CET Karnataka Medical CET - Karnataka Medical Solved Paper-2006

  • 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 \[50~J{{K}^{-1\text{ }}}mo{{l}^{-1}}\]respectively.  The temperature at which the above reaction attains equilibrium is about:

    A)  \[400K\]         

    B)  \[500K\]

    C)  \[273K\]

    D)  \[373K\]

    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  \[\Delta {{H}^{o}}=-40kJ=-40,000J\] \[\Delta {{G}^{o}}=\Delta {{H}^{o}}-T\Delta {{S}^{o}}\] At equilibrium, \[\Delta {{G}^{o}}=0\] \[\therefore \] \[\Delta {{H}^{o}}=T\Delta {{S}^{o}}\] or \[T=\frac{\Delta {{H}^{o}}}{\Delta {{S}^{o}}}=\frac{40,000}{80}=500K\]


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