/ News & Press / Video / 18.2e Calculating reaction free energy under nonstandard conditions
18.2e Calculating reaction free energy under nonstandard conditions
WEBVTT Kind: captions Language: en
00:00:00.740 this question says a chemist fills a 00:00:03.50000:00:03.510 reaction vessel with 9.6 atmospheres of 00:00:06.01900:00:06.029 methane 0.699 atmospheres of oxygen 8.28 00:00:11.87000:00:11.880 atmospheres of carbon dioxide and 1.74 00:00:14.95900:00:14.969 atmospheres of hydrogen at a temperature 00:00:18.17000:00:18.180 of 25 degrees C under these conditions 00:00:20.77900:00:20.789 calculate the reaction free energy Delta 00:00:23.87000:00:23.880 G for the following chemical reaction 00:00:26.39000:00:26.400 and it asks us to use the thermodynamic 00:00:29.00000:00:29.010 information the Ark's data tab and round 00:00:31.13000:00:31.140 our answer to the nearest kilojoules so 00:00:33.17000:00:33.180 essentially we have methane plus oxygen 00:00:34.70000:00:34.710 yields carbon dioxide and hydrogen so 00:00:40.01000:00:40.020 this is the reaction that we're going 00:00:41.99000:00:42.000 after and we need to find Delta G for 00:00:44.11900:00:44.129 the energy of it well there's a couple 00:00:47.00000:00:47.010 things we need to do in order to solve 00:00:48.38000:00:48.390 this the first thing is we need to 00:00:50.20900:00:50.219 realize that Delta G is equal to Delta G 00:00:57.35000:00:57.360 naught under standard conditions plus RT 00:01:02.59000:01:02.600 ln Q and in this case we can look at Q 00:01:09.17000:01:09.180 as the current conditions so this is 00:01:11.63000:01:11.640 going to be products over reactants and 00:01:14.35000:01:14.360 it's going to be under current 00:01:16.58000:01:16.590 conditions as opposed to under standard 00:01:18.80000:01:18.810 conditions so this is how we relate the 00:01:20.99000:01:21.000 standard change in free energy with the 00:01:24.89000:01:24.900 non standard under these current 00:01:26.71900:01:26.729 conditions of this pressures of these 00:01:30.67900:01:30.689 different gases but before we can do any 00:01:33.05000:01:33.060 of this we need to find out what the 00:01:35.60000:01:35.610 standard table in free energy is and to 00:01:39.59000:01:39.600 do that we do it just like the previous 00:01:40.91000:01:40.920 problem on this playlist it's just the 00:01:42.80000:01:42.810 products minus reactants so Delta G 00:01:46.21900:01:46.229 naught equals the summation of n times 00:01:50.42000:01:50.430 the Delta G of formation it's an F for 00:01:56.89900:01:56.909 the products minus the summation of n 00:02:01.12000:02:01.130 Delta G formation of the reactants 00:02:06.24000:02:06.250 and as I said there's a previous problem 00:02:08.55900:02:08.569 on the playlist where we do exactly this 00:02:09.96900:02:09.979 and you've also done this for entropy 00:02:11.44000:02:11.450 and enthalpy so Delta G nought equals 00:02:15.18000:02:15.190 well let's look at our first product at 00:02:17.55900:02:17.569 co2 and there's only one of them so I'm 00:02:20.55900:02:20.569 the table it's minus three hundred and 00:02:22.05900:02:22.069 ninety four point four kilojoules 00:02:24.58000:02:24.590 multiplied by one plus well the next 00:02:28.80900:02:28.819 product is hydrogen that's an element so 00:02:31.12000:02:31.130 it has a free energy of formation of 00:02:32.86000:02:32.870 zero so we're not going to include it - 00:02:36.24000:02:36.250 the first product is methane and if you 00:02:39.16000:02:39.170 look up on the table there's one of them 00:02:41.08000:02:41.090 so it's 1 times minus 50 point 5 00:02:45.15000:02:45.160 kilojoules and again the second reactant 00:02:49.30000:02:49.310 is an element so it's free energy of 00:02:52.53900:02:52.549 formation is 0 so Delta G naught under 00:02:56.58900:02:56.599 standard conditions here is equal to 00:02:59.38000:02:59.390 minus three hundred and forty three 00:03:01.63000:03:01.640 point nine kilojoules now we can use 00:03:06.43000:03:06.440 this in order to find it under the non 00:03:09.49000:03:09.500 standard conditions given in the problem 00:03:11.62000:03:11.630 with Q well let's take a look at how 00:03:14.83000:03:14.840 we're gonna do that the first thing is 00:03:16.69000:03:16.700 are so are in joules is eight point 00:03:23.19900:03:23.209 three one joules per K but there's a 00:03:26.53000:03:26.540 problem 00:03:27.00900:03:27.019 Delta G naught is in kilojoules so 00:03:30.87900:03:30.889 instead of using eight point three one 00:03:32.62000:03:32.630 we want to divide this by a thousand and 00:03:34.80900:03:34.819 use zero point zero zero eight three one 00:03:38.64000:03:38.650 kilojoules per K for R so this is gonna 00:03:43.62900:03:43.639 be our R this is the standard one this 00:03:47.05000:03:47.060 is in kilojoules T we're just gonna add 00:03:51.49000:03:51.500 273 point one five to our twenty five 00:03:54.52000:03:54.530 because s be in Kelvin so it's going to 00:03:56.50000:03:56.510 be 298.15 cat so now we have R and we 00:04:02.68000:04:02.690 have T what about Q so Q is equal to 00:04:08.68000:04:08.690 products over reactants in this case 00:04:11.02000:04:11.030 they're all gasses and we're given 00:04:12.49000:04:12.500 pressures so it's the pressure of co2 00:04:15.75000:04:15.760 times the pressure 00:04:19.90000:04:19.910 I guess we don't need the time sign 00:04:21.58000:04:21.590 pressure of hydrogen squared because 00:04:24.79000:04:24.800 there's two of them divided by the 00:04:27.28000:04:27.290 pressure of ch4 times the pressure of o2 00:04:33.55000:04:33.560 so products over reactants accept its 00:04:36.73000:04:36.740 non-standard conditions so we're gonna 00:04:38.41000:04:38.420 use these atmospheres I'm gonna leave 00:04:40.42000:04:40.430 the unit's out because Q is typically 00:04:42.40000:04:42.410 given as a unitless so here 00:04:45.37000:04:45.380 the first one is co2 we find its 00:04:47.71000:04:47.720 pressure up here which is eight point 00:04:50.41000:04:50.420 two eight times hydrogen we find it up 00:04:55.21000:04:55.220 here which is one point seven four since 00:04:58.00000:04:58.010 there are two hydrogen's we need to 00:04:59.86000:04:59.870 square that divided by ch4 which is nine 00:05:05.74000:05:05.750 point six times o2 which is point six 00:05:12.49000:05:12.500 nine nine when you plug those all in you 00:05:17.26000:05:17.270 find that Q equals three point seven 00:05:21.31000:05:21.320 three five eight and I'm not gonna worry 00:05:24.19000:05:24.200 about sig figs because it says round 00:05:25.63000:05:25.640 your answer to the nearest kilojoule so 00:05:28.06000:05:28.070 this is basically the value of Q when we 00:05:30.70000:05:30.710 solve this problem well now we have all 00:05:33.07000:05:33.080 of our variables we have Delta G naught 00:05:35.05000:05:35.060 we found that from products minus 00:05:36.55000:05:36.560 reactants we have R that's a constant we 00:05:39.15900:05:39.169 just converted it to kilojoules per K 00:05:40.63000:05:40.640 instead of joules per K we have T we 00:05:43.12000:05:43.130 found the absolute temperature in Kelvin 00:05:44.56000:05:44.570 and we have Q which is just products 00:05:46.72000:05:46.730 over reactants so now Delta G equals 00:05:51.39000:05:51.400 Delta G naught which is minus three 00:05:54.37000:05:54.380 hundred and forty three point nine 00:05:56.61000:05:56.620 kilojoules plus R in kilojoules 0.0083 00:06:02.32000:06:02.330 one kilojoules per k times the 00:06:06.31000:06:06.320 temperature 298 0.15 K times the Ln of Q 00:06:13.00000:06:13.010 which is the Ln of three point seven 00:06:17.29000:06:17.300 three five eight I'm just gonna do that 00:06:21.43000:06:21.440 so there's a clear distinction between 00:06:22.83000:06:22.840 the numbers and that's a seven even 00:06:25.65900:06:25.669 though it's kind of hard to see 00:06:28.48000:06:28.490 now we just solve this we use a 00:06:31.46000:06:31.470 calculator to take the hour and then we 00:06:32.87000:06:32.880 get Delta G equals minus three forty 00:06:35.63000:06:35.640 three point nine kilojoules plus three 00:06:42.35000:06:42.360 point two six five four kilojoules so 00:06:46.67000:06:46.680 when we take the Ln of this number 00:06:47.87000:06:47.880 multiply it by the 298 point one five 00:06:49.88000:06:49.890 times the point zero zero eight three 00:06:51.50000:06:51.510 one and we find Delta G for these 00:06:55.64000:06:55.650 non-standard conditions is equal to two 00:06:59.33000:06:59.340 the nearest kilojoule minus 341 00:07:02.77000:07:02.780 kilojoules so this is how we can find 00:07:05.75000:07:05.760 Delta G on non-standard conditions
Office location
Engineering company LOTUS®
Russia, Ekaterinburg, Lunacharskogo street, 240/12