Question

(b) If equilibrium constant for formation of 1 mole ammonia from hydrogen and nitrogen is X. What will be equilibrium constant for dissociation of 2 moles ammonia?​

Answer 1

Explanation:

The equilibrium constant of a reaction depends only on temperature and the equation that you write for the reaction. It does not depend on the amounts of reactants.

So, whatever be the amount of ammonia that you start your reaction with, the equilibrium constant will have the same value.

You can write the equation for dissociation of ammonia as

(1) NH3 <=> 1/2 N2 + 3/2 H2

(2) 2 NH3 <=> N2 +3 H2

or any multiple , say nth multiple, of (1)

(3) n NH3 <=> n/2 N2 + 3*n/2 H2

Eq.(2) = 2* Eq.(1) , Eq.(3) = n* Eq.(1)

At any temperature T if K1, K2 and K3 are the equilibrium constant s respectively for (1), (2) and (3) then they are related as follows:

K2 = K1^2, K3 = K1^n

Why so?

Write the expressions for the equilibrium constants of (1), (2) .

K1 ={ [N2]^1/2} *{[H2]^3/2}/ [NH3]

K2 = [N2]*[H2]^3/ [NH3]

The equilibrium molarities ( or the partial pressures) of N2, H2 and NH3 in the two expressions are the same because they don't depend on how you write the equation.

You can easily see that K2 is the square of K1. Since Eq(2) = 2* Eq.(1) , K2 = K1^2

Similarly Eq.(3) = n*Eq.(1)

So K3 = K1^n

Hope this helps.