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Question

Chemistry Question on Equilibrium

A(g)B(g)+C2(g)A_{(g)} \rightleftharpoons B_{(g)} + \frac{C}{2}_{(g)} The correct relationship between KPK_P, α\alpha, and equilibrium pressure PP is:

A

KP=α1/2P1/2(2+α)1/2K_P = \frac{\alpha^{1/2} P^{1/2}}{(2 + \alpha)^{1/2}}

B

KP=α3/2P1/2(2+α)1/2(1α)K_P = \frac{\alpha^{3/2} P^{1/2}}{(2 + \alpha)^{1/2}(1 - \alpha)}

C

KP=α1/2P3/2(2+α)3/2K_P = \frac{\alpha^{1/2} P^{3/2}}{(2 + \alpha)^{3/2}}

D

KP=α1/2P1/2(2+α)3/2K_P = \frac{\alpha^{1/2} P^{1/2}}{(2 + \alpha)^{3/2}}

Answer

KP=α3/2P1/2(2+α)1/2(1α)K_P = \frac{\alpha^{3/2} P^{1/2}}{(2 + \alpha)^{1/2}(1 - \alpha)}

Explanation

Solution

Consider the reaction at equilibrium:

A(g)B(g)+12C(g)A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2}C_{(g)}

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

PA=(1α)P,PB=α2+αP,PC=α2(2+α)PP_A = (1 - \alpha)P, \quad P_B = \frac{\alpha}{2 + \alpha}P, \quad P_C = \frac{\alpha}{2(2 + \alpha)}P

The expression for KPK_P is given by:

KP=PBPC1/2PA=α3/2P1/2(2+α)1/2(1α)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)}