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NEET Sample Paper NEET Sample Test Paper-7

  • 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, \[\frac{1}{2}{{X}_{2}}+\frac{3}{2}{{Y}_{2}}\xrightarrow{{}}X{{Y}_{3}},\] \[\Delta H=-30\,kJ\]to be at equilibrium, the temperature will be

    A)  \[1250\text{ }K\]                    

    B)  \[500\,K\]

    C)  \[750\text{ }K\]                      

    D)  \[1000\text{ }K\]

    Correct Answer: C

    Solution :

    \[\frac{1}{2}{{X}_{2}}+\frac{3}{2}{{Y}_{2}}\xrightarrow{{}}X{{Y}_{3}}\] \[\Delta {{S}_{reaction}}={{S}_{product}}-{{S}_{r\text{eactant}}}\] \[\Delta {{S}_{reaction}}=50-\left( \frac{3}{2}\times 40+\frac{1}{2}\times 60 \right)\] \[=-40\,Jmo{{l}^{-1}}\] \[\Delta G=\Delta H-T.\Delta S\] At equilibrium, \[\Delta G=0\]                 \[\Delta H=T\Delta S\]                 \[T=\frac{\Delta H}{\Delta S}=\frac{30\times {{10}^{3}}}{40}=750\,K\]


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