Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) 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: $$ \begin{aligned} 2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-1452.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

A standard enthalpy of formation involves the transition of an element from its pure form in its standard state to a compound. Methanol, \(\mathrm{CH}_3\mathrm{OH}\), is composed of Carbon (C), Hydrogen (H), and Oxygen (O). Therefore, its formation reaction from its pure elements can be written as follows: \[ \mathrm{C}(s) + \frac{1}{2}\mathrm{O}_2(g) + \frac{4}{2}\mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_3\mathrm{OH}(l) \] Notice that the stoichiometric coefficients are such that exactly 1 mole of CH3OH is produced, thus making the reaction a formation reaction.

We are given the reaction \[2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\] with its associated enthalpy change of \(\Delta H_{\mathrm{rxn}}^{\circ}= -1452.8 \mathrm{~kJ/mol} \). We will use Hess's law to connect this reaction with the formation reaction in Step 1. According to Hess’s law, the enthalpy change of a chemical reaction is independent of the route by which the chemical reaction takes place. This means we can calculate the enthalpy of methanol by breaking down the given reaction into steps that include the formation reaction of methanol. The reverse of the formation reaction happens twice in the given reaction, but also two formation reactions of CO2 and four formation reactions of H2O happen. If we take the enthalpy of formation of CO2 to be -393.5 kJ/mol and that of H2O to be -285.8 kJ/mol, we can write \[ -2\Delta H_f(\mathrm{CH}_3\mathrm{OH}) = -1452.8 \mathrm{kJ/mol} - 2*(-393.5 \mathrm{kJ/mol}) - 4*(-285.8 \mathrm{kJ/mol}) \]

You can now solve the equation obtained in Step 2 for \(\Delta H_f(\mathrm{CH}_3\mathrm{OH})\), the standard enthalpy of formation of methanol, as follows: \[ -2\Delta H_f(\mathrm{CH}_3\mathrm{OH}) = -1452.8 \mathrm{kJ/mol} - 2*(-393.5 \mathrm{kJ/mol}) - 4*(-285.8 \mathrm{kJ/mol}) \] \[ \Delta H_f(\mathrm{CH}_3\mathrm{OH}) = \frac{-1452.8 \mathrm{kJ/mol} - 2*(-393.5 \mathrm{kJ/mol}) - 4*(-285.8 \mathrm{kJ/mol})}{-2} \]