The dissociation equilibrium of a gas AB_{2} | vedclass

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(D) The reaction is $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$.Initial moles: $1$ for $AB_2$,$0$ for $AB$,$0$ for $B_2$.At equilibrium: $2(1

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$(2K_P/P)^{1/3}$

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(D) The reaction is $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$.Initial moles: $1$ for $AB_2$,$0$ for $AB$,$0$ for $B_2$.At equilibrium: $2(1-x)$ for $AB_2$,$2x$ for $AB$,$x$ for $B_2$.Total moles at equilibrium $= 2-2x+2x+x = 2+x$.Partial pressures are $p_{AB_2} = \frac{2(1-x)P}{2+x}$,$p_{AB} = \frac{2xP}{2+x}$,and $p_{B_2} = \frac{xP}{2+x}$.$K_P = \frac{(p_{AB})^2 (p_{B_2})}{(p_{AB_2})^2} = \frac{(\frac{2xP}{2+x})^2 (\frac{xP}{2+x})}{(\frac{2(1-x)P}{2+x})^2}$.Simplifying,$K_P = \frac{4x^2 P^2 \cdot xP}{(2+x)^3} \cdot \frac{(2+x)^2}{4(1-x)^2 P^2} = \frac{x^3 P}{(2+x)(1-x)^2}$.Since $x \ll 1$,we approximate $2+x \approx 2$ and $(1-x) \approx 1$.Thus,$K_P \approx \frac{x^3 P}{2}$,which gives $x = (2K_P/P)^{1/3}$.

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