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EAMCET Medical EAMCET Medical Solved Paper-1997

  • question_answer
    The value of \[{{\text{K}}_{\text{p}}}\] for the reaction \[2{{H}_{2}}S(g)\rightleftharpoons 2{{H}_{2}}(g)+{{S}_{2}}(g)\]is \[1.2\times {{10}^{-2}}\]at \[\text{1065}{{\,}^{\text{o}}}\text{C}\text{.}\]The value of \[{{\text{K}}_{\text{c}}}\]for this reaction is:

    A)  \[1.2\times {{10}^{-2}}\]                             

    B)  \[<1.2\times {{10}^{-2}}\]

    C)  83                                         

    D)  \[>1.2\times {{10}^{-2}}\]

    Correct Answer: B

    Solution :

                     \[2{{H}_{2}}S(g)\rightleftharpoons 2{{H}_{2}}(g)+{{S}_{2}}(g)\] \[{{K}_{p}}=1.2\times {{10}^{-12}}\] \[T=1065+273=1338\,K\] \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\] where \[\Delta n=\]number of moles of products - number of moles of reactants \[\Delta n=3-2=1\] \[R=0.0821\]                  \[2\times {{10}^{-2}}={{K}_{e}}{{(0.0821\times 1338)}^{1}}\]                 \[{{K}_{c}}=\frac{1.2\times {{10}^{-2}}}{0.0821\times 1338}=\frac{1.2\times {{10}^{-12}}}{109.8}\] \[=0.0109\times {{10}^{-2}}\]                 It means that                                 \[{{K}_{c}}<1.2\times {{10}^{-2}}\]


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