Consider the reaction X_2Y_{(g)} rightlefthar | vedclass

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(A) For the reaction: $X_2Y_{(g)} \rightleftharpoons X_{2(g)} + \frac{1}{2}Y_{2(g)}$Initial moles: $1, 0, 0$Moles at equilibrium: $(1-x), x, \frac{x}{

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$x = \sqrt[3]{\frac{2 Kp^2}{p}}$

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(A) For the reaction: $X_2Y_{(g)} \rightleftharpoons X_{2(g)} + \frac{1}{2}Y_{2(g)}$Initial moles: $1, 0, 0$Moles at equilibrium: $(1-x), x, \frac{x}{2}$Total moles at equilibrium: $1-x+x+\frac{x}{2} = 1+\frac{x}{2}$Partial pressures: $P_{X_2Y} = \frac{1-x}{1+x/2} P$,$P_{X_2} = \frac{x}{1+x/2} P$,$P_{Y_2} = \frac{x/2}{1+x/2} P$$Kp = \frac{P_{X_2} \cdot (P_{Y_2})^{1/2}}{P_{X_2Y}} = \frac{[x/(1+x/2)] P \cdot [x/2(1+x/2) P]^{1/2}}{[(1-x)/(1+x/2)] P}$Since $x \ll 1$,we approximate $1-x \approx 1$ and $1+x/2 \approx 1$.$Kp \approx \frac{x \cdot (x/2)^{1/2} \cdot P^{1/2}}{1} = \frac{x^{3/2} \cdot P^{1/2}}{\sqrt{2}}$$Kp^2 = \frac{x^3 \cdot P}{2} \implies x^3 = \frac{2 Kp^2}{P}$$x = \left( \frac{2 Kp^2}{P} \right)^{1/3}$

Consider the reaction $X_2Y_{(g)} \rightleftharpoons X_{2(g)} + \frac{1}{2}Y_{2(g)}$. The equation representing the correct relationship between the degree of dissociation $(x)$ of $X_2Y_{(g)}$ with its equilibrium constant $Kp$ is. . . . . . . Assume $x$ to be very very small.

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