Question

The enthalpy change for a chemical reaction is the sum of the energy consumed in breaking bonds and the energy released during bond formation. One way to determine the overall energy change for a chemical reaction is to apply Hess’s law to add together a group of reactions which can be arranged such that the chemical equations, when combined, give the overall equation we are trying to characterize.

part 1

Write a balanced chemical equation for the combustion of gaseous propane in gaseous oxygen to produce gaseous carbon dioxide and liquid water.

part 2

Combine the following equations to determine the enthalpy change for the combustion of 1 mole of propane. Assume that solid carbon is graphite.

Part 3

The average propane cylinder for a residential grill holds approximately 18 kg of propane. How much energy (in kJ) is released by the combustion of 18.50 kilograms of propane in sufficient oxygen?

Answer 1

1.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

2.
("C(s)" is graphite throughout what follows.)
You need "C3H8(g)" on the left. The only given equation that mentions C3H8 is the first one. So write the first given equation backwards:
C3H8(g) → 3 C(s) + 4 H2(g), ΔH = +103.8 kJ/mol
You also need "3 CO2(g)" on the right. The only given equation that mentions CO2 is the second one. So multiply the second given equation by 3:
3 C(s) + 3 O2(g) → 3 CO2(g), ΔH = -1180.5 kJ/mol
You also need "4 H2O(g)" on the right. The only given equation that mentions H2O is the third one. So multiply the third given equation by 4:
4 H2(g) + 2 O2(g) → 4 H2O(g), ΔH = -1143.2 kJ/mol
Add the last three equations here:
C3H8(g) + 3 C(s) + 3 O2(g) + 4 H2(g) + 2 O2(g) →
3 C(s) + 4 H2(g) + 3 CO2(g) + 4 H2O(g), ΔH = +103.8 kJ/mol -1180.5 kJ/mol -1143.2 kJ/mol
Cancel like amounts on opposite sides of the arrow, combine O2 on the left, and do the arithmetic for ΔH:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g), ΔH = -2219.9 kJ/mol

3.
(15000 g C3H8) / (44.0956 g C3H8/mol) x (2219.9 kJ/mol) = 755143 = 7.551 x 10^5 kJ