At temperature T,the dissociation equilibrium | vedclass

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Question

(A) The initial moles are $2$ moles of $AB_2$. Let the degree of dissociation be $x$.For the reaction: $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2

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$Px^3 / 2$

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(A) The initial moles are $2$ moles of $AB_2$. Let the degree of dissociation be $x$.For the reaction: $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$Initial moles: $2, 0, 0$Moles at equilibrium: $2(1-x), 2x, x$Total moles at equilibrium $= 2-2x+2x+x = 2+x$.Partial pressures at equilibrium:$p_{AB_2} = \frac{2(1-x)}{2+x} P \approx P$ (since $x \ll 1$)$p_{AB} = \frac{2x}{2+x} P \approx xP$$p_{B_2} = \frac{x}{2+x} P \approx \frac{x}{2} P$$K_p = \frac{(p_{AB})^2 (p_{B_2})}{(p_{AB_2})^2} = \frac{(xP)^2 (xP/2)}{P^2} = \frac{Px^3}{2}$

At temperature $T$,the dissociation equilibrium of $AB_{2(g)}$ is given as: $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$. If the degree of dissociation $x$ is very small compared to $1$,then the expression for the equilibrium constant $K_p$ in terms of $x$ and total pressure $P$ is:

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