The combustion of benzene(/) gives `CO_2(g) and H_2O(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. 3260
B. `-3 2 6 7 . 6`
C. `4152.6`
D. `-4 5 2 . 4 6`

Answer

B

Explanation

`C_(6)H_(6)(l) + 15/2 O_(2)(g) rarr 6CO_(2)(g) + 3H_(2)O(l)`
`Delta _(g) = -3/2`
`Delta H = Delta U + Deltan_((g)) RT = -3263.9 -(1.5 xx 8.314 xx 298)/(1000) = -3267.6 kJ//mol`.