The combustion of benzene (l) gives `CO_(2)(g)` and `H_(2)O(l)`. Given that heat of combustion of benzene at constant volume is `–3263.9 kJ mol^(–1)` at `25^(@)C`, heat of combustion (in kJ` mol^(–1)`) of benzene at constant pressure will be
(R = 8.314 JK–1 mol–1)

Options

A. `-452.46`
B. `3260`
C. `-3267.6`
D. `4152.6`

Answer

C

Explanation

Benzene undergoes combustion as:
`C_(6)H_(6)(l)+(15)/(2)O_(2)(g)to6CO_(2)(g)+3H_(2)O(l)`
`DeltaU^(@)=` Heat of combustion at constant volume ltbr `=-3263.9kJ//mol`
`Deltan_(g)=6-(15)/(2)=-(3)/(2)`
Heat of combustion at constant pressure i.e., `DeltaH^(@)` may be calculated as,
`DeltaH^(@)=DeltaU^(@)+Deltan_(g)RT`
`=-3263.9+(-(3)/(2))xx8.314xx10^(-3)xx298`
`=-3267.6kJ//mol`