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VIT Engineering VIT Engineering Solved Paper-2010

  • question_answer
    For the reaction, \[2A(g)+{{B}_{2}}(g)2A{{B}_{2}}(g)\] the equilibrium constant, \[{{K}_{p}}\] at 300 K is 16.0. The value of \[{{K}_{p}}\] for \[A{{B}_{2}}(g)A(g)+1/2{{B}_{2}}(g)\]

    A)  \[8\]             

    B)  \[0.25\]

    C)  \[0.125\]          

    D)  \[32\]

    Correct Answer: B

    Solution :

     For the reaction, \[2A+{{B}_{2}}2A{{B}_{2}}\], Equilibrium constant,  \[{{K}_{P}}=\frac{P_{A{{B}_{2}}}^{2}}{P_{A}^{2}.{{P}_{{{B}_{2}}}}}=16\]  ....(i) For the reaction, \[A{{B}_{2}}A+1/2{{B}_{2}},\] \[K_{P}^{}=\frac{{{p}_{A}}.p_{{{B}_{2}}}^{1/2}}{{{p}_{A{{B}_{2}}}}}\]                  ???(ii) On squaring Eq. (ii), we get \[{{(K_{P}^{})}^{2}}=\frac{p_{A}^{2}{{p}_{{{B}_{2}}}}}{p_{A{{B}_{2}}}^{2}}\]            ???(iii) From Eq. (i) and (iii)^ we get \[{{K}_{p}}.{{(K_{p}^{})}^{2}}=1\] \[16{{(K_{p}^{})}^{2}}=1\] \[{{(K_{p}^{})}^{2}}=\frac{1}{16}\] \[K_{p}^{}=\frac{1}{4}=0.25\]


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