Question

$\text{The following concentrations were observed at 500 K for the formation of NH}_3 \text{ from N}_2 \text{ and H}_2\text{. At equilibrium: [N}_2] = 2 \times 10^{-2} \, \text{M, [H}_2] = 3 \times 10^{-2} \, \text{M, and [NH}_3] = 1.5 \times 10^{-2} \, \text{M. Equilibrium constant for the reaction is \_\_\_\_\_\_.}$

Answer 1

Correct Answer: 417

The equilibrium reaction is: $$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$$. The equilibrium constant expression \(K_c\) is given by: $$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$$. The provided concentrations are: $$[\text{N}_2] = 2 \times 10^{-2} \, \text{M}$$, $$[\text{H}_2] = 3 \times 10^{-2} \, \text{M}$$, and $$[\text{NH}_3] = 1.5 \times 10^{-2} \, \text{M}$$. Substituting these values into the expression yields: $$K_c = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2})(3 \times 10^{-2})^3}$$. The numerator is $$(1.5 \times 10^{-2})^2 = 2.25 \times 10^{-4}$$. The denominator is $$(2 \times 10^{-2})(3 \times 10^{-2})^3 = 2 \times 10^{-2} \times 27 \times 10^{-6} = 54 \times 10^{-8} = 5.4 \times 10^{-7}$$. Therefore, $$K_c = \frac{2.25 \times 10^{-4}}{5.4 \times 10^{-7}}$$, which calculates to $$K_c \approx 416.67$$. This calculated value of \(K_c \approx 416.67\) aligns with the expected range of 417,417.