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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    In the reaction : \[{{H}_{2}}+{{I}_{2}}=2HI\] In a 2 L 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 :

                     \[\underset{\text{At}\,\text{equilibrium}\,=0.15}{\mathop{\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.4-0.25}{\mathop{\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.4}{\mathop{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{H}_{2}}}}\,}}\,}}\,\,\,\,\,\,\,+\underset{=\,0.15}{\mathop{\underset{0.4-0.25}{\mathop{\underset{0.4}{\mathop{{{I}_{2}}}}\,}}\,}}\,=\,\,\,\,\,\,\,\,\,\,\underset{0.05}{\mathop{\underset{0}{\mathop{2HI}}\,}}\,\] \[{{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}\] \[=\frac{{{\left( \frac{0.50}{2} \right)}^{2}}}{\left( \frac{0.50}{2} \right)\left( \frac{0.50}{2} \right)}=\frac{0.5\times 0.5}{0.15\times 0.15}=11.11\]


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