Areaction a(aq)+b(aq)↽−−⇀c(aq) has a standard free‑energy change of −4.20 kj/mol at 25 °c. what are the concentrations of a, b, and c at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 m, 0.40 m, and 0 m, respectively?

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,

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

the answer is na (sodium)