IB Chemistry HL Topic 4 — Measuring Enthalpy Change Paper 1 & 2 Core skill ~10 min read

Energy Profile Diagrams

An enthalpy level diagram tells you where a reaction starts and finishes. An energy profile adds the bit in between — the hill the reactants have to climb before anything happens at all. This one diagram explains why petrol does not ignite in the tank, and it is drawn in almost every exam session.

📘 What you need to know

What the axes actually mean

The vertical axis is potential energy (you will also see it labelled enthalpy, H) — the energy stored in the bonds and arrangement of the particles.

The horizontal axis is the reaction coordinate. This is not time, and it is not concentration. It is simply “how far along the process of turning into products are these particles”. Left-hand end: untouched reactants. Right-hand end: finished products. Everything in between is the messy middle where old bonds are stretching and new ones are forming.

Do not read the horizontal axis as a stopwatch. A reaction can sit at the left of the profile for years and then cross the whole diagram in a millisecond.

The transition state and the hill

Two particles that collide do not slide smoothly into being products. First they must be forced into an awkward, unstable arrangement where the old bonds are stretched almost to breaking and the new ones are only half made. That arrangement is the transition state, and it sits at the top of the hill.

Getting there costs energy, and colliding particles pay for it out of their own kinetic energy. If a collision does not bring enough energy, the particles simply bounce apart unchanged. The minimum needed is the activation energy.

Definition to learn word for word Ea = the minimum energy that colliding particles must have for a successful collision
Energy profile for an exothermic reaction Uphill first, then further downhill than it started. potential energy Ea (forward) ΔH negativereactants products transition state reaction coordinateActivation energy is measured from the reactants, not from the axis. The height of the hill controls the rate; the drop across it controls delta H.
Two completely separate quantities live on one diagram. Confusing them is the most common error on this topic, so check the start of every arrow you draw.

Endothermic profiles

Nothing about the shape changes except where the right-hand plateau ends up. For an endothermic reaction the products finish above the reactants, so the reaction is uphill overall as well as uphill on the way.

Because the products are already high up, they only need a small push to get back over the peak. That is why endothermic reactions have a large forward activation energy and a small reverse one.

Energy profile for an endothermic reaction The products never come back down as far as they started. potential energy Ea (forward) ΔH positivereactants products transition state reaction coordinateSame hill, different landing height. That is the only difference. Here the reverse activation energy is the small one, since products start high.
If you can draw one of these two profiles from memory you can draw the other — only the height of the right-hand plateau moves.

Reading numbers off a profile

Exam questions rarely just ask you to draw the diagram. They give you a sketch with a couple of energy values on it and ask you to extract Ea or ΔH. Both come out of the same three levels.

🧩 Extracting values from any profile

  1. Write down the three heights: reactants, peak, products.
  2. Ea forward = peak − reactants. Always positive.
  3. ΔH = products − reactants. Sign tells you exo or endo.
  4. Ea reverse = peak − products. Also always positive.
  5. Check your work: Ea(reverse) should equal Ea(forward) − ΔH.
WORKED EXAMPLE

On an energy profile the reactants lie at 50 kJ mol−1, the transition state at 185 kJ mol−1 and the products at 20 kJ mol−1. Find Ea for the forward reaction, ΔH, and Ea for the reverse reaction.

Step 1: Activation energy forward = peak − reactants 185 − 50 = +135 kJ mol⁻¹ Step 2: Enthalpy change = products − reactants 20 − 50 = −30 kJ mol⁻¹ negative, so exothermic − as expected, products are lower Step 3: Activation energy reverse = peak − products 185 − 20 = +165 kJ mol⁻¹ Step 4: Check 135 − (−30) = 165 ✓ Eₐ fwd = 135, ΔH = −30, Eₐ rev = 165 kJ mol⁻¹
WORKED EXAMPLE

A reaction has ΔH = +52 kJ mol−1 and a forward activation energy of 90 kJ mol−1. Deduce the activation energy of the reverse reaction and state which direction is faster at a given temperature.

Step 1: Use the relationship Eₐ(rev) = Eₐ(fwd) − ΔH = 90 − (+52) = +38 kJ mol⁻¹ Step 2: Compare the two barriers the reverse barrier is much lower, so more collisions can clear it Eₐ(rev) = +38 kJ mol⁻¹, reverse reaction is faster an endothermic forward reaction always has the taller barrier

What a catalyst does — and does not do

A catalyst provides an alternative pathway with a lower transition state. More of the colliding particles now have enough energy to get over the top, so the rate goes up.

What a catalyst absolutely cannot do is move the reactants or products levels. Those are fixed by the bonds in the substances themselves. So ΔH is completely unchanged, and both the forward and reverse activation energies fall by the same amount.

A catalyst lowers the hill, not the destination potential energy without catalyst with catalyst smaller Ea larger Ea ΔH is identicalreactants products reaction coordinateBoth routes start and finish at exactly the same two levels.
A common exam trap is a diagram where the catalysed curve also ends lower. That is wrong: a catalyst cannot change how much energy the bonds store.
Why this matters for exothermic reactions. An exothermic reaction is downhill overall, but the hill still has to be climbed first. That is precisely why a mixture of petrol and air sits harmlessly in a fuel tank: the activation energy has not been supplied. A spark supplies it to a few molecules, those release enough energy to activate their neighbours, and the reaction runs away.
WORKED EXAMPLE

The reaction from the first worked example (Ea = 135 kJ mol−1, ΔH = −30 kJ mol−1) is repeated with a catalyst that lowers the forward activation energy to 78 kJ mol−1. State the new ΔH and calculate the new reverse activation energy.

Step 1: What does a catalyst change? only the height of the peak, so the two levels stay put ΔH = −30 kJ mol⁻¹ (unchanged) Step 2: New reverse barrier Eₐ(rev) = 78 − (−30) = +108 kJ mol⁻¹ Step 3: Sense-check both barriers dropped by 57 kJ mol⁻¹, the same amount, exactly as expected ΔH unchanged at −30; Eₐ(rev) = +108 kJ mol⁻¹

Drawing one under exam pressure

Marks for sketching a profile are handed out for very specific features. Include all of these and the marks are hard to lose.

💡 Exam tip

⚠ Common mix-up

Up next: Standard Enthalpy Changes — before we can compare any two ΔH values fairly, we need to agree on the conditions they were measured under.

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