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MGIMS WARDHA MGIMS WARDHA Solved Paper-2004

  • question_answer
    In the reaction: \[{{H}_{2}}+{{I}_{2}}\]    \[2HI\] In a 2 litre flask 0.4 moles of each\[{{H}_{2}}\]and\[{{I}_{2}}\]are taken. At equilibrium 0.5 moles of HI are formed. What will be the value of equilibrium constant,\[{{K}_{c}}\]?

    A)  20.2                                      

    B)  25.4

    C)   0.284                                  

    D)  11.1

    Correct Answer: D

    Solution :

                     
                                   \[\underset{0.4}{\mathop{{{H}_{2}}}}\,+\underset{0.4}{\mathop{{{I}_{2}}}}\,\underset{0.50}{\mathop{2HI}}\,\]
    At equilibrium 0.4-0.25 0.4-0.25 0.50
    =0.15 =0.15
    \[{{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}=\frac{{{\left( \frac{0.50}{2} \right)}^{2}}}{\left( \frac{0.15}{2} \right)\left( \frac{0.15}{2} \right)}\]                 \[=\frac{0.5\times 0.5}{0.15\times 0.15}=11.11\]


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