Enter your answer in the provided box. Methanol (CH3OH) is an organic solvent and is also used as a fuel in some automobile engines. From the following data, calculate the standard enthalpy of formation of methanol. 2CH3OH (l) + 3O2 (g) → 2CO2 (g) + 4H2O (l) ΔH o rxn = - 1452.8 kJ/mol

-477.4 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 CH₃OH (l) + 3 O₂ (g) → 2 CO₂ (g) + 4 H₂O (l) ΔH°rxn = - 1452.8 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) using the following expression.

ΔH°rxn = 2 mol * ΔH°f (CO₂ (g)) + 4 mol * ΔH°f (H₂O (l)) - 2 mol * ΔH°f (CH₃OH (l)) - 3 mol * ΔH°f (O₂ (g))

2 mol * ΔH°f (CH₃OH (l)) = 2 mol * ΔH°f (CO₂ (g)) + 4 mol * ΔH°f (H₂O (l)) - ΔH°rxn - 3 mol * ΔH°f (O₂ (g))

2 mol * ΔH°f (CH₃OH (l)) = 2 mol * (-393.5 kJ/mol) + 4 mol * (-285.8 kJ/mol) - (-1452.8 kJ) - 3 mol * 0 kJ/mol

ΔH°f (CH₃OH (l)) = - 477.4 kJ/mol