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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2008

  • question_answer
    What is the equation for the equilibrium constant \[({{K}_{c}})\] for  the following reaction? \[\frac{1}{2}A(g)+\frac{1}{3}B(g)\frac{2}{3}C(g)\]

    A)                  \[{{K}_{c}}=\frac{{{[A]}^{1/2}}{{[B]}^{1/3}}}{{{[C]}^{3/2}}}\]

    B)                  \[{{K}_{c}}=\frac{{{[C]}^{3/2}}}{{{[A]}^{2}}{{[B]}^{3}}}\]

    C)                  \[{{K}_{c}}=\frac{{{[C]}^{2/3}}}{{{[A]}^{1/2}}{{[B]}^{1/3}}}\]

    D)                  \[{{K}_{c}}=\frac{{{[C]}^{2/3}}}{{{[A]}^{1/2}}+{{[B]}^{1/3}}}\]

    Correct Answer: C

    Solution :

    \[\frac{1}{2}A(g)+\frac{1}{3}B(g)\frac{2}{3}C(g)\] Equilibrium constant, \[{{K}_{c}}=\frac{Rate\,\,product}{Rate\,of\,\text{reactant}}\]                                 \[{{K}_{c}}=\frac{{{[C]}^{2/3}}}{{{[A]}^{1/2}}{{[B]}^{1/3}}}\]


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