The heat of combustion of benzene at $27°\mathrm{C}$ found by bomb calorimeter i.e., for the Reaction ${\mathrm{C}}_6{\mathrm{H}}_{6 \mathrm{m}}+7\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{a}\right)}⟶6{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}$ is $-780\mathrm{K}\mathrm{.}\mathrm{C}\mathrm{a}\mathrm{l}{\mathrm{mol}}^{-1}$. The heat evolved on burning $39\mathrm{ }\mathrm{g}$ of benzene in an open vessel will be

 Thermodynamic properties - the heat of combustion Detailed solution:

 $\Delta H=\Delta E+\Delta n_{g}RT;$ also written as $\Delta E=\Delta H-\Delta n_{g}RT$

 $\Delta n_g=$ no. of moles of gas products - no. of moles of gas reactants

 $\Delta n_g=6-7.5=-1.5;\mathrm{T}=27°\mathrm{C}=27+273=300 \mathrm{K},\Delta \mathrm{H}=-780 \mathrm{K} \mathrm{cal}/\mathrm{mol}$

 By substituting all the values

 $\Delta E=-780-1.5\times 2\times 300\times {10}^{-3}=-780.9\mathrm{Kcal}/\mathrm{mole}$

 For $39\mathrm{ }\mathrm{g}$ of benzene no. of moles $\frac{39}{78}=\frac{1}{2}=0.5$

 $\Delta H=-\frac{780.9}{2}=-390.45\mathrm{Kcal}$

Hence the correct option is (C).