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. 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 13.50 kilograms of propane in sufficient oxygen?

Answer 1

first write the balance combution reaction

C3H8 (g) + 5O2 (g)-----> 3CO2 (g) + 4H2O (l)

first we calculate molar heat of combution of propane

ΔH of C3H8 = -103.8 KJ/mol

ΔH of O2 = 0.0 KJ/mol

ΔH of CO2 = -393.5 KJ/mol ---> 3x (-393.5 KJ/mol ) = -1180.5 KJ/mol

ΔH of H2O =  -285.8 KJ/mol ---> 4 x (-285.8 KJ/mol ) = -1143.2 KJ/mol

ΔH°(rxn) = Σ ΔH°f products - Σ ΔH°f reactants

ΔH°(rxn) = [-1143.2 + (-1180.5)] - [-103.8]

ΔH° (rxn) = -2219.9 kJ/mol

energy released from 1 mol combution of propane = 2219.9 KJ { - tive sign indicate heat released only }

now find for 13.5 Kg propane combution

1 mol of propane = 44.1 g so,

2219.9 KJ / 44.1 g = 50.3378 KJ/g

for 13.5 Kg propane = 50.3378 KJ/g x 13500 g = 679561.225 KJ