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NEET Sample Paper NEET Sample Test Paper-91

  • question_answer
    Standard entropy of \[{{X}_{2}},\]\[{{Y}_{2}}\] and X \[{{Y}_{3}}\] are 60, 40 and \[50\,J\,{{K}^{-1}}mo{{l}^{-1}},\] respectively. For the reaction, \[\frac{1}{2}{{X}_{2}}+\frac{3}{2}{{Y}_{2}}\to X{{Y}_{3}},\]\[\Delta H=-30kJ,\] to be at equilibrium, the temperature will be

    A) 1250 K             

    B) 500 K

    C) 1000 K 

    D)        750 K

    Correct Answer: D

    Solution :

    [d] For a reaction to be at equilibrium \[\Delta G=0\] Since \[\Delta G=\Delta H-T\Delta S\] so at equilibrium \[\Delta H-T\Delta S=0\] or \[\Delta H=T\Delta S\] For the reaction \[\frac{1}{2}{{X}_{2}}+\frac{3}{2}{{Y}_{2}}\xrightarrow{{}}X{{Y}_{3}};\] \[\Delta H=-30kJ\](given) Calculating \[\Delta S\] for the above reaction, we get \[\Delta S=50-\left[ \frac{1}{2}\times 60+\frac{3}{2}\times 40 \right]J{{K}^{-1}}\]      \[=50-(30+60)J{{K}^{-1}}=-40J{{K}^{-1}}\] At equilibrium, \[T\Delta S=\Delta H\] \[\therefore \] \[T\times (-40)=-30\times 1000\]\[[\because 1kJ=1000J]\] or \[T=\frac{-30\times 1000}{-40}\] or \[750\,K\]


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