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JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank Heat of reaction

  • question_answer
    The heat change for the following reaction at \[{{298}^{o}}K\] and at constant pressure is \[+7.3\,kcal\] \[{{A}_{2}}B(s)\,\to \,2A(s)+1/2\,{{B}_{2}}(g)\], \[\Delta H=+7.3\,kcal\] The heat change at constant volume would be [DCE 2000]

    A)                 7.3 kcal

    B)                 More than 7.3

    C)                 Zero      

    D)                 None of these

    Correct Answer: D

    Solution :

           \[\Delta H=\Delta \text{E}+\Delta \text{nRT}\] or\[\Delta E=\Delta H-\Delta nRT\]                 \[\therefore \Delta E=+7.3-\frac{1}{2}\times 0.002\times 298\]\[=7.3-0.298\]= 7 kcal.


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