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

Answer:

ΔH°c = -2219.9 kJ

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