Question

For the equilibrium at 500 K, \( \text{N}_2 (\text{g}) + 3\text{H}_2 (\text{g}) \rightleftharpoons 2\text{NH}_3 (\text{g}) \), the equilibrium concentrations of \(\text{N}_2 (\text{g})\), \(\text{H}_2 (\text{g})\) and \(\text{NH}_3 (\text{g})\) are respectively 4.0 M, 2.0 M, and 2.0 M. The \(K_c\) for the formation of \(\text{NH}_3\) at 500 K is:

• A. \( \frac{1}{16} \, {mol}^{-2} \, {dm}^6 \)
• B. \( \frac{1}{32} \, {mol}^{-2} \, {dm}^6 \)
• C. \( \frac{1}{8} \, {mol}^{-2} \, {dm}^6 \)
• D. \( \frac{1}{4} \, {mol}^{-2} \, {dm}^6 \)
• E. \( \frac{1}{2} \, {mol}^{-2} \, {dm}^6 \)

Hint:

For equilibrium constants, remember that concentrations of products are raised to the power of their coefficients, and the same for reactants.

Answer 1

The Correct Option is C

The equilibrium constant \(K_c\) for the reaction is given by the formula: \[ K_c = \frac{[{NH}_3]^2}{[{N}_2][{H}_2]^3} \] Substituting the given concentrations: \[ K_c = \frac{(2.0)^2}{(4.0)(2.0)^3} = \frac{4.0}{4.0 \times 8.0} = \frac{1}{8} \, {mol}^{-2} \, {dm}^6 \]