Worked example 1
Find ΔH for CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) given ΔHf: CH₄(g) = -74.6, CO₂(g) = -393.5, H₂O(l) = -285.8 kJ/mol.
Try it first: Note which species has a formation enthalpy of zero before adding anything up.
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Thermochemistry & ThermodynamicsEnthalpy of FormationContent level: Core 22 min
Standard Enthalpies of Formation
What you'll be able to do: Calculate a reaction enthalpy from tabulated standard enthalpies of formation.
Best after: Hess's Law and Combining Reactions
Introduction
Hess's law works, but hunting for the right set of equations is slow. Formation enthalpies package the same idea into one table and one subtraction.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- State the definition and conditions for a standard enthalpy of formation
- Explain why elements in their standard states have ΔHf of zero
- Compute a reaction enthalpy using the products-minus-reactants relationship
- Write correct formation equations for a given compound
Lesson
The definition
A formation reaction makes exactly one mole of a compound from its elements in the form they naturally take at 1 bar and the stated temperature. Oxygen is O₂(g), carbon is graphite, bromine is Br₂(l). Fractional coefficients on the element side are expected, because the product side is fixed at one mole.
C(graphite) + 2 H₂(g) → CH₄(g), ΔHf = -74.6 kJ/mol
Why elements are zero
Forming an element from itself involves no change, so the enthalpy change is zero by definition. This is a chosen reference point rather than a claim that elements contain no energy, and it works because only differences in enthalpy are ever measured.
Only the standard state form is zero. O₃(g) and diamond both have non-zero ΔHf even though ozone and diamond are elemental forms.
Products minus reactants
Imagine breaking every reactant down into its elements and rebuilding the products from them. The first stage is the negative of the reactant formation enthalpies, the second is the product formation enthalpies. Adding the two stages gives the standard relationship.
ΔH(rxn) = S n ΔHf(products) - S n ΔHf(reactants)
Coefficients are not optional
Each formation enthalpy is per mole, so it must be multiplied by the coefficient in the balanced equation. A combustion producing 2 mol of water needs twice the water value.
Reading the sign
A strongly negative ΔHf means the compound is enthalpically stable relative to its elements. A positive ΔHf, as for NO(g) or C₂H₂(g), signals a compound that stores energy relative to its elements and often burns or decomposes vigorously.
Key ideas
Definition
Standard enthalpy of formation
Enthalpy change forming one mole of a substance from its elements in standard states, in kJ/mol.
Rule
Elements are zero
ΔHf equals zero for an element in its standard state, and only for that form.
Rule
Products minus reactants
Multiply each ΔHf by its coefficient, sum for products, sum for reactants, then subtract.
Definition
Standard state
The pure substance at 1 bar, with solutions at 1 M, usually tabulated at 298 K.
Key concept
Interpreting ΔHf
A large negative value indicates a compound that is enthalpically stable relative to its elements.
Equation
Reaction enthalpy from formation data
ΔH(rxn) = S n ΔHf(products) - S n ΔHf(reactants)
- n = stoichiometric coefficient of the species
- ΔHf = standard enthalpy of formation in kJ/mol
Worked examples
Worked example 2
Write the formation equation for ethanol, C₂H₅OH(l).
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Worked example 3
For 2 NO(g) + O₂(g) → 2 NO₂(g), ΔHf(NO) = +91.3 and ΔHf(NO₂) = +33.2 kJ/mol. Find ΔH.
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Common mistakes
Computing reactants minus products.
Why it's wrong: It reverses the sign of the entire answer.
Check instead: Products come first in the subtraction, every time.
Ignoring the stoichiometric coefficients.
Why it's wrong: Formation enthalpies are per mole and must be scaled.
Check instead: Multiply each value by its coefficient before summing.
Assigning zero to any elemental form.
Why it's wrong: Only the standard state form is zero; ozone and diamond are not.
Check instead: Check that the form given is the standard state before using zero.
Writing a formation equation with two moles of product.
Why it's wrong: The definition fixes the product at exactly one mole.
Check instead: Use fractional coefficients on the element side instead.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
What you should now know
The standard enthalpy of formation is the enthalpy change when one mole of a substance forms from its elements in their standard states at 1 bar and a stated temperature, normally 298 K. Elements in their standard states are defined as zero. For any reaction, ΔH = sum of n ΔHf(products) minus sum of n ΔHf(reactants), with each value multiplied by its coefficient. The two things that go wrong are forgetting the coefficients and reversing the subtraction.
- ΔHf forms one mole of compound from elements in standard states
- Elements in their standard states have ΔHf of zero
- ΔH = products minus reactants, each scaled by its coefficient
- Formation equations must produce exactly one mole
- A negative ΔHf indicates enthalpic stability relative to the elements
Sources and further reading
- Chemistry 2e, Section 5.3: EnthalpyPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 5.3View source
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License