Given the heat of formation of the following compounds:
CO₂(g) ΔH°f = -393.5 kJ/mol
H₂O(l) ΔH°f = -285.9 kJ/mol
CH₃OH(l) ΔH°f = -238.6 kJ/mol
What is the value of ΔH for the reaction:
CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2 H₂O(l)

Question

Quizplus · Accepted Answer

The enthalpy change for the reaction can be calculated using the formula: ΔH = ΣΔH°f(products) - ΣΔH°f(reactants). For the given reaction, ΔH = [(-393.5) + 2(-285.9)] - [(-238.6)] = -726.7 kJ.

Given the Heat of Formation of the Following Compounds