Question

At equilibrium for the reaction N$_2$(g) + 3H$_2$(g) $\rightleftharpoons$ 2NH$_3$(g), the concentrations are [N$_2$] = 0.20 M, [H$_2$] = 0.60 M and [NH$_3$] = 0.80 M. The value of K$_c$ is:

• A. 7.4
• B. 14.8
• C. 22.2
• D. 44.4

Hint:

Always pay close attention to the stoichiometric coefficients in the chemical equation.
These coefficients become the exponents in the $K_c$ expression.
A common mistake is to forget to raise the concentration to its respective power (e.g., squaring $[\text{NH}_3]$ and cubing $[\text{H}_2]$).

Answer 1

The Correct Option is B

Step 1: Kc expression.
$K_c = \dfrac{[NH_3]^2}{[N_2][H_2]^3}$
Step 2: Substitute.
$K_c = \dfrac{(0.80)^2}{(0.20)(0.60)^3} = \dfrac{0.64}{0.0432}$
\[ \boxed{K_c \approx 14.8} \]