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


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