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BCECE Medical BCECE Medical Solved Papers-2007

  • question_answer
    For the reaction, \[{{H}_{2}}+{{I}_{2}}2HI\], the equilibrium concentration of \[{{H}_{2}},\,{{I}_{2}}\] and HI are 8.0, 3.0 and 28.0 mol/L respectively. The equilibrium constant is

    A)  28.34         

    B)  32.66

    C)  34.78          

    D)  38.88

    Correct Answer: B

    Solution :

    Given    \[[{{H}_{2}}]=8.0\,mol/L\]                 \[[{{I}_{2}}]=3.0\,mol/L\]                 \[[HI]=28\,\,mol/L\]                     \[K=?\] \[{{H}_{2}}+{{I}_{2}}2HI\] \[\therefore \] \[K=\frac{{{[HI]}^{2}}}{[{{H}_{2}}]\,[{{I}_{2}}]}=\frac{{{(28)}^{2}}}{(8)\times (3)}\]                 \[=\frac{28\times 28}{24}\] = 32.66


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