4.2.5Additional lessons
Calculate Enthalpy Changes Using ΔHc⦵ (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 77 of 172
Part 1 of 5Standard Enthalpy of Combustion
Combustion data suits organic chemistry because every organic compound burns to the same small set of products.
Definition
- Definition ΔHc⦵ is the enthalpy change when one mole of a substance burns completely in excess oxygen under standard conditions.
Sign
- Sign Complete combustion always releases energy, so every ΔHc⦵ value is negative in kJ mol⁻¹.
Products
- Products Carbon becomes CO₂(g), hydrogen becomes H₂O(l) and other elements form their stated oxides.
Why use it
- Why use it Many compounds are hard to make from elements directly, yet burning them is easy to measure in a calorimeter.
Part 2 of 5Building the Hess Cycle
Notice that the subtraction order is the opposite of the formation-enthalpy method, so check which data you are using.
1
Write the target equation and identify all reactants and products with state symbols.
2
Draw a combustion cycle in which reactants and products are each burned to the same common combustion products.
3
The route through the reactants' combustion has an enthalpy equal to the sum of their ΔHc⦵ values, each multiplied by its coefficient.
4
The route through the products' combustion is the sum of the products' ΔHc⦵ values, also scaled by coefficients.
5
Equate the two routes to obtain ΔH⦵ = ΣΔHc⦵(reactants) − ΣΔHc⦵(products).
Part 3 of 5Comparing the Two Methods
Comparing the Two Methods
ΔHf⦵ method
- ΔHf⦵ method Products minus reactants, using data for forming each compound from its elements.
ΔHc⦵ method
- ΔHc⦵ method Reactants minus products, using data for burning each substance completely.
Non-combustible species
- Non-combustible species CO₂ and H₂O(l) have ΔHc⦵ = 0 because they cannot burn any further, so ignore them.
Elements
- Elements C(graphite) and H₂(g) have ΔHc⦵ values equal to ΔHf⦵ of CO₂ and H₂O(l), respectively.
Part 4 of 5Worked Example
Question
- Calculate ΔHf⦵ of ethane for 2C(s) + 3H₂(g) → C₂H₆(g).
- ΔHc⦵ values (kJ mol⁻¹): C(s) = −393.5, H₂(g) = −285.8, C₂H₆(g) = −1560.
- Reactants2(−393.5) + 3(−285.8) = −787.0 + (−857.4) = −1644.4 kJ mol⁻¹.
- ProductsC₂H₆(g) = −1560 kJ mol⁻¹.
- ΔH⦵ = ΣΔHc⦵(reactants) − ΣΔHc⦵(products) = −1644.4 − (−1560) = −84.4 kJ mol⁻¹.
Answer ΔHf⦵ of ethane = −84.4 kJ mol⁻¹
Part 5 of 5Exam Checklist & Tips
4.2.5 – Key Points for the Exam
- ΔHc⦵ Enthalpy change when one mole of a substance burns completely in oxygen under standard conditions.
- Complete combustion Burning in excess oxygen so that carbon forms only CO₂ and hydrogen forms only H₂O.
- Hess cycle A diagram linking reactants and products through a common set of substances so enthalpies can be added.
- Calorimeter Apparatus used to measure the heat released by a reaction such as combustion.
Common Mistakes
- Using products minus reactants with combustion data, which gives the wrong sign.
- Including ΔHc⦵ values for CO₂ and H₂O, which are zero and add nothing useful.
- Leaving out coefficients so a value for one mole is used for several moles.
- Quoting a positive combustion enthalpy, forgetting that burning always releases energy.
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