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  3. JEE Main & Advanced

JEE Main & Advanced AIEEE Solved Paper-2005

  • question_answer
    A schematic plot of In\[{{K}_{eq}}\]versus inverse of temperature for a reaction is shown below \[\log {{K}_{eq}}=\frac{-\Delta {{H}^{o}}}{2.303RT}+\frac{-\Delta {{S}^{o}}}{R}\] The reaction must be     AIEEE  Solved  Paper-2005

    A) highly spontaneous at ordinary temperature

    B) one with negligible enthalpy change

    C) endothermic

    D) exothermic

    Correct Answer: D

    Solution :

    Variation of\[{{K}_{eq}}\]with temperature T is given by van't Hoff equation. \[\log {{K}_{eq}}=-\frac{\Delta {{H}^{o}}}{2.303RT}+\frac{\Delta {{S}^{o}}}{R}\]                                 A             B Slope of the given line is +ve indicating that term A is positive thus\[\Delta {{H}^{o}}\]is -ve. Thus reaction is exothermic.


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