2. Solved Papers
  3. MGIMS WARDHA

MGIMS WARDHA MGIMS WARDHA Solved Paper-2008

  • 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 \[50\text{ }J{{K}^{-1}}mo{{l}^{-1}}\]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 :

     \[{{X}_{(g)}}+3Y(g)2Z(g)\] \[\Delta {{s}^{o}}=2\Delta {{s}^{o}}(Z)-\{{{s}^{o}}(X)+3{{s}^{o}}(Y)\}\]                 \[=2\times 50-\{60+3\times 40\}\]                 \[=100-180\]                 \[=-80\,J{{K}^{-1}}mo{{l}^{-1}}\] 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           \[T=\frac{\Delta {{H}^{o}}}{\Delta {{s}^{o}}}\]                 \[=\frac{40,000}{80}\] \[=500K\]


You need to login to perform this action.
You will be redirected in 3 sec spinner