Calculate the enthalpy for the combustion of propane shown using the given bond dissociation energies:
C₃H₈ + 5 O₂ → 3 CO₂ + 4H₂O
Number and types of bonds in each molecule:
C₃H₈ (2 C-C,8 C-H bonds) ;O₂ (1 O=O) ; (2 C=O bonds) ; (2 O-H bonds) .
Bond dissociation energies:
C-C (347 kJ/mol) ;C-H (413 kJ/mol) ;O=O (498 kJ/mol;C=O (799 kJ/mol) ;O-H (467 kJ/mol)

Question

Quizplus · Accepted Answer

First, we need to calculate the total energy required to break all the bonds in the reactants:
2 C-C x 347 kJ/mol = 694 kJ/mol
8 C-H x 413 kJ/mol = 3304 kJ/mol
1 O=O x 498 kJ/mol = 498 kJ/mol
2 C=O x 799 kJ/mol = 1598 kJ/mol
2 O-H x 467 kJ/mol = 934 kJ/mol
Total bond dissociation energy = 7028 kJ/mol

Next, we need to calculate the total energy required to form all the bonds in the products:
3 C=O x 799 kJ/mol = 2397 kJ/mol
4 H-O x 467 kJ/mol = 1868 kJ/mol
Total bond formation energy = 4265 kJ/mol

Finally, we need to subtract the bond formation energy from the bond dissociation energy to get the overall enthalpy change for the reaction:
Enthalpy change = (bond formation energy) - (bond dissociation energy)
Enthalpy change = 4265 kJ/mol - 7028 kJ/mol
Enthalpy change = -2763 kJ/mol

Since the enthalpy change is negative, the reaction is exothermic, meaning it releases heat. The closest answer choice is C, which is -2042 kJ, and since -2042 kJ is also negative, it is the best choice.

Calculate the Enthalpy for the Combustion of Propane Shown Using