Answer
$h_{c}= −1,559,850\text{ kJ}$
Work Step by Step
The stoichiometric equation for this reaction is
$$
\mathrm{C}_2 \mathrm{H}_6+3.5\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}(\ell)+13.16 \mathrm{~N}_2
$$ Both the reactants and the products are at the standard reference state of $25^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$. Also, $\mathrm{N}_2$ and $\mathrm{O}_2$ are stable elements, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of $\mathrm{C}_2 \mathrm{H}_6$ becomes $$
h_C=H_P-H_R=\sum N_P \bar{h}_{f, P}^{\circ}-\sum N_R \bar{h}_{f, R}^{\circ}=\left(N \bar{h}_f^{\circ}\right)_{\mathrm{CO}_2}+\left(N \bar{h}_f^{\circ}\right)_{\mathrm{H}_2 \mathrm{O}}-\left(N \bar{h}_f^{\circ}\right)_{\mathrm{C}_2 \mathrm{H}_{\mathrm{o}}}
$$ Using $\bar{h}_f^{\circ}$ values from Table A-26, $$
\begin{aligned}
h_C= & (2 \mathrm{kmol})(-393,520 \mathrm{~kJ} / \mathrm{kmol})+(3 \mathrm{kmol})(-285,830 \mathrm{~kJ} / \mathrm{kmol}) \\
& -(1 \mathrm{kmol})(-84,680 \mathrm{~kJ} / \mathrm{kmol}) \\
& =-1,559,850 \mathrm{~kJ}\left(\text { per } \mathrm{kmolC}_2 \mathrm{H}_6\right)
\end{aligned}
$$
