Back to the on-screen lesson ·
Industrial catalytic cycles traced step by step, the rate-determining step, and measuring a catalyst by its turnover number, turnover frequency and loading.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to trace a catalytic cycle and find a catalyst's turnover number, turnover frequency and loading from a run.
Lesson 30 named the elementary steps, oxidative addition, reductive elimination, migratory insertion and beta-hydride elimination, and tracked a metal's oxidation state and electron count through each. From general chemistry you know that a catalyst speeds a reaction without being used up. This lesson strings the steps into complete cycles and measures how much work a catalyst does before it stops.
| Term | What it means |
|---|---|
| Catalytic cycle | A closed sequence of elementary steps that turns reactants into products and regenerates the catalyst. |
| Turnover | One trip round the cycle by one catalyst molecule, making one molecule of product. |
| Turnover number, TON | Moles of product made per mole of catalyst before the catalyst is spent. |
| Turnover frequency, TOF | Turnovers per unit time, usually per hour or per second: how fast the catalyst works. |
| Catalyst loading | Moles of catalyst as a percentage of moles of substrate. |
| Rate-determining step | The slowest step of the cycle, which sets how fast the whole cycle turns. |
| Catalyst deactivation | Any process, such as decomposition or poisoning, that stops a catalyst from turning over. |
In a catalytic cycle the metal passes through a series of elementary steps and comes back to where it started, having turned reactants into product on the way. Each trip round the loop is one turnover. Because the catalyst is regenerated, one metal center can make thousands or millions of product molecules, which is why a few grams of rhodium can run a chemical plant.
Every cycle has the same shape. The metal takes up a reactant, usually by oxidative addition or by binding a molecule to an open site. It builds the new bond while the pieces sit on the metal, often by migratory insertion. And it releases the product, usually by reductive elimination, which returns the metal to its starting oxidation state. The Monsanto process for acetic acid, for instance, runs: rhodium(I) adds methyl iodide; the methyl migrates onto carbon monoxide; a fresh carbon monoxide fills the open site; acetyl iodide is eliminated, and water turns it into acetic acid.
Industry judges a catalyst by two numbers. The turnover number is how many molecules of product each catalyst molecule makes before it dies:
$$\text{TON} = \frac{\text{moles of product}}{\text{moles of catalyst}}.$$
The turnover frequency is how fast it works:
$$\text{TOF} = \frac{\text{TON}}{\text{time}}.$$
If $0.001$ mol of Wilkinson's catalyst hydrogenates $2.0$ mol of an alkene in $95$ percent yield over two hours, it makes $1.9$ mol of product, a turnover number of $1{,}900$ and a turnover frequency of $950$ per hour. A related number, the catalyst loading, is the catalyst as a percentage of the substrate, here $0.05$ mole percent. A lower loading means each catalyst molecule must turn over more times, and costs less metal.
Another way: picture
Think of the catalyst as a worker on an assembly line who takes a part, fits a piece to it and hands on the finished item, then turns back for the next part. The turnover frequency is how many items one worker finishes each hour; the turnover number is how many they finish before they have to stop for good. A good worker is both fast and tireless.
Another way: steps
Hydrogenation with Wilkinson's catalyst. $\mathrm{RhCl(PPh_3)_3}$, rhodium(I) with sixteen electrons, loses a phosphine to open a site, adds hydrogen by oxidative addition to become rhodium(III), binds the alkene, inserts it into a rhodium-hydrogen bond to make an alkyl, and eliminates the alkane by reductive elimination, returning to rhodium(I).
Acetic acid by the Monsanto process. $\mathrm{[Rh(CO)_2I_2]^-}$ adds methyl iodide, the methyl migrates onto a carbon monoxide to form an acetyl, another carbon monoxide binds, and acetyl iodide is eliminated. The rate-determining step is the oxidative addition of methyl iodide, so the reaction runs faster the more methyl iodide there is.
A Suzuki coupling. Palladium(0) adds an aryl bromide by oxidative addition; the second carbon group is transferred from a boron compound onto the palladium, a step called transmetalation; the two carbon groups move next to each other; and reductive elimination releases them joined, regenerating palladium(0). In each cycle the oxidation state goes up by two and comes back down by two, exactly as the bookkeeping of lesson 30 requires.
No catalyst lasts forever. Metal centers clump together into inactive metal particles, ligands decompose or react with the substrate, and impurities in the feed poison the metal by binding more strongly than the reactant. Sulfur compounds are notorious poisons for palladium and platinum, the soft-soft binding of lesson 18, which is why feedstocks for catalytic processes are cleaned of sulfur first.
The turnover number measures how long a catalyst survives all this, and the turnover frequency how quickly it works while it does. Both matter economically. A fast catalyst that dies after a few hundred turnovers can cost more than a slower one that lasts a million, because the metal, often rhodium, palladium or iridium, is expensive. Industrial processes that use precious metals also recover the metal from the spent catalyst and recycle it, which can make even a modest turnover number affordable.
The turnover frequency is set by the slowest step in the cycle, the rate-determining step. Speeding up any other step makes no difference, just as widening one stretch of a road does not help if traffic is stuck at a single narrow bridge. Finding the rate-determining step tells a chemist what to change.
If oxidative addition is slowest, an electron-richer metal or more reactive substrate helps: that is why aryl iodides couple faster than aryl bromides, and bromides faster than chlorides, and why chemists developed special electron-rich phosphines to make cheap aryl chlorides usable. If reductive elimination is slowest, bulkier ligands that crowd the two groups together help. Kinetic measurements, how the rate depends on each reactant's concentration, reveal which step is slowest: in the Monsanto process the rate depends on methyl iodide but not on carbon monoxide, pointing to the oxidative addition.
The catalysts of this unit are homogeneous: they dissolve in the reaction mixture, every metal center is the same, and each can be studied as a single molecule. That is why their cycles can be traced step by step and their ligands tuned one atom at a time for speed and selectivity. Their drawback is separation: the catalyst ends up mixed with the product and must be recovered, which matters when the metal is rhodium or the product is a medicine.
Heterogeneous catalysts are solids, such as platinum on alumina in a car's catalytic converter or iron in the Haber process for ammonia. The reaction happens at the surface, the catalyst stays behind when the gases or liquids flow past, and separation is free. But the surface has many kinds of site, some far more active than others, so a turnover number is averaged over sites that may behave very differently. Chemists increasingly borrow ideas across the divide, anchoring well-defined molecular catalysts onto solid supports to get the precision of one kind and the easy recovery of the other, a field known as surface organometallic chemistry.
Three checks catch most slips. First, the product is the substrate times the yield as a fraction, not the substrate itself; using the substrate gives a turnover number that is too high whenever the yield is below one hundred percent. Second, the turnover number is a pure number, product over catalyst, and for a working catalyst it is large, usually hundreds or more; a turnover number below one means the catalyst was used up before making even one molecule each, which is not catalysis at all.
Third, the units of frequency: a turnover number divided by hours gives turnovers per hour, and by seconds turnovers per second, so the two differ by a factor of $3{,}600$. And the catalyst loading and the turnover number are linked: at full conversion, the turnover number is one hundred divided by the loading in mole percent, a quick check that the two numbers belong to the same run.
Acetic acid is made on an enormous scale, more than fifteen million tonnes a year worldwide, for vinyl acetate paints and adhesives, for the plastic PET and for solvents. Most of it comes from methanol and carbon monoxide by metal-catalyzed carbonylation: first the Monsanto process with rhodium, introduced in the 1970s, and now largely the Cativa process with iridium, introduced by BP in the 1990s.
Iridium replaced rhodium partly on turnover. The iridium catalyst is more stable at the low water concentrations that make the process cheaper to run, so it lasts longer and loses less metal, and it produces fewer byproducts. Choosing between the two came down to comparing their turnover numbers and turnover frequencies under plant conditions, the same comparison this lesson makes on a small scale.
Olefin metathesis cuts two carbon-carbon double bonds and rejoins the pieces the other way round, a way of swapping partners between alkenes. Yves Chauvin worked out its mechanism, a metal carbene reacting with an alkene through a four-membered ring, and Robert Grubbs and Richard Schrock made practical catalysts from ruthenium and molybdenum. The three shared the 2005 Nobel Prize in Chemistry.
Grubbs's ruthenium catalysts tolerate air and water and work at loadings of a mole percent or less, and they are used to make medicines, agricultural chemicals and tough plastics. Pharmaceutical chemists routinely report the loading and turnover number of each metathesis step, because the ruthenium left in a drug must be kept below strict limits, and a catalyst that does more turnovers leaves less metal behind to remove.
Textbook definitions say a catalyst is not used up, and that is true of each turn of the cycle: the metal ends where it began. But real catalysts die. Metal atoms clump together, ligands break down, and impurities bind the metal and block it. The turnover number is the count of how many cycles a catalyst survives, and for an industrial process it is one of the most important numbers there is.
A second error is to confuse the turnover number with the amount of product. A run that makes more product is not necessarily using a better catalyst: it may simply have used more catalyst. Only product per mole of catalyst compares catalysts fairly.
Find the moles of product.
$2.0 \times 0.95 = 1.9\ \text{mol}$
Substrate times the fraction converted.
Find the turnover number.
$1.9 \div 0.001 = 1{,}900$
Product per mole of rhodium.
Find the turnover frequency.
$1{,}900 \div 2 = 950\ \text{h}^{-1}$
Over the two-hour run.
Find the loading.
$0.001 \div 2.0 \times 100 = 0.05\ \text{mol}\ \%$
One rhodium for every two thousand alkene molecules.
Check the link.
$100 \div 0.05 = 2{,}000 \approx 1{,}900 \div 0.95$
Full conversion would give a turnover number of two thousand.
Start with the catalyst.
$\mathrm{[Rh(CO)_2I_2]^-}: \ \mathrm{Rh^{I}}, \ 16 \text{ electrons}$
Square-planar rhodium(I).
Add methyl iodide.
$\mathrm{Rh^{I}} \to \mathrm{Rh^{III}}, \ 16 \to 18$
Oxidative addition, the rate-determining step.
Move the methyl onto carbon monoxide.
$18 \to 16$
Migratory insertion makes an acetyl and opens a site.
Fill the site with carbon monoxide.
$16 \to 18$
A fresh carbon monoxide binds.
Eliminate acetyl iodide.
$\mathrm{Rh^{III}} \to \mathrm{Rh^{I}}, \ 18 \to 16$
Reductive elimination closes the cycle.
Make the product.
$\mathrm{CH_3COI} + \mathrm{H_2O} \to \mathrm{CH_3COOH} + \mathrm{HI}$
Water turns the acetyl iodide into acetic acid.
Write the target.
$40{,}000\ \text{mol of butanal per batch}$
Hydroformylation of propene.
Write the catalyst's turnover number.
$\text{TON} = 8{,}000$
Measured for this catalyst under plant conditions.
Find the moles of catalyst.
$40{,}000 \div 8{,}000 = 5\ \text{mol}$
Each rhodium makes eight thousand molecules.
Convert to grams of rhodium.
$5 \times 102.91 = 514.55\ \text{g}$
Rhodium's molar mass is $102.91$ g/mol.
See what a better catalyst saves.
$\text{TON } 16{,}000 \Rightarrow 257.3\ \text{g}$
Doubling the turnover number halves the metal.
Say why it matters.
$\text{rhodium costs thousands of dollars an ounce}$
Turnover number translates directly into cost.
Find the moles of product.
$0.50 \times 0.88 = 0.44\ \text{mol}$
Substrate times the fraction converted.
Divide by the catalyst.
$0.44 \div 0.0005$
Product per mole of palladium.
Write the turnover number.
Put the steps of the catalytic cycle for a Suzuki coupling in order.
Number the steps in order (write the number in the box):
Complete the worked solution: the product, turnover number and turnover frequency for hydroformylating propene with $\mathrm{HRh(CO)(PPh_3)_3}$.
Multiply the substrate by the fraction converted.
$\text{product} =$ p $\text{mol}$
What the run actually made.
Divide the product by the catalyst.
$\text{turnover number} =$ n
Molecules of product per molecule of catalyst.
Divide the turnover number by the time.
$\text{turnover frequency} =$ r $\text{per hour}$
How fast each catalyst molecule works.
Three runs: $0.02$ mol of $\mathrm{HRh(CO)(PPh_3)_3}$ gives $96$ mol of product; $0.001$ mol of $\mathrm{RhCl(PPh_3)_3}$ gives $1.9$ mol; $0.0005$ mol of $\mathrm{Pd(PPh_3)_4}$ gives $0.44$ mol. Which catalyst has the highest turnover number?
In making acetic acid from methanol, $0.005$ mol of catalyst converts $50$ mol of substrate in $99$ percent yield over $5$ hours; in an alkene metathesis, $0.002$ mol converts $0.20$ mol in $90$ percent yield over $1.5$ hours. For each, in that order, fill in the moles of product, the turnover number and the turnover frequency per hour.
| product (mol) | turnover number | turnover frequency (per hour) | |
|---|---|---|---|
| the first run | |||
| the second run |
In hydrogenating an alkene, $0.001$ mol of catalyst is used with $2.0$ mol of substrate. What is the catalyst loading in mole percent?
Answer: mole percent catalyst
In hydrogenating an alkene, $0.001$ mol of $\mathrm{RhCl(PPh_3)_3}$ converts $2.0$ mol of substrate in $95$ percent yield over $2$ hours. What is the turnover frequency, in turnovers per hour?
Answer: turnovers per hour
A plant manager budgets the precious metal each batch needs. For butanal from propene, $40000$ mol per batch with a catalyst of turnover number $8000$ and a metal of $102.91$ g/mol; for a drug intermediate, $500$ mol, $2500$, $106.42$ g/mol; for a fragrance compound, $2000$ mol, $50000$, $101.07$ g/mol. In that order, fill in the moles of catalyst and the grams of metal each batch needs.
| catalyst (mol) | metal (g) | |
|---|---|---|
| the first batch | ||
| the second batch | ||
| the third batch |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
In an asymmetric hydrogenation, $0.0002$ mol of catalyst converts $10$ mol of substrate in $100$ percent yield over $10$ hours; in an alkene metathesis, $0.002$ mol converts $0.20$ mol in $90$ percent yield over $1.5$ hours. For each, in that order, fill in the moles of product, the turnover number and the turnover frequency per hour.
| product (mol) | turnover number | turnover frequency (per hour) | |
|---|---|---|---|
| the first run | |||
| the second run |
You can measure a catalyst. Explain the difference between turnover number and turnover frequency, and why a higher turnover number saves precious metal.
16. Your turn: a Suzuki coupling uses $0.0005$ mol of palladium to convert $0.50$ mol of substrate in $88$ percent yield. What is the turnover number?, step 3
$880$
Each palladium made 880 molecules.