Enthalpy change and activation-energy diagrams
Label the two different vertical energy differences and use the correct sign for the overall change.
Exam scope and assessment notes
Extended only (Supplement) · Chemistry 0971, 2026–2028. Cambridge states that IGCSE 0971 has the same subject content as IGCSE 0620, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Interpret exothermic and endothermic endpoint levels.
- Subtract signed energy values and distinguish an endpoint from a maximum.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For an exothermic reaction, what sign does ΔH have?
What you will learn
- State the signs of enthalpy change and define activation energy.
- Draw and label a reaction pathway using supplied reactant, product and peak energies.
Enthalpy change is the overall difference
The enthalpy change of a reaction, ΔH, describes its thermal energy transfer. It is negative for an exothermic reaction and positive for an endothermic reaction. On the supplied reaction pathway model, calculate product energy minus reactant energy. Use the units and reaction quantity stated in the question, commonly kJ/mol of reaction as written; do not attach molar units to an arbitrary uncalibrated sketch.
A downward arrow from reactant to product level shows a negative overall change, while an upward arrow shows a positive one. The diagram must label reactants, products and ΔH, and its horizontal axis is reaction progress. A negative value does not mean energy itself is negative or that the reaction has no starting barrier.
Activation energy measures the starting barrier
Activation energy Eₐ is the minimum energy that colliding particles must have to react. In a forward reaction pathway diagram, its vertical arrow runs from the reactant level up to the peak. It is not the entire distance from the baseline to the peak unless the reactant level happens to be that baseline. Particles also need a suitable collision arrangement; merely calling every collision a reaction ignores the barrier.
When drawing a pathway from data, position reactants and products at the stated relative levels, draw a curve through the given higher peak, then label the forward Eₐ and ΔH arrows separately. For the reverse reaction, the activation barrier is measured from the original product level to the same peak. A catalyst can lower the barrier for a reaction route without changing the starting and final energy levels, a connection developed in the reaction-rate lessons.
Pause and explain
Where does a forward Eₐ arrow begin on a reaction pathway?
Worked example
A supplied molar-energy pathway has reactants at 50 kJ/mol, products at 20 kJ/mol and a peak at 110 kJ/mol. Find ΔH and forward Eₐ, classify it and describe the arrows.
Show the worked solution
- ΔH=20−50=−30 kJ/mol, so the reaction is exothermic.
- Forward Eₐ=110−50=60 kJ/mol.
- Draw ΔH down from level 50 to 20 and Eₐ up from level 50 to the peak at 110; label both endpoints and reaction progress.
Answer ΔH=−30 kJ/mol; forward Eₐ=60 kJ/mol; exothermic.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Draw an endothermic pathway using reactants 25, products 55 and peak 85 kJ/mol. Label reactants, products, ΔH and forward Eₐ.
Give me a hint
The final level is higher, but the peak is higher still.
Compare my reasoning
- Place reactants at 25 and products at 55, and draw a curve rising to 85 before reaching 55.
- Draw ΔH up between 25 and 55, labelled +30 kJ/mol.
- Draw forward Eₐ up between 25 and 85, labelled 60 kJ/mol.
Endothermic; ΔH=+30 kJ/mol; forward Eₐ=60 kJ/mol.
Look for these in your work
- All required pathway labels are present.
- The two energy differences have different endpoints.
A pathway has reactants 80, products 35 and peak 140 kJ/mol. Find forward and reverse Eₐ and forward ΔH.
Give me a hint
The reverse reaction starts at the original product level.
Compare my reasoning
- Forward Eₐ=140−80=60 kJ/mol.
- Reverse Eₐ=140−35=105 kJ/mol.
- Forward ΔH=35−80=−45 kJ/mol; the reverse barrier exceeds the forward by 45 kJ/mol.
Forward Eₐ=60; reverse Eₐ=105; forward ΔH=−45 kJ/mol.
Look for these in your work
- Each barrier uses its own starting level.
- The barrier difference is consistent with the endpoint difference.
For reactants 50 and products 20 kJ/mol, one route peaks at 110 and another at 90 kJ/mol. Compare forward barriers and overall ΔH. Explain what the lower peak changes.
Give me a hint
Only the route maximum changes; the endpoints stay fixed.
Compare my reasoning
- The first forward barrier is 110−50=60 kJ/mol; the second is 90−50=40 kJ/mol.
- Both overall changes are 20−50=−30 kJ/mol.
- The lower-peak route has a smaller starting barrier; it does not change the overall reactant-to-product energy difference.
Eₐ decreases from 60 to 40 kJ/mol; ΔH remains −30 kJ/mol.
Look for these in your work
- Barrier and overall energy change are calculated separately.
- The changed route is not treated as extra energy released.
Common mistakes
- Using peak minus products as the forward activation energy.
- Using the baseline-to-peak height when reactants are above the baseline.
- Giving exothermic ΔH a positive sign.
- Drawing ΔH to the peak instead of between endpoint levels.
What can you explain now?
Reactants are at 30, products at 70 and the peak at 100 kJ/mol on supplied molar-energy data. Find ΔH and forward Eₐ.
Compare with the explanation
ΔH=+40 kJ/mol; forward Eₐ=70 kJ/mol.
Products minus reactants is 70−30=+40; peak minus reactants is 100−30=70.
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Make it stick
Draw a numerical pathway from memory and label both energy arrows. Reverse the reaction and calculate its barrier and overall change without confusing the starting levels.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.