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

`C_(6)H_(6("liq"))+7.5O_(2(g))rarr6CO_(2(g))+3H_(2)O_("liq")`
`Deltan_((g))=6-7.5=-15`
`DeltaH=DeltaE+Deltan_((g))RT`
`DeltaH=-3263.9kJ-(1.5xx8.314xx298)/(1000)kJ=-3267.6 kJ`