The standard heat of combustion of propane is –2220.1 kJ/mol. The standard heat of vaporization of l — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

The combustion reaction to standard state products is $\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \to 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}$ with $\Delta H = -2220.1\text{ kJ}$. Vaporization of 4 moles of water requires $4 \times 44 = 176\text{ kJ}$. Total $\Delta H = -2220.1 + 176 = -2044.1\text{ kJ}$.