## Q. 13.2

Calculating the Equilibrium Constant $K_c$

The following concentrations were measured for an equilibrium mixture at 500 K: $[N_2]$ = 3.0 × $10^{-2}$ M; $[H_2]$ = 3.7 × $10^{-2}$ M; $[NH_3]$ = 1.6 × $10^{-2}$ M. Calculate the equilibrium constant at 500 K for each of the reactions in Worked Example 13.1.

STRATEGY

To calculate the value of the equilibrium constant, substitute the equilibrium concentrations into the equilibrium equation.

## Verified Solution

(a) $K_{\mathrm{c}}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}=\frac{\left(1.6 \times 10^{-2}\right)^2}{\left(3.0 \times 10^{-2}\right)\left(3.7 \times 10^{-2}\right)^3}=1.7 \times 10^2$

(b) $\text { (b) } K_{\mathrm{c}}{ }^{\prime}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^3}{\left[\mathrm{NH}_3\right]^2}=\frac{\left(3.0 \times 10^{-2}\right)\left(3.7 \times 10^{-2}\right)^3}{\left(1.6 \times 10^{-2}\right)^2}=5.9 \times 10^{-3}$

Note that $K_{c}^{ʹ}$ is the reciprocal of $K_c$. That is,

$5.9 \times 10^{-3}=\frac{1}{1.7 \times 10^2}$