The power of the equation ΔG° = –RT ln(K) is that K can be determined from tabulated values of ΔG°f. Use the tabulated values of the Gibb's energy to determine K for the reaction, 2AB(l) ↔ 2A(g) + B2(g), at 298K.

Question

(R=8.314 J/K mol or R=0.008314 kJ/K mol)

ΔGfº (kJ/mol)
A(g)=53
B2(g)=-114
AB(l)=162
2AB(l) ↔ 2A(g) + B2(g)
I have been working on this for awhile and i can't seem to get the right answer. I have tried finding the sum of the products minus the sum of the reactants to find what ΔG. Then taking e^(-ΔG/RT) to find the value for K.

Answer 1

From your description I can't tell what you did wrong. If you post your work I will find the error.

Answer 2

A common student error is to use kJ for dGo. You must use dGo in J (unless of course you change R).

Answer 3

ΔG= (2x53 + -114) - (2x162)= -332
K=e^(-332/(.008314x298)= 6.36 e-59
I have also tried using 8.314 as the value for R

Answer 4

Isn't that e^(-dG/RT). If dG is -332,000 then the rxn should be spontaneous and K should be a large number and not a small number.
dGo = -RTlnK
-dGo = RTlnK
(332000/8.314*298)= lnK