Answer to Question #90237 in General Chemistry for daseem

Question

How does heat combustion of paraffin compare to heat combustion of propane?

Answer 1

Both paraffin and propane are saturated hydrocarbons with a general formula C_nH_2n+2. Let's estimate first heat combustion of 1 mole each hydrocarbon assuming n=25 for averaged paraffin. General equation of combustion is following:

$C_nH_{2n+2}(g) +\frac{3n+1}{2}O_2(g) \rightarrow nCO_2(g) + (n+1)H_2O(g)$

Heat of combustion can be estimated as a difference of bond energies of products formed and reactants (bond energies are tabulated values: E_C=O=192.0 kcal/mol, E_H-O=110.6 kcal/mol, E_C-C=82.6 kcal/mol, E_C-H=98.7 kcal/mol, E_O=O=118.9 kcal/mol).

For propane (n=3):

$Q(propane \ per \ mole)=6E_{C=O} + 8E_{H-O} - 2E_{C-C}-8E_{C-H}-5E_{O=O} = 487.5 \frac{kcal}{mol}$

For paraffin (n=25):

$Q(paraffin \ per \ mole)=50E_{C=O} + 52E_{H-O} - 24E_{C-C}-52E_{C-H}-38E_{O=O} = 3718.2 \frac{kcal}{mol}$

Now heat of combustion per gram can be calculated:

$Q(propane \ per \ g) = \frac{Q(propane \ per \ mole)}{M(propane)} = \frac{487.5 \frac{kcal}{mol}}{44\frac{g}{mol}}=11.1 \frac{kcal}{g}$

$Q(paraffin \ per \ g) = \frac{Q(paraffin \ per \ mole)}{M(paraffin)} = \frac{3718.2 \frac{kcal}{mol}}{352\frac{g}{mol}}=10.6 \frac{kcal}{g}$

Thus 1 g of propane produces a bit more heat than the same amount of gaseous paraffin. As paraffins are usually liquids or solids, extra energy should be spent to transfer paraffin to gaseous phase that makes heat of combustion of condensed paraffin even less.