SolutionsCombustion
1. Balance the combustions
Balance the complete combustion reaction of:
- Propane, C3H8
- Octane, C8H18
- Cyclohexane, C6H12
- C3H8 + 5 O2 → 3 CO2 + 4 H2O
- C8H18 + 25/2 O2 → 8 CO2 + 9 H2O (or 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O)
- C6H12 + 9 O2 → 6 CO2 + 6 H2O
2. Which isomer is more stable?
The heats of combustion of butane and 2-methylpropane are −687.4 and −685.4 kcal/mol. Which of the two is more stable? By how much?
2-Methylpropane, by 2 kcal/mol: both give the same products (4 CO2 + 5 H2O), and the one that releases less heat started from a lower energy content.
Branched alkanes are usually slightly more stable than their linear isomers.
3. Heat of formation
Calculate the heat of formation of butane from its heat of combustion (−687.4 kcal/mol, with liquid water) and these data: C(graphite) + O2 → CO2, −94.1; H2 + ½ O2 → H2O(g), −57.8; H2O(g) → H2O(l), −10.52 kcal/mol.
Combustion of the elements that make up butane (4 C + 5 H2) to 4 CO2 + 5 H2O(l): 4 × (−94.1) + 5 × (−57.8) + 5 × (−10.52) = −376.4 − 289.0 − 52.6 = −718.0 kcal/mol.
ΔH°f = −718.0 − (−687.4) = −30.6 kcal/mol.
The heat of formation is negative: butane is more stable than its separate elements.
4. Energy per carbon
If the combustion of an alkane releases on average about 165 kcal/mol per carbon, estimate the heat released by one mole of octane.
8 × 165 ≈ 1320 kcal/mol. The experimental value is about 1300 kcal/mol.