2. Solved Papers
  3. EAMCET Medical

EAMCET Medical EAMCET Medical Solved Paper-1997

  • question_answer
    In the complete combustion of butanol \[{{\text{C}}_{\text{4}}}{{\text{H}}_{\text{9}}}\text{OH(I)}\] if \[\text{ }\!\!\Delta\!\!\text{ H}\] is enthalpy of combustion and \[\text{ }\!\!\Delta\!\!\text{ E}\] is the heat of combustion at constant volume, then:

    A)  \[\Delta H <\Delta E\]

    B) \[\Delta H =\Delta E\]

    C) \[\Delta H >\Delta E\]             

    D)   \[\Delta H ,\Delta E\] relation cannot to be predicted

    Correct Answer: C

    Solution :

    Combustion reaction of butanol is : \[{{C}_{4}}{{H}_{9}}OH+6{{O}_{2}}\xrightarrow{{}}4C{{O}_{2}}+5{{H}_{2}}O\] Total no. of moles product = 4 + 5 = 9 Total no. of moles reactant =1+ 6 = 7 \[\Delta n=\]number of moles of products - number of moles of reactants = 9 - 7 = 2means that \[\Delta n\] have +ve value. From the equation \[\Delta H=\Delta E+\Delta n\,RT\] If             \[\Delta n=\,0\]then \[\Delta H =\Delta E\] If             \[\Delta n=\,-ve\]then \[\Delta H<\Delta E\] If             \[\Delta n=+ve\]then \[\Delta H>\Delta E\] In the question \[\Delta n\]has +ve value so, AH>AE. \[\Delta H>\Delta E.\]


You need to login to perform this action.
You will be redirected in 3 sec spinner