IB Chemistry SL Topic 5 — The Rate of Reaction Paper 1 & 2 Core idea ~11 min read

Activation Energy

Even a reaction that releases enormous energy has to spend some first. Bonds must be stretched and broken before new ones can form, and that upfront cost is why petrol sits safely in a tank until something lights it.

📚 What you need to know

Reading an energy profile

Activation energy the minimum energy that colliding particles must have
for a reaction to take place
THE SAME HILL, TWO LANDSCAPESEXOTHERMICreactantsproductstransition stateEₐΔH −ENDOTHERMICreactantsproductstransition stateEₐΔH +REACTION COORDINATEEₐ is always measured upwards from the reactants to the peak
Two quantities, two different arrows. Ea goes reactants-to-peak; ΔH goes reactants-to-products and ignores the peak entirely.

The two arrows answer different questions, and the exam relies on you keeping them apart:

A reaction can be strongly exothermic and glacially slow. Diamond turning into graphite releases energy, but the activation energy is so enormous that a diamond will outlast everything you own. ΔH tells you whether a reaction can release energy; Ea tells you whether it will, in any useful timeframe.
WORKED EXAMPLE

On an energy profile, the reactants lie at 50 kJ mol–1, the peak at 190 kJ mol–1 and the products at 110 kJ mol–1. State whether the reaction is exothermic or endothermic, and find Ea and ΔH.

Step 1 — which way round? Products (110) are HIGHER than reactants (50), so energy has been absorbed. endothermic Step 2 — activation energy, reactants to peak 190 − 50 = 140 Eₐ = +140 kJ mol⁻¹ Step 3 — enthalpy change, reactants to products 110 − 50 = +60 ΔH = +60 kJ mol⁻¹ Always subtract in the order products minus reactants for ΔH, and peak minus reactants for Eₐ.

Going the other way

Every reaction can in principle run backwards, and the reverse reaction has to climb to the same peak — it just starts from a different level. For an exothermic reaction the products sit lower down, so the return journey is a bigger climb.

GOING BACK UP THE OTHER SIDE60120220reactantsproducts100160−60Eₐ forwardEₐ reverseΔHEₐ(reverse) = Eₐ(forward) − ΔH = 100 − (−60) = 160 kJ mol⁻¹
One peak, two climbs. The gap between the two activation energies is exactly ΔH.
The reverse activation energy Ea(reverse) = Ea(forward) – ΔH

The formula handles both cases without you having to think about signs. For an exothermic reaction ΔH is negative, so subtracting it adds to Ea and the reverse reaction is harder. For an endothermic reaction ΔH is positive, so the reverse is easier.

WORKED EXAMPLE

For a reaction, Ea(forward) = 258 kJ mol–1 and ΔH = –92 kJ mol–1. Calculate the activation energy of the reverse reaction.

Step 1 — substitute Eₐ(rev) = 258 − (−92) Step 2 — mind the double negative = 258 + 92 Eₐ(reverse) = 350 kJ mol⁻¹ Sensible? The forward reaction is exothermic, so the products are more stable and harder to push back uphill. A larger reverse Eₐ is exactly what you would expect.
WORKED EXAMPLE

A student says: “Heating the reaction lowers the activation energy, which is why it goes faster.” Identify the error and give the correct explanation.

The error Eₐ is a property of the reaction pathway. Temperature does not change it at all — the hill stays exactly the same height. What actually changes Heating changes the PARTICLES, not the barrier. A greater proportion of them now have energy ≥ Eₐ, and they collide more often. the barrier is fixed; the particles change Only a catalyst changes Eₐ, and it does so by offering a different pathway rather than by lowering the original one.

💡 Exam tip

⚠️ Common mix-up

Up next: Energy Profiles With and Without Catalysts — the one thing that genuinely does change the height of the hill, and exactly what it leaves untouched.

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