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Manipal Engineering Manipal Engineering Solved Paper-2008

  • question_answer
    For a reversible reaction : \[X(g)+3Y(g)2Z(g);\,\,\Delta H=-40kJ\] the standard entropies of\[X,\,\,Y\]and\[Z\]are 60, 40 and 50 \[J{{K}^{-1}}mo{{l}^{-1}}\]respectively. The temperature at which the above reaction attains equilibrium is about

    A) \[400\,\,K\]                       

    B) \[500\,\,K\]

    C) \[273\,\,K\]       

    D)        \[373\,\,K\]

    Correct Answer: B

    Solution :

    \[X(g)+3Y(g)2Z(g)\] \[\Delta {{S}^{o}}=2{{S}^{o}}(Z)-\{{{S}^{o}}(X)+3{{S}^{o}}(Y)\}\]                 \[=2\times 50-\{60+3\times 40\}\]                 \[=100-180=-80\,\,J\,\,{{K}^{-1}}\,\,mo{{l}^{-1}}\]                 \[\Delta G=\Delta H-T\Delta S\] At equilibrium,                 \[\Delta G=0\] \[\therefore \]  \[\Delta H=T\Delta S\]                 \[T=\frac{\Delta H}{\Delta S}\]                 \[=\frac{-40\times {{10}^{3}}J}{-80J{{K}^{-1}}mo{{l}^{-1}}}\]                 \[=500K\]


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