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)).

The combustion of benzene (l) gives CO_(2)(g) and H_(2)O(l).Given that

1.	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)).
Answer» 4152.6 `-452.46` 3260 `-3267.6` SOLUTION :`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`.