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JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank 2nd and 3rd law of thermodynamics and Entropy

  • question_answer
    The entropy values (in JK-1 mol-1) of H2(g) = 130.6, Cl2(g) = 223.0 and HCl(g) = 186.7 at 298 K and 1 atm pressure. Then entropy change for the reaction                 \[{{H}_{2(g)}}+C{{l}_{2(g)}}\to 2HC{{l}_{(g)}}\] is                                               [BHU 2005]

    A)                 + 540.3 

    B)                 + 727.3

    C)                 ? 166.9 

    D)                 + 19.8

    Correct Answer: D

    Solution :

            \[\Delta {{S}^{o}}=2S_{HCl}^{o}-(S_{{{H}_{2}}}^{o}+S_{C{{l}_{2}}}^{o})\]                                 \[=2\times 186.7-(130.6+223.0)=19.8\,J{{K}^{-1}}mo{{l}^{-1}}\]


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