For the reactions (1) and (2) :A rightlefthar | vedclass

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(B) For reaction $(1)$: $A \rightleftharpoons B + C$Initial moles: $1, 0, 0$Equilibrium moles: $(1 - \alpha), \alpha, \alpha$Total moles at equilibriu

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) For reaction $(1)$: $A \rightleftharpoons B + C$Initial moles: $1, 0, 0$Equilibrium moles: $(1 - \alpha), \alpha, \alpha$Total moles at equilibrium: $1 + \alpha$$K_{P_1} = \frac{p_B \cdot p_C}{p_A} = \frac{(\frac{\alpha}{1+\alpha} P_1)(\frac{\alpha}{1+\alpha} P_1)}{\frac{1-\alpha}{1+\alpha} P_1} = \frac{\alpha^2}{1-\alpha^2} P_1$For reaction $(2)$: $D \rightleftharpoons 2E$Initial moles: $1, 0$Equilibrium moles: $(1 - \alpha), 2\alpha$Total moles at equilibrium: $1 + \alpha$$K_{P_2} = \frac{p_E^2}{p_D} = \frac{(\frac{2\alpha}{1+\alpha} P_2)^2}{\frac{1-\alpha}{1+\alpha} P_2} = \frac{4\alpha^2}{1-\alpha^2} P_2$Given $\frac{K_{P_1}}{K_{P_2}} = \frac{9}{1}$.Substituting the expressions: $\frac{\frac{\alpha^2}{1-\alpha^2} P_1}{\frac{4\alpha^2}{1-\alpha^2} P_2} = \frac{P_1}{4P_2} = 9$Therefore,$\frac{P_1}{P_2} = 36 : 1$.

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