Octane (C8H18) undergoes combustion according to the following thermochemical equation:2C8H18 (l) + 25O2 (g) → 16CO2 (g) + 18H2O (l) ΔH°rxn = - 1.0940 * 104 kJ/mol What is the standard enthalpy of formation of liquid octane? ΔH°f (CO2 (g)) = - 393.5 kJ/mol and ΔH°f (H2O (l)) = - 285.8 kJ/mol

-250.40 kJ = ΔH°f (C8H18) (l)

Explanation:

The enthalpy change for a given chemical reaction is given by the sum of the standard heats of formation of products multiplied by their respective coefficients in the balanced equation minus the sum of the standard heat of formation of reactants again multiplied by their coefficients.

For this particular reaction,

2C8H18 (l) + 25O2 (g) → 16CO2 (g) + 18H2O (l)

the ΔH°rxn will be given by

ΔH°rxn = 16ΔH°f (CO2 (g)) + 18ΔH°f (H2O (l)) - (2ΔH°f (C8H18) (l) + 25 ΔH°fO2 (g))

Notice we are given the information for the ΔH°rxn and all the ΔH°fs except for O2 which is zero and the ΔH° (C8H18) (l) which is what we are looking for.

So plugging our values and solving:

-1.0940 * 10⁴ kJ = 16mol x (-393.5 kJ/mol) + 18mol x (-285.8 kJ/mol) - 2 (ΔH°f (C8H18) (l)

-1.094 x10⁴ kJ = - 6296 kJ + (-5144.40 kJ) - 2 (ΔH°f (C8H18) (l)

-500.40 kJ/2 = ΔH°f (C8H18) (l)

-250.40 kJ = ΔH°f (C8H18) (l)