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(B) For the reaction $A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)}$:Initial moles: $A=1, B=0, C=0$At equilibrium: $A=1-\alpha, B=\alpha, C=\alpha$Tota

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When $p$ increases,$\alpha$ decreases

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(B) For the reaction $A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)}$:Initial moles: $A=1, B=0, C=0$At equilibrium: $A=1-\alpha, B=\alpha, C=\alpha$Total moles $= (1-\alpha) + \alpha + \alpha = 1+\alpha$Partial pressures: $P_A = \frac{1-\alpha}{1+\alpha} p, P_B = \frac{\alpha}{1+\alpha} p, P_C = \frac{\alpha}{1+\alpha} p$$K_p = \frac{P_B \cdot P_C}{P_A} = \frac{(\frac{\alpha}{1+\alpha} p) (\frac{\alpha}{1+\alpha} p)}{\frac{1-\alpha}{1+\alpha} p} = \frac{\alpha^2 p}{1-\alpha^2}$Rearranging for $\alpha$: $\alpha^2 p = K_p - K_p \alpha^2$ $\Rightarrow \alpha^2(p+K_p) = K_p$ $\Rightarrow \alpha = \sqrt{\frac{K_p}{p+K_p}}$This expression shows that as the total pressure $p$ increases,the degree of dissociation $\alpha$ decreases.

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