Direction of Reversible Reactions: Study Guide (Chemistry 7.2)

Two rates, one net direction

In Topic 7.1 you saw that a reversible reaction runs both ways at once. Particles of reactant turn into product (the forward reaction) while particles of product turn back into reactant (the reverse reaction). What you actually observe is the net result of the two.

The net rate is the difference between them:

net rate = forward rate − reverse rate

That gives three cases.

Comparison of rates	Net rate	What you observe
forward rate > reverse rate	positive	reactants are used up, products build up: net change to the right
reverse rate > forward rate	negative	products are used up, reactants build up: net change to the left
forward rate = reverse rate	zero	no net change: the system is at equilibrium

So the direction of a reversible reaction is not fixed by the equation. It depends on which of the two rates is larger at that moment.

How the rates change on the way to equilibrium

Each rate depends on the concentrations of the species that react in that direction. The forward rate depends on the reactants; the reverse rate depends on the products. As the net reaction runs, those concentrations change, and so do the rates.

Suppose the net change is to the right:

• Reactants are being used up, so the forward rate falls.
• Products are building up, so the reverse rate rises.
• The gap between them shrinks until the two rates are equal. That is equilibrium.

The same happens in the other direction. Figure 1 shows a fictional reaction, E(aq) ⇌ F(aq), started with only F present: [F] = 0.80 M and [E] = 0. Each direction is a single elementary step (Topic 5.4), so forward rate = k_f[E] and reverse rate = k_r[F], with the invented values k_f = 0.060 s⁻¹ and k_r = 0.020 s⁻¹.

Read Figure 1 like this:

• At t = 0 there is no E, so the forward rate is zero. The reverse rate is k_r[F] = 0.020 × 0.80 = 0.016 M s⁻¹, its largest value.
• While reverse > forward, F is used up and E builds up. The net change is to the left.
• As this happens the reverse rate falls (less F) and the forward rate rises (more E).
• At equilibrium [E] = 0.20 M and [F] = 0.60 M, and both rates are 0.012 M s⁻¹.

The equation is written E ⇌ F, but here the reaction runs “backwards” until equilibrium is reached. Direction depends on the rates, not on which side of the equation you call the reactants.

Calculating the two rates for an elementary step

When a reversible reaction is a single elementary step, you can write the rate law for each direction from its molecularity, just as in Topic 5.4:

• A ⇌ B: forward rate = k_f[A], reverse rate = k_r[B].
• 2A ⇌ A₂: forward rate = k_f[A]², reverse rate = k_r[A₂].
• A + B ⇌ C + D: forward rate = k_f[A][B], reverse rate = k_r[C][D].

Then substitute the concentrations, compare the two rates, and read off the direction. For a reaction that is not a single step, you cannot write the rate laws from the equation, but the rule is the same: whichever measured rate is larger decides the direction.

Worked example 1: which way will each mixture go?

Question. Use the reaction E(aq) ⇌ F(aq) from Figure 1, with k_f = 0.060 s⁻¹ and k_r = 0.020 s⁻¹ at the temperature of the experiment. Three separate mixtures are prepared:

Mixture	[E] (M)	[F] (M)
1	0.30	0.50
2	0.10	0.50
3	0.20	0.60

For each mixture, calculate the forward and reverse rates and state the direction of net change.

1. Mixture 1. Forward rate = 0.060 s⁻¹ × 0.30 M = 0.018 M s⁻¹. Reverse rate = 0.020 s⁻¹ × 0.50 M = 0.010 M s⁻¹. Forward > reverse, so the net change is to the right: E is converted to F. Net rate = 0.018 − 0.010 = 0.008 M s⁻¹.
2. Mixture 2. Forward rate = 0.060 × 0.10 = 0.006 M s⁻¹. Reverse rate = 0.020 × 0.50 = 0.010 M s⁻¹. Reverse > forward, so the net change is to the left: F is converted to E, at a net rate of 0.004 M s⁻¹.
3. Mixture 3. Forward rate = 0.060 × 0.20 = 0.012 M s⁻¹. Reverse rate = 0.020 × 0.60 = 0.012 M s⁻¹. The rates are equal, so mixture 3 is at equilibrium. Its concentrations will not change.

Check. Mixtures 1 and 2 contain the same [F] but different [E]. Only the forward rate differs, and that alone is enough to reverse the direction.

Interpretation. Mixture 1 holds 0.80 M in total. As it reacts, [E] falls and [F] rises until the rates match. That happens when 0.060[E] = 0.020[F], that is when [F] is three times [E]: [E] = 0.20 M and [F] = 0.60 M, which is mixture 3. So mixture 1 ends up as mixture 3, and Figure 1 (which also holds 0.80 M in total) ends up there too, from the other side. Mixture 2 (0.60 M in total) settles at [E] = 0.15 M and [F] = 0.45 M. Equilibrium mixtures for this reaction always have [F] three times [E] at this temperature; Topic 7.3 turns that fixed ratio into the equilibrium constant.

Particle models of direction

A particle diagram, or “snapshot”, shows the particles in a small region at one instant. Snapshots are a good way to connect what particles do with what you measure. They also have limits, and the exam may ask you to say what a model does and does not show.

Figure 2 shows four snapshots of a fictional gas reaction, 2Z(g) ⇌ Z₂(g), in a sealed container at constant temperature.

Worked example 2: reading and judging a particle model

Question. Use Figure 2.

(a) Show that the snapshots are consistent with the equation. (b) State the direction of net change between snapshots 1 and 2, and compare the forward and reverse rates in that interval. (c) What can you conclude from snapshots 3 and 4? (d) Give two things these snapshots do not show about the reaction.

(a) Count the Z units. Snapshot 1: 12. Snapshot 2: 6 + 2 × 3 = 12. Snapshots 3 and 4: 4 + 2 × 4 = 12. The total is conserved, and each Z₂ uses 2 Z, which matches 2Z ⇌ Z₂.

(b)

1. Z falls from 12 to 6 and Z₂ rises from 0 to 3. Six Z have become three Z₂.
2. The net change is to the right (forward).
3. So in this interval the forward rate (2Z → Z₂) is greater than the reverse rate (Z₂ → 2Z). At the very start, with no Z₂ present, the reverse rate is zero.

(c) The counts are the same at 20 s and 30 s (4 Z, 4 Z₂), while the particles are in different positions. The amounts are no longer changing, so the system has probably reached equilibrium by 20 s. At equilibrium the forward and reverse rates are equal. Note that 4 Z and 4 Z₂ are equal counts here by coincidence; equal amounts are not what defines equilibrium.

(d) Any two of:

• Rates are not shown. A single snapshot shows how many particles there are, not how fast they react. You can only infer which rate is larger by comparing snapshots.
• The dynamic process is hidden. In snapshots 3 and 4, Z₂ molecules are still forming and breaking up. The diagrams cannot show which particular Z₂ broke apart and re-formed between 20 s and 30 s.
• Too few particles. A real sample contains an enormous number of particles. With only 12 units, random variation could change the counts from one snapshot to the next even at equilibrium.
• Timing. You cannot tell exactly when equilibrium began; it could be any time between 10 s and 20 s.

Interpretation. The model is good for showing conservation of atoms and the direction of net change. To show that equilibrium is dynamic, you would need extra information, for example arrows on individual particles, or a labelled-particle experiment like the one in Topic 7.1.

Preview: what if you upset an equilibrium?

The rate rule also explains what happens when you disturb a system at equilibrium, which you will study in Topics 7.9 and 7.10. Take mixture 3 from Worked example 1 and suddenly add more F. The reverse rate, k_r[F], jumps up at once, but the forward rate, k_f[E], has not changed yet. Now reverse > forward, so there is a net change to the left until the rates match again. You do not need to calculate this now; just notice that the same rule, “the larger rate wins”, decides the direction.

Common misconceptions

• “Forward means the reaction always goes left to right.” The net direction depends on which rate is larger. A mixture with too much product goes right to left.
• “At the start, the reaction only goes forward.” Only if no product is present. If both reactants and products are there, both directions run from the start.
• “The direction depends on the rate constants alone.” In Worked example 1, the same k_f and k_r gave a net change to the right for mixture 1 and to the left for mixture 2. Concentrations matter as much as the constants.
• “A larger k_f means the forward rate is always faster.” Rate = k × concentration term. A large k_f with a tiny reactant concentration can still give the smaller rate.
• “The faster direction stays faster until the reactants run out.” The faster direction slows and the slower one speeds up until they are equal. Neither rate reaches zero.
• “Two snapshots with equal counts prove nothing is happening.” Equal counts at two times show no net change, not no reaction.
• “One snapshot is enough to tell the direction.” One snapshot shows amounts only. Direction needs a change over time or rate information.

Where this leads

Next, Topic 7.3 introduces the reaction quotient Q and the equilibrium constant K. Comparing Q with K gives the direction of net change straight from concentrations, with no rate constants needed. It is a shortcut for the reasoning on this page. Look back at Topic 7.1 if the signs of equilibrium are not secure. Try the practice questions now, then use the revision notes and the checklist to consolidate.