The heat of combustion of benzene at 27°C  for the reaction  ${\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right)+7\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to 6{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ is 780 kcal/mole as determined by a bomb calorimeter. The heat evolved on burning 39g of benzene in an open vessel is

$$\begin{gathered} \mathrm{\Delta n}=-1.5\text{ and }\mathrm{\Delta H}=\mathrm{\Delta E}+\mathrm{\Delta nRT} \\ \therefore \mathrm{\Delta H}=-780+(-1.5)\left(2\times {10}^{-3}\right)\left(300\right) \\ =-780-\left(3\times 3\times {10}^{-1}\right) \end{gathered}$$

=-780.9 or 780.9 is the heat of combustion of 1 mol, i.e., 78 g of benzene. Thus, for 39 g, that is, half mole of benzene heat evolved is 390.45 kcal.