00:01
In this problem, we're asked to use data from appendix 4, calculate delta g for the reaction i'm writing down right here.
00:16
I've got to make sure this is the right one.
00:22
Yes, it looks like the right equation.
00:35
Okay, and for these conditions.
00:45
Okay, 298 kelvin, p of no equals 1 .00 times 10 to the minus 6 atmospheres.
00:57
The pressure of o3 will equal 2 .00 times 10 to the minus 6 atmospheres.
01:07
The pressure of no2 is 1 .00 times 10 to the minus 7 atmospheres.
01:19
And po2 equals 1 .00 times 10 to the minus 3 atmospheres.
01:29
Okay, so from a previous problem, we know that the delta g0 equals negative 198 kelvin or kilojoules.
01:48
Okay, let's go ahead and set up our problem.
01:51
So our delta g will equal this plus rt times the ln of pno2 times p of o2 over p of no, where all my coefficients went, yes, and p of o3...