At temperature T,compound AB_{2(g)} dissociat | vedclass

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(C) The dissociation reaction is: $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$Initial moles: $1, 0, 0$Moles at equilibrium: $(1-x),

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$\sqrt[3]{\frac{2 K_p^2}{p}}$

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(C) The dissociation reaction is: $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$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: $P_{AB_2} = \frac{1-x}{1+x/2}P$,$P_{AB} = \frac{x}{1+x/2}P$,$P_{B_2} = \frac{x/2}{1+x/2}P$Since $x \ll 1$,we approximate $1+x/2 \approx 1$ and $1-x \approx 1$.Thus,$P_{AB_2} \approx P$,$P_{AB} \approx xP$,$P_{B_2} \approx \frac{x}{2}P$.$K_p = \frac{P_{AB} \cdot (P_{B_2})^{1/2}}{P_{AB_2}} = \frac{(xP) \cdot (xP/2)^{1/2}}{P} = x \cdot (xP/2)^{1/2} = x^{3/2} \cdot (P/2)^{1/2}$.Squaring both sides: $K_p^2 = x^3 \cdot \frac{P}{2}$.Solving for $x$: $x^3 = \frac{2 K_p^2}{P} \Rightarrow x = \left( \frac{2 K_p^2}{P} \right)^{1/3}$.

At temperature $T$,compound $AB_{2(g)}$ dissociates as $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$ having degree of dissociation $x$ (small compared to unity). The correct expression for $x$ in terms of $K_p$ and $p$ is

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