Question

Calculare ∆G°' for the following values where Keq = 1 x 10^4, 1, and 1 x 10^-6.

Here is my work:
∆G°' = -RT ln Keq

1) ∆G°' = - (8.31 J/mol)(298 K) ln (1x10^4) = -22808 J/mol = -22 kJ/mol
2) ∆G°' = - (8.31 J/mol)(298 K) ln (1) = 0 kJ/mol
3) ∆G°' = - (8.31 J/mol)(298 K) ln (1x10^-6) = 34212 J/mol = 34 kJ/mol

The answers are 20 kJ/mol, 0 kJ/mol, and +30 kJ/mol

Answer 1

I haven't calculated the specific numbers, but it looks very correct. While temperature is not considered a part of the standard conditions I think it is save to assume the temperature being 298 K.

Answer 2

@Frostbite I just don't understand why I get 22 and the answer is 20 and then 34, but the answer is 30

Answer 3

Let me fast try do the calculations then. :)

Answer 4

okay, thank you

Answer 5

I get the very same as you.... 34.3088 kJ / mol for the Keq = 10^(-6)

Answer 6

What if you reduce the temperature?

Answer 7

okay, awesome. It must just be how the book rounds

Answer 8

Wondering if the calculations are done at 0 degree C.

Answer 9

The question didn't say, I typed exactly what it asked

Answer 10

Even if it was, then it still don't fit. Assuming the equilibrium constant is at 0 degrees it would be 31.3 kJ/mol

Answer 11

It must be the book :S