LearnChem · Part IV — What makes a reaction go further, or faster?
High in the mountains the air is still there. You can feel it on your face. Yet every step costs breath. Something in your blood has been pushed off balance.
“Why do you get out of breath high in the mountains, when there is still plenty of air?”
Push a mixture at balance, and find out where it settles.
Module 17 gave three results. A reaction stops early because the reverse catches up. Q measures where the mixture sits right now. K marks where it lands. At balance Q equals K. Module 14 gave the referee with two terms. ΔG = ΔG° + RT ln Q, and the position moves until that sum is zero. Module 16 gave the energy barrier and the rate constant. All three come back here.
Climb high enough and every step costs breath. The air is still there. Each breath fills your lungs as before. So why does the body struggle? Before you read on, make a guess about the air itself.
What about the air changes as you climb?
Answer from your gut. The reveal follows.
The share of oxygen in the air holds steady. At sea level oxygen is 20.9% of the air. High on a mountain it is still 20.9%. What falls is the pressure. At sea level the partial pressure of oxygen is about 21 kilopascals. Partial pressure is the push from one gas alone in a mixture. A kilopascal is a unit of pressure. At about 5,000 metres that push is about half. At the summit of Everest it is about one third.
These figures come from Wikipedia. Wikipedia is a tertiary source. A tertiary source collects other people's findings and measures nothing itself. The figures are ordinary and widely repeated. Still, this page tells you where they come from.
So here is the crux. The share of oxygen did not change. The pressure did. Anything in the body that depends on the share of oxygen is untouched. Anything that depends on the pressure of oxygen must change. The next step is to find the thing that depends on pressure.
That thing is in your blood. Haemoglobin is the protein in red blood cells that carries oxygen. It picks oxygen up in the lungs. It lets oxygen go in the muscles. Picking up and letting go are the forward and reverse of one reaction. In the simplest picture the reaction is written as one balance. Hb(aq) + 4 O2(g) ⇌ Hb(O2)4(aq) Hb stands for haemoglobin. Hb(O2)4 is haemoglobin carrying four oxygen molecules. The double arrow says both directions run. In the lungs the oxygen pressure is high. There the forward direction outruns the reverse. In the muscles the oxygen pressure is low. There the reverse outruns the forward. Now go up the mountain. The oxygen pressure in the lungs falls. The balance has been pushed. Less haemoglobin is loaded.
The equation above is the simplest possible picture. Real blood is more complicated than one balance. The binding is cooperative. Cooperative means the first oxygen makes the second easier to bind. The curve of loading against oxygen pressure is S-shaped, rather than the simple curve this one equation gives. Acid, carbon dioxide and a molecule called bisphosphoglycerate all shift it. This module names those complications and teaches none of them. One more thing about the source. The simple equation comes from a LibreTexts page. That page is student coursework from a university writing project. No chemistry in this module rests on its authority. The equation is used only as a picture of a balance being pushed.
Here is the whole idea of the module in one line. A balance can be pushed. The push moves it off balance. Then it settles somewhere new. Module 17 gave you two numbers for this. Q is the ratio of products to reactants in the mixture right now. K is the value that ratio takes at balance. Keep both in mind. Every push in this module lands either on Q or on K. Telling those two apart is most of the module.
The oxygen pressure in the lungs falls. Which way is the haemoglobin balance pushed, and what happens to loaded haemoglobin?
— The share of oxygen holds steady with height, and only the pressure falls.
Start with the simplest balance there is. A turns into B, and B turns into A. Write it as A ⇌ B. Q is the amount of B divided by the amount of A. K is the value of Q at balance. Suppose the mixture sits at balance, so Q equals K. Now tip in more A. Nothing else changes. The bench below offers two reactions. One has one molecule on each side. The other has one molecule on the left and two on the right. The difference between them matters later. For now, make a guess before the bench.
You add more A to a mixture at balance. What happens next, overall?
Answer from your gut. Then drive the bench.
Think about Q. Q is the amount of B divided by the amount of A. You raised the amount of A. So Q fell. Q is now smaller than K. The mixture is no longer at balance. The forward direction now outruns the reverse. A turns into B faster than B turns into A. So some of the added A becomes B.
Watch what that does to Q. As A falls and B rises, Q climbs. It climbs until it reaches K again. Then the two directions match and the net change stops. Nothing here is a new rule. It is module 17's Q against K, run after a poke. The bench below lets you do the poke yourself.
That is the whole mechanism. A poke moves Q away from K. The reaction then runs net in whichever direction brings Q back to K. Add product, and Q rises above K. Then the reverse direction outruns the forward. Add reactant, and Q falls below K. Then the forward outruns the reverse. Check the direction rule below.
A mixture of A ⇌ 2B holds a lot of B and almost no A. So Q sits far above K. Which way does it run?
This pattern has a name. Le Chatelier's principle is the usual name for what you just watched. Textbooks state it as a rule. It is not a rule. It is what Q against K does after a poke. If you know Q and K, you never need the slogan.
— A poke moves Q away from K, and the reaction runs whichever way brings it back.
Go back to the bench. You added A and watched it settle. Now look harder at the numbers. The amount of A started somewhere. Your poke sent it up. Then the reaction pulled it back down. The question is where it stops. Guess before you look.
You add A to a mixture at balance and wait. Where does the amount of A settle?
Answer from your gut. The reveal follows.
It settles between the two. It ends above where it started. It ends below where your poke put it. The shift undoes part of your push. It never undoes all of it. Here is why. Suppose A did fall all the way back to its start. The added material has to be somewhere. So B would sit above its start. Then Q, which is B over A, would be above K. The reverse would run and push A back up. So A cannot rest at its start.
The arithmetic behind bench 1 was run over 22 additions. The fraction of the push given back ran from 0.0625 to 0.9375. It was never 1. The exact fraction depends on the reaction. The rule does not. Below is one case in full.
Picture a seesaw resting level. Lean on one end. It tips and takes up a new position. While you keep leaning, it stays there. An equilibrium is like that. Your poke tips it. It settles somewhere new. It stays there for as long as the disturbance lasts.
The picture breaks in three places. First, stop the disturbance and both return. Let the volume back out, or cool the mixture down, and the balance comes back. But you cannot un-add a substance. For an addition there is nothing to stop. Second, a seesaw gives back nothing while you lean. An equilibrium gives back part of your push while you keep pushing. That is the thing the seesaw cannot show. Third, leaning does not change the seesaw. Heating a reaction changes K itself. That third point is the next section. The arithmetic below shows the second point in numbers.
This worked case uses the simple one-molecule-each-side reaction, A ⇌ B, not the bench's default. The bench's A ⇌ B preset does not use this K, so do not look for these numbers there. Take a mixture of A and B sitting at balance with 0.2000 mol/L of A. For this mixture K is 4, so at balance B is 4 times A. So B starts at 4 times 0.2000, which is 0.8000 mol/L of B. Now add enough A to take its amount at once up to 0.7000 mol/L. The push is therefore 0.7000 minus 0.2000, which is 0.5000 mol/L of A added.
The reaction now runs forward and turns some of the added A into B. Turning A into B does not change the total, so A plus B is conserved. After the push, A plus B is 0.7000 plus 0.8000, which is 1.5000 mol/L. At balance B is K times A, so the total is A plus 4 times A. That is 5 times A, so A settles at 1.5000 divided by 5, which is 0.3000 mol/L. That is above the start of 0.2000 and below the push of 0.7000. The amount given back is 0.7000 minus 0.3000, which is 0.4000 mol/L of A. The fraction given back is 0.4000 divided by 0.5000, which is 0.8000. The arithmetic behind bench 1, run on this case, lands on the same 0.3000 mol/L.
For the simple one-molecule-each-side reaction A ⇌ B, that working gives a formula. Turning A into B never changes the total, so the total after the push is fixed. At balance B is K times A, so the total is A times (1 + K). So A at balance is the total divided by (1 + K), for any K. The give-back fraction then works out as K divided by (1 + K). Here K is 4, so the fraction is 4 divided by 5, which is 0.8000. That formula is proved here for A ⇌ B only. For other reactions the fraction differs, and bench 1 shows that, but it is never 1.
For A ⇌ B with K equal to 3, what fraction of an added push comes back?
So the settled point is always between the start and the push. The bench numbers say it. The formula says it. Le Chatelier's principle is often taught as "the system pushes back". It pushes back partly. That is the sentence at the top of this module, now earned.
— The shift gives back part of a push and never all of it.
On the bench you had two more buttons. They were squeeze and heat. Both look like pushes. Are they the same kind of push? Start with the squeeze. Squeezing a gas mixture raises its pressure. Make a guess about what that does to the position.
You squeeze a gas mixture at balance into a smaller volume. What happens to its position?
Answer from your gut. The reveal follows.
Squeezing raises every concentration at once. What matters is what that does to Q. Take A ⇌ 2B first. Two molecules of B are made for each one of A, so B enters Q twice. So Q is the amount of B multiplied by itself, divided by the amount of A. Now halve the volume. Every amount doubles. The top of Q doubles twice over, so it goes up four times. The bottom doubles once. So Q goes up two times. Q is now above K. The reverse must run to bring it down. That is why the side with fewer molecules wins. It is arithmetic, not a slogan.
Now take A ⇌ B. Q is the amount of B divided by the amount of A. Halve the volume and both amounts double. The top doubles once and the bottom doubles once. So Q does not move. It still equals K. Nothing shifts. You can run this on bench 1. Squeeze both reactions and watch the readout. It scales the volume change back out and prints the leftover shift. For A ⇌ B that shift came out as zero, to the limit of the arithmetic. Notice what this means. This is not a small give-back. It is none at all. A push that does not move Q is not a push on the position. So there is nothing to undo. Pressure is not a special push. It is a push on Q, and only when the two sides are unequal.
Now the heat button. Heating is different in kind. Adding a species moves Q and leaves K where it was. Squeezing moves Q and leaves K where it was. Heating moves K itself. To see why, go back to module 14's referee.
Module 14 gave a referee with two terms, ΔG = ΔG° + RT ln Q. A change happens while ΔG is negative, and the position moves until ΔG reaches zero. At that point Q has become K, so ΔG° + RT ln K equals zero. That equation fixes K from ΔG° alone, at one temperature. Adding a species changes Q but leaves ΔG° untouched, so K keeps its value. The sum is no longer zero, so the position moves until Q returns to K. Squeezing does the same thing, because it also changes only Q.
ΔG° is a property of the reaction at one temperature, and it changes with temperature. Module 14 split ΔG° into two parts, an enthalpy part and a temperature times entropy part. Enthalpy is the heat given out or taken in, and entropy is the measure of spread. For ammonia synthesis the forward direction gives out heat, so the enthalpy part is negative. Exothermic means the forward direction gives out heat as it runs. It also turns four gas molecules into two, so the entropy part is negative as well. Raising the temperature makes the temperature times entropy term count for more. Because the entropy part is negative, that pushes ΔG° upward. K is fixed by ΔG°, so a higher ΔG° means a lower K. So heating lowers K, and a hot mixture holds less ammonia at balance. Cooling does the reverse, so a cold mixture holds more ammonia at balance.
A reaction takes in heat going forward and turns two gas molecules into four. You heat it. What happens to K?
So there are two kinds of push. One kind moves Q and leaves K alone. The other kind moves K. A standard textbook says the same thing about a temperature change.
“When an equilibrium shifts in response to a temperature change, however, it is re-established with a different relative composition that exhibits a different value for the equilibrium constant.”
OpenStax, Chemistry 2e, §13.3, “Shifting Equilibria: Le Châtelier's Principle” · openstax.org · CC BY-NC-SA 4.0
— Concentration and volume move Q and leave K alone, but temperature moves K itself.
Now put the rule to work on a reaction that matters. Nitrogen and hydrogen make ammonia. Ammonia makes fertiliser. The reaction is a balance. Its yield depends on temperature, on pressure and on the mix of gases fed in. Before the bench, guess where the yield is largest.
Nitrogen and hydrogen make ammonia. Four gas molecules become two, and the forward direction gives out heat. Where is the share of ammonia at balance largest?
Answer from your gut. Then drive the bench.
Cold and at high pressure. The forward direction gives out heat, so cooling raises K. That is the §4 result. Four gas molecules become two, so squeezing raises the bottom of Q faster than the top. Q falls below K, and the forward direction runs. The equation and the arithmetic are set out below.
There is no middle setting to hunt for. Cooler is better at every pressure. Higher pressure is better at every temperature. So the best corner is a corner. The bench below shows how large the yield gets there.
Here is the reaction. Nitrogen and hydrogen are fed in together. Three hydrogen molecules go in for each nitrogen. The balance is written as follows. N2(g) + 3 H2(g) ⇌ 2 NH3(g) Write Q the same way as in §4. Ammonia is made two at a time, so it enters twice on top. Below the line, nitrogen enters once and hydrogen enters three times. Count the factors. There are two above the line and four below. Now squeeze so that every amount doubles. The top doubles twice over, so it goes up four times. The bottom doubles four times over, so it goes up sixteen times. So Q falls to a quarter of what it was. Q is below K, and the forward direction runs. That is the same arithmetic you did in §4, with more factors. The forward direction also gives out heat. So cooling raises K. The bench below turns those two facts into numbers.
First, the ammonia grid is computed and ideal gas. It was checked against no published grid. The table of record is Table 10 of Max Appl's "Ammonia" chapter in Ullmann's Encyclopedia of Industrial Chemistry. Its pages are scanned images this course could not read, so the page computes its own grid instead. Second, the real gas is not ideal at these pressures. The source is El-Gharbawy, Gad and Shehata (2021), Egyptian Journal of Petroleum, volume 30, pages 11 to 15. Its DOI is 10.1016/j.ejpe.2020.12.003, and it is open access. It reports that the equilibrium constant for this reaction rises with pressure. That sounds like a contradiction of §4. It is not. The constant that drifts is the one built out of pressures. The thermodynamic constant §4 talked about is built from ΔG° and depends only on temperature. So §4 stands.
What does the drift mean for you? The real yield at the cold high-pressure corner is higher than the bench says. No source gives the size, so this page gives none. So the sacrifice a plant makes is at least as large as the page's figure, not smaller. Third, no recycle ratio or purge fraction is given. No source gives one. Fourth, the bench says nothing about speed. It gives the share at balance and not the time taken to reach it. That is why §6 needs a separate table of what plants do.
Read the corners. At 450 °C and 200 bar the computed ideal-gas share is 25.3 mol %. At 450 °C and 1 bar it is 0.2 mol %, computed and ideal gas. At 300 °C and 300 bar it is 67.7 mol %, computed and ideal gas. At 200 °C and 200 bar it is 85.4 mol %, computed and ideal gas. Now drag both sliders to their far ends, 200 °C and 300 bar. There the share is 87.9 mol %, computed and ideal gas. That is the largest anywhere the bench can reach. Cooler wins at every pressure. Higher pressure wins at every temperature. So the best temperature and the best pressure sit at the edge of any range you offer. There is nothing to discover there. Widen the range and the best moves with it. Is the feed ratio the same? Guess before the next passage.
Temperature and pressure are both best at an edge. Is the feed ratio best at an edge too?
Answer from your gut. The reveal follows.
It is best in the middle. Too little hydrogen and there is not enough to build the ammonia. Too much and the surplus hydrogen simply dilutes the ammonia in the mixture. Both ends lose. So the best sits inside the range, at three hydrogen per nitrogen. That is the ratio in the equation itself.
Notice the shape of that answer. For temperature and pressure the answer was always an edge. For the feed ratio it is a peak with a fall on each side. That difference in shape is the thing worth keeping. The numbers below show the peak.
Hold the temperature at 450 °C and the pressure at 200 bar, and vary the feed. One hydrogen per nitrogen gives a computed ideal-gas share of 17.9 mol %. Three hydrogen per nitrogen gives a computed ideal-gas share of 25.3 mol %. Six hydrogen per nitrogen gives a computed ideal-gas share of 21.7 mol %. The share rises from one to three, and then it falls from three to six. So the best feed sits at three, which is the ratio in the equation itself. Inert gases are gases that take no part in the reaction, and they dilute the mixture too. With 0% inerts the computed ideal-gas share is 25.3 mol %, and with 5% it is 22.9 mol %. With 10% inerts it is 20.6 mol %, computed and ideal gas, so the loss is steady.
The feed goes from three to four hydrogen per nitrogen, at the same temperature and pressure. What happens to the share?
One more check on the bench. Start from almost pure ammonia instead of the synthesis gases. The synthesis gases are the nitrogen and hydrogen fed in. At 450 °C and 200 bar it lands on the same 25.3 mol %, computed and ideal gas. Same answer from both ends. That is what a balance means. It does not matter which side you start from.
— Temperature and pressure are best at an edge, and the feed ratio is best in the middle.
The bench says the yield is largest cold and at high pressure. At 200 °C and 300 bar the computed ideal-gas share was 87.9 mol %. No plant runs there. Every plant runs far hotter. Before you see the real numbers, guess why.
Why do plants not run at the cold, high-pressure corner where the computed ideal-gas yield is largest?
Answer from your gut. The reveal follows.
Cold gas is slow. Module 16 gave the reason. Every reaction has an energy barrier. Cooling lowers the fraction of collisions that clear it. So the rate falls. At the cold corner the balance holds the most ammonia. But the mixture would take far too long to get there. A plant cannot wait.
So a plant trades yield for speed. It runs hot enough for the gas to reach balance in the time it spends in the reactor. It runs at high pressure to win some yield back. And it uses a catalyst. A catalyst is a substance that lowers the barrier without being used up. Here the catalyst is finely divided iron. It is made in place by reducing magnetite. Magnetite is an ore of iron. A few per cent of promoters are added. Promoters are small additions that make the catalyst work better. The table below shows what real plants do.
| What | What is published | Who says so |
|---|---|---|
| Pressure | above 100 atm; 150 bar in the case worked there | Humphreys, Lan & Tao 2021 · peer-reviewed, open access |
| Pressure, across sources | 150–300 bar · 250–350 bar · 100–300 bar · 70–120 bar for low-pressure designs | Royal Society · Ullmann's via Wikipedia · a peer-reviewed review · Ullmann's via an unofficial mirror — the sources disagree on the edges |
| Temperature | 425–450 °C | Humphreys, Lan & Tao 2021 · peer-reviewed, open access |
| Temperature, across sources | 350–500 °C · 450–550 °C · 400–600 °C · 400–450 °C | Royal Society · Wikipedia · a peer-reviewed review · LibreTexts — again the edges disagree |
| Catalyst | finely divided iron, made in place by reducing magnetite | Ullmann's · Humphreys et al. · Wikipedia, agreeing |
| Promoters | Al2O3, K2O, CaO, at a few per cent by weight | Wikipedia · Ullmann's · Humphreys et al., agreeing |
| Conversion in one pass | around 10–15% | Humphreys, Lan & Tao 2021 · a reactor number, not an equilibrium one |
| Energy per tonne | 28 GJ/t for the best gas-route plants; about 41 GJ/t as the world average | IEA, Ammonia Technology Roadmap · an intergovernmental report |
| Plants running ruthenium instead of iron | about ten worldwide · or three, counting only Trinidad | Humphreys et al. 2021 · IIP and Ullmann's — not reconciled |
One of the sources puts the trade in a single sentence.
“Although the theoretical ammonia equilibrium concentration can be close to 100% at low temperatures and high pressures, the ammonia formation rate is extremely slow and not suitable for production purposes.”
Humphreys, Lan & Tao (2021), “Development and Recent Progress on Ammonia Synthesis Catalysts for Haber–Bosch Process,” Advanced Energy and Sustainability Research, DOI 10.1002/aesr.202000043 · open access, via Warwick · peer-reviewed
Now put two numbers side by side. The largest share the bench can reach is at 200 °C and 300 bar, at 87.9 mol %, computed and ideal gas. Now take 450 °C and 200 bar as a plant condition. That pressure is this page's own choice. It sits inside the 150–300 bar and 100–300 bar ranges that sources give. The source of record's own worked case uses 150 bar. At 450 °C and 200 bar the bench gives 25.3 mol %, computed and ideal gas. The gap is 62.6 percentage points. That is the yield given up for speed. Keep one more number strictly apart. The 10–15% is conversion in one pass. Conversion in one pass is the share of the feed turned into ammonia in one trip through the reactor. It is what the reactor actually achieves in the contact time it has. Contact time is the time the gas spends in the reactor. The 25.3 mol % is the thermodynamic ceiling at those conditions. A thermodynamic ceiling is the most the balance allows there. The reactor falls short of the ceiling. The two are different kinds of number.
Why did it take until 1909? Nobody could hold that pressure at that temperature until someone built a vessel that would. That is what Bosch's 1931 prize was for. Its citation reads "in recognition of their contributions to the invention and development of chemical high-pressure methods". Now the date itself. The usually-taught date for the bench demonstration is 1909, at Karlsruhe. That date comes from a biography, by way of Wikipedia. Wikipedia is a tertiary source.
Now read a primary source. A primary source is a record from the time and the people involved. Haber's Nobel Prize in Chemistry for 1918 was "for the synthesis of ammonia from its elements". The Nobel Foundation's own award ceremony speech does not give 1909 at all. It says the study began in 1904. It says construction of the first large factory at Oppau began in 1910. It says the key paper was published in 1913. So the primary source and the secondary source do not agree on a date. A secondary source is a later account built from primary ones. Neither is wrong on its own terms. A biography can hold a detail a ceremony speech left out. But you should know which kind of source each date rests on.
A plant has a computed ideal-gas ceiling of 25.3 mol % and a 12% one-pass conversion. What is the recycle loop for?
The unreacted gas is not thrown away. The ammonia is taken out and the rest is sent round again. Inert gases, the gases that take no part in the reaction, build up as the gas goes round. So a bleed is needed to let them out. A bleed is a small stream drawn off the loop. No source gives a recycle ratio or a purge fraction. So this page gives none. Energy is the other cost. The IEA gives 28 GJ per tonne for the best gas-route plants. A GJ is a gigajoule, and the figure is the energy spent for each tonne of ammonia made. Gas-route means the plant starts from natural gas. The IEA gives about 41 GJ per tonne as the world average.
— A plant gives up yield for speed, because the cold corner delivers its ammonia too slowly to use.
1. From module 16.
2. From module 14.
3.
4.
1.
The question gives the computed ideal-gas share at 450 °C and 100 bar. Doubling the pressure takes the mixture from 100 bar to 200 bar. Higher pressure is better at every temperature, so the share must rise. There is no rule that doubles the share when the pressure doubles. Pressure moves Q and leaves K alone, so the new share must be read, not scaled. Read the bench at 450 °C and 200 bar. There it shows 25.3 mol % of ammonia, computed and ideal gas. That is the answer, and it is larger than the figure the question gave.
2.
Before the addition, the mixture holds 0.40 mol/L of A and 0.60 mol/L of B. K is 0.60 divided by 0.40, which is 1.50. Adding 0.50 mol/L of A pushes A from 0.40 up to 0.90 mol/L. The total of A and B is now 0.90 plus 0.60, which is 1.50 mol/L. At balance A is the total divided by (1 + K), and 1 + K is 2.50. So A settles at 1.50 divided by 2.50, which is 0.60 mol/L. The amount given back is 0.90 minus 0.60, which is 0.30 mol/L. The fraction given back is 0.30 divided by 0.50, which is 0.60. As a check, K divided by (1 + K) is 1.50 divided by 2.50, which is 0.60.
Back to the mountain. The share of oxygen in the air held steady. The pressure fell. At about 5,000 metres the partial pressure of oxygen is about half. Oxygen sits on the bottom of Q for the haemoglobin balance. Lower the pressure and Q rises above K. The reverse direction runs. Less haemoglobin is loaded with oxygen.
In a flask, the shift would give part of the push back. The total is conserved there. The lung is different. The atmosphere holds the oxygen pressure down, and nothing inside the blood can raise it. So the push cannot be given back at all. The balance settles where less haemoglobin is loaded. It stays there for as long as you stay high. That is why you gasp, with the air still there. The push landed on Q. Adding and squeezing move Q, and removing is the obvious extension. Heating moves K.
Push an equilibrium so that Q moves, and it settles somewhere new that partly undoes your push.
Balances appear again in the modules ahead. Each time the question will be the same. Did the push move Q, or did it move K? Everything else follows from where Q sits against K.