Punjab Medical Punjab - MET Solved Paper-2008

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

    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 \[{{K}_{c}}\text{=}\frac{\text{rate}\,\,\text{of}\,\,\text{product}}{\text{rate}\,\,\text{of}\,\text{reactant}}\] \[{{K}_{c}}=\frac{{{[C]}^{2/3}}}{{{[A]}^{1/2}}{{[B]}^{1/3}}}\]


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