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AP Chemistry · Unit 5 · Topic 5.6

Reaction Energy Profile & Activation Energy

L3 · Exam Readiness⏱ 25 minSP1ENE-1.D

By the end you can…

  • ✓Read a reaction-coordinate diagram and identify reactants, products, the transition state, E_a and \Delta H.
  • ✓Explain, at the particle level, why a minimum energy is required for a reaction to occur.
  • ✓Predict how a catalyst changes the energy profile — and what it does NOT change.

Why this matters

Every reaction that keeps you alive — and every one that spoils your food — runs at a rate set by a single hill: the activation energy. Understand that hill and you can explain why a spark ignites petrol, why the fridge slows decay, and why enzymes make impossible reactions routine.

The reaction as a journey over a hill

Macroscopic — what you observe

Mix two reactants and often nothing visible happens until you add heat or a catalyst. A reaction that releases energy overall can still refuse to start — as if the reactants are stuck at the bottom of a valley with a hill in the way.

Particulate — atoms, ions & electrons

For a reaction, particles must collide with enough energy and the correct orientation to reach a strained, high-energy arrangement called the transition state. Most collisions bounce off; only the energetic ones make it over.
Watch: toggle catalyst → the hill (E_a) shrinks; toggle exo/endo → products drop below (\Delta H<0) or rise above (\Delta H>0) the reactants. A catalyst lowers E_a but never changes \Delta H.

Make a prediction

Adding a catalyst lowers the hill. What happens to \Delta H (reactants → products)?

Reading the diagram

Symbolic — the mathematics

On a reaction-coordinate diagram, the peak height above the reactants is the activation energy E_a = E_{\text{transition state}} - E_{\text{reactants}} and the level difference between products and reactants is the enthalpy change \Delta H = E_{\text{products}} - E_{\text{reactants}}. These are independent: a big hill can sit in front of an exothermic drop.

Derivation

Why temperature speeds things up (Arrhenius):

  1. 1.The fraction of collisions with energy \geq E_a follows a Boltzmann factor e^{-E_a/RT}.
  2. 2.So the rate constant is k = A\,e^{-E_a/RT}.
  3. 3.Raising T shrinks the exponent magnitude, so e^{-E_a/RT} grows — dramatically more collisions clear the hill.
≈9% exceed E_a (approx)
Raise T  the curve flattens and shifts right — far more molecules clear E_a, so the reaction speeds up.

Raise T → the shaded fraction past Eₐ grows → faster reaction.

Worked examples

Worked example

L2 · Curriculum Application

Q: A reaction has E_a = 50 kJ/mol and \Delta H = -80 kJ/mol. Sketch-wise, what is the activation energy of the REVERSE reaction?

  1. 1.Forward barrier from reactants to peak: 50 kJ/mol.
  2. 2.Products lie 80 kJ/mol BELOW reactants (\Delta H<0).
  3. 3.Reverse barrier = peak minus product level = E_a - \Delta H = 50 - (-80).
Answer: E_{a,\text{rev}} = 130 kJ/mol

Faster method: Reverse E_a = E_{a,\text{fwd}} - \Delta H (watch the sign).

Worked example

L4 · Advanced Reasoning

Q: Two reactions have equal \Delta H. Reaction A: E_a=75 kJ/mol. Reaction B has a catalyst giving E_a=25 kJ/mol, both at 300 K. By what factor is B faster (assume equal A)? Use R=8.314 J/mol·K.

  1. 1.Ratio \dfrac{k_B}{k_A} = e^{-(E_{a,B}-E_{a,A})/RT}.
  2. 2.E_{a,B}-E_{a,A} = (25-75)\times10^3 = -5.0\times10^4 J/mol.
  3. 3.Exponent = -(-5.0\times10^4)/(8.314\times300) = +20.05.
  4. 4.k_B/k_A = e^{20.05} \approx 5\times10^{8}.
Answer: \approx 5\times10^{8} times faster

Faster method: A 50 kJ/mol lower barrier at room T is worth ~8 orders of magnitude — why catalysts feel like magic.

Common misconception

A catalyst does not change \Delta H, the equilibrium position, or how much product is possible — it only lowers E_a (offers a new path), speeding both forward and reverse equally. Also: E_a \neq \Delta H. A large \Delta H release tells you nothing about how fast the reaction goes.

On the AP exam

The AP exam loves asking you to (1) label E_a and \Delta H on a diagram, (2) compare forward vs reverse E_a, and (3) justify — in words — that a catalyst lowers E_a without changing \Delta H. Always tie rate to "fraction of collisions with energy \geq E_a."

Concept map

CollisionsActivation energy EₐTransition stateRateCatalystTemperature
  • Collisions → reach if E ≥ Eₐ → Transition state
  • Activation energy Eₐ → higher → slower → Rate
  • Temperature → higher → faster → Rate
  • Catalyst → lowers → Activation energy Eₐ

Concept check

3 questions — answer before moving on.