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 standard combustion of propane produces liquid water: $C_3H_8$(g) + 5$O_2$(g) -> 3$CO_2$(g) + 4$H_2O$(l), δ H1 = -2220.1 kJ/mol. The vaporization of water is: $H_2O$(l) -> $H_2O$(g), δ $H_2$ = 44 kJ/mol. For the reaction producing gaseous water: $C_3H_8$(g) + 5$O_2$(g) -> 3$CO_2$(g) + 4$H_2O$(g), we need to add 4 times the vaporization reaction to the combustion reaction: δ H° = -2220.1 + 4 * 44 = -2220.1 + 176 = -2044.1 kJ.