Question

Butane (C4 H10(g), mc031-1.jpgHf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , mc031-2.jpgHf = –393.5 kJ/mol ) and water (H2 O, mc031-3.jpgHf = –241.82 kJ/mol) according to the equation below. mc031-4.jpg What is the enthalpy of combustion (per mole) of C4H10 (g)? Use mc031-5.jpg. –2,657.5 kJ/mol –5315.0 kJ/mol –509.7 kJ/mol –254.8 kJ/mol

Answer 1

The balanced chemical equation for the combustion of butane is:

`2C_(4)H_(10)(g) +13 O_(2)(g)-->8CO_(2)(g)+10H_(2)O(g)`

Δ
`H_(reaction)^(0)` = Σ
`n_(products)`Δ
`H_(f)^(0)_((products))`-Σ
`n_(reactants)`Δ
`H_(f)^(0)_((reactants))`

=
`[{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]`=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]

= -5315 kJ/mol

Calculating the enthalpy of combustion per mole of butane:

`1mol C_(4)H_(10)*( (-5315kJ)/(2mol C_(4)H_(10) ))=-2657.5 (kJ)/(molC_(4)H_(10))`

Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol

Correct answer: -2657.5 kJ/mol