UNIT 2 CHEMISTRY · PEARSON AAQ · D1.3.4
Hess’s Law and Energy Cycles
FREE CHEMISTRY REVISION · TOPIC 41
Hess’s Law and Energy Cycles
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · D1.3.4
What you will learn
Use Hess’s Law to connect direct and indirect routes, read the directions of energy-cycle arrows, and combine chemical equations while adjusting their enthalpy changes. This lesson develops cycle logic and introductory equation arithmetic; comprehensive standard-enthalpy calculations belong to D1.3.5.
1. Hess’s Law: same start and finish
The enthalpy change of a reaction is independent of the pathway, provided the initial and final states are the same. Enthalpy is a state function. The states include composition, amounts, physical states, temperature and pressure. Compare one direct route with a sequence of reactions connecting exactly the same states.
This is consistent with conservation of energy. If two routes between identical states had different enthalpy changes, following one route and reversing the other would return to the start with a nonzero net enthalpy change. That would make a closed cycle an energy source or sink without a change of state. The algebraic sum of enthalpy changes around a closed cycle is therefore zero.
The route can be hypothetical; every individual arrow must still represent a defined transformation. Hess’s Law concerns the overall change, not the time taken or activation barriers. A catalyst can alter rate and pathway while leaving the same overall ΔH.
2. Reading a directed cycle
If A → B has unknown change x, A → C has change b and C → B has change c, then x = b + c. Add changes along the route actually followed. A positive arrow label means that transformation absorbs enthalpy; a negative label means it releases enthalpy. Arrow direction alone does not indicate the sign.
When both known arrows point towards a common state C, follow A → C and then C → B. The second step opposes the B → C arrow, so its sign reverses: x = b − c. Do not simply add every printed number on a triangle.
3. Three operations on equations
| Operation | Equation change | Enthalpy change |
|---|---|---|
| Reverse | Swap all reactants and products | Multiply ΔH by −1. |
| Scale | Multiply every coefficient by the same factor | Multiply ΔH by that factor. |
| Add | Add equations, then cancel identical species on opposite sides | Add the adjusted ΔH values. |
A species can cancel only if its formula and physical state match. H₂O(l) cannot cancel against H₂O(g) without an explicit vaporisation or condensation step. Keep spectator substances when constructing complete cycle states so each route conserves atoms.
The value belongs to the equation as written. If you halve a reaction, halve its associated change. Express final molar reaction data in kJ mol⁻¹ for the stated reaction; a supplied total for a particular amount remains a total in kJ. All numerical values below are supplied or rounded teaching data for consistent states and conditions, not precision experimental results.
4. Worked examples: from arrows to equations
Example 1 — add two arrows
A → C has b = −40 kJ mol⁻¹ and C → B has c = −25 kJ mol⁻¹ on a consistent reaction basis. A → B therefore has x = −40 + (−25) = −65 kJ mol⁻¹. Both arrows are followed in their written directions. The closed loop A → C → B → A gives −40 −25 +65 = 0.
Example 2 — reverse one arrow
A → C is −120 and B → C is −75 kJ mol⁻¹, with consistent states and amounts. To reach B from A, take A → C, then reverse B → C. Thus x = −120 − (−75) = −45 kJ mol⁻¹. Recheck x + (B → C) = −45 −75 = −120, matching A → C.
Example 3 — reverse and halve a balanced equation
Given 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = −572 kJ mol⁻¹ for that reaction, obtain H₂O(l) → H₂(g) + ½O₂(g). First reverse: +572. Then halve every coefficient and the change: +286 kJ mol⁻¹. Reversing changes the sign; halving changes the magnitude.
Example 4 — build an equation by cancellation
Given C(s, graphite) + ½O₂(g) → CO(g), ΔH₁ = −110.5 kJ mol⁻¹, and CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283.0 kJ mol⁻¹, add the equations. CO(g) appears on opposite sides and cancels. The net equation is C(s, graphite) + O₂(g) → CO₂(g), with ΔH = −110.5 −283.0 = −393.5 kJ mol⁻¹.
Example 5 — find the missing carbon monoxide step
Use C(s, graphite) + O₂(g) → CO₂(g), −393.5 kJ mol⁻¹, and CO(g) + ½O₂(g) → CO₂(g), −283.0 kJ mol⁻¹. Reverse the second equation and add it to the first. CO₂ cancels, and ½O₂ cancels from both sides. Net: C(s, graphite) + ½O₂(g) → CO(g). ΔH = −393.5 +283.0 = −110.5 kJ mol⁻¹.
Example 6 — scale before cancelling an intermediate
Given N₂(g) + O₂(g) → 2NO(g), ΔH₁ = +180 kJ mol⁻¹, and NO(g) + ½O₂(g) → NO₂(g), ΔH₂ = −57 kJ mol⁻¹. Target: N₂(g) + 2O₂(g) → 2NO₂(g). Multiply the second equation by 2 so that 2NO cancels: its change becomes −114. Add: ΔH = +180 −114 = +66 kJ mol⁻¹. Adding +180 and −57 without scaling would leave one NO uncancelled.
Example 7 — reverse a combined result
For the reverse target 2NO₂(g) → N₂(g) + 2O₂(g), reverse Example 6. The answer is −66 kJ mol⁻¹. If instead the target is NO₂(g) → ½N₂(g) + O₂(g), halve that reverse equation and its change: −33 kJ mol⁻¹.
Example 8 — repair a state mismatch
One supplied step produces H₂O(g), but the target requires H₂O(l). Include H₂O(g) → H₂O(l). If its supplied change is −44.0 kJ mol⁻¹ and the first step forms one mole of gaseous water, add −44.0 once. For two moles, add −88.0. Do not cancel liquid water against steam or treat condensation as zero. The number is illustrative at specified conditions.
5. A reliable construction method
- Write the target equation first, including states and coefficients.
- For each supplied equation, identify whether it contains a useful target species or intermediate.
- Reverse and scale each equation until target coefficients and intermediate cancellations match.
- Write its adjusted ΔH alongside it immediately.
- Add equations; cancel only identical species on opposite sides.
- Check the final equation against the target, then add the adjusted changes.
- Check atoms, direction, units and the meaning of one mole of reaction.
For a cycle, trace a continuous route from the target’s starting node to its finishing node. Write a plus sign for a step along its arrow and a minus sign for a step against its arrow. Node placement in a Hess cycle is usually a layout choice, unlike a reaction profile’s vertical energy axis. Do not infer an enthalpy sign from an upward or downward cycle arrow unless an actual energy scale is stated.
6. Scope and useful limits
Hess’s Law lets us obtain an inaccessible overall change from other known changes, provided the state bookkeeping is consistent. It does not prove that the direct reaction is fast, occurs by the drawn intermediate route, or can be performed safely in one step.
This topic covers the law, directed cycles and elementary equation manipulation. The systematic calculations from tables of formation or combustion enthalpies, ΔH = ΔU + pΔV and calorimetry using ΔQ = mcΔT are specified separately in D1.3.5. They are not taught as part of this lesson.
Common misconceptions
- Every number in the cycle is added: choose a directed route and reverse signs where needed.
- Reverse an arrow without changing ΔH: reversing multiplies ΔH by −1.
- Scale coefficients but not ΔH: the enthalpy change scales with the reaction amount.
- Cancel the same formula in different states: include the state-change step.
- Cycle heights show activation energies: a Hess cycle is not a reaction profile.
- Hess’s Law says the mechanism is the drawn cycle: the cycle is thermodynamic bookkeeping.
Exam tips
- Show each adjusted equation and its enthalpy before adding.
- Use brackets when subtracting a negative value.
- Recheck cancellation before trusting arithmetic.
- Write the final balanced equation with its ΔH.
- For a closed loop, verify that the signed sum is zero.
- Keep the required target direction visible throughout.
Check your understanding
Choose one answer, select Check Answer, then read the explanation. Try Again resets the question. Use Tab between controls, arrow keys within a radio group, and Enter or Space on buttons. Answers are not saved.