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

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Question

(B) The reaction is $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$.Initially,let the moles of $AB_2$ be $1$.At equilibrium,the moles are: $AB_2

Vedclass · Accepted Answer

$\sqrt[3]{\frac{2K_P}{P}}$

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(B) The reaction is $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$.Initially,let the moles of $AB_2$ be $1$.At equilibrium,the moles are: $AB_2 = 1-x$,$AB = x$,$B_2 = x/2$.Total moles at equilibrium $= 1-x+x+x/2 = 1+x/2 \approx 1$ (since $x \ll 1$).The partial pressures are $P_{AB_2} = (1-x)P$,$P_{AB} = xP$,and $P_{B_2} = (x/2)P$.$K_P = \frac{(P_{AB})^2 (P_{B_2})}{(P_{AB_2})^2} = \frac{(xP)^2 (xP/2)}{(1-x)^2 P^2} = \frac{x^3 P}{2(1-x)^2}$.Since $x \ll 1$,$(1-x) \approx 1$,so $K_P \approx \frac{x^3 P}{2}$.Thus,$x^3 = \frac{2K_P}{P}$,which gives $x = \sqrt[3]{\frac{2K_P}{P}}$.Therefore,option $B$ is correct.

At temperature $T$,a compound $AB_{2(g)}$ dissociates according to the reaction $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$ with a degree of dissociation $x$,which is very small compared to unity. The value of $x$ is:

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