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,$\frac{P_1}{P_2} = 36 : 1$.

Similar Questions

For the reaction $2AB_{(g)} \rightleftharpoons 2A_{(g)} + B_{2(g)}$,$AB$ dissociates. If the initial pressure of $AB$ is $500 \, mm$ and the total pressure at equilibrium is $625 \, mm$,calculate $K_p$ for the reaction. Assume constant volume.

$A$ mixture of $0.3 \ mol$ of $H_2$ and $0.3 \ mol$ of $I_2$ is allowed to react in a $10 \ L$ evacuated flask at $500 \ ^oC$. The reaction is $H_2 + I_2 \rightleftharpoons 2HI$,and the equilibrium constant $K_c$ is found to be $64$. The amount of unreacted $I_2$ at equilibrium is $...... \ mol$.

In which of the following reactions is ${K_p} > {K_c}$?

The relation between equilibrium constant $K_p$ and $K_c$ is

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