --> 00:09:49.130 But anyway, I just wanted to do this calculation for you to 00:09:49.130 --> 00:09:51.360 show you that there's nothing too abstract here. 00:09:51.360 --> 00:09:53.950 You can look up everything on the web, and then figure out 00:09:53.950 --> 00:09:56.320 if something is going to be spontaneous.