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RAJASTHAN PMT Rajasthan - PMT 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 :

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


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