Question

A student needs to calculate the standard enthalpy change of combustion for propane using Hess's Law. They have the following information: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH° = -2220 kJ/mol 2H2(g) + O2(g) → 2H2O(l) ΔH° = -484 kJ/mol 3C(s) + 2H2(g) → C3H8(g) ΔH° = +51 kJ/mol What is the standard enthalpy change of combustion for propane?

Answer 1

To find the standard enthalpy change of combustion for propane using Hess's Law, we need to manipulate the given reactions to create the combustion reaction for propane. The combustion reaction for propane is:C3H8 g  + 5O2 g   3CO2 g  + 4H2O l We have the following reactions:1. C3H8 g  + 5O2 g   3CO2 g  + 4H2O l  H = -2220 kJ/mol2. 2H2 g  + O2 g   2H2O l  H = -484 kJ/mol3. 3C s  + 2H2 g   C3H8 g  H = +51 kJ/molFirst, we need to reverse reaction 3 to get C3H8 g  on the reactant side:3'. C3H8 g   3C s  + 2H2 g  H = -51 kJ/molNow, we need to multiply reaction 2 by 2 to get the correct amount of H2O l  in the product side:2'. 4H2 g  + 2O2 g   4H2O l  H = -968 kJ/molNow, we can add reactions 3' and 2' to get the combustion reaction for propane:C3H8 g  + 5O2 g   3CO2 g  + 4H2O l The standard enthalpy change of combustion for propane can be calculated by adding the H values of reactions 3' and 2':H_combustion = H 3'  + H 2' H_combustion = -51 kJ/mol +  -968 kJ/mol H_combustion = -1019 kJ/molTherefore, the standard enthalpy change of combustion for propane is -1019 kJ/mol.