The heat of combustion of benzene determined in a bomb calorimeter is $-870\mathrm{kcal}/\mathrm{mol}$ at $298\mathrm{ }\mathrm{K}$. The value of $\mathrm{\Delta E}$ for the reaction is

In the bomb calorimeter $\text{ΔH}$ is determined by the change in internal energy.

Here, $\mathrm{\Delta H}=870\mathrm{kcal}/\mathrm{mol}$

According to bomb calorimetry,

$\text{ΔH=ΔU}$

Thus,

$\mathrm{\Delta H}=-870\mathrm{kcal}/\mathrm{mol}$

Hence, option (C) is correct.