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:

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