Combustion of butane is expressed as C4H10 + 6.5 O2(g) = 4CO2(g) + 5 H2O(l) The
ID: 699937 • Letter: C
Question
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.)
Explanation / Answer
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.