• # question_answer The dissociation equilibrium of a gas $A{{B}_{2}}$can be represented as $2A{{B}_{2}}(g)\rightleftharpoons 2AB(g)+{{B}_{2}}(g)$ The degree of dissociation is $x$and is small com- pared to 1. The expression relating the degree of dissociation$(x)$with equilibrium constant ${{K}_{p}}$ and total pressure P is A) ${{(2{{K}_{p}}/P)}^{1/3}}$B) ${{(2{{K}_{p}}/P)}^{1/2}}$C) $({{K}_{p}}/P)$D) $(2{{K}_{p}}/P)$

$\underset{1-x}{\mathop{\underset{1}{\mathop{2A{{B}_{2}}(g)}}\,}}\,\rightleftharpoons \underset{x}{\mathop{\underset{0}{\mathop{2AB(g)}}\,}}\,+\underset{x,2}{\mathop{\underset{0}{\mathop{{{B}_{2}}(g)}}\,}}\,$ $\therefore$ ${{K}_{P}}=\frac{{{x}^{2}}.x}{2(1-x)}.\left[ \frac{P}{1+\frac{x}{2}} \right]=\frac{{{x}^{3}}.P}{2}$ $(1-x\approx 1and\,1+\frac{x}{2}\approx 1,\sin ce\,x<<1)$ or $P=\sqrt[3]{\frac{2{{K}_{p}}}{P}}$