What is dynamic equilibrium — Chemistry

Before this

This leads to

A reaction that stopped two-thirds of the way

In 1861, two French chemists, Marcellin Berthelot and Léon Péan de Saint-Gilles, mixed equal amounts of ethanoic acid and ethanol in sealed flasks. The two react to make an ester, ethyl ethanoate, and water. Every few weeks they tested how much acid was left. They kept going for more than a year.

Berthelot and Péan de Saint-Gilles's results at room temperature, for equal amounts of ethanoic acid and ethanol (from their 1862 memoir). The reaction slowed down long before the acid ran out. Heated runs, and runs started from ester and water, settled at the same limit, about 67%.

The reaction was slow at room temperature, so they also ran it hot, where it is much faster. Either way, the result was the same kind of surprise. The reaction never used up all the acid. It approached a limit, with about two-thirds of the acid turned into ester, and then went no further. They published this in 1862.

Your experience so far says that a reaction goes until a reactant runs out. Here, there was plenty of acid and plenty of ethanol left, and nothing changed. Why did it stop? The answer is that it did not stop at all. It only looks stopped. This lecture shows what is really going on.

Reactions that run both ways

Berthelot and Péan de Saint-Gilles also tried the reaction backwards. They mixed the ester with water. The ester slowly broke down into acid and ethanol. This is called hydrolysis. So the reaction can run in both directions:

CH_{3} COOH(l) + C_{2} H_{5} OH(l) ⇌ CH_{3} COOC_{2} H_{5} (l) + H_{2} O(l)

A reaction that can go both ways is called a reversible reaction. The arrow ⇌ says exactly that. The direction from left to right is the forward reaction. The direction from right to left is the reverse reaction. The double arrow was first used in this sense by the Dutch chemist J. H. van 't Hoff, in a book of 1884.

Why the reaction seems to stop

Now follow the two rates, one step at a time.

At the start, the flask holds only acid and ethanol. The forward reaction runs at its fastest. There is no ester yet, so the reverse reaction cannot happen at all.

As time passes, acid and ethanol are used up, so the forward reaction slows down. This is because fewer reactant molecules means fewer successful collisions each second, as in the factors-affecting-rate lecture. At the same time, ester and water build up. So the reverse reaction starts and speeds up.

Eventually, the falling forward rate and the rising reverse rate become equal. From then on, every second, the forward reaction makes as many ester molecules as the reverse reaction breaks down. The amount of each substance stops changing.

The forward rate falls as reactants are used up; the reverse rate rises as products build up. Once they are equal, the amounts stop changing. (A model curve, not measured data.)

That is it. That is the whole idea. The reaction has not stopped. Both reactions are still running, as fast as each other. Because they cancel, the amounts stay constant. This state is called dynamic equilibrium. "Dynamic" means moving. On the outside, nothing changes. On the inside, molecules are reacting all the time.

Dynamic equilibrium

A reversible reaction in a closed system is at dynamic equilibrium when the forward and reverse reactions happen at the same rate.

Both reactions continue, but the concentrations of all the substances stay constant.

Watch it happen in the model below. It is not a real reaction, but it follows the same rules: each molecule has a fixed chance of changing each way.

Simulation — molecules that keep changing (invented model)

Each dot is one molecule, either X (dark) or Y (green). In every tick, each X has a 2% chance of turning into Y, and each Y has a 1% chance of turning back into X. Press Play. Watch the counts level off while the changes go on. Then start again from the other side, and try the catalyst.

Add a catalyst (both changes twice as likely)

This table shows one run of an invented model with 150 molecules. In each tick, each X has a 2% chance of becoming Y, and each Y a 1% chance of becoming X.

Tick	Start all X: X / Y	Start all Y: X / Y
0	150 / 0	0 / 150
25	93 / 57	28 / 122
50	73 / 77	43 / 107
100	63 / 87	48 / 102
200	52 / 98	49 / 101
400	43 / 107	49 / 101

From either start, the counts settle near 50 X and 100 Y. After that, about the same number of molecules change each way, so the counts stay steady while the molecules keep changing. With a catalyst (both chances doubled), the counts settle at the same values in about half the time.

The model shows three things at once. First, the counts level off at about 50 X and 100 Y, even though the dots never stop changing colour. Second, starting from all Y gives the same final mixture as starting from all X. Third, the catalyst makes the counts settle sooner, at the same values.

Four features of every dynamic equilibrium

1. It needs a closed system. A closed system is one where no substance can enter or leave. If the ester were evaporating out of an open flask, the reverse reaction would keep losing its reactant, and the balance could never be reached. Energy may still pass in or out, which is how the flask stays at a fixed temperature.

2. Both reactions continue. Equilibrium is not the end of the reaction. It is the point where the two directions cancel.

3. The concentrations are constant, but usually not equal. In Berthelot's flask at equilibrium, there is about twice as much ester as acid. In other reactions, almost everything is products, or almost everything is reactants. "Constant" means not changing. It says nothing about the amounts being the same.

4. The same mixture is reached from either direction, as long as the flask holds the same atoms in the same volume at the same temperature. Mixing 1 mol of acid with 1 mol of ethanol gives the same final mixture as mixing 1 mol of ester with 1 mol of water. Both start with exactly the same atoms, so both settle at the same balance. Test this idea below.

Predict — two flasks, two starting points

Flask 1 holds 1.00 mol of ethanoic acid and 1.00 mol of ethanol. Flask 2 holds 1.00 mol of ethyl ethanoate and 1.00 mol of water. Both are sealed, with the same volume, at the same temperature, and left until nothing changes. Which flask ends with more ester? Choose one answer.

This figure shows the amount of ester in each flask over time. Flask 1 rises from zero and flask 2 falls from 1.00 mol. Both level off at 0.67 mol.

Stop and think

The catalysts lecture said that a catalyst would "help a reaction reach equilibrium sooner, but not change where the equilibrium lies". Explain why, using the two rates.

Show the answer

The catalyst gives a new route with a lower activation energy. The reverse reaction uses the same route backwards, so its barrier is lowered too, and both rates rise by the same factor. The forward and reverse rates therefore become equal at the same mixture as before. The only difference is that the rates are bigger, so that mixture is reached sooner. You saw this in the simulation: the counts settled at about 50 and 100 with or without the catalyst.

How do we know it is dynamic?

A fair question. If the amounts never change, how can anyone tell that molecules are still reacting? A still flask and a busy flask look identical.

The answer came from radioactive tracers. In the early twentieth century, George de Hevesy shook solid lead sulfate with a saturated solution of lead sulfate. Some of the lead in the solution was radioactive, so it could be tracked. The amount of dissolved lead did not change. Yet radioactive lead turned up in the solid, and ordinary lead from the solid turned up in the solution. Hevesy described a constant exchange between the solid and the solution. Lead was dissolving and crystallising at the same rate, all the time. He later received the 1943 Nobel Prize in Chemistry for developing tracer methods.

That example also shows that dynamic equilibrium is not only about chemical reactions. Two more examples are physical changes.

Water in a sealed bottle. Some water evaporates. Some vapour condenses back. When the two rates are equal, the amount of liquid stops changing.
A saturated solution with crystals at the bottom. Solid dissolves and dissolved particles crystallise, at equal rates. The crystals change shape slowly, but their total mass stays the same.

Worked examples

Worked Example 1 — The amounts at equilibrium

1.00 mol of ethanoic acid is mixed with 1.00 mol of ethanol in a sealed flask. At equilibrium there is 0.67 mol of ester. Find the amount of every substance at equilibrium.

Think first. Use the equation as a ratio: 1 acid + 1 ethanol → 1 ester + 1 water.

Making 0.67 mol of ester uses 0.67 mol of acid and 0.67 mol of ethanol, and makes 0.67 mol of water. So at equilibrium:

Substance	Start (mol)	Change (mol)	Equilibrium (mol)
Ethanoic acid	1.00	−0.67	0.33
Ethanol	1.00	−0.67	0.33
Ethyl ethanoate	0	+0.67	0.67
Water	0	+0.67	0.67

Check. The total is $0.33 + 0.33 + 0.67 + 0.67 = 2.00$ mol, the same as at the start. This must be so, because the equation has two molecules on each side. ✓ Notice that the amounts are constant but not equal.

Worked Example 2 — Starting from the other side

1.00 mol of ethyl ethanoate is mixed with 1.00 mol of water, at the same temperature and in the same volume. How much ester breaks down before equilibrium is reached?

Think first. 1.00 mol of ester + 1.00 mol of water contains exactly the same atoms as 1.00 mol of acid + 1.00 mol of ethanol. So it must reach the same equilibrium mixture: 0.67 mol of ester.

The ester falls from 1.00 mol to 0.67 mol, so $1.00 - 0.67 = 0.33$ mol of ester breaks down. This makes 0.33 mol of acid and 0.33 mol of ethanol, and leaves 0.67 mol of water.

Check. The final amounts, 0.33, 0.33, 0.67 and 0.67 mol, match Worked Example 1 exactly. ✓

Worked Example 3 — A gas equilibrium you can see

Dinitrogen tetroxide, N₂O₄, is colourless. Nitrogen dioxide, NO₂, is brown. In a sealed flask they reach equilibrium very quickly:

N_{2} O_{4} (g) ⇌ 2 NO_{2} (g)

A flask starts with 0.100 mol of N₂O₄. At equilibrium it contains 0.040 mol of NO₂. Find the amount of N₂O₄ left, and the total amount of gas. (These numbers are invented.)

Think first. The ratio is 1 N₂O₄ → 2 NO₂, so the N₂O₄ used is half the NO₂ made.

N₂O₄ used $= 0.040 \div 2 = 0.020$ mol. N₂O₄ left $= 0.100 - 0.020 = 0.080$ mol. Total gas $= 0.080 + 0.040 = 0.120$ mol.

Check. Count nitrogen atoms. At the start: $0.100 \times 2 = 0.200$ mol. At equilibrium: $0.080 \times 2 + 0.040 \times 1 = 0.200$ mol. ✓ The colour of the flask stays the same pale brown, but N₂O₄ molecules keep splitting and NO₂ molecules keep joining.

Three traps

Trap 1 — "At equilibrium, the reaction has stopped"

Both reactions are still happening, at equal rates. Only the overall amounts have stopped changing. An exam answer that says "the reaction stops" loses the mark. Write "the forward and reverse rates are equal".

Trap 2 — "At equilibrium, the concentrations are equal"

They are constant, not equal. In Berthelot's flask there was twice as much ester as acid. Equal concentrations at equilibrium would be a coincidence.

Trap 3 — Every reaction is an equilibrium

Many reactions go essentially to completion, and are written with →. A strong acid such as hydrochloric acid ionises almost completely in water. A weak acid such as ethanoic acid is the one that sets up an equilibrium, with most of its molecules still whole. Use ⇌ only when both directions really matter.

Try it yourself

Work each one on paper first. Write the balanced equation before any calculation. Open a hint only when you are stuck.

A student writes: "At equilibrium the reaction stops, so the amounts of reactants and products are equal." Correct both parts of the sentence.

Hint 1
What are the two reactions doing at equilibrium?

Hint 2
Think about Berthelot's flask: 0.33 mol of acid and 0.67 mol of ester.

Full answer
"At equilibrium the forward and reverse reactions continue at equal rates, so the amounts of reactants and products stay constant. They are usually not equal." Check: in Berthelot's flask, the amounts were constant (0.33 and 0.67 mol) but not equal. ✓
Hydrogen and iodine react reversibly: H₂(g) + I₂(g) ⇌ 2HI(g). A sealed flask starts with 1.00 mol of H₂ and 1.00 mol of I₂. At equilibrium there is 1.56 mol of HI. Find the amounts of H₂ and I₂ at equilibrium. (These numbers are invented.)

Hint 1
2 mol of HI is made from 1 mol of H₂ and 1 mol of I₂.

Hint 2
H₂ used = 1.56 ÷ 2.

Full answer
H₂ used $= 1.56 \div 2 = 0.78$ mol, and the same for I₂. So H₂ left $= 1.00 - 0.78 = 0.22$ mol and I₂ left $= 0.22$ mol. Check: hydrogen atoms at the start, $1.00 \times 2 = 2.00$ mol; at equilibrium, $0.22 \times 2 + 1.56 = 2.00$ mol. ✓
The same flask, at the same temperature, is filled instead with 2.00 mol of HI and nothing else. Predict the amounts of H₂, I₂ and HI at equilibrium, and explain your answer.

Hint 1
Compare the atoms in 2.00 mol of HI with the atoms in 1.00 mol of H₂ + 1.00 mol of I₂.

Hint 2
The same atoms in the same volume at the same temperature reach the same equilibrium.

Full answer
2.00 mol of HI contains 2.00 mol of H atoms and 2.00 mol of I atoms, exactly the same as 1.00 mol of H₂ + 1.00 mol of I₂. So the flask reaches the same equilibrium as in question 2: 0.22 mol of H₂, 0.22 mol of I₂ and 1.56 mol of HI. To get there, $2.00 - 1.56 = 0.44$ mol of HI breaks down. Check: 0.44 mol of HI gives 0.22 mol of H₂ and 0.22 mol of I₂. ✓
Solid silver chloride sits at the bottom of a beaker of its saturated solution. A chemist adds a small amount of dissolved chloride ions made with a radioactive isotope of chlorine. One day later, she filters off the solid and finds it is radioactive, although the mass of solid has not changed. Explain what this shows. What would she have found if the equilibrium were static, with nothing dissolving or crystallising?

Hint 1
How can radioactive chloride ions from the solution end up inside the solid?

Hint 2
If ions are crystallising into the solid, but the mass does not change, what else must be happening at the same rate?

Full answer
Radioactive chloride ions from the solution have become part of the solid, so ions are crystallising. The mass of solid has not changed, so ions must also be dissolving from the solid at the same rate. Both processes continue at equal rates: the equilibrium is dynamic, as in Hevesy's lead experiments. If the equilibrium were static, no ions would move between the solid and the solution, so the solid would not be radioactive. Check: the conclusion uses both pieces of evidence, the radioactivity (exchange happens) and the constant mass (the rates are equal). ✓

The bigger idea

The Gibbs free energy lecture said that a positive ΔG⦵ means the reactants are favoured, and promised that the equilibrium lectures would make this precise. Here is the first step. A reaction with a negative ΔG⦵ does not go 100% of the way, and one with a positive ΔG⦵ does not go 0% of the way. Each settles at an equilibrium mixture, and ΔG⦵ tells you which side that mixture favours. The equilibrium constant lecture turns this into a number.

Dynamic equilibrium is everywhere once you look for it. The oxygen carried by your blood is picked up and released at equilibria that shift as conditions change. The carbon dioxide in a fizzy drink is held in equilibrium with the gas above it, until the bottle is opened and the system stops being closed. Many systems that look still are balanced between two busy processes. The next lecture asks what happens when you disturb that balance.

Key concepts in this lesson

A reversible reaction can go in both directions, written with ⇌, and in a closed system it reaches dynamic equilibrium.
At dynamic equilibrium the forward and reverse rates are equal, so both reactions continue but the concentrations stay constant.
The equilibrium concentrations are constant but usually not equal.
The same equilibrium mixture is reached from either direction, as long as the same atoms are in the same volume at the same temperature.
A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached sooner but at the same mixture.