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DUMET Medical DUMET Medical Solved Paper-2004

  • question_answer
    The enthalpy change for the reaction of 50.00 mL of ethylene with 50.00 mL of H^ at 1.5 atm pressure is \[\Delta H=-0.31\,kJ\]. The value of \[\Delta E\]will be:

    A)  0.235 kJ

    B)  0.3024 kJ

    C)  2.567 kJ

    D)  0.0076 kJ

    Correct Answer: A

    Solution :

    Key Idea: Use the following formula to find the value of \[\Delta E\]. \[\Delta H-\Delta E+P\Delta V\] where \[\Delta H=-0.31\,kJ\,mo{{l}^{-1}}\] \[P=1.5\,atm\] \[\Delta V=-50\,mL=-0.050\,L\] \[\Delta E=?\] \[\Delta H=\Delta E+P\Delta V\] or \[-0.31=\Delta E+(1.5\times -0.05)\] or \[-0.31=\Delta E-0.075\] or \[-0.31+0.075=\Delta E\] \[\Delta E=0.235\]


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