Activation energy

Mix hydrogen and oxygen gas and nothing happens — even though burning them releases a huge amount of energy. Strike a match near the same mixture and it explodes. Why the wait? Every reaction, however energetically favourable, must first climb an energy hill before it can roll down to products. The height of that hill is the activation energy, $E_a$ — the extra energy reactant molecules need above their own average energy before old bonds can break and new bonds can form. This article builds the energy-diagram picture behind $E_a$ : threshold energy, the fleeting activated complex at the top of the hill, how exothermic and endothermic reactions differ, and why a catalyst is really just a lower, easier hill rather than a different destination. The ideas here are the qualitative foundation for collision theory and the Arrhenius equation later in Chemical Kinetics.

The core insight

Activation energy is the extra energy — above the reactants' average energy — that colliding molecules must acquire to reach the unstable activated complex at the top of the energy barrier before they can become products.

The core idea

Threshold energy is the minimum total energy a colliding pair of reactant molecules must possess for their collision to have any chance of producing products. At ordinary temperatures, molecular energies spread over a wide range, and only a fraction of molecules ever carry enough energy at a given instant to clear this threshold — most collisions are simply too gentle to react.

Activation energy $E_a$ measures the same barrier from the reactants' side: it is the extra energy reactants must gain above their own average energy to reach the threshold.

E_a = E_{\text{threshold}} - E_{\text{reactants (average)}}

where: $E_{\text{threshold}}$ is the minimum energy needed for reaction, and $E_{\text{reactants (average)}}$ is the average energy the reactant molecules already carry. Graphically, $E_a$ is simply the height of the hump on a potential-energy-versus-reaction-progress diagram, measured from the reactants' energy level up to the peak. At that peak sits the activated complex (also called the transition state) — a strained, high-energy arrangement of atoms in which old bonds are half-broken and new bonds are half-formed. It is not a real, isolable substance; it exists only for the instant of collision and can fall either forward to products or backward to reactants.

Exothermic and endothermic reactions look different on this diagram, even though both need $E_a$ to start. In an exothermic reaction the products sit lower than the reactants ( $\Delta H < 0$ ): the reaction still has to climb the forward barrier $E_{a(f)}$ , but once over the top it falls further than it climbed, releasing net energy overall. The reverse reaction, going from the lower products back up to the same peak, therefore has to climb higher: $E_{a(\text{reverse})} = E_{a(\text{forward})} + |\Delta H|$ . In an endothermic reaction the products sit higher than the reactants ( $\Delta H > 0$ ), so the forward climb must itself be at least as large as $\Delta H$ , since the reaction absorbs net energy.

Figure

Energy diagram showing reactants climbing an activation energy hill to a peak activated complex then descending further to lower energy products in an exothermic reaction

Effect of a catalyst. A catalyst opens up an alternative reaction pathway with a lower activation energy than the uncatalysed route, without touching the energy levels of the reactants or products. Because the hill is lower, a larger fraction of molecules clears it at the same temperature, so the rate rises. The catalyst speeds up the forward and reverse reactions equally, so it never changes $\Delta H$ and never shifts the equilibrium — it only gets the system to equilibrium faster.

Derivation ARelating the reverse activation energy to the forward one

The goal is to show, using nothing more than reading heights off an energy diagram, that for any exothermic reaction the reverse activation energy exceeds the forward one by exactly the magnitude of the enthalpy change, $E_{a(\text{reverse})} = E_{a(\text{forward})} + |\Delta H|$ , so that either quantity can be recovered from the other two without redrawing the whole diagram each time.

Step 1. Fix a common zero. Place the reactants' average energy at height $0$ on the energy axis. The activated complex, the peak of the hill, then sits at height $E_{a(f)}$ above this zero, by the very definition of forward activation energy.

Step 2. Locate the products. For an exothermic reaction the products release net energy, so they sit below the reactants by $|\Delta H|$ — that is, the products are at height $-|\Delta H|$ on the same axis.

Step 3. Measure the reverse climb. The reverse reaction starts at the products' height, $-|\Delta H|$ , and must climb up to the same activated-complex peak at height $E_{a(f)}$ , since forward and reverse reactions pass through the identical transition state. The size of that climb is the difference in heights: $E_{a(\text{reverse})} = E_{a(f)} - (-|\Delta H|) = E_{a(f)} + |\Delta H|$ .

Step 4. Read off the result. Because $|\Delta H| \geq 0$ always, this proves $E_{a(\text{reverse})} \geq E_{a(\text{forward})}$ for every exothermic reaction — the downhill direction always has the smaller barrier, and the uphill (reverse) direction always has the larger one, with the gap between them fixed exactly by $|\Delta H|$ .

Worked example

Question. For a certain exothermic reaction, $\Delta H = -120\ \text{kJ mol}^{-1}$ and the forward activation energy $E_{a(f)} = 60\ \text{kJ mol}^{-1}$ . Find the activation energy for the reverse reaction.

Step 1. Take the reactants' average energy as the zero of the diagram. The activated complex then sits $E_{a(f)} = 60\ \text{kJ mol}^{-1}$ above the reactants, by definition.

Step 2. Since the reaction is exothermic with $\Delta H = -120\ \text{kJ mol}^{-1}$ , the products sit $120\ \text{kJ mol}^{-1}$ below the reactants, i.e. at height $-120\ \text{kJ mol}^{-1}$ .

Step 3. The reverse reaction climbs from the products (height $-120$ ) up to the same activated complex (height $60$ ), a climb of $60-(-120)=180\ \text{kJ mol}^{-1}$ .

Step 4. Check against the derived relation: $E_{a(\text{reverse})} = E_{a(\text{forward})} + |\Delta H| = 60 + 120 = 180\ \text{kJ mol}^{-1}$ , which matches. The reverse activation energy is $\mathbf{180\ \text{kJ mol}^{-1}}$ .

Competitive stretch

A reaction has $E_a = 75\ \text{kJ mol}^{-1}$ uncatalysed and $\Delta H = -40\ \text{kJ mol}^{-1}$ throughout. Adding a catalyst drops the forward $E_a$ to $50\ \text{kJ mol}^{-1}$ . What happens to the reverse activation energy, and does the catalyst change the equilibrium constant $K$ ? Uncatalysed, $E_{a(\text{reverse})} = 75 + 40 = 115\ \text{kJ mol}^{-1}$ . Catalysed, $E_{a(\text{reverse})} = 50 + 40 = 90\ \text{kJ mol}^{-1}$ — the reverse barrier drops by the same $25\ \text{kJ mol}^{-1}$ as the forward one, keeping the gap fixed at $|\Delta H|=40\ \text{kJ mol}^{-1}$ . Since $\Delta H$ (and hence $K$ ) depends only on reactant and product energies, not on the pathway between them, the catalyst leaves $\Delta H$ and $K$ unchanged; it only makes both directions faster and reaches equilibrium sooner.

Common traps

Treating $E_a$ as the total energy reactants need to react, rather than the extra energy above their own average energy.
Assuming a catalyst changes $\Delta H$ — it only lowers $E_a$ for both directions equally; reactant and product energy levels never move, so $\Delta H$ and $K$ are untouched.
Believing the activated complex is a real, isolable substance rather than a fleeting high-energy arrangement that instantly proceeds forward or falls back.
Forgetting that $E_{a(\text{reverse})} = E_{a(\text{forward})} + |\Delta H|$ applies to exothermic reactions (reverse climbs further); for endothermic reactions the forward barrier is the larger one instead.

Ask the tutor: if you want a step-by-step sketch of the energy-profile diagram for an endothermic reaction of your choice, ask your AI tutor to walk through it.

Why do reactants need extra energy at all if the reaction releases energy overall?

Because reacting is not just about the final energy difference — it is about physically breaking old bonds before new ones can form, and breaking bonds always costs energy up front. Even a strongly exothermic reaction, like hydrogen burning in oxygen, must first pay this bond-breaking cost to reach the activated complex before it can collect the larger payoff of forming new, more stable bonds. That upfront cost is $E_a$ ; the net energy released afterward is $\Delta H$ . They are two separate quantities describing two separate stages of the same reaction.

Does a catalyst let a reaction happen that would otherwise be impossible?

No. A catalyst cannot make an energetically unfavourable reaction favourable, because it never changes $\Delta H$ or the relative energies of reactants and products. What it does is provide an alternative route with a lower $E_a$ , so a reaction that was already thermodynamically possible but painfully slow now proceeds at a usable rate. If a reaction has no accessible lower-energy pathway available under the given conditions, a catalyst will not conjure one out of nothing — it only exploits pathways that exist.

If most molecules do not have enough energy to react, why does the reaction still happen at ordinary temperature?

Because molecular energies are spread over a wide range at any instant, and even though only a small fraction of molecules sit above the threshold energy at a given moment, that fraction is constantly being replenished as molecules collide and exchange energy. So while any single molecule may lack enough energy right now, some molecules always do, and reaction proceeds continuously through this ever-refreshed high-energy tail of the population — just more slowly than it would at a higher temperature, where a larger fraction clears the threshold.

Activation energy

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads