Question

The enthalpy of combustion for octane (C8H18(l)), a key component of gasoline, is –5074 kJ/mol. The reaction equation is: C8H18(l) + 12.5O2(g) es001-1.jpg 8CO2(g) + 9H2O(g).
What is the es001-2.jpgHf for this reaction?
16CO2(g) + 18H2O(g) es001-3.jpg 2C8H18(l) + 25O2(g)

Answer 1

Use ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants). For the combustion

C8H18(l) + 12.5 O2(g) → 8 CO2(g) + 9 H2O(g), ΔHcomb = −5074 kJ mol−1,

and with ΔH°f(CO2,g) = −393.5 kJ mol−1, ΔH°f(H2O,g) = −241.8 kJ mol−1 (O2 = 0):

−5074 = [8(−393.5) + 9(−241.8)] − ΔH°f(C8H18)

So
ΔH°f(C8H18,l) = 8(−393.5) + 9(−241.8) − (−5074)
= −5324.2 + 5074
= −250.2 kJ mol−1 ≈ −250 kJ mol−1.

The shown reverse reaction (16 CO2 + 18 H2O → 2 C8H18 + 25 O2) is just the reverse of 2 combustions, so its ΔH = +2(5074) = +10 148 kJ.