How a catalyst speeds up a reaction

Hydrogen peroxide breaks down into water and oxygen on its own, but slowly. Add iodide ions and it goes faster. Add catalase, an enzyme you can extract from lettuce, and it goes faster still. Nothing in the flask has been given extra energy. The reaction has been handed a lower hill to climb.

What a catalyst is

The OpenStax chemistry text defines a catalyst as “a substance that can increase the reaction rate without being consumed in the reaction.” It works, in the same text's words, “by providing an alternative reaction mechanism with a lower activation energy”. The reactants still end up as the same products. They simply take a different route between them.

Why it is worth knowing

Much of chemistry runs on this. OpenStax lists ammonia, nitric acid, sulfuric acid and methanol as industrial products made with solid catalysts, and the platinum–rhodium surface in a car's catalytic converter as another. Your own cells run on enzymes, catalysts that are usually proteins. Cheung reports that one molecule of catalase can turn 4 × 10⁷ molecules of hydrogen peroxide into products every second under the best conditions. The same idea explains harm, too: a single chlorine atom in the upper atmosphere, OpenStax notes, “can break down thousands of ozone molecules.”

The hill, and who gets over it

Molecules react when they collide hard enough. The minimum energy a collision needs is the activation energy, written E_a. Draw the energy of the reacting system as it goes from reactants to products and you get a hill: E_a is the height of the top above the starting point.

The Arrhenius equation turns the height into a rate: k = A e^−E_a/RT. Here k is the rate constant, R = 8.314 J/mol/K, T is the temperature in kelvin, and A reflects how often molecules meet in a useful orientation. The exponential factor is, as Cheung puts it, the proportion of molecules with at least the activation energy. That share is tiny for a tall hill. For uncatalysed hydrogen peroxide, with E_a = 75 kJ/mol at 25 °C, Cheung prints 6.9 × 10⁻¹⁴; recomputing his expression gives 7.1 × 10⁻¹⁴ (our arithmetic). Either way it is well under one collision in ten trillion.

One reaction, three hills

Cheung gathers literature values for this single reaction. Uncatalysed, the hill is 75 kJ/mol. With iodide it is 56. With catalase it is about 18, and his students measured 19 using catalase from lettuce. He warns that the enzyme's value depends on where the catalase came from, so treat it as “about 18–19”.

Because the hill sits in an exponent, a modest drop multiplies the rate enormously. With A held fixed, the rate goes up by e^ΔE_a/RT. Cheung's summary: lowering the barrier from 75 to 55 kJ/mol raises the rate about 10³ times, and catalase beats iodide by about 10⁶. Our arithmetic at 25 °C agrees: 75 to 55 gives 3.2 × 10³, and 55 to 19 gives 2.0 × 10⁶.

What a catalyst does not do

It does not change where the reaction ends up. OpenStax says the catalysed and uncatalysed curves “begin and end at the same energies”, so the energy released or absorbed overall is the same. Since the product level does not move, the hill seen from the products side drops by exactly as much as the forward hill. The reverse reaction speeds up by the same factor, and the balance point between the two, the equilibrium, stays put (our reasoning from the diagram). A catalyst gets you there sooner; it does not get you more.

“Not consumed” does not mean “not involved”. In OpenStax's ozone example, nitric oxide is a reactant in the first step and a product in the last, so it comes out as it went in, ready to go again. The new route can even have more steps. OpenStax's own diagram shows a catalysed path with two humps and a dip between them, where an intermediate briefly exists. What matters is the highest hump, the slowest step. In one of their worked exercises that first step is 80 kJ without the catalyst and 70 kJ with it, a 10 kJ drop worth a factor of about 57 at 25 °C (our arithmetic; the book gives no temperature).

Heat does the same arithmetic

Temperature sits in the same exponent. OpenStax says the exponential term describes the effect of activation energy and also the effect of temperature. At 25 °C, each 5.7 kJ/mol taken off the hill multiplies the rate by ten, and taking 1 kJ/mol off the 75 kJ/mol hill does what warming it by 4 °C would do (our arithmetic). Small drops trade for small temperature changes. The large drops do not: matching catalase by heating alone would take the uncatalysed reaction to nearly 970 °C on paper (our arithmetic).

In short

A catalyst opens a different route from the same reactants to the same products, with a lower highest hill. The share of collisions energetic enough to cross depends exponentially on that height, so taking 20 kJ/mol off makes hydrogen peroxide break down about a thousand times faster. The catalyst comes back unchanged, and the equilibrium does not move.