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Manipal Medical Manipal Medical Solved Paper-2009

  • question_answer
    The heat change for the. following reaction at 298 K and constant pressure is\[+7.3\text{ }kcal\]. \[{{A}_{2}}B(s)\xrightarrow[{}]{{}}2A(s)+\frac{1}{2}{{B}_{2}}(g);\]\[\Delta H=+7.3\text{ }kcal\] The heat change at constant volume would be

    A)  \[+7.3\text{ }kcal\]

    B)  more than 7.3 kcal

    C)  less than 7.3 kcal

    D)  zero

    Correct Answer: C

    Solution :

     \[\Delta H=\Delta E+\Delta {{n}_{g}}RT\] Or \[\Delta E=\Delta H-\Delta {{n}_{g}}RT\] \[\Delta {{n}_{g}}=\frac{1}{2}\] \[\Delta E=2.3\times {{10}^{3}}-\frac{1}{2}\times 2\times 298\] \[=7300\text{ }cal-298\] \[=7002\,cal\] \[\approx 7.00\,kcal\]


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