Calculate the enthalpy of combustion

Question

Calculate the enthalpy of combustion (kJ/mol) of cetane (C₁₆H₃₄), an analogous component in diesel fuel to octane in gasoline, given the following standard enthalpies of formation: CO₂(g) = -393.5 kJ/mol H₂O(l) = -285.8 kJ/mol C₁₆H₃₄(s) = -458.3 kJ/mol

Answer 1

Hi Josh W.!

To calculate the enthalpy of reaction given the values for the enthalpies of formation, we can simply apply Hess's Law. However, we must first balance our equation:

Δ

C₁₆H_34(s) + (49/2)O_2(g) -------> 16CO_2(g) + 17H₂O_(l)

Now that we know the stoichiometric coefficients, we can simply apply them to Hess's Law:

ΣH°_rxn = ΣH°_f Products - ΣH°_f Reactants

ΣH°_rxn = [16ΔH°_fCO_2(g) + 17ΔH°_f H₂O_(l)] - [ΔH°_f C₁₆H_34(s) + (49/2)ΔH°_f O₂]

=> ΔH°_f O₂ = 0 kJ*mol^-1 because O₂ is an element in most stable form

ΣH°_rxn = [16ΔH°_f CO_2(g) + 17ΔH°_f H₂O_(l)] - [ΔH°_f C₁₆H_34(s)]

<=> [(16)(-393.5 kJ*mol^-1) + (17)(-285.8 kJ*mol^-1)] - [-458.3 kJ*mol^-1]

ΣH°_rxn = -10,696.3 kJ*mol^-1

Hope this helps!

Cheers