Question

Propane (C3 H8 (g), Hf = –103.8 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O, Hf = –241.82 kJ/mol) according to the equation below. What is the enthalpy of combustion (per mole) of C3 H8 (g)? Use . –2,044.0 kJ/mol –531.5 kJ/mol 531.5 kJ/mol 2,044.0 kJ/mol

Answer 1

Given:

Enthalpy of formation data

Hf(C3H8) = -103.8 KJ/Mol

Hf(CO2) = -393.5 kJ/mol

Hf (H2O) = -241.82 kJ/mol

To determine:

Enthalpy of combustion of C3H8

Explanation:

The combustion reaction is:

C3H8 + 5O2 → 3CO2 + 4H2O

ΔHcombustion = ∑nHf(products) - ∑nHf(reactants)

ΔHc = [3(Hf CO2) + 4(Hf H2O)] - [Hf(C3H8) + 5Hf(O2)]

       = [3(-393.5) + 4(-241.82)] -[-103.8 + 5(0)] = -2044 kJ/mol

Ans: (a) -2044 kJ/mol

Answer 2

Answer:

A) –2,044.0

Explanation

Its right on edge