For the complete combustion of ethanol, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ the amount of heat produced as measured in bomb calorimeter is 1364.47 kJ ${\mathrm{mol}}^{-1}$ at ${25}^0\mathrm{C}.$ Assuming ideality, the enthalpy of combustion, $∆{\mathrm{H}}_{\mathrm{C}},$ for the reaction will be

$\left[\mathrm{R}=8.314 {\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1}\right]$

${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Amount of heat produced in bomb calorimeter,

$∆\mathrm{U}=-1364.47 \mathrm{kJ} {\mathrm{mol}}^{-1}$

Enthalpy of a combustion reaction is

$$\begin{gathered} ∆\mathrm{H}=∆\mathrm{U}+∆{\mathrm{n}}_{\mathrm{g}}\mathrm{RT} \\ \mathrm{where} ∆\mathrm{U}=\mathrm{internal} \mathrm{energy} \\ ∆{\mathrm{n}}_{\mathrm{g}} = \mathrm{moles} \mathrm{of} \mathrm{gas} \left(\mathrm{products} - \mathrm{reactants}\right) \\ \mathrm{R}=\mathrm{Gas} \mathrm{constant} \\ \mathrm{T}=\mathrm{Temperature} \mathrm{in} \mathrm{K} \\ \mathrm{As} \mathrm{per} \mathrm{equation}, \\ ∆{\mathrm{n}}_{\mathrm{g}} =2 - 3 = -1 \\ \mathrm{T} = {25}^0\mathrm{C}= 25+273 \mathrm{K} \\ \Rightarrow \mathrm{T}=298 \mathrm{K} \\ \mathrm{Thus}, ∆\mathrm{H}= -1364.47 + \left[\frac{\left(-1\right) \times 8.314 \times 298}{1000}\right] \\ ∆\mathrm{H}= -1364.47 - 2.477 = -1366.647 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

Hence, the enthalpy of combustion, $∆{\mathrm{H}}_{\mathrm{C}},$ for the given reaction will be $-1366.947 \mathrm{kJ} {\mathrm{mol}}^{-1}.$