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AP® Chemistry · Unit 5 · Topic 5.5

Activation Energy and Temperature

The Arrhenius equation links the rate constant k to temperature and activation energy, showing that rate jumps sharply with temperature because far more molecules collide with enough energy to react.

Activation energy, Ea, is the energy barrier a collision must clear before reactants can become products. At any temperature, molecular speeds and energies spread out over a range described by the Maxwell-Boltzmann distribution. Only the molecules in the high-energy tail, past Ea, react on collision. Raising the temperature shifts the whole distribution slightly, but it stretches that tail a lot. A small bump in average kinetic energy can multiply the fraction of molecules above Ea several times over, which is why rate constants often double or triple for just a 10 K rise even though average speed barely changes.

The Arrhenius equation, k = A e−Ea/RT, makes this precise. Taking the natural log gives ln k = -Ea/R times (1/T) plus ln A, a straight line when you plot ln k against 1/T. The slope of that line is -Ea/R, so measuring k at several temperatures and plotting lets you pull Ea straight out of the graph. With only two data points, the two-point form does the same job algebraically: Ea = R ln(k2/k1) divided by (1/T1 - 1/T2).

This sits right after the rate law and the idea of a rate-determining step, because Ea and A describe that slow step specifically, not the overall reaction. A catalyst speeds things up by lowering Ea for that step, which you can see directly as a change in the slope of the Arrhenius plot, not a change in concentration dependence.

Key terms

Activation energy (Ea)
The minimum energy a collision needs for reactants to convert to products.
Arrhenius equation
k = A e−Ea/RT, the equation relating the rate constant to temperature and activation energy.
Maxwell-Boltzmann distribution
The spread of molecular kinetic energies at a given temperature, showing what fraction exceed any energy value.
Pre-exponential factor (A)
The constant in the Arrhenius equation accounting for collision frequency and orientation, independent of temperature's effect on energy.
Arrhenius plot
A graph of ln k versus 1/T whose slope equals -Ea/R, used to extract activation energy experimentally.

Watch it explained

  1. How it worksHow does the Maxwell-Boltzmann tail past the activation energy explain the jump in rate?
  2. Worked exampleGraph ln k against 1 over T and show how the slope gives the activation energy.
  3. Worked exampleFind the activation energy from rate constants of 0.0025 at 300 K and 0.0400 at 350 K.
  4. Common mistakeWhy isn't the rate constant a constant if it is called a constant?

Check yourself

Your turn1 of 2
Find the activation energy, in kJ/mol, given rate constants of 7 x 10−3 at 306 K and 85 x 10−3 at 344 K (same units for both). (Use the two-point Arrhenius equation, R = 8.314 J/(mol K).)

End of topic.