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) to 6CO_(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 - 3.716`
` = - 3267.678`.