Topic 5.11: Catalysis

Catalysis: a new path on platinum

Inside a car’s catalytic converter, every bit of the exhaust squeezes through thousands of tiny channels, and if we zoom into one channel wall, past its rough coating and onto one of the metal specks, we land on a surface made of platinum atoms. Chemists have known since the 1820s that platinum can make gases react without being used up, and it was a German chemist, Gerhard Ertl, who worked out the actual steps on its surface, part of the work that won him the 2007 Nobel Prize in Chemistry.

First, a carbon monoxide molecule lands carbon end down on this platinum atom and sticks there, held by a real bond to the surface. Then an oxygen molecule lands across two free platinum atoms that sit right next to each other, and it falls apart into two oxygen atoms, each one bonded to the surface.

On platinum that happens even at minus a hundred and eight degrees Celsius, colder than the coldest day ever recorded anywhere on Earth, while pulling an oxygen molecule apart in the gas, with nothing to hold onto, takes nearly five hundred kilojoules per mole. Now a bound carbon monoxide and a bound oxygen atom slide together, carbon dioxide forms and leaves the surface right away, and our platinum atom is left exactly the way it started, free for the next one.

And then it happens again, and again, the same atom grabbing a molecule, letting go of the product, and coming back as good as new every time. The carbon monoxide and the oxygen atoms stuck to the metal are brand-new intermediates that don’t exist anywhere in the gas, so the platinum gives the reaction a different set of steps, a whole new mechanism.

Here’s what happens if we swap the metal for bare ceramic, with the same gas at the same temperature. The molecules hit the wall and bounce right off, and nothing forms at all. Same gas, same temperature, and the only difference is the route the metal opens up for the molecules. Now I’ll hand you an oxygen molecule, and you can run the surface yourself. Land this oxygen molecule somewhere it can split apart, and let’s watch the surface make a carbon dioxide.

Same start, same finish, lower climbs

Let’s keep that surface in mind and draw what it just did. Ertl’s lab measured the energies of these steps, and this is the real energy profile from his Nobel lecture, for carbon monoxide and oxygen on platinum.

When the carbon monoxide and the oxygen stick to the metal, the energy drops by two hundred sixty kilojoules per mole, and then the surface step climbs a hundred to its transition state. Carbon dioxide forms while it’s still stuck to the metal, it takes twenty-one more to pull it off into the gas, and from start to finish the energy drops by two hundred eighty-three kilojoules per mole.

Here’s one for you about Ertl’s picture, so how high is the transition state of the surface step, compared with the starting gases, in kilojoules per mole? The top of the hill on the platinum route sits below where the gases started. That doesn’t make the climb free, though. Sticking hands those two hundred sixty to the metal as heat.

From down in that valley, the molecules still climb a hundred. That’s far less than they’d climb without the metal. You might be tempted to add up the climbs, a hundred plus twenty-one, and call that the activation energy, but the climbs aren’t a bill the molecules pay all at once, because each one is measured from the valley it starts in.

Now here’s the route without the metal, where carbon monoxide and oxygen have to react in the gas on their own, and its climb is so much higher that it runs right off the top of the board. But look at where the two routes start and finish, because it’s exactly the same place, so the energy change is minus two hundred eighty-three either way. A catalyst changes the route a reaction takes, never its start or its finish.

At any temperature, the molecules have a spread of energies, a few very fast ones and lots of slower ones, and only the collisions with enough energy to get over the climb can react, which is this thin sliver out here. Give the reaction a lower climb, and at the very same temperature a much bigger share of the collisions clears it, so more of the collisions are effective, and the reaction runs faster.

Holding the molecules in place on the surface, already lined up, helps too, because more of their meetings happen the right way round. So ‘a catalyst adds energy’ gets crossed out, because the molecules have exactly the same energies they had before. And ‘a catalyst lowers the energy of the products’ is wrong too, because the finish never moved. A catalyst can’t change where a reaction ends up, only how fast it gets there, and it speeds up the reverse reaction just as much, which you’ll see again in Unit 7.

Here are four drawings, each with the original route in grey underneath. Which of these four drawings shows the same reaction, at the same temperature, with a catalyst that gives it a lower-energy route?

Used up, then given back

Chemists write surface steps with a star for a free spot on the metal, so let’s write Ertl’s steps that way. Carbon monoxide plus a free site gives bound carbon monoxide, oxygen plus two free sites gives two bound oxygen atoms, and a bound carbon monoxide plus a bound oxygen gives carbon dioxide and two free sites back.

Each oxygen molecule makes two bound oxygens, so we use the first and third steps twice and the second step once, and add everything up. The bound carbon monoxides cancel, the bound oxygens cancel, and four free sites on the left cancel four on the right, which leaves two CO plus O two gives two CO two, with no platinum anywhere in the overall equation.

Both the stars and the bound molecules cancel, but they play opposite roles. The free sites get used up first and handed back later, and that’s the catalyst, while the bound carbon monoxide and oxygen get made first and used up later, and those are the intermediates. Cancelling can’t tell them apart, because both cancel, so you have to look at the order, which one shows up on the left first and which one shows up on the right first.

A catalyst that binds a reactant works like this, grabbing the reactant to make a new intermediate and coming back out in a later step, so the total amount of catalyst stays the same, even though it’s used up in one step, very often the slow one, and regenerated in a later step, and that’s what ‘never used up’ really means for the metal in a car’s converter. That’s also why showing up on both sides isn’t enough to make something a catalyst. Back in our exhaust mechanism, NO two showed up on both sides too, but two of them got used and only one came back, so NO two is just a reactant.

Here’s a mechanism from the old way of making sulfuric acid, where sulfur dioxide gets turned into sulfur trioxide. In this two-step mechanism, which species is the catalyst, and which one is the intermediate?

Enzymes and protons

A metal surface isn’t the only thing that can grab a molecule and give it a new route. The carbon dioxide coming out of that converter is the same gas your own cells make all day long, and to carry it away, your blood relies on an enzyme packed inside every red blood cell, called carbonic anhydrase.

Most enzymes are catalysts made of protein, like the ones in the bombardier beetle’s spray chamber earlier in this unit, and each molecule of this one turns carbon dioxide and water into bicarbonate close to a million times every second. Deep inside the protein sits a zinc ion holding onto a hydroxide, and a carbon dioxide slides into a little pocket right beside it, held so that its carbon faces the hydroxide.

Scientists have actually caught carbon dioxide sitting in exactly that spot, in a crystal of the enzyme, and the enzyme holding its carbon dioxide like that is a new intermediate, with the catalyst bound to the reactant. Then the hydroxide’s oxygen bonds to the carbon, and now we have bicarbonate stuck to the zinc, a second new intermediate, this time with the product still bound to the catalyst.

A water molecule pushes the bicarbonate off, and then that water hands one of its protons away along a little chain of water molecules, which turns it back into a hydroxide, ready for the next carbon dioxide. That proton hand-off is acid–base catalysis, a step where something gains or loses a proton, and it’s a step that only exists because the enzyme is there.

Out in plain water, carbon dioxide and water bump into each other every which way, and most of those collisions don’t work, because they happen the wrong way round or without enough energy, so the molecules just bounce apart. The enzyme fixes that by holding the carbon dioxide just right, and now you get to do the holding.

Turn this carbon dioxide so the enzyme can react with it. And when that blood reaches your lungs, the very same enzyme runs the reaction backwards, turning bicarbonate back into carbon dioxide for you to breathe out, because a catalyst speeds up a reaction in both directions.

The sentence that earns the point

That proton hand-off has a simpler cousin called acid catalysis, where a hydronium ion hands a proton to an oxygen on the reactant and makes a brand-new O–H bond, that protonated molecule is a new intermediate that reacts more easily, and later a water molecule takes a proton back, which gives you the hydronium ion again. On the 2025 AP exam, students got this proposed two-step mechanism for making glycolic acid, and the question said that a student claims H three O plus is a catalyst, then asked whether you agree, with a justification based on the mechanism.

Hydronium is on the left in step one, so it gets used up, and it’s on the right in step two, so it comes back, while this ion, C two H five O three plus, gets made in step one and used up in step two, so it’s the intermediate. So which of these justifications would earn the point for ‘H three O plus is a catalyst’?

And the sentence that earns the point is this one, agree, because H three O plus is consumed in step one and regenerated in step two. Only about half the students earned this point, and the graders noted that answers like ‘it lowers the activation energy’ or ‘it speeds up the reaction’ weren’t enough on their own, because those describe catalysts in general, and the question asked for evidence from this mechanism, which is the order of the steps.

Why the metal is still there

So let’s go back to that platinum surface one last time. The metal gives carbon monoxide and oxygen a new route, with intermediates stuck to its surface and much smaller climbs, and the platinum atoms get used in one step and handed back in the next, over and over, for years, while the reaction still starts and finishes in exactly the same place. That’s how a few grams of metal can keep cleaning a car’s exhaust for years without being used up, and it’s why the metal was still there, worth stealing, when the thief crawled under that car.