1. Tardigrade
  2. Question
  3. Chemistry
  4. In the complete combustion of butanol textC text4 textH text9 textOH(I) if text ! !Δ ! ! text H is enthalpy of combustion and text ! !Δ ! ! text E is the heat of combustion at constant volume, then:

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Q. 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:

EAMCETEAMCET 1997

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. $