Chapter 13

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.


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}