Question

Calculate the enthalpy of combustion of propane, C₃H₈(g), for the formation of H₂O(g) and CO₂(g). The enthalpy of formation of propane is -104 kJ/mol. What is the enthalpy of combustion of propane?

Answer 1

The enthalpy of combustion of propane is calculated using the standard enthalpies of formation for CO2, H2O, and propane, resulting in -2204.7 kJ/mol.

Step-by-step explanation:

The enthalpy of combustion of a substance is the heat released when one mole of the substance completely reacts with oxygen under standard conditions. To calculate the enthalpy of combustion of propane, C3H8(g), for the formation of H2O(g) and CO2(g), we use the standard enthalpies of formation of the products and reactants involved in the reaction.

The balanced equation for the complete combustion of propane is:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

The enthalpy change for this reaction can be calculated using the formula:

ΔHcombustion = [ΣΔHf(products)] - [ΣΔHf(reactants)]

The standard enthalpies of formation (ΔHf) are as follows:

• ΔHf of CO2(g) = -393.5 kJ/mol
• ΔHf of H2O(g) = -241.8 kJ/mol (not given in problem but can be found in thermodynamics tables)
• ΔHf of C3H8(g) = -104 kJ/mol (given)

We can now plug in the values into the formula:

ΔHcombustion = [3(-393.5 kJ/mol) + 4(-241.8 kJ/mol)] - [-104 kJ/mol]

ΔHcombustion = -2204.7 kJ/mol

Therefore, the enthalpy of combustion of propane is -2204.7 kJ/mol.

Please note that the enthalpy of formation of water in the gaseous state was not provided in the problem, but is readily available in standard thermodynamic tables.