(i) X(g) rightleftharpoons Y(g) + Z(g), K_{p1 | vedclass

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(A) For reaction $(i): X(g) \rightleftharpoons Y(g) + Z(g)$   $Initial \ moles$ $n$ $0$ $0$   $Equilibrium \ moles$ $n(1-\alpha)$ $n\alpha$ $n\alpha$

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) For reaction $(i): X(g) ightleftharpoons Y(g) + Z(g)$   $Initial \ moles$ $n$ $0$ $0$   $Equilibrium \ moles$ $n(1-\alpha)$ $n\alpha$ $n\alpha$   Total moles at equilibrium $= n(1-\alpha) + n\alpha + n\alpha = n(1+\alpha)$. Partial pressures: $p_X = \frac{1-\alpha}{1+\alpha} p_1$,$p_Y = \frac{\alpha}{1+\alpha} p_1$,$p_Z = \frac{\alpha}{1+\alpha} p_1$. $K_{p1} = \frac{p_Y p_Z}{p_X} = \frac{\alpha^2 p_1}{1-\alpha^2} = 3$ For reaction $(ii): A(g) ightleftharpoons 2B(g)$   $Initial \ moles$ $n$ $0$   $Equilibrium \ moles$ $n(1-\alpha)$ $2n\alpha$   Total moles at equilibrium $= n(1-\alpha) + 2n\alpha = n(1+\alpha)$. Partial pressures: $p_A = \frac{1-\alpha}{1+\alpha} p_2$,$p_B = \frac{2\alpha}{1+\alpha} p_2$. $K_{p2} = \frac{p_B^2}{p_A} = \frac{4\alpha^2 p_2}{1-\alpha^2} = 1$ Dividing the two expressions: $\frac{K_{p1}}{K_{p2}} = \frac{3}{1} = \frac{\alpha^2 p_1 / (1-\alpha^2)}{4\alpha^2 p_2 / (1-\alpha^2)} = \frac{p_1}{4p_2}$ Therefore,$\frac{p_1}{p_2} = 3 	imes 4 = 12$. Thus,$x = 12$.

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