Question

Butane C4 H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Hf = –393.5 kJ/mol), and H2O(g) (Hf = –241.82) in the reaction: What is the enthalpy of combustion, per mole, of butane? Use

Answer 1

Following is the balanced reaction for combustion of butane,

2 C4H10 + 13 O2 → 8 CO2 + 10H2O

Given, Heat of formation of C4H10 = = -125.7 kJ/mol

Heat of formation of water = = -241.82 kJ/mol

Heat of formation of CO2 = = -393.5 kJ/mol

Thus, enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants

= 8 + 10 - 2

= (-393.5X8) + (-241.82X10) - (-125.7X2)

= -5314.8 kJ/mol

This energy would result in combustion of 2 moles of butane

∴ For 1 mole, energy required for combustion of butane = -5314.8/2 = -2657.4 kJ/mol

Answer 2

The combustion reaction, assuming complete combustion and water is gaseous, is
C4 H10 + 13/2 O2 ---> 4 CO2 + 5 H2O

The enthalpy of combustion is
4(–393.5) + 5(–241.82) - ( –125.7 + (13/2)(0)) = -2657.4 kJ/mol