The equilibrium constants K_{p1} and K_{p2} f | vedclass

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(A) For reaction $X \rightleftharpoons 2Y$:Initial moles: $a$,$0$At equilibrium: $a(1 - \alpha)$,$2a\alpha$Total moles $= a(1 + \alpha)$$K_{p1} = \fra

Vedclass · Accepted Answer

$1 : 36$

Vedclass · Answer

(A) For reaction $X ightleftharpoons 2Y$:Initial moles: $a$,$0$At equilibrium: $a(1 - \alpha)$,$2a\alpha$Total moles $= a(1 + \alpha)$$K_{p1} = \frac{(2a\alpha)^2}{a(1 - \alpha)} 	imes \left(\frac{P_{T1}}{a(1 + \alpha)}ight)^1 = \frac{4\alpha^2 P_{T1}}{1 - \alpha^2}$For reaction $Z ightleftharpoons P + Q$:Initial moles: $b$,$0$,$0$At equilibrium: $b(1 - \alpha)$,$b\alpha$,$b\alpha$Total moles $= b(1 + \alpha)$$K_{p2} = \frac{(b\alpha)(b\alpha)}{b(1 - \alpha)} 	imes \left(\frac{P_{T2}}{b(1 + \alpha)}ight)^1 = \frac{\alpha^2 P_{T2}}{1 - \alpha^2}$Given $\frac{K_{p1}}{K_{p2}} = \frac{1}{9}$,we have:$\frac{4\alpha^2 P_{T1} / (1 - \alpha^2)}{\alpha^2 P_{T2} / (1 - \alpha^2)} = \frac{1}{9}$$\frac{4 P_{T1}}{P_{T2}} = \frac{1}{9}$$\frac{P_{T1}}{P_{T2}} = \frac{1}{36}$

The equilibrium constants $K_{p1}$ and $K_{p2}$ for the reactions $X \rightleftharpoons 2Y$ and $Z \rightleftharpoons P + Q$ respectively are in the ratio of $1 : 9$. If the degree of dissociation of $X$ and $Z$ be equal,then the ratio of total pressures at these equilibria is:

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