Question

$A_{(g)} \rightleftharpoons B_{(g)} + \frac{C}{2}_{(g)}$ The correct relationship between $K_P$, $\alpha$, and equilibrium pressure $P$ is:

• A. $K_P = \frac{\alpha^{1/2} P^{1/2}}{(2 + \alpha)^{1/2}}$
• B. $K_P = \frac{\alpha^{3/2} P^{1/2}}{(2 + \alpha)^{1/2}(1 - \alpha)}$
• C. $K_P = \frac{\alpha^{1/2} P^{3/2}}{(2 + \alpha)^{3/2}}$
• D. $K_P = \frac{\alpha^{1/2} P^{1/2}}{(2 + \alpha)^{3/2}}$

Answer 1

The Correct Option is B

Consider the reaction at equilibrium:

$$A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2}C_{(g)}$$

At equilibrium, let the fraction dissociated be $\alpha$. Then:

$$P_A = (1 - \alpha)P, \quad P_B = \frac{\alpha}{2 + \alpha}P, \quad P_C = \frac{\alpha}{2(2 + \alpha)}P$$

The expression for $K_P$ is given by:

$$K_P = \frac{P_B \cdot P_C^{1/2}}{P_A} = \frac{\alpha^{3/2}P^{1/2}}{(2 + \alpha)^{1/2}(1 - \alpha)}$$