Introduction to Equilibrium: Study Guide (Chemistry 7.1)

Reversible processes are everywhere

Up to now you have mostly written reactions with a single arrow, →, as if they ran in one direction until a reactant was used up. Many processes do not work like that. They are reversible: the products can turn back into the reactants. Chemists show this with a double arrow, ⇌.

The process from left to right is the forward process. The process from right to left is the reverse process. Reversible processes can be physical or chemical.

Type	Forward process	Reverse process
Phase change	water evaporates: H₂O(l) → H₂O(g)	water vapour condenses: H₂O(g) → H₂O(l)
Gas and liquid	CO₂ dissolves in water (absorption)	dissolved CO₂ escapes as gas (desorption)
Salt and solution	a salt dissolves into ions	the ions come together and precipitate as solid
Acid–base (proton transfer)	an acid gives H⁺ to a base	the conjugate acid gives the H⁺ back
Redox (electron transfer)	one species gives electrons to another	the electrons transfer back

A sealed bottle of fizzy drink is a good everyday example. Carbon dioxide gas sits above the liquid and dissolved CO₂ sits in it. Some CO₂ molecules leave the liquid while others enter it. Both processes happen at once.

What you see when a system reaches equilibrium

Take a reversible process in a closed system: no matter can get in or out, and the temperature is held constant. Start it off and watch. At first things change: a colour deepens, a pressure rises, a solid dissolves. After a while the changes stop. The system has reached equilibrium.

At equilibrium you observe three things:

1. No observable change. Colour, pressure, mass of solid and pH all stay the same.
2. Reactants and products are all present. The reaction does not run to completion. Some of every species remains.
3. Concentrations and partial pressures are constant. They are not changing with time. They are usually not equal to each other.

Point 3 is where many students slip. “Constant” means each concentration has stopped changing. It does not mean the reactant and product concentrations have the same value.

Equilibrium is dynamic

If nothing observable changes, has the reaction stopped? No. At the particle level, the forward and reverse processes are still going on. They happen at equal rates, so every particle made in one direction is matched by a particle used up in the other. The result is no net change.

Think of the sealed bottle of water. Molecules keep escaping from the liquid surface, and vapour molecules keep hitting the surface and sticking. Once the two rates are equal, the number of molecules in the vapour stays constant, and so does the water level. Yet individual molecules are always moving between the phases.

How do we know it is dynamic? Labelled-atom experiments give direct evidence. Suppose a solution is saturated with a salt and some undissolved solid sits at the bottom. If you add a little extra solid made with a radioactive isotope, the solution soon becomes radioactive, even though its concentration does not change and the total mass of solid stays the same. Labelled solid has dissolved, and an equal amount of unlabelled ions has precipitated. That only makes sense if dissolving and precipitating are both still happening.

How equilibrium is established: graphs against time

Equilibrium does not appear instantly. It builds up as the forward and reverse rates change. Graphs against time show this clearly. The figures below use a fictional gas-phase reaction, A(g) ⇌ 2B(g). The flask starts with A at 1.00 atm and no B, at constant temperature and volume. The data are invented.

Read Figure 1 like this:

• Before about 45 s: both lines are still sloping. The system is not yet at equilibrium.
• After about 45 s: both lines are flat. This is equilibrium. P_A = 0.70 atm and P_B = 0.60 atm. Both gases are present, and the pressures are different.
• Stoichiometry check: A fell by 1.00 − 0.70 = 0.30 atm, and B rose by 0.60 atm, which is 2 × 0.30. That matches the 1 : 2 coefficients.

Now look at the rates of the two processes over the same time.

The story behind Figure 2:

• At the start, only A is present. The forward rate is at its highest (0.020 atm s⁻¹) and the reverse rate is zero, because there is no B to react.
• As time passes, A is used up, so the forward rate falls. B builds up, so the reverse rate rises.
• At equilibrium the two rates are equal, 0.014 atm s⁻¹ each. A is used up exactly as fast as it is made, so P_A stops changing, and the same is true for B.

So the two graphs tell one story. Flat concentration or pressure lines (Figure 1) and equal rates (Figure 2) are two views of the same equilibrium state.

Worked example 1: reading a concentration table

Question. A fictional reaction C₂(g) + D₂(g) ⇌ 2CD(g) is started in a sealed flask with [C₂] = [D₂] = 0.500 M and no CD. The temperature is constant. The invented data are below ([D₂] is always equal to [C₂]).

Time (min)	0	5	10	15	20	30	40	50	60	70	80
[C₂] (M)	0.500	0.419	0.368	0.338	0.321	0.307	0.302	0.301	0.300	0.300	0.300
[CD] (M)	0.000	0.162	0.264	0.324	0.358	0.386	0.396	0.398	0.400	0.400	0.400

(a) At which time in the table can you first say the system is at equilibrium? (b) A student says: “Equilibrium was reached between 15 and 20 minutes, because that is when [C₂] and [CD] became equal.” Evaluate the claim. (c) Show that the data are consistent with the balanced equation.

(a)

1. Look for the first time after which every concentration stays the same.
2. [C₂] is still falling at 50 min (0.301 M) and reaches 0.300 M at 60 min. After that it does not change. [CD] behaves the same way.
3. So the system is at equilibrium by 60 min (it may have got there a little before, between 50 and 60 min).

(b)

1. Between 15 and 20 min the [C₂] line and the [CD] line do cross: [C₂] goes from above [CD] (0.338 vs 0.324) to below it (0.321 vs 0.358).
2. But both concentrations are still changing after 20 min, for example [CD] rises by a further 0.028 M between 20 and 30 min.
3. Equilibrium is defined by constant concentrations, not equal ones. The claim is incorrect: equal concentrations happened during the approach, and equilibrium came later, with [CD] = 0.400 M and [C₂] = 0.300 M, which are not equal.

(c)

1. Change in [C₂] from 0 to 60 min: 0.500 − 0.300 = 0.200 M (a decrease).
2. Change in [CD]: 0.400 − 0.000 = 0.400 M (an increase).
3. Ratio 0.200 : 0.400 = 1 : 2, which matches the coefficients 1 C₂ : 2 CD. The same check works at every row, for example at 10 min: 2 × (0.500 − 0.368) = 0.264 M.

Interpretation. Only 40% of the C₂ has reacted (0.200 of 0.500 M) when the mixture stops changing. Both reactants and the product are present at equilibrium, as expected for a reversible reaction.

Worked example 2: approaching equilibrium from both sides

Question. Two identical flasks are held at the same temperature. Flask 1 starts with A(g) at 1.00 atm and no B. Flask 2 starts with B(g) at 2.00 atm and no A. The reaction is A(g) ⇌ 2B(g), the one in Figures 1 and 2. After a long time, flask 1 contains P_A = 0.70 atm and P_B = 0.60 atm.

(a) Show that flask 2 contains the same total amount of material as flask 1. (b) Flask 2 settles at P_A = 0.70 atm. Find P_B in flask 2. (c) Explain what the result tells you about the reaction.

(a) 2.00 atm of B is the amount that 1.00 atm of A would make (each A gives 2B). So both flasks hold the same atoms in the same volume. They differ only in which side of the equation they start from.

(b)

1. In flask 2 the net change is B → A. Making 0.70 atm of A uses 2 × 0.70 = 1.40 atm of B.
2. P_B = 2.00 − 1.40 = 0.60 atm.

So both flasks end with P_A = 0.70 atm and P_B = 0.60 atm, total 1.30 atm.

(c)

1. In flask 2 the reverse process, 2B → A, clearly happens, because A appears where there was none. So the reaction is reversible.
2. Flask 1 moves forward and flask 2 moves in reverse, yet both stop at the same mixture. At that mixture the forward and reverse rates are equal, so there is no net change either way.
3. The total pressure gives an easy way to follow this. Flask 1 rises from 1.00 atm to 1.30 atm; flask 2 falls from 2.00 atm to 1.30 atm. Once the total pressure is steady, the flask is at equilibrium.

Check. In flask 2 at the start, there is no A, so the forward rate is zero and only the reverse process can occur. That is why the net change goes from right to left.

Common misconceptions

• “At equilibrium, the concentrations of reactants and products are equal.” They are constant, not equal. In Worked example 1, [CD] = 0.400 M and [C₂] = 0.300 M at equilibrium.
• “At equilibrium, the reaction has stopped.” The forward and reverse processes both continue, at equal rates. Labelled-atom experiments show particles still moving between reactants and products.
• “The forward reaction finishes, then the reverse reaction starts.” Both run at the same time from the moment some product exists. The reverse rate just starts at zero and grows.
• “Equilibrium means the rate constants are equal.” It is the rates that are equal. The rate constants can be very different; the concentrations adjust until the rates match (you will see this in Topic 7.2).
• “Where the lines cross on a concentration graph is equilibrium.” A crossing only means two concentrations are momentarily equal. Look for where the lines go flat.
• “Equilibrium only applies to chemical reactions.” Evaporation in a sealed container and a saturated solution with solid present are physical equilibria.
• “A system in an open container will reach equilibrium.” If a product escapes, for example a gas leaving an open beaker, the reverse process cannot keep up and the system does not settle. Equilibrium needs a closed system.

Where this leads

Topic 7.2 looks more closely at the rates: what decides the direction of net change when the forward and reverse rates are not equal. Read Direction of Reversible Reactions next. Later topics in Unit 7 give equilibrium a number, K, and you will meet proton-transfer equilibria again in Unit 8 and electron transfer again in Unit 9. If you want to look back at the last topic of Unit 6, see Hess’s law. Try the practice questions now, then use the revision notes and the checklist to consolidate.