Standard Enthalpy of Formation

Hess’s law is powerful, but manipulating multiple reactions can be tedious. There is a shortcut: if you know the standard enthalpy of formation (ΔHf°) for every compound in a reaction, you can calculate ΔH° in a single step using a simple “products minus reactants” formula.

What Is a Formation Reaction?

A formation reaction produces exactly one mole of a compound from its elements in their standard states (the most stable form at 25 C and 1 atm).

Examples:

• C(graphite) + O₂(g) → CO₂(g), ΔHf° = -393.5 kJ/mol
• H₂(g) + $\frac{1}{2}$ O₂(g) → H₂O(l), ΔHf° = -285.8 kJ/mol
• $\frac{1}{2}$ N₂(g) + $\frac{1}{2}$ O₂(g) → NO(g), ΔHf° = +90.3 kJ/mol

The Critical Rule: Elements Have ΔHf° = 0

Elements in their standard states are the reference point. Their ΔHf° is defined as zero:

• ΔHf°[O₂(g)] = 0
• ΔHf°[N₂(g)] = 0
• ΔHf°[C(graphite)] = 0
• ΔHf°[Fe(s)] = 0

This makes sense: you do not need to “form” an element from itself.

Calculating ΔH°rxn from ΔHf° Values

Worked Example

Calculate ΔH° for the combustion of methane:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l)

Given ΔHf° values (kJ/mol): CH₄(g) = -74.8, O₂(g) = 0, CO₂(g) = -393.5, H₂O(l) = -285.8

Products: (1)(-393.5) + (2)(-285.8) = -393.5 + (-571.6) = -965.1 kJ

Reactants: (1)(-74.8) + (2)(0) = -74.8 kJ

ΔH°rxn = -965.1 - (-74.8) = -965.1 + 74.8 = -890.3 kJ

The reaction is strongly exothermic, which makes sense - methane combustion powers your stove.

Standard Enthalpy of Combustion

The standard enthalpy of combustion (ΔH°comb) is the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions. Combustion reactions are always exothermic (ΔH°comb < 0).

Common values:

• Methane (CH₄): -890 kJ/mol
• Glucose (C₆H₁₂O₆): -2,803 kJ/mol
• Ethanol (C₂H₅OH): -1,367 kJ/mol