Enthalpy changes, calorimetry and Hess's law
Enthalpy change, ΔH, is the heat energy transferred in a reaction at constant pressure. You can measure it in a calorimeter, or calculate it from other enthalpy changes with Hess's law when the reaction cannot be measured directly.
Part 1 of 3: Learn it
In short
- Negative ΔH means exothermic; positive ΔH means endothermic.
- Calorimetry: q = mcΔT, then ΔH = −q ÷ n.
- Hess's law: the enthalpy change is the same whatever route is taken from reactants to products.
Where this is in your specification
Spec points: AQA 7405 3.1.4.1 to 3.1.4.4, OCR A H432 3.2.1, Edexcel 9CH0 Topic 8
| Board | Topic: Energetics and thermodynamics |
|---|---|
| AQA 7405 | 3.1.4 and 3.1.8 |
| Edexcel 9CH0 | Topics 8 and 13 |
| OCR H432 | 3.2.1, 5.2.1 and 5.2.2 |
| Higher C813 76 | 3(c) Chemical energy |
Standard enthalpy changes
Standard conditions are 100 kPa and a stated temperature, usually 298 K, with substances in their standard states. Two definitions you must be able to write:
- Standard enthalpy of formation, ΔfH°: one mole of a compound is formed from its elements in their standard states.
- Standard enthalpy of combustion, ΔcH°: one mole of a substance is burned completely in oxygen.
By definition, ΔfH° of an element in its standard state is zero.
Calorimetry
m is the mass of the water or solution being heated (not of the fuel), c is its specific heat capacity (4.18 J g⁻¹ K⁻¹ for water) and ΔT the temperature change. Divide q by the moles of the reactant that was limiting, convert to kJ, and give the sign. Heat escaping to the air and incomplete combustion make experimental values less exothermic than data-book values.
Hess's law
- From enthalpies of formation: ΔH = ΣΔfH(products) − ΣΔfH(reactants).
- From enthalpies of combustion: ΔH = ΣΔcH(reactants) − ΣΔcH(products).
- Bond enthalpies: ΔH = Σ(bonds broken) − Σ(bonds made). Mean bond enthalpies are averages over many compounds, so answers differ slightly from measured values.
What are standard conditions for enthalpy changes?
Show the answer
100 kPa and a stated temperature, usually 298 K.
Part 2 of 3: See it worked
Worked examples
Example 1
Burning 0.32 g of methanol (Mᵣ = 32.0) raises the temperature of 100 g of water by 15.0 K. Find the enthalpy of combustion.
- q = 100 × 4.18 × 15.0 = 6270 J = 6.27 kJ
- n = 0.32 ÷ 32.0 = 0.0100 mol
- ΔH = −6.27 ÷ 0.0100
Answer: −627 kJ mol⁻¹ (far less exothermic than the data-book value because of heat loss).
Example 2
Use enthalpies of combustion (kJ mol⁻¹): C −394, H₂ −286, CH₄ −890 to find ΔfH of methane, C + 2H₂ → CH₄.
- ΔH = ΣΔcH(reactants) − ΣΔcH(products)
- = [−394 + 2 × (−286)] − (−890)
- = −966 + 890
Answer: −76 kJ mol⁻¹.
Common mistakes
- Using the mass of fuel as m in q = mcΔT.
- Leaving ΔH positive for an exothermic reaction.
- Mixing up the formation and combustion versions of the Hess's law shortcut.
- Forgetting to multiply by the number of moles in the balanced equation.
What is ΔfH° of O₂(g)?
Show the answer
Zero: it is an element in its standard state.
Part 3 of 3: Test yourself
Check yourself
Answer each one in your head or on paper first, then open it to check.
What are standard conditions for enthalpy changes?
100 kPa and a stated temperature, usually 298 K.
What is ΔfH° of O₂(g)?
Zero: it is an element in its standard state.
Using ΔfH values (kJ mol⁻¹) C₂H₄ +52.4 and C₂H₆ −84.7, find ΔH for C₂H₄ + H₂ → C₂H₆.
−84.7 − (52.4 + 0) = −137.1 kJ mol⁻¹.
Jobs that use this
- Chemical engineer (opens a new tab)
- Chemist (opens a new tab)
- Renewable energy engineer (opens a new tab)
Each link opens the job profile on the National Careers Service (England). In the rest of the UK: My World of Work (Scotland), Careers Wales, nidirect careers (Northern Ireland).
Full lessons and marked practice for this course are coming soon to Brainlag Learn. See courses