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=8.314J*K^(-1)*mol^(-1))`-

Options

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

Answer

D

Explanation

`C_(6)H_(6)(l)+(15)/(2)O_(2)(g)to6CO_(2)(g)+3H_(2)O(l)`
`Deltan=6-(15)?(2)=-(3)/(2)`
`DeltaH=DeltaU+DeltanRT`
`DeltaH=[-3263.9-(3)/(2)xx8.314xx10^(-3)xx298]kJ*mol^(-1)`
`=-3267.6kJ*mol^(-1)`.