When a question shows two energy profiles for the same reaction, compare three levels: the reactants, the peak and the products. The peak decides activation energy, and the products decide the enthalpy change.
This lesson is part of energy evidence and industrial chemistry reasoning. The measuring skills come from activation-energy diagrams.
What do you compare first?
Compare the levels in this order and write each down.
- The reactant level in both profiles.
- The product level in both profiles.
- The peak level in both profiles.
If the first two match and only the peak differs, only the activation energy has changed.
Worked example with numbers
Profile P: reactants at 60 kJ, peak at 150 kJ, products at 30 kJ. Profile Q: reactants at 60 kJ, peak at 110 kJ, products at 30 kJ.
| Quantity | Profile P | Profile Q | Changed? |
|---|---|---|---|
| Activation energy | 150 − 60 = 90 kJ | 110 − 60 = 50 kJ | Yes, lower in Q |
| Enthalpy change ΔH | 30 − 60 = −30 kJ | 30 − 60 = −30 kJ | No |
A full answer to “What has changed from P to Q?” is: Claim: Q has a lower activation energy and the same ΔH. Evidence: the peak is at 110 kJ instead of 150 kJ while the products stay at 30 kJ. Reason: this pattern is what a catalyst does, because it offers a pathway with a lower activation energy without changing the reactants and products.
The mistake that costs marks
The common slip is to say the catalyst “reduces the energy change” or “makes more heat”. The heat released is given by ΔH, which is −30 kJ in both profiles.
| Statement | Verdict |
|---|---|
| “The catalyst makes the reaction release more energy.” | Wrong: ΔH is unchanged |
| “The catalyst lowers the activation energy.” | Right: the peak is lower |
Another slip is writing that the temperature changed the diagram. The diagram is the same at any temperature. Temperature changes the fraction of particles that can climb the peak.
Check yourself
Profile R has reactants at 25 kJ, peak at 100 kJ and products at 70 kJ. Profile S has reactants at 25 kJ, peak at 80 kJ and products at 70 kJ. State what changed and what stayed the same from R to S, in claim, evidence and reason form.
Answer
Activation energy: R = 100 − 25 = 75 kJ and S = 80 − 25 = 55 kJ. ΔH: R and S are both 70 − 25 = +45 kJ.
Claim: S has a lower activation energy but the same ΔH.
Evidence: the peak is at 80 kJ instead of 100 kJ, while reactants and products are at the same levels.
Reason: a lower peak with unchanged reactant and product levels matches a catalysed pathway, which changes activation energy but not the enthalpy change.
What to study next
Move to distinguishing a faster reaction from a larger equilibrium yield, which uses the same separation between speed and amount. The thermochemistry overview shows where this fits.
If you want a teacher to give you unseen profile pairs, see online one-to-one Chemistry tuition. The mistake log and paper-error review tool helps you track which quantity you confuse.