Question

Methanol is an alcohol used as a fuel for some race cars. The standard molar enthalpy of formation of methanol CH₂OH() is -238.4 kJ/mol. For the balanced reaction equation

of methanol use:
CH₂OH(I) +2O₂ --> CO₂ + 2H₂O.
Methanol is a liquid, but all the other chemicals in this reaction are gases. What is the enthalpy change of this reaction in kJ/mol?

Answer 1

-488.2 kJ/mol

Step-by-step explanation:

To calculate the enthalpy change of the reaction, we can use the standard molar enthalpies of the formation of the products and reactants, which are given in the problem.

ΔH = Σ(nΔHf° products) - Σ(nΔHf° reactants)

where ΔH is the enthalpy change of the reaction, n is the stoichiometric coefficient, and ΔHf° is the standard molar enthalpy of formation.

Substituting the values given in the problem, we get:

ΔH = [(1 mol)(-393.5 kJ/mol) + (2 mol)(-241.8 kJ/mol)] - [(1 mol)(-238.4 kJ/mol) + (2 mol)(0 kJ/mol)]

= -726.6 kJ/mol + 238.4 kJ/mol

= -488.2 kJ/mol

The negative sign indicates that the reaction is exothermic, which means that it releases heat. Therefore, the enthalpy change of the reaction is -488.2 kJ/mol.