## Q. 13.1

Writing Equilibrium Equations for Gas-Phase Reactions

Write the equilibrium equation for each of the following reactions:

(a) $N_2$(g) + 3 $H_2$(g) $\rightleftharpoons$ 2 $NH_3$(g)

(b) 2 $NH_3$(g) $\rightleftharpoons$ $N_2$(g) + 3 $H_2$(g)

STRATEGY

Put the concentrations of the reaction products in the numerator of the equilibrium-constant expression and the concentrations of the reactants in the denominator. Then raise the concentration of each substance to the power of its coefficient in the balanced chemical equation.

## Verified Solution

$\text { (a) } K_{\mathrm{c}}=\frac{\left[\mathrm{NH}_3\right]^{2 \ \leftarrow \ Coefficient \ of \ NH_3}}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^{3 \ \leftarrow \ Coefficient \ of \ H_2}}$

(b) $K_{\mathrm{c}}{ }^{\prime}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^{3 \ \leftarrow \ Coefficient \ of \ H_2}}{\left[\mathrm{NH}_3\right]^{2 \ \leftarrow \ Coefficient \ of \ NH_3}} \ \ \ \ \ \ \ \ K_c{ }^{\prime}=\frac{1}{K_c}$

Because the balanced equation in part (b) is the reverse of that in part (a), the equilibrium-constant expression in part (b) is the reciprocal of the expression in part (a) and the equilibrium constant $K_{c}^{ʹ}$ is the reciprocal of $K_c$