Question

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
• B. $\Delta H =\Delta E$
• C. $\Delta H >\Delta E$
• D. $\Delta H ,\Delta E$ relation cannot to be predicted

Answer 1

Answer: (C)

$\Delta H >\Delta E$

Answer 2

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 $\Delta H\Delta E$ In the question $\Delta n$ has +ve value so, AH>AE. $\Delta H>\Delta E.$