The heat of combusion of benzene in a bomb calorimeter (i.e constant volume) was found to be `3263.9kJ mo1^(-1)` at `25^(@)C` Calculate the heat of combustion of benzene at constan pressure .

Explanation

The reaction in
`C_(6)H_(6)(l)+7(1)/(2)O_(2)(g)rarr 6CO_(2)(g) +3H_(2)O(l)`
In this reaction, `O_(2)` is the only gaseous reactant and `CO_(2)` is the only gaseous product.
`:. Deltan_(g) =n_(p) - n_(r) =6 -7 (1)/(2) =-1(1)/(2) =-3//2`
Also, we are given
`DeltaU = (or q_(V)) =- 3163.9 kJ mol^(-1)`
`T = 25^(@)C = 298K,R = 8.314 J K^(-1) mol^(-1)`
or `R = 8.313//1000 kJ K^(-1) mol^(-1)`
`DeltaH (orq_(p)) = DeltaU +Deltan_(g) RT`
`=- 3263.9 + (-3//2) (8.314//1000) (298)` ltbRgt `=- 3263.9 - 3.7 =- 3267.6 kJ mol^(-1)`