Answer
🔥 Thermochemistry of Propene Combustion (C₃H₆)
Part (a): Enthalpy of Combustion
Given Data:
- ΔHf [C₃H₆(g)] = +53.3 kJ/mol
- ΔHf [CO₂(g)] = –393.5 kJ/mol
- ΔHf [H₂O(l)] = –286.0 kJ/mol
Balanced equation:
C₃H₆(g) + 4.5 O₂(g) → 3 CO₂(g) + 3 H₂O(l)
Apply Hess’s Law:
ΔHcomb = [3×(–393.5) + 3×(–286)] – [53.3] = –2091.8 kJ/mol
✅ Enthalpy of combustion of propene = –2091.8 kJ/mol
Part (b): Effect on O₂ Partial Pressure (Rigid Container)
Reaction:
C₃H₆(g) + 4.5 O₂(g) ⇌ 3 CO₂(g) + 3 H₂O(l)
- (i) Increase in CO₂ → shift left → ↑ O₂
- (ii) Increase in H₂O(l) → no effect (liquid not in Kp)
- (iii) Increase in temp → exothermic → shift left → ↑ O₂
- (iv) Add catalyst → no change in equilibrium → no effect on O₂
✅ O₂ partial pressure increases in (i) and (iii); no change in (ii) and (iv)
Part (c): Expression for Kp
Note: H₂O(l) is not included in the expression.
Kp = (PCO₂)³ / [PC₃H₆ × (PO₂)4.5]
✅ Kp = (PCO₂)³ / [PC₃H₆ × (PO₂)9/2]
Part (d): Determining Δn
Δn = moles of gaseous products – moles of gaseous reactants
Reactants = 1 mol C₃H₆ + 4.5 mol O₂ = 5.5 mol
Products = 3 mol CO₂ (H₂O is liquid and excluded)
Products = 3 mol CO₂ (H₂O is liquid and excluded)
Δn = 3 – 5.5 = –2.5
✅ Δn = –2.5