Question

Butane C4 H10 (g),(mc016-1.jpgHf = –125.7), combusts in the presence of oxygen to form CO2 (g) (mc016-2.jpgHf = –393.5 kJ/mol), and H2 O(g) (mc016-3.jpgHf = –241.82) in the reaction:

mc016-4.jpg

What is the enthalpy of combustion, per mole, of butane?

Answer 1

Following is the balanced reaction for combustion of butane,

2 C4H10 + 13 O2 → 8 CO2 + 10H2O

Given, Heat of formation of C4H10 =
`Hf_(C4H10)` = -125.7 kJ/mol
Heat of formation of water =
`Hf_(H2O)` = -241.82 kJ/mol
Heat of formation of CO2 =
`Hf_(CO2)` = -393.5 kJ/mol

Thus, enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants
= 8
`Hf_(H2O)` + 10
`Hf_(CO2)` - 2
`Hf_(C4H10)`
= (-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