Question

Attached is my question! Thanks so much!

Answer 1

delta G = -4.51 KJ/mol
= -4510 KJ/mol
T = 25oC = 298 K
use:
delta G = -R*T*ln Kc
-4510 = - 8.314*298*ln Kc
kc = 6.174

A + B <----> C
0.30 0.40 0
0.30-x 0.40-x x

Kc = [C] / [A][B]
6.174 = (x)/(0.30-x)(0.40-x)
6.174*(0.30-x)(0.40-x) = (x)
6.174 * (0.12 - 0.70x + x^2) = x
0.7409 - 4.3218 x + 6.174*x^2 = x
6.174*x^2 - 5.3218 x + 0.7409 = 0
Solving for x,
x = 0.687 and x=0.174
x can't be greater than 0.30
so,
x = 0.17 M

at equilibrium,
[A] = 0.30-x = 0.30 - 0.17 = 0.13 M
[B] = 0.40-x = 0.40 - 0.17 = 0.23 M
[C] = x = 0.17 M

if delta G is positive, k will be small
Then [C] will be small
[A] and [B] will be more
Answer: there would be more A and B but less C