Question

The combustion of propane, C₃H₈, occurs via the reaction C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g) with heat of formation values given by the following table: Substance ΔH°f (kJ/mol) C₃H₈(g) -104.7 CO₂(g) -393.5 H₂O(g) -241.8. Calculate the enthalpy for the combustion of 1 mole of propane.

Answer 1

The enthalpy of combustion of 1 mole of propane is calculated using the enthalpies of formation, resulting in -2044.1 kJ/mol.

Step-by-step explanation:

The enthalpy of combustion of propane can be calculated using the standard heat of formation values for the reactants and products in the balanced combustion reaction for propane: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g). The enthalpy change for the reaction (ΔH°ₓ) can be found with the equation:

ΔH°ₓ = ∑ΔH°ₓ(products) - ∑ΔH°ₓ(reactants)

Using the provided enthalpies of formation:

• ΔH°ₓf for C₃H₈(g) = -104.7 kJ/mol
• ΔH°ₓf for CO₂(g) = -393.5 kJ/mol
• ΔH°ₓf for H₂O(g) = -241.8 kJ/mol

The enthalpy for the combustion of 1 mole of propane is calculated as:

ΔH°ₓ = [3(-393.5) + 4(-241.8)] - [-104.7] = -2044.1 kJ/mol

Therefore, the enthalpy of combustion of 1 mole of propane is -2044.1 kJ/mol.