1.8.4Physical Chemistry
Hess Cycles — Edexcel AS Chemistry Revision Notes & Topical Questions
5 min read 6 parts Lesson 36 of 42
Part 1 of 6Hess Law
Statement
- The enthalpy change of a reaction is the same whichever route is taken, provided the start and end conditions are the same.
Reason
- Enthalpy is a state function, so it depends only on the initial and final states.
Use
- It lets us calculate enthalpy changes that cannot be measured directly.
Cycles
- Enthalpy cycles show direct and indirect routes between reactants and products.
Part 2 of 6Cycles Using Formation Data
Cycles Using Formation Data
Route
- Reactants break down into elements, and the elements form the products.
Equation
- Use ΔrH = ΣΔfH(products) − ΣΔfH(reactants), with every value multiplied by its coefficient.
Multiples
- Multiply each ΔfH by the coefficient of that substance in the equation.
Elements
- Elements in standard states have ΔfH = 0 and do not appear in the calculation.
Part 3 of 6Cycles Using Combustion Data
1
Write the equation for the reaction with all reactants and products.
2
Draw arrows from reactants and from products down to the combustion products.
3
Label each arrow with ΔcH multiplied by the right number of moles.
4
Apply the rule: ΔrH = ΣΔcH(reactants) − ΣΔcH(products).
5
Check the sign: a more exothermic combustion value means more energy released by that substance.
| Data given | Formula |
|---|---|
| Formation | ΔrH = ΣΔfH(products) − ΣΔfH(reactants) |
| Combustion | ΔrH = ΣΔcH(reactants) − ΣΔcH(products) |
Part 4 of 6Avoiding Mistakes
Direction of arrows
- Follow an arrow in its direction to add the value and against it to subtract the value.
Coefficients
- Multiply each enthalpy by the number of moles shown in the balanced equation.
States
- Use the correct state for water, as ΔfH of H₂O(l) differs from H₂O(g).
Units
- Give the final answer in kJ mol⁻¹ with the correct sign.
Part 5 of 6Worked Example
Question
- Calculate ΔcH of methane given ΔfH values (kJ mol⁻¹): CH₄(g) = −75, CO₂(g) = −394, H₂O(l) = −286.
- CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).
- Sum of products = (−394) + 2(−286) = −966.
- Sum of reactants = (−75) + 0 = −75, because O₂ is an element.
- ΔrH = −966 − (−75) = −891 kJ mol⁻¹.
Answer ΔcH = −891 kJ mol⁻¹
Part 6 of 6Exam Checklist & Tips
1.8.4 – Key Points for the Exam
- Hess law Total enthalpy change is independent of the route.
- State function A property that depends only on the initial and final states.
- Enthalpy cycle Diagram showing alternative routes between reactants and products.
- Standard enthalpy of formation Enthalpy change forming one mole of a compound from its elements in standard states.
Common Mistakes
- Adding formation values for reactants instead of subtracting them.
- Forgetting to multiply by the stoichiometric coefficient.
- Including ΔfH for elements as non-zero values.
- Reversing the sign of ΔfH when applying combustion data.
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