Reversible Reactions and Equilibria

Chemistry revision notes · Edexcel · 4CH1

Back Flashcards

1. Reversible reactions

What "reversible" means

A can proceed in both directions: the products can themselves combine, or break apart, to give the original reactants back again
The turns reactants into products
The reverse (backward) reaction turns products back into reactants
A reversible equation uses a special two-headed arrow, ⇌, instead of the single arrow → used for one-way reactions
- The top half-arrow points right (forward); the bottom half-arrow points left (reverse)
Whether the reaction settles toward the products side or toward the reactants side depends on the conditions — temperature, pressure, concentration

Common exam question

What the ⇌ symbol means

Question: State what the ⇌ symbol in an equation represents, or give its meaning (1 mark).

The mark is for reversible: the reaction can go in both directions, so the products can react to re-form the reactants. Writing that the reaction goes both ways is also accepted. Asked in 3 of the 23 papers, once on Paper 1 and twice on Paper 2, always attached to an equation that the later parts of the question then use.

Do not write "equilibrium" on its own: all three mark schemes ignore it. The symbol tells you the reaction is reversible; whether it has actually reached equilibrium depends on the container, not on the arrow.

Two reversible reactions to know

Heating hydrated copper(II) sulfate drives off its water of crystallisation: the blue crystals turn white as CuSO₄·5H₂O becomes anhydrous CuSO₄. Adding water reverses the change, turning the white solid blue again and releasing heat, which is why anhydrous copper(II) sulfate is used as a test for water: CuSO₄·5H₂O ⇌ CuSO₄ + 5H₂O
Heating ammonium chloride decomposes the white solid into ammonia and hydrogen chloride gas; the two gases recombine on the cooler upper walls of the tube to reform solid ammonium chloride: NH₄Cl ⇌ NH₃ + HCl

Reversible reaction of copper(II) sulfate: heating blue hydrated CuSO4.5H2O is endothermic and drives off water to leave white anhydrous CuSO4; adding water reverses the change and is exothermic — Source: Reversible reactions by Save My Exams

2. Dynamic equilibrium

What equilibrium looks like

Place a reversible reaction in a sealed container with no escape route for any chemical — a
At first only the forward reaction occurs (no products to react in the reverse direction yet); its rate is highest
As the products build up, the reverse reaction starts and speeds up
The forward rate falls (reactants becoming scarcer) and the reverse rate rises (products becoming more abundant)
Eventually the two rates become equal: the system is at

Graph of reaction rate against time: the forward rate starts high and decreases while the reverse rate starts at zero and increases, until the two lines meet and level off where equilibrium is reached with both rates equal — Source: Dynamic equilibrium by Save My Exams

Two defining features

The rate of the equals the rate of the
The concentrations of reactants and products stay constant from this point onward (provided temperature and pressure do not change)

Common exam question

Stating the characteristics of dynamic equilibrium

Question: State what is meant by dynamic equilibrium, or give one or two characteristics of a reaction at dynamic equilibrium (1–2 marks).

Asked in 4 of the 23 papers, all on Paper 2. There are two credited points, one mark each: the forward and reverse reactions are taking place at the same rate, and the reactant and product concentrations stay constant. Three of the four schemes also accept amounts or moles in place of concentrations, and for a one-mark version either point on its own is enough.

The standard slip is to say the reactant and product concentrations are "the same" or "equal": every one of the four schemes rejects it. Constant means they stop changing over time, not that there is as much product as reactant. Nor have the reactions stopped; both keep running, which is what "dynamic" means.

Why "dynamic", not "static"

At equilibrium the reactions have not stopped — both still happen at full speed
Each forward conversion of a reactant into a product is matched by a reverse conversion of a product back into a reactant
The chemicals interchange constantly; only the overall amounts stay fixed

Closed-system requirement

Equilibrium can only be reached when nothing can leak out (or leak in)
Open systems lose products as they form (a gas escaping into the air, for instance), so the reverse reaction never has enough material to keep up and equilibrium is never reached

Iodine in an open versus a closed test tube: in the open system iodine vapour escapes so the change is one-way, I2(s) → I2(g); in the sealed closed system the vapour cannot leave and recondenses, so a reversible equilibrium I2(s) ⇌ I2(g) is set up — Source: Dynamic equilibrium by Save My Exams

3. Position of equilibrium and Le Chatelier's principle

What "position of equilibrium" means

The position of equilibrium describes how much of each species is present once dynamic equilibrium has been reached
The position lies "to the right" if products are favoured (more product, less reactant at equilibrium)
The position lies "to the left" if reactants are favoured (less product, more reactant)
Changing the conditions of the system shifts where this balance sits

Le Chatelier's principle

states that if a closed system at equilibrium is disturbed, the equilibrium position shifts so as to oppose the change imposed on it
Le Chatelier's principle predicts the shift direction caused by:
- Changing the temperature
- Changing the pressure (gases)
- Changing the concentration of a species (not assessed numerically, but the direction is)

Exam tip

Get the yield direction right before the reason

In the two-mark questions on the effect of a change on the yield, the first mark is for the direction and the second for the reason. In three of the four papers that set these, the reason mark depends on the direction being correct or left blank: a wrong direction loses both marks, while a reason alone can still earn one.

Naming Le Chatelier's principle earns nothing: three schemes ignore it explicitly, and two ignore remarks about rate. Give the direction, then the specific reason: the sign of ΔH, the moles of gas, or equal rates for a catalyst.

Effect of temperature

Identify whether the forward reaction is exothermic or endothermic; the reverse direction is then the opposite
Increasing temperature shifts the equilibrium in the direction of the reaction (the system "absorbs" the extra heat to oppose the rise)
Decreasing temperature shifts the equilibrium in the direction of the reaction (the system releases heat to oppose the fall)
Worked example. In the Haber process N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), the forward reaction is exothermic.
- Raising the temperature drives the equilibrium to the left — ammonia decomposes — so the yield of NH₃ falls
- Industry compromises by running the reaction at about 450 °C: hot enough for a fast rate, cool enough that the yield is still useful

Flow diagram of the Haber process: nitrogen (from air) and hydrogen (from methane and steam) are compressed and passed over a catalyst at about 450 °C and high pressure, where they reach equilibrium; the ammonia is cooled and liquefied and the unreacted gases are recycled — Source: Haber Process by Wikimedia Commons

Common exam question

Effect of temperature on the yield at equilibrium

Question: Explain the effect on the yield of the product at equilibrium when the temperature is raised or lowered at constant pressure (2 marks).

Asked in 3 of the 23 papers, all on Paper 2, and each time the equation is printed with its ΔH. The first mark is for the direction of the yield change. The second is for the reason: that the forward reaction is exothermic (or endothermic), which you read straight from the sign of ΔH. Saying instead that the reverse reaction is endothermic (or exothermic) is accepted, and you may add that the equilibrium shifts in the endothermic direction on heating.

So a negative ΔH means the forward reaction is exothermic: raising the temperature moves the equilibrium in the endothermic, reverse direction and the yield falls, while lowering it raises the yield. A positive ΔH reverses both conclusions.

Effect of pressure (gaseous reactions only)

Count the number of moles of gas on each side of the equation; only species in the gaseous state count
Increasing pressure shifts the equilibrium toward the side with fewer moles of gas — that side occupies a smaller volume, opposing the pressure rise
Decreasing pressure shifts the equilibrium toward the side with more moles of gas
If the two sides have the same number of moles of gas, changing the pressure has no effect on the position of equilibrium (it does still alter the rate at which equilibrium is reached)
Worked example. In the Contact process 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g):
- Left side has 3 mol of gas (2 SO₂ + 1 O₂); right side has 2 mol of gas (2 SO₃)
- Raising the pressure shifts the equilibrium to the right, producing more SO₃
- The industrial reaction runs at slightly above atmospheric pressure because the yield is already high; the extra cost of higher-pressure equipment is not worth the small gain

Common exam question

Effect of pressure on the yield at equilibrium

Question: Explain, give a reason for, or predict with a reason the effect on the yield of the product at equilibrium when the pressure is increased or decreased at constant temperature (1–2 marks).

Asked in 5 of the 23 papers, all on Paper 2. For two marks the first is for the direction of the yield change and the second for the reason, a count of the moles of gas on each side; when the paper states the direction itself and asks only for a reason, the moles comparison alone earns the single mark. The comparison ("fewer moles of gas on the right") or the actual numbers are both credited, and "molecules" is accepted for moles.

Raising the pressure favours the side with fewer moles of gas; lowering it favours the side with more. One scheme wants the direction of the shift alongside the yield change for the first mark, so for a pressure decrease the safe answer reads: the yield falls, because the equilibrium shifts to the left, the side with more moles of gas (for an increase, reverse both halves).

Effect of concentration

Adding more of a reactant shifts the equilibrium toward the products to consume the added material — position shifts to the right
Adding more of a product shifts the equilibrium toward the reactants — position shifts to the left
Removing a product as it forms (e.g. condensing it out of a gas mixture) drains material from the right side, so the system makes more product to compensate — equilibrium keeps shifting to the right
This last trick is heavily used in industry: continuously removing the product is a way to push reactions that would otherwise stop at a moderate yield

Effect of a catalyst

A speeds up both the forward and the reverse reactions by the same factor
It lets equilibrium be reached faster, but it does not change the position once equilibrium has been reached
The amounts of reactant and product at equilibrium are the same with or without the catalyst
Industrially, catalysts are still used because reaching a useful yield in minutes instead of hours saves enormous amounts of energy

Rate against time graph comparing a reaction with and without a catalyst: the dashed catalysed forward and backward rates meet sooner than the solid uncatalysed rates, so equilibrium is reached faster, but both reach the same final equilibrium rate so the position of equilibrium is unchanged — Source: The position of equilibrium by Save My Exams

Common exam question

Effect of a catalyst on the yield at equilibrium

Question: Explain the effect, if any, of adding a catalyst on the yield at equilibrium, or give the reason why a catalyst is used (1–2 marks).

Asked in 5 of the 23 papers, all on Paper 2: four ask for its effect on the yield or the position of equilibrium, one for the reason it is used. The credited reason is that a catalyst speeds up the forward and the reverse reaction equally. For two marks, the first is usually for no effect on the yield and the second for that reason; one scheme gives its second mark for the word "equally" alone. Name both directions and say equally, or that both rates are affected the same.

Asked why a catalyst is used, say that it increases the rate, or lets equilibrium be reached sooner. "To increase the yield" was rejected, and the explanation about an alternative pathway with a lower activation energy was ignored.

Change made	Position of equilibrium shifts	Why
Increase temperature	Toward the endothermic side	System absorbs the added heat
Decrease temperature	Toward the exothermic side	System releases heat to oppose the fall
Increase pressure (gases)	Toward the side with fewer moles of gas	Reduces total moles → reduces pressure
Decrease pressure (gases)	Toward the side with more moles of gas	Increases total moles → opposes the drop
Add more reactant	To the right (products)	Consumes the excess reactant
Remove a product	To the right (products)	Replaces the missing product
Add a catalyst	No change in position; equilibrium reached faster	Catalyst speeds both rates equally