Catalysts and reaction mechanism

The converter in a car

The exhaust of a car holds carbon monoxide and unburnt fuel. Both are harmful. A box in the exhaust pipe, the catalytic converter, turns them into carbon dioxide and water within a second or two. If the same gases were left alone in the air, the change would take years.

The box does not get used up. After thousands of kilometres it is still there and still working. This is what a catalyst does.

A catalyst is a substance that increases the rate of a reaction and is not used up in the reaction. Small amounts are often enough.

How a catalyst works

A catalyst does not give molecules more energy. It gives them a different road. The new path has a lower activation energy.

Look at the picture. The blue curve is the path without the catalyst. The barrier is $98\ \text{kJ/mol}$. The dashed green curve is the path with the catalyst, and its barrier is $62\ \text{kJ/mol}$. The molecules at the start and the end are the same in both cases, so those two levels are the same.

A lower barrier means a larger share of collisions with enough energy. The Arrhenius equation tells us how much. If the pre-exponential factor stays the same,

where $\Delta E_a$ is how much the barrier was lowered.

Example: the speed-up

Here $\Delta E_a = 98 - 62 = 36\ \text{kJ/mol}$. At $340\ \text{K}$:

So $\dfrac{k_{cat}}{k} = e^{12.7} = 3.4 \times 10^{5}$. The reaction goes about three hundred thousand times faster. A drop of $36\ \text{kJ/mol}$ is a large gain, and this is why catalysts are so useful.

What a catalyst changes, and what it does not

Think about the profile again. A catalyst does not alter the reactants or the products. It only changes the hill between them. So:

• It lowers the barrier in both directions by the same amount. For the reaction above, $\Delta H = -31\ \text{kJ/mol}$. The reverse barrier is $98 + 31 = 129\ \text{kJ/mol}$ without the catalyst and $62 + 31 = 93\ \text{kJ/mol}$ with it. Both fell by $36\ \text{kJ/mol}$.
• It does not change $\Delta H$. The energy given out or taken in depends on the reactants and the products only.
• It does not change the equilibrium. Forward and reverse reactions are sped up equally, so the equilibrium constant stays the same. The catalyst helps a reaction get to equilibrium sooner, and not to a different one.
• It cannot start a reaction that cannot happen. If $\Delta G$ is positive for a reaction, no catalyst will make it run.
• It is not used up. It may change in form, for example a powder may become coarser. Its amount stays the same.
• It can be specific. Different catalysts can send the same reactant to different products. Ethanol passed over hot copper gives ethanal, and over hot alumina it gives ethene.

Kinds of catalysis

In homogeneous catalysis the catalyst is in the same phase as the reactants. The chlorine atoms that break down ozone in the upper atmosphere are an example, because both are gases.

In heterogeneous catalysis the catalyst is in a different phase, usually a solid with gas or liquid reactants. Examples:

• Iron in the making of ammonia, $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$.
• Vanadium(V) oxide in the making of sulphur trioxide for sulphuric acid.
• Platinum and rhodium in the catalytic converter.
• Finely divided nickel in the hydrogenation of vegetable oils.

On a solid catalyst, the reaction goes in steps.

1. The reactant molecules stick to the surface. This is adsorption.
2. The bonds in the adsorbed molecules weaken, and the molecules react on the surface.
3. The products leave the surface. This is desorption, and it frees the place for new molecules.

Only the spots where reactants can stick, the active sites, matter. A finely divided catalyst has far more surface than a lump of the same mass, so it works better.

Two more terms are useful. A promoter is a substance that makes a catalyst work better. Molybdenum improves the iron in ammonia production. A poison is a substance that stops a catalyst from working, by sticking to its active sites. Lead compounds poison the converter, which is why petrol for modern cars has no lead in it.

Enzymes and autocatalysis

Living things run on enzymes, which are catalysts made of protein. Each enzyme fits its own reactant, in the way that a key fits a lock. The enzyme urease breaks urea into ammonia and carbon dioxide, and nothing else. Enzymes work at the temperature of the body, about $310\ \text{K}$, and at a suitable pH.

The rate of an enzyme reaction rises with temperature at first, like any reaction. Then it falls. At a higher temperature the protein loses its shape, and the enzyme stops working. This is why the Arrhenius equation applies to an enzyme only in the range where the enzyme is intact.

In autocatalysis, one product of the reaction is a catalyst for it. A reaction of potassium permanganate with oxalic acid starts slowly. The manganese(II) ions that form speed it up, so after a while the colour fades quickly.

The catalyst in a mechanism

A catalyst takes part in the steps, even though it does not show in the overall equation. It is used in one step and made again in a later one. Here is a known example, with chlorine atoms and ozone:

• Step 1, slow: $\text{Cl} + \text{O}_3 \rightarrow \text{ClO} + \text{O}_2$
• Step 2, fast: $\text{ClO} + \text{O} \rightarrow \text{Cl} + \text{O}_2$

Add the two steps. $\text{Cl}$ and $\text{ClO}$ both cancel. What remains is $\text{O}_3 + \text{O} \rightarrow 2\text{O}_2$.

• $\text{Cl}$ is the catalyst. It is there at the start and is given back at the end.
• $\text{ClO}$ is an intermediate. It is made in step 1 and used up in step 2.

The slow step decides the rate, so the rate law is $\text{rate} = k[\text{Cl}][\text{O}_3]$. The catalyst appears in the rate law, though not in the overall equation. This is how a catalyst can change the rate: it is part of the slow step.

Three names are easy to confuse. A catalyst is present at the start and the end. An intermediate is made and used up. An activated complex is the peak of a barrier and is never separated.

In the simulation, the starting level and the ending level stay in place whatever you do to the barrier. Check that the reverse barrier falls by the same amount as the forward one.

Practise with new numbers

Try it in Python

The program works out the speed-up for three catalysts, in the same reaction, at one temperature. It also finds the reverse barrier of each.

Common mistakes

• Saying a catalyst changes $\Delta H$ or the equilibrium. It changes neither. It only changes how fast the equilibrium is reached.
• Lowering only the forward barrier. The reverse barrier falls by the same amount.
• Thinking a catalyst supplies energy. It does not. It gives a path with a lower hill.
• Calling a catalyst a reactant, or listing it in the overall equation. It cancels between the steps.
• Mixing up a catalyst and an intermediate. A catalyst is used and then made again. An intermediate is made and then used.

Remember these

• A catalyst gives a path with a lower activation energy and is not used up.
• $\dfrac{k_{cat}}{k} = e^{\Delta E_a/RT}$ when $A$ stays the same.
• Both barriers fall by the same amount. $\Delta H$ and the equilibrium constant do not change.
• A catalyst cannot make a reaction happen that is impossible by $\Delta G$.
• Solid catalysts work by adsorption on active sites. Poisons block them, and promoters help them.
• Enzymes are protein catalysts. They are specific and work in a narrow range of temperature.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A catalyst is added to a reaction. Does the enthalpy change of the reaction alter?

2. A catalyst lowers the forward barrier by $30\ \text{kJ/mol}$. By how much does it lower the reverse barrier?

3. Can a catalyst make a reaction with a positive $\Delta G$ run?

4. Why is a catalyst in the form of a fine powder more effective than the same mass in one lump?

5. A reaction goes by two steps. Step 1: $\text{A} + \text{X} \rightarrow \text{AX}$. Step 2: $\text{AX} + \text{B} \rightarrow \text{C} + \text{X}$. Name the catalyst and the intermediate, and write the overall equation.