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

For the reaction:

\[ A(g) \rightleftharpoons B(g) + \frac{1}{2} C(g) \]

Step 1: Initial moles

Initial moles: \[ n_A = n, \quad n_B = 0, \quad n_C = 0 \]

Step 2: Equilibrium moles

Equilibrium moles: \[ n_A = n(1 - \alpha), \quad n_B = n\alpha, \quad n_C = \frac{n\alpha}{2} \]

Step 3: Total moles at equilibrium

Total moles: \[ n_{\text{total}} = n\left(1 + \alpha\right)/2 \]

Step 4: Equilibrium pressure expressions

Equilibrium pressure for each component: \[ P_A = \frac{(1 - \alpha)P}{1 + \alpha/2}, \quad P_B = \frac{\alpha P}{1 + \alpha/2}, \quad P_C = \frac{(\alpha/2) P}{1 + \alpha/2} \]

Step 5: Expression for \( K_p \)

The equilibrium constant \( K_p \) is given by: \[ K_p = \frac{\alpha P}{1 + \alpha/2} \times \left( \frac{\alpha P}{(2 + \alpha)} \right)^{1/2} \]

Step 6: Simplification

Simplifying the expression for \( K_p \): \[ K_p = \frac{\alpha P}{1 + \alpha/2} \times \frac{\alpha P^{1/2}}{(2 + \alpha)^{1/2}} \]

Final expression for \( K_p \):

\[ K_p = \frac{\alpha^{3/2} P^{1/2}}{(2 + \alpha)^{1/2} (1 - \alpha)} \]

Answer 2

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:

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