Question

Propane has the structure H₃C–CH₂ –CH₃ . Calculate the change in enthalpy for the reaction:

C₃H₈ (g) + 5 O₂ (g) → 3 CO₂ (g) + 4 H₂O (g)

Given that average bond enthalpies are:

C–C C–H C = O O = O O–H

347 414 741 498 464 kJ mol^–1

Answer 1

ΔH = (2e_C–C + 8e_C–H + 5e_O=O) – (6e_C=O + 8e_O–H) 

= 2 (347) + 8 (414) + 5 (498) – 6 (741) – 8 (464)

= – 1662 kJ mol^–1