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 given by: $$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$$ The equilibrium constant expression \(K_c\) is: $$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$$ Substituting the given concentrations, we have: $$[\text{N}_2] = 2 \times 10^{-2} \, \text{M}$$ $$[\text{H}_2] = 3 \times 10^{-2} \, \text{M}$$ $$[\text{NH}_3] = 1.5 \times 10^{-2} \, \text{M}$$ Plug these values into the equation: $$K_c = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2})(3 \times 10^{-2})^3}$$ First, calculate the numerator: $$(1.5 \times 10^{-2})^2 = 2.25 \times 10^{-4}$$ Next, calculate the denominator: $$(2 \times 10^{-2})(3 \times 10^{-2})^3 = 2 \times 10^{-2} \times 27 \times 10^{-6} = 54 \times 10^{-8} = 5.4 \times 10^{-7}$$ Thus, $$K_c = \frac{2.25 \times 10^{-4}}{5.4 \times 10^{-7}}$$ $$K_c \approx 416.67$$ The calculated equilibrium constant \(K_c \approx 416.67\) is indeed within the expected range of 417,417.

Answer 2

The equilibrium constant expression \( K_c \) for the formation of ammonia (\( \text{NH}_3 \)) from nitrogen (\( \text{N}_2 \)) and hydrogen (\( \text{H}_2 \)) is:  
\(K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}\)

Substituting the given concentrations:  
\(K_c = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2})(3 \times 10^{-2})^3}\)

Calculating the result:  
\(K_c = 417\)

The Correct Answer is:417