Question

What is the enthalpy of formation of butane (C₄H₁₀) if the enthalpy of combustion for butane is -2876.9 kJ/mol (water is produced as a liquid)? Use the appendix in your textbook, do not enter units, and answer with 3 significant digits.

Answer 1

Combustion of butane is expressed as

C4H10 + 6.5 O2(g) = 4CO2(g) + 5 H2O(l)

The heat of combustion for one mole of butane is -2876.9 kJ/mol

Heats of formation from literature are
CO₂ = -393.5 kJ/mole,
H₂O(l) = -285.8 kJ/mole              (But, H2O(g) = -241.8 kJ/mole)

Heat of combustion of butane = sum heats of formation products minus heats of formation reactants.
or, -2877 Kj = [4 (-393.5 kJ) + 5 (-285.8 kJ)] minus heat or enthalpy of formation butane.
(since, the enthalpy formation oxygen an element is assigned a value of zero)

-2876.9 kJ = -1574 kJ + (-1429 kJ) - (heat or enthalpy of formation butane).
-2876.9 kJ = -3003 kJ - (heat or enthalpy of formation butane)
heat or enthalpy of formation butane = - 3003 kJ + 2876.9 kJ
heat or enthalpy of formation butane = -126.1 kJ
​ = - 126 kJ (upto 3 sig. fig.)