AP Chemistry 6.6: Enthalpy of Formation

The standard enthalpy of formation, written ΔH°f, is the enthalpy change when 1 mole of a substance is formed from its constituent elements in their standard states.

• Khan Academy's current Unit 6 uses this definition and the products-minus-reactants equation for calculating reaction enthalpy.

The ° symbol indicates standard-state conditions.

In typical AP Chemistry data tables, standard-state values are commonly tabulated around 298 K, and gases are referenced at standard pressure.

• The important idea is that every substance is being compared under a consistent set of standard conditions.

A standard state is the most stable form of an element under the specified standard conditions.

• Examples include:
• O₂(g) for oxygen
• H₂(g) for hydrogen
• N₂(g) for nitrogen
• C(s, graphite) for carbon

Be careful: not every form of an element is its standard state.

• For example:
• O₂(g) is the standard state of oxygen.
• O₃(g), ozone, is not.
• Graphite is the standard reference form of carbon under ordinary standard conditions.
• Diamond is not assigned ΔH°f = 0.

By convention:

• ΔH°f = 0
• for an element in its standard state.
• Examples:
• ΔH°f[O₂(g)] = 0
• ΔH°f[H₂(g)] = 0
• ΔH°f[N₂(g)] = 0
• ΔH°f[C(s, graphite)] = 0
• This does not mean these substances contain no energy. It means their standard enthalpies of formation are defined as the reference value zero.

The standard formation reaction for CO₂(g) is:

• C(s, graphite) + O₂(g) → CO₂(g)
• This produces exactly:
• 1 mol CO₂
• from elements in their standard states.
• Therefore the enthalpy change is:
• ΔH°f of CO₂(g)

Suppose the formation reaction is for H₂O(l).

• Correct:
• H₂(g) + ½O₂(g) → H₂O(l)
• because exactly one mole of H₂O is produced.

Fractions are completely acceptable in formation equations.

• Writing:
• 2H₂ + O₂ → 2H₂O
• would describe twice the standard molar enthalpy of formation rather than ΔH°f for one mole.

The main equation is:

• ΔH°rxn = ΣnΔH°f(products) − ΣnΔH°f(reactants)
• Think:
• products minus reactants

The coefficient n from the balanced equation multiplies each substance's ΔH°f.

Consider:

• CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
• Suppose:
• ΔH°f CH₄(g) = −74.8 kJ/mol
• ΔH°f O₂(g) = 0
• ΔH°f CO₂(g) = −393.5 kJ/mol
• ΔH°f H₂O(l) = −285.8 kJ/mol

Products:

• [1(−393.5)] + [2(−285.8)]
• = −393.5 − 571.6
• = −965.1 kJ

Reactants:

• [1(−74.8)] + [2(0)]
• = −74.8 kJ

Now:

• ΔH°rxn = products − reactants
• ΔH°rxn = −965.1 − (−74.8)
• ΔH°rxn = −890.3 kJ/molrxn
• Because the result is negative, the reaction is exothermic.

The enthalpy-of-formation equation is essentially a convenient application of Hess's law.

• Instead of manually combining many formation reactions, the formula automatically subtracts the reactant formation values from the product formation values.
• That is why Unit 6 topics connect so strongly.

If a reaction contains:

• 3H₂O(l)
• you use:
• 3 × ΔH°f[H₂O(l)]

Do not ignore coefficients.

ΔH°f for:

• H₂O(l)
• is different from:
• H₂O(g)
• because vaporizing water requires energy.

Always choose the value corresponding to the correct state.

Do not assume every element has ΔH°f = 0.

Only an element in its standard state gets zero.

Do not forget stoichiometric coefficients.

Do not reverse the formula. It is:

products − reactants

not reactants − products.

Do not ignore physical states.

For formation enthalpies:

• ΔH°rxn = products − reactants
• and:
• elements in their standard states = 0