Why do reactions go faster when it’s hot?

The spread of energies

The molecules in a gas don’t all have the same energy. Collisions constantly pass energy from one to another, so at any moment a few are almost still, most have middling energy, and a few have a lot. The Maxwell–Boltzmann distribution is the shape of that spread.

• The curve starts at zero: no molecule has exactly zero energy.
• It rises to a peak, the most likely energy, then falls away in a long tail.
• The tail never quite reaches the axis. There is no upper limit, just fewer and fewer molecules.
• The area under the curve is the total number of molecules.

The activation energy Eₐ sits out in the tail. Only the molecules in the shaded area beyond it have enough energy to react.

What heating does to the curve

At a higher temperature the average energy is higher, so the peak moves to the right. The number of molecules is the same, so the area must stay the same. The curve therefore gets lower and wider. Try it with the slider and watch the shaded tail.

The tail is where the action is. For a reaction with Eₐ = 50 kJ mol⁻¹, the share of molecules with at least that energy is about 9 in a billion at 298 K and about 17 in a billion at 308 K. A rise of 10 K, about 3% in absolute temperature, nearly doubles the share in the tail. That is the origin of the old rule of thumb that rates double every 10 °C, and why it only works for activation energies around that size.

The Arrhenius equation

The rate constant k captures how temperature and activation energy together set the rate:

R is the gas constant, 8.314 J K⁻¹ mol⁻¹, and T is in kelvin. The factor e^−Eₐ/RT is, roughly, the share of collisions with enough energy. A, the pre-exponential factor, includes how often molecules collide and how often they meet the right way round.

Taking natural logarithms gives a straight line:

Plot ln k against 1/T and the gradient is −Eₐ/R. Multiply the gradient by −8.314 to get Eₐ in J mol⁻¹, then divide by 1000 for kJ mol⁻¹. Getting the units of R and Eₐ to match is where most marks go.

Check the rule of thumb with it. For Eₐ = 50 000 J mol⁻¹, going from 298 K to 308 K multiplies k by e^{(50000/8.314)(1/298 − 1/308)} = e^0.655 ≈ 1.9.

Catalysts on the same graph

A catalyst doesn’t change the curve at all: the molecules have the same energies. It provides a route with a lower activation energy, which moves the Eₐ line to the left, so a bigger share of the same molecules is now in the tail. Use the Eₐ slider to see how strongly the share depends on it.