Question

Calculate the standard enthalpy changes for the combustion of a mole of methane and the combustion of a mole of methanol.

Answer 1

The equation for combustion of methane is written as

CH₄(g) + O₂(g) 2H₂O (l) + CO₂ (g)

The standard enthalpy of reaction is given by the equation

H = sum(enthalpy of formation of products) - sum( enthalpy of formation of reactants)

H = 2 x Δ_fH ^owater + Δ_fH ^oCO₂ - Δ_fH ^omethane - Δ_fH ^ooxygen

H = 2(-285.83) - 393.5 + 74.8 + 0 = -890.36 kJ/mol

this is the standard enthalpy changes for the combustion of a mole of methane

For methanol:

2CH₃OH(g) + 3O₂(g) 2CO₂(g) + 4H₂O (l)

CH₃OH(l) + 3/2O₂(g) CO₂(g) + 2H₂O (l)

H = sum(enthalpy of formation of products) - sum( enthalpy of formation of reactants)

H = 2 x Δ_fH ^owater + Δ_fH ^oCO₂ - Δ_fH ^omethanol - 3/2Δ_fH ^ooxygen

H = 2(-285.83) - 393.5 + 238.4 + 0 = -726.7 kJ/mol

this is the standard enthalpy changes for the combustion of a mole of methanol