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JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

  • question_answer
                    In reaction\[\operatorname{A}+2B\rightleftharpoons 2C+\operatorname{D},\] initial concentration of B was 1.5 tunes of \[\left[ \operatorname{A} \right],\]but at equilibrium the concentrations of A and B became equal. The equilibrium constant for the reaction is:      JEE Main Online Paper (Held On 09 April 2013)             

    A)                 8                

    B)                 4                

    C)                 12                

    D)                 6                

    Correct Answer: D

    Solution :

                    Let degree of dissociation \[=x\]                                            \[A\,\,\,\,\,+\,\,\,\,2B\,\underset{{}}{\leftrightarrows}\,2C\,\,+\,\,\,D\]                 Initial cone.      1        1.5             0        0                 At equilibrium \[(1-x)\] \[(1.5-2x)\] \[2x\]            \[x\]                 Given,                   \[(1-x)=(1.5-2x)\]                 \[1=1.5-2x+x\]                 \[1=1.5-x\]                 \[x=1.5-1=0.5\]                 Equilibrium constant for the reaction                 \[{{K}_{C}}=\frac{{{[C]}^{2}}[D]}{[A]{{[B]}^{2}}}=\frac{{{(2x)}^{2}}\,(x)}{(1-x)\,(1.5-2x)}\] \[\because \]     \[x=0.5\]                 \[{{K}_{C}}=\frac{{{(2\times 0.5)}^{2}}\,(0.5)}{(1-0.5)\,{{(1.5-2\times 0.5)}^{2}}}\]                 \[=\frac{(1)\times (0.5)}{(0.5)\,{{(0.5)}^{2}}}=\frac{0.5}{0.5\times 0.25}=\frac{0.5}{0.125}=4\]                


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