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:

$1mol C_{4}H_{10}*( \frac{-5315kJ}{2mol C_{4}H_{10} })=-2657.5 \frac{kJ}{molC_{4}H_{10}}$

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

Correct answer: -2657.5 kJ/mol