Using $Δ_{f} H^{\circ} / kJ mo l^{- 1}$ : $C X_{3} H X_{8} (g) = - 103.8$ , $CO X_{2} (g) = - 393.5$ , $H X_{2} O (l) = - 285.8$ , calculate the standard enthalpy of combustion of propane, $C X_{3} H X_{8} (g) + 5 O X_{2} (g) 3 CO X_{2} (g) + 4 H X_{2} O (l)$ , in $kJ mo l^{- 1}$ (nearest integer).

Numerical answer type, answer in kJ/mol, tolerance 0.5.

Answer: -2219.9 kJ/mol

Δ_{c} H^{\circ} = \sum Δ_{f} H^{\circ} (products) - \sum Δ_{f} H^{\circ} (reactants) = 3 (- 393.5) + 4 (- 285.8) - (- 103.8) = - 2219.9 kJ mo l^{- 1}

(oxygen is an element, so its term is zero).