NEET Sample Paper NEET Sample Test Paper-36

  • 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) \[(2{{K}_{p}}/P)\]    

    Correct Answer: A

    Solution :

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

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