Question

If $K_c$ for the formation of ammonia is 2 $mo{l}^{-2}{L}^2,K_c$ 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

The correct option is A $0.5mo{l}^2/L^2$

If the equilibrium constant for the forward reaction is K, then the equilibrium constant for the reverse reaction is $\frac{1}{K}$

For the formation of ammonia, $K_c=2{\text{mol}}^{-2}{\text{L}}^2$

For the decomposition of ammonia, $K_c=\frac{1}{2{\text{mol}}^{-2}{\text{L}}^2}=0.5{\text{mol}}^2{\text{L}}^{-2}$