For a reaction at equilibrium A(g) B(g) + 12C(g), the relation between dissociation constant (K), degree of…

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

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

Note: $1 + \alpha/2 = (2+\alpha)/2$

Step 2 — Partial pressures: $P_A = \frac{(1-\alpha)p}{(2+\alpha)/2} = \frac{2(1-\alpha)p}{2+\alpha} P_B = \frac{2\alpha p}{2+\alpha}, \quad P_C = \frac{\alpha p}{2+\alpha}$

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

Answer: Option (1)