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AMU Medical AMU Solved Paper-2015

  • question_answer
    The \[{{K}_{p}}\] value for the reaction,\[{{H}_{2}}(g)+{{I}_{2}}(g)2HI(g)\] at \[{{460}^{o}}C\] is 49. If the partial pressure of \[{{H}_{2}}\] and \[{{I}_{2}}\] is 0.5 atm, respectively. What will be the partial pressure of HI at equilibrium?

    A)  0.389                   

    B)  0.111

    C)  0.788                   

    D)  0,222

    Correct Answer: C

    Solution :

                                                    \[{{H}_{2}}+\]    \[{{I}_{2}}\]        \[\]         \[2\,\,HI\] Initial pressure  5.0          5.0                          0 At equilibrium   (5.0 - p) (5.0 - p)                2p                 \[{{K}_{p}}=\frac{{{({{p}_{HI}})}^{2}}}{{{p}_{{{H}_{2}}}}\times {{p}_{{{I}_{2}}}}}=\frac{{{(2\,p)}^{2}}}{(0.5-p)\,(0.5-p)}\] or            \[49=\frac{{{(2\,p)}^{2}}}{{{(0.5-p)}^{2}}}\] or            \[\sqrt{49}=\frac{2\,p}{0.5-p}\] or            \[7=\frac{2\,p}{0.5-p}\] or            \[2p=7\,(0.5-p)\]              or            \[2p=3.5-7p\] or            \[9p=3.5\] or            \[9p=3.5\] or            \[p=\frac{3.5}{9}=0.388\] Hence, \[{{p}_{HI}}=2p=2\times 0.388=0.788\]atm


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