Answer : The concentration of
at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.
Explanation :
The given chemical reaction is,

First we have to calculate the equilibrium constant for the reaction.
The relation between the equilibrium constant and standard free‑energy is:

where,
= standard free‑energy change = -4.20 kJ/mole
R = universal gas constant = 8.314 J/mole.K
k = equilibrium constant = ?
T = temperature = 
Now put all the given values in the above relation, we get:


Now we have to calculate the concentrations of A, B, and C at equilibrium.
The given equilibrium reaction is,

Initially 0.30 0.40 0
At equilibrium (0.30-x) (0.40-x) x
The expression of equilibrium constant will be,
![k=\frac{[C]}{[A][B]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BC%5D%7D%7B%5BA%5D%5BB%5D%7D)

By solving the term x, we get

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.
The value of x will be, 0.168 M
The concentration of
at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M
The concentration of
at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M
The concentration of
at equilibrium = x = 0.168 M