4.2.4Additional lessons
Calculate Enthalpy Changes Using ΔHf⦵ (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 76 of 172
Part 1 of 5Standard Enthalpy of Formation
Formation enthalpies give every substance a common reference point, which is the set of elements it could be made from.
Definition
- ΔHf⦵ is the enthalpy change when one mole of a compound forms from its elements in their standard states under standard conditions.
Standard conditions
- Pressure of 100 kPa, a stated temperature (usually 298 K) and 1 mol dm⁻³ for any solutions involved in the process.
Elements
- The ΔHf⦵ of any element in its standard state, such as O₂(g) or C(graphite), is defined as exactly zero kJ mol⁻¹.
Data source
- Values are taken from the data booklet or the question, and each is quoted per mole of the compound formed.
Part 2 of 5The Calculation Route
Hess's law says the overall enthalpy change is independent of route, so the path through the elements is valid.
1
Write a fully balanced equation for the reaction and include state symbols, because the state changes the enthalpy of formation value.
2
Imagine each reactant being broken down into its elements, which is the reverse of formation, so each reactant contributes the opposite sign of its ΔHf⦵.
3
Imagine the elements recombining to give the products, so each product contributes its own ΔHf⦵ value.
4
Multiply every ΔHf⦵ by the stoichiometric coefficient from the balanced equation before adding anything together.
5
Apply ΔH⦵ = ΣΔHf⦵(products) − ΣΔHf⦵(reactants) and quote the answer in kJ mol⁻¹ with the correct sign.
Part 3 of 5Interpreting the Answer
Interpreting the Answer
Negative result
- An exothermic reaction, because the products are more stable than the reactants relative to the elements.
Positive result
- An endothermic reaction, where the products lie higher in energy than the reactants.
Units
- The answer is in kJ mol⁻¹ and refers to the amounts shown in the balanced equation as written.
Scaling
- Doubling every coefficient in the equation doubles the enthalpy change, and reversing it changes the sign.
| Species | ΔHf⦵ / kJ mol⁻¹ |
|---|---|
| CH₄(g) | −74.8 |
| CO₂(g) | −393.5 |
| H₂O(l) | −285.8 |
| O₂(g) | 0 |
Part 4 of 5Worked Example
Question Calculate ΔH⦵ for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) using ΔHf⦵ values of CH₄ = −74.8, CO₂ = −393.5 and H₂O(l) = −285.8 kJ mol⁻¹.
- ProductsΣΔHf⦵ = (−393.5) + 2(−285.8) = −965.1 kJ mol⁻¹.
- ReactantsΣΔHf⦵ = (−74.8) + 2(0) = −74.8 kJ mol⁻¹.
- ΔH⦵ = (−965.1) − (−74.8) = −890.3 kJ mol⁻¹.
Answer ΔH⦵ = −890.3 kJ mol⁻¹ (exothermic)
Part 5 of 5Exam Checklist & Tips
4.2.4 – Key Points for the Exam
- Standard state The most stable physical form of a substance at 100 kPa and the stated temperature.
- Hess's law The enthalpy change of a reaction is the same whichever route is taken between the same start and end points.
- ΔHf⦵ Enthalpy change when one mole of a compound forms from its elements in standard states.
- Stoichiometric coefficient The number multiplying each formula in a balanced equation, used to scale enthalpy values.
Common Mistakes
- Subtracting products from reactants instead of reactants from products, which reverses the sign of the answer.
- Forgetting to multiply each ΔHf⦵ value by its coefficient in the balanced equation.
- Using ΔHf⦵ of an element such as O₂(g) as a non-zero number instead of zero.
- Using the value for H₂O(g) when the equation shows H₂O(l), so the state symbol is ignored.
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