RAJASTHAN PMT Rajasthan - 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,                 \[{{K}_{c}}=\frac{\text{Rate}\,\,\text{product}}{\text{Rate}\,\,\text{of}\,\,\text{reactant}}\]                 \[{{K}_{c}}=\frac{{{[C]}^{2/3}}}{{{[A]}^{1/2}}{{[B]}^{1/3}}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner