Question

The combustion of propane is represented by the following chemical equation. C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(l) The standard enthalpies of formation ( ΔH∘f ) for C3H8(g), CO2(g), and H2O(l) are −103.8 kJ/mol, −393.5 kJ/mol, and −285.8 kJ/mol respectively. What is the enthalpy of combustion for propane at 25 °C and 1 atm?

Answer 1

The enthalpy of combustion for propane at 25 °C and 1 atm is -2418.7 kJ/mol.

Step-by-step explanation:

The enthalpy of combustion for propane at 25 °C and 1 atm can be calculated using the standard enthalpies of formation of the reactants and products. The enthalpy change for the combustion reaction is equal to the sum of the enthalpies of the formation of the products minus the sum of the enthalpies of the formation of the reactants. Given the standard enthalpies of formation for C3H8(g), CO2(g), and H2O(l) as -103.8 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol respectively, we can substitute these values into the equation:

ΔH = (3 × -393.5 kJ/mol) + (4 × -285.8 kJ/mol) - (-103.8 kJ/mol)

Simplifying the equation gives:

ΔH = -2418.7 kJ/mol

Therefore, the enthalpy of combustion for propane at 25 °C and 1 atm is -2418.7 kJ/mol.

Answer 2

ΔH°c = -2219.9 kJ

Step-by-step explanation:

Let's consider the combustion of propane.

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)

We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.

ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]

ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]

ΔH°c = -2219.9 kJ