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BVP Medical BVP Medical Solved Paper-2002

  • question_answer
     If the equilibrium constant for the reaction \[{{N}_{2}}+3{{H}_{2}}2N{{H}_{3}}\], then \[{{K}_{C}}\] the equilibrium constant for the reaction \[N{{H}_{3}}\frac{1}{2}{{N}_{2}}+\frac{3}{2}{{H}_{2}}\] is:

    A)  \[\frac{1}{K_{C}^{2}}\]                                

    B)  \[\sqrt{{{K}_{C}}}\]

    C)  \[\frac{1}{{{K}_{C}}}\]                                  

    D)  \[\frac{\sqrt{1}}{{{K}_{C}}}\]

    Correct Answer: D

    Solution :

                    \[{{N}_{2}}+3{{H}_{2}}2N{{H}_{3}},{{K}_{C}}=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]{{[{{H}_{2}}]}^{3}}}\] ?.(i) \[N{{H}_{3}}\frac{1}{2}{{N}_{2}}+\frac{3}{2}{{H}_{2}},K_{_{C}}^{}=\frac{{{[{{N}_{2}}]}^{1/2}}{{[{{H}_{2}}]}^{3/2}}}{[N{{H}_{3}}]}\]?..(ii) From equation (i) and (ii),  \[K_{_{C}}^{}=\frac{\sqrt{1}}{{{K}_{C}}}\]


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