2. Solved Papers
  3. MGIMS WARDHA

MGIMS WARDHA MGIMS WARDHA Solved Paper-2003

  • question_answer
    If equilibrium constant for reaction\[2AB~{{A}_{2}}+{{B}_{2}},\]is 49, then the equilibrium constant   for   reaction\[AB\frac{1}{2}{{A}_{2}}+\frac{1}{2}{{B}_{2}},\]will be;

    A)  7                                            

    B)  280

    C)  49                                         

    D)  0.01

    Correct Answer: A

    Solution :

                     For the reaction, \[2AB{{A}_{2}}+{{B}_{2}}\]                 \[{{K}_{c}}=\frac{[{{A}_{2}}][{{B}_{2}}]}{[AB]}\]        ...(i) for reaction, \[AB\frac{1}{2}{{A}_{2}}+\frac{1}{2}{{B}_{2}}\]                                 \[{{K}_{c}}'=\frac{{{[{{A}_{2}}]}^{1/2}}{{[{{B}_{2}}]}^{1/2}}}{[AB]}\]         ??(ii) Hence, from equation (i) and (ii)                 \[{{K}_{c}}'=\sqrt{{{K}_{c}}}=\sqrt{49}=7\]


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