Answer:
[tex]\large \boxed{\text{-2043.96 kJ/mol}}[/tex]
Explanation:
Assume the reaction is the combustion of propane.
Word equation: propane plus oxygen produces carbon dioxide and water
Chemical eqn: C₃H₈(g) + O₂(g) ⟶ CO₂(g) + H₂O(g)
Balanced eqn: C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)
(a) Table of enthalpies of formation of reactants and products
[tex]\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}[/tex]
(b)Total enthalpies of reactants and products
[tex]\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\= \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}[/tex]
ΔᵣH° is negative, so the reaction is exothermic.
