00:01
So this problem is going off of reactions shown in the previous problem, and then wants you to calculate delta g not using the values from your appendix of delta g formation.
00:16
And so delta g not equals the sum of n of products time the formation of products minus the sum of n reactants, which is the mule, the stoichiometry coefficients, times delta g -not of reactants.
00:45
So basically you're doing products minus reactants and accounting for the stoichiometry.
00:52
So we can do this for a.
00:57
Your reactants are methane, ch4, as a gas.
01:06
Going to c2h6.
01:12
As a gas plus hydrogen gas.
01:24
And so you can look up these values and your textbook and so we get delta g not equals minus 32 from c2h6 plus 0 for hydrogen minus 2 to count for the 2 in front of ch4 comes minus 50 .5 and this equals 69 kilojoules.
02:03
Next reaction to an h3 as a gas going to n2h4 plus h2 gas.
02:28
So you look up these values in your textbook and delta g equals 159 .4 plus zero from hydrogen minus 2 times negative 16 .4.
02:55
And that equals 192 .2 kilojoules.
03:05
All right for c, you have n2 plus o2, both as a gas, going to 2 and 0...