The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are: -2220.0 kJ mol⁻¹, -393.5 kJ mol⁻¹ and -285.8 kJ mol⁻¹ respectively. The magnitude enthalpy of formation of propane (C₃H₈) is........ kJ mol⁻¹. (Nearest integer)
Answer: 104
Solution
The enthalpy of formation of propane is calculated using Hess's law. The formation reaction is:
3C(graphite) + 4H₂(g) → C₃H₈(g)
Using the combustion reactions:
Applying Hess's law:
ΔH°f(C₃H₈) = [3 × ΔH°c(C) + 4 × ΔH°c(H₂)] - ΔH°c(C₃H₈)
= [3(-393.5) + 4(-285.8)] - (-2220.0)
= [-1180.5 - 1143.2] + 2220.0
= -2323.7 + 2220.0
= -103.7 kJ/mol
Magnitude = 104 kJ/mol (nearest integer)
3C(graphite) + 4H₂(g) → C₃H₈(g)
Using the combustion reactions:
- C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), ΔH°c = -2220.0 kJ/mol
- C(graphite) + O₂(g) → CO₂(g), ΔH°c = -393.5 kJ/mol
- H₂(g) + ½O₂(g) → H₂O(l), ΔH°c = -285.8 kJ/mol
Applying Hess's law:
ΔH°f(C₃H₈) = [3 × ΔH°c(C) + 4 × ΔH°c(H₂)] - ΔH°c(C₃H₈)
= [3(-393.5) + 4(-285.8)] - (-2220.0)
= [-1180.5 - 1143.2] + 2220.0
= -2323.7 + 2220.0
= -103.7 kJ/mol
Magnitude = 104 kJ/mol (nearest integer)
JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Integer answer