A_{(g)} rightleftharpoons B_{(g)} + frac{1}{2 | vedclass

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(A) For the reaction: $A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$Initial moles: $1, 0, 0$Moles at equilibrium: $(1-\alpha), \alpha, \fr

Vedclass · Accepted Answer

$K_P = \frac{\alpha^{3/2} P^{1/2}}{(1-\alpha)(2+\alpha)^{1/2}}$

Vedclass · Answer

(A) For the reaction: $A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$Initial moles: $1, 0, 0$Moles at equilibrium: $(1-\alpha), \alpha, \frac{\alpha}{2}$Total moles at equilibrium: $n_{total} = 1 - \alpha + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2} = \frac{2+\alpha}{2}$Partial pressures are given by $P_i = x_i \cdot P$:$P_A = \frac{1-\alpha}{(2+\alpha)/2} \cdot P = \frac{2(1-\alpha)P}{2+\alpha}$$P_B = \frac{\alpha}{(2+\alpha)/2} \cdot P = \frac{2\alpha P}{2+\alpha}$$P_C = \frac{\alpha/2}{(2+\alpha)/2} \cdot P = \frac{\alpha P}{2+\alpha}$$K_P = \frac{P_B \cdot (P_C)^{1/2}}{P_A} = \frac{(\frac{2\alpha P}{2+\alpha}) \cdot (\frac{\alpha P}{2+\alpha})^{1/2}}{\frac{2(1-\alpha)P}{2+\alpha}}$Simplifying the expression:$K_P = \frac{2\alpha P}{2+\alpha} \cdot \frac{\alpha^{1/2} P^{1/2}}{(2+\alpha)^{1/2}} \cdot \frac{2+\alpha}{2(1-\alpha)P} = \frac{\alpha^{3/2} P^{1/2}}{(1-\alpha)(2+\alpha)^{1/2}}$

$A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$. The correct relationship between $K_P$,$\alpha$,and equilibrium pressure $P$ is:

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