Using Hess's Law, calculate the heat of formation of propane ( extC3 extH8) given the following reactions and their enthalpy changes: ... [list of reactions]

1. $\text{C(s)} + \text{O}_2\text{(g)} \longrightarrow \text{CO}_2\text{(g)}$; $\Delta H_1 = -393 \text{ kJ/mol}$
2. $\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \longrightarrow \text{H}_2\text{O(l)}$; $\Delta H_2 = -286 \text{ kJ/mol}$
3. $\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \longrightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}$; $\Delta H_3 = -2209 \text{ kJ/mol}$

• To get $3\text{C(s)}$ on the reactant side, multiply Reaction 1 by 3: $$3\text{C(s)} + 3\text{O}_2\text{(g)} \longrightarrow 3\text{CO}_2\text{(g)}; \quad \Delta H_{1'} = 3 \times (-393 \text{ kJ/mol}) = -1179 \text{ kJ/mol}$$
• To get $4\text{H}_2\text{(g)}$ on the reactant side, multiply Reaction 2 by 4: $$4\text{H}_2\text{(g)} + 2\text{O}_2\text{(g)} \longrightarrow 4\text{H}_2\text{O(l)}; \quad \Delta H_{2'} = 4 \times (-286 \text{ kJ/mol}) = -1144 \text{ kJ/mol}$$
• To get $\text{C}_3\text{H}_8\text{(g)}$ on the product side, reverse Reaction 3. This changes the sign of its enthalpy change: $$3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)} \longrightarrow \text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)}; \quad \Delta H_{3'} = -(-2209 \text{ kJ/mol}) = +2209 \text{ kJ/mol}$$

Adding the three manipulated reactions:
\begin{align*} 3C(s) + 3O_2(g) &\longrightarrow 3CO_2(g) \ 4H_2(g) + 2O_2(g) &\longrightarrow 4H_2O(l) \ 3CO_2(g) + 4H_2O(l) &\longrightarrow C_3H_8(g) + 5O_2(g) \ \hline 3C(s) + 4H_2(g) + (3+2)O_2(g) + 3CO_2(g) + 4H_2O(l) &\longrightarrow 3CO_2(g) + 4H_2O(l) + C_3H_8(g) + 5O_2(g) \end{align*}
Cancel out common species ($5\text{O}_2\text{(g)}$, $3\text{CO}_2\text{(g)}$, $4\text{H}_2\text{O(l)}$) from both sides to get the target equation:

The heat of formation of propane is $\text{-114 kJ/mol}$.