Question

C₆H₆($\mathcal{l}$) + $\frac{15}{2}$ O₂(g) $\to$ 6CO₂(g) + 3H₂O($\mathcal{l}$) 
$∆$n_g = 6 - 7.5 = - 1.5 (change in gaseous mole)
$∆$U or $∆$E = - 3263.9 kJ
$∆$H = $∆$U + $∆$n_gRT
$∆$n_g = - 1.5
R = 8.314 JK^-1 mol^-1
T = 298 K
So $∆$H = - 3263.9 + (-1.5) 8.314 $\times$ 10^-3 $\times$ 298
= - 3267.6 kJ
$∆$H = Heat at constant pressure
$∆$U / $∆$E = Heat at constant volume
R = gas constant