A(g) B(g) + C2(g). The correct relationship between KP, and equilibrium pressure P is

Step 1 — Set up ICE table (start with 1 mol A):

| | A | B | C/2 | Total | |--|--|--|--|--| | Initial | 1 | 0 | 0 | 1 | | Change | $-\alpha$ | $+\alpha$ | $+\alpha/2$ | | | Equil. | $1-\alpha$ | $\alpha$ | $\alpha/2$ | $1 + \alpha/2$ |

Step 2 — Mole fractions and partial pressures: $x_A = \frac{1-\alpha}{1+\alpha/2}, \quad x_B = \frac{\alpha}{1+\alpha/2}, \quad x_C = \frac{\alpha/2}{1+\alpha/2}$

Multiply by $P$: $P_A = \frac{(1-\alpha)P}{1+\alpha/2}$, $P_B = \frac{\alpha P}{1+\alpha/2}$, $P_C = \frac{\alpha P/2}{1+\alpha/2}$

Step 3 — Write $K_P$: $K_P = \frac{P_B \cdot P_C^{1/2}}{P_A} = \frac{\frac{\alpha P}{1+\alpha/2} \cdot \left(\frac{\alpha P/2}{1+\alpha/2}\right)^{1/2}}{\frac{(1-\alpha)P}{1+\alpha/2}} = \frac{\alpha \cdot (\alpha/2)^{1/2} \cdot P^{1/2}}{(1-\alpha)(1+\alpha/2)^{1/2}}$

With $1 + \alpha/2 = (2+\alpha)/2$: $K_P = \frac{\alpha^{3/2} P^{1/2}}{2^{1/2}(1-\alpha) \cdot \left(\frac{2+\alpha}{2}\right)^{1/2}} = \frac{\alpha^{3/2} P^{1/2}}{(1-\alpha)(2+\alpha)^{1/2}}$

Answer: Option (2)