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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

  • question_answer
    For a reversible reaction: \[X(g)+3Y(g)2Z(g);\,\,\Delta H=-40\,kJ\], 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 :

    \[X(g)+3Y(g)2Z(g)\] \[\Delta {{S}^{o}}=2{{S}^{o}}(Z)-\{{{S}^{o}}(X)+3{{S}^{o}}(Y)\}\]                 \[=2\times 50-\{60+3\times 40\}\]                                 \[=100-180=-80J\,{{K}^{-1}}mo{{l}^{-1}}\] Given \[\Delta {{H}^{o}}=-40\,kJ=-40,000\,J\]                 \[\Delta {{G}^{o}}=\Delta {{H}^{o}}-T\Delta {{S}^{o}}\] At equilibrium, \[\Delta {{G}^{o}}=0\] \[\therefore \]                  \[\Delta {{H}^{o}}=T\Delta {{S}^{o}}\] or            \[T=\frac{\Delta {{H}^{o}}}{\Delta {{S}^{o}}}=\frac{40,000}{80}=500\,K\]


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