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

Answer

D

Explanation

`C_(6)H_(6)(l) +(15)/(2)O_(2)(g) rarr6CO_(2)(g)+3H_(2)O(l)`
`Deltan_(g)= 6 - ( 15)/( 2) = - (3)/(2)`
`DeltaH = DeltaU+ Deltan_(g) RT`
`= - 3263.9 + ( - (3)/(2)) xx ( 8.314 xx 10^(-3)) xx ( 298)`
`= - 3263.9 + ( -3.71)= - 3267 .6 kJ mol^(-1)`