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BHU PMT BHU PMT (Screening) Solved Paper-2005

  • question_answer
    The entropy values (in\[J{{K}^{-1}}mo{{l}^{-1}}\]) of\[{{H}_{2}}(g)=130.6,\text{ }C{{l}_{2}}(g)=223.0\]and \[HCl(g)=\] 186.7 at 298 K and 1 atm pressure. Then entropy change for the reaction. \[{{H}_{2}}(g)+C{{l}_{2}}(g)\xrightarrow[{}]{{}}2HCl(g)\]is:

    A)  +540.3                                 

    B)  +727.3

    C)   \[-166.9\]                         

    D)  +19.8

    Correct Answer: D

    Solution :

                     Key Idea: \[\Delta S={{S}_{p}}-{{S}_{R}}\] \[{{H}_{2}}(g)+C{{l}_{2}}(g)\xrightarrow[{}]{{}}2HCl(g)\] Given    \[{{S}^{o}}H(g)=130.6J{{K}^{-1}}/mol\] \[{{S}^{o}}C{{l}_{2}}(g)=223.0\text{ }J{{K}^{-1}}/mol\] \[{{S}^{o}}HCl(g)=186.7\text{ }J{{K}^{-1}}/mol\] \[\Delta S=2{{S}^{o}}HCl-({{S}^{o}}{{H}_{2}}+{{S}^{o}}C{{l}_{2}})\] \[=2\times 186.7-(130.6+223.0)\] \[=373.4-353.6=19.8\text{ }J{{K}^{-1}}mo{{l}^{-1}}\]


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