Question

At 25°C, the standard enthalpy of combustion of gaseous propane (C3H8) is –2219.0 kJ per mole of propane, and the standard enthalpy of combustion of gaseous propylene (C3H6) is –2058.3 kJ per mole of propylene. What is the standard enthalpy change for the following reaction at 25°C? C3H6(g) + H2(g) → C3H8(g) Substance ∆H°f (kJ/mol) CO2(g) –393.5 H2O(l) –285.8 a) +160.7 kJ b) –160.7 kJ c) +104.7 kJ d) –20.4 kJ e) –125.1 kJ

Answer 1

Write the equation for combustion of propane:

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

Enthalpy of combustion, ΔH⁰ = Σn.ΔH_f⁰ (products) – Σn.ΔH_f⁰(reactants) (n denotes number of moles of species and ΔH_f⁰ is the standard enthalpy of formation at 25˚C)

(-2219.0 kJ = [(3 mol CO₂)*ΔH_f⁰(CO₂, g) + (4 mol H₂O)*ΔH_f⁰(H₂O, l)] – [(1 mol C₃H₈)*ΔH_f⁰(C₃H₈, g)] (enthalpy of formation of diatomic gases like H₂ and O₂ is 0).

===> -2219.0 kJ= [(3 mol)*(-393.5 kJ/mol) + ( 4 mol)*(-285.8)] – ΔH_f⁰ (C₃H₈)

===> -2219.0 kJ = -2323.7 kJ – ΔH_f⁰ (C₃H₈)

===> ΔH_f⁰ (C₃H₈) = (-2323.7 + 2219.0) kJ = -104.7 kJ

Since we started with 1 mole of propane, the enthalpy of formation of propane is -104.7 kJ/mol.

Write the balanced equation for combustion of propylene:

C₃H₆ (g) + 9/2 O₂ ----> 3 CO₂ (g) + 3 H₂O (l)

Enthalpy of combustion, ΔH⁰ = Σn.ΔH_f⁰ (products) – Σn.ΔH_f⁰(reactants)

====> -2058.3 kJ = [(3 mol)*(-393.5 kJ/mol) + (3 mol)*(-285.8 kJ/mol)] – ΔH_f⁰ (C₃H₆)

====> ΔH_f⁰ (C₃H₆) = -2037.9 kJ + 2058.3 kJ = 20.4 kJ

As before, since we started with 1 mole of C₃H₆, the enthalpy of formation of propylene is 20.4 kJ/mol.

Now consider our required reaction:

C₃H₆ (g) + H₂ -----> C₃H₈ (g)

ΔH⁰ = Σn.ΔH_f⁰ (products) – Σn.ΔH_f⁰(reactants) = [(1 mol)*(-104.7 kJ/mol)] – [(1 mol)*(20.4 kJ/mol)] = -104.7 kJ – 20.4 kJ = -125.1 kJ (ans)

Ans: (e) -125.1 kJ