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

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