Question

If ${K_c}$for the formation of ammonia is $2mo{l^{ - 2}}{L^2}$, ${K_c}$ for the decomposition of ammonia is:
A. $0.5mo{l^{ - 2}}{L^2}$
B. $0.25mo{l^{ - 2}}{L^2}$
C. $0.75mo{l^{ - 2}}{L^2}$
D. None of these

Answer 1

To solve this problem we need to understand the concept of equilibrium. ${K_c}$of the decomposition reaction gets reciprocal when the equilibrium is reversed i.e., formation reaction.

Complete step-by-step solution:
Let $K$ be the equilibrium constant for formation of $N{H_3}$.
Reaction of formation of $N{H_3}:$ ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$, Equilibrium constant,${K_c} = 2mo{l^{ - 2}}{L^2}$

$${K_f} = \dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}} \\\ 2mo{l^{ - 2}}{L^2} = \dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}} - - (i) \\\$$

Reaction of decomposition of $N{H_3}:$ $2N{H_3} \rightleftharpoons 2{N_2} + 3{H_2}$, Equilibrium constant, ${K_c} = ?$
${K_d} = \dfrac{{\left[ {{N_2}} \right]{{\left[ H \right]}^3}}}{{{{\left[ {N{H_3}} \right]}^2}}} - - (ii)$
Substituting equation$(i)$within equation$(ii)$we get:

$$\dfrac{{\left[ {{N_2}} \right]{{\left[ H \right]}^3}}}{{{{\left[ {N{H_3}} \right]}^2}}} = \dfrac{1}{2} \\\ \therefore \dfrac{{\left[ {{N_2}} \right]{{\left[ H \right]}^3}}}{{{{\left[ {N{H_3}} \right]}^2}}} = 0.5mo{l^{ - 2}}{L^2} \\\$$

Final answer: $$
The ${K_c}$of decomposition of ammonia is option(A)$0.25mo{l^{ - 2}}{L^2}$

Note: Characteristics of equilibrium constant are as follows:
It has a specific value for every chemical reaction at a particular temperature.
It changes with change at temperature.
It is independent of pressure, volume and concentrations of reactants and products.
It is not affected by the introduction catalyst.
It depends on the nature of reaction.