Chemical equilibrium

In a nutshell

Many reactions are reversible: products can react to re-form the reactants. In a closed system they settle into a dynamic equilibrium, where both reactions keep going at the same rate so nothing seems to change.

This subtopic is about predicting how the position of equilibrium shifts when you change concentration, pressure or temperature (le Chatelier's principle), why industry picks compromise conditions, and how to write and calculate the equilibrium constant, K_c.

Assumed knowledge: Enthalpy changes: the sign of ΔH; Reaction rates: collision theory and catalysts; Amount of substance: concentration in mol dm^-3.

Core content

Dynamic equilibrium: what it means

A reversible reaction is written with the ⇌ sign. For example, the Haber process:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

The left-to-right reaction is the forward reaction. The right-to-left reaction is the reverse reaction.

Dynamic equilibrium: exists in a closed system when the rate of the forward reaction is equal to the rate of the reverse reaction and the concentrations of reactants and products do not change.

Each part of that definition earns its place:

• Closed system: nothing is added or removed, so no reactant or product can escape.
• Rates equal: both reactions are still happening. That is why it is called dynamic.
• Concentrations do not change: they are constant, not equal. An equilibrium mixture can be mostly products or mostly reactants.

Here is how it gets there, for a simple reaction A ⇌ B that starts with only A:

Read the two graphs together:

1. At the start there is only A, so the forward rate is high and the reverse rate is zero.
2. As A is used up, the forward reaction slows. As B builds up, the reverse reaction speeds up.
3. When the two rates become equal, A is made as fast as it is used up, so the concentrations stop changing. That is dynamic equilibrium.

Notice that at equilibrium [B] is about twice [A]. The concentrations are constant, not equal.

The position of equilibrium describes the relative amounts of reactants and products in the equilibrium mixture:

• Position to the right: the mixture is mostly products.
• Position to the left: the mixture is mostly reactants.

Still don't get it? · dynamic equilibrium

Picture a busy shop. People walk in through one door and walk out through another. If five people come in every minute and five leave every minute, the number of people inside stays the same, even though the actual people keep changing.

Now rebuild the chemistry one step at a time:

1. The people walking in are the forward reaction turning A into B.
2. The people walking out are the reverse reaction turning B back into A.
3. Both doors are always open: neither reaction ever stops.
4. When "in" and "out" happen at the same rate, the count inside (the concentration) stops changing.
5. The count inside does not have to equal the count outside. It only has to stay the same.
6. The doors must be the only way in or out. If people could climb out of a window, the count would never settle. That is why the system must be closed.

Exam version: in a closed system, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products do not change. Never write that the reaction has stopped, or that the concentrations are equal.

Le Chatelier's principle: the idea

Le Chatelier's principle lets you predict which way the position of equilibrium moves when you change a condition. OCR does not ask you to define it, but you must use it.

The working version: if a closed system at equilibrium is changed, the position of equilibrium shifts to oppose (minimise) the change.

At AS you apply it to homogeneous equilibria: equilibria where all the species are in the same phase (for example, all gases, or all in aqueous solution).

Every le Chatelier answer needs three things:

1. Which way the position of equilibrium shifts (to the left or right).
2. Why, in terms of the specific change (fewer moles of gas, the endothermic direction, removing the added substance).
3. The effect asked for (yield of a product, colour, and so on).

Le Chatelier's principle: changing concentration

• Increase the concentration of a reactant: the position shifts to the right, to use up the added reactant. More products form.
• Decrease the concentration of a product (remove it): the position shifts to the right, to replace the product.
• Increasing a product's concentration, or removing a reactant, does the opposite: the position shifts to the left.

The graph shows extra A being added to the A ⇌ B mixture at 40 s:

1. [A] jumps up when A is added.
2. The position of equilibrium shifts to the right: some of the added A turns into B.
3. [A] falls (but stays above its old value) and [B] rises, until the concentrations are constant again.

Le Chatelier's principle: changing pressure

Pressure only matters for equilibria with gases. Count the moles of gas on each side of the balanced equation.

• Increase pressure: the position shifts towards the side with fewer moles of gas, to reduce the pressure.
• Decrease pressure: the position shifts towards the side with more moles of gas, to increase the pressure.
• Same number of moles of gas on each side: changing pressure has no effect on the position of equilibrium.

Le Chatelier's principle: changing temperature

The ΔH value quoted with an equilibrium always refers to the forward reaction. The reverse reaction has the same size of ΔH with the opposite sign.

• Increase temperature: the position shifts in the endothermic direction, to absorb the extra heat energy and lower the temperature.
• Decrease temperature: the position shifts in the exothermic direction, to release heat energy and raise the temperature.

For the Haber process, ΔH = -92 kJ mol^-1, so the forward reaction is exothermic:

Still don't get it? · which way does the equilibrium shift?

Think of a stubborn cat on a sofa. Push it to the left and it leans to the right. Whatever you do to it, it pushes back the other way. It never fully undoes your push, but it always fights it.

Now rebuild it for each kind of change, one step at a time:

1. Ask: what did I change, and in which direction? (More of something? More pressure? Hotter?)
2. Ask: what would undo that change?
   • Added a substance: use it up.
   • Raised the pressure: make fewer gas molecules, because fewer molecules push less.
   • Raised the temperature: take in heat, which is what an endothermic reaction does.
3. Ask: which direction of the reaction does that? Look at the equation: which side has fewer moles of gas, or which direction has a positive ΔH?
4. That direction is where the position of equilibrium shifts.

Lowering something works the same way, just reversed: the equilibrium shifts to put back what you took away.

Exam version: "The position of equilibrium shifts to the right / left to oppose the change, because that is the side with fewer moles of gas / the endothermic direction / the direction that removes the added substance." Then state the effect on the yield.

Le Chatelier's principle: adding a catalyst

A catalyst increases the rate of both the forward and reverse reactions by the same amount. So the position of equilibrium is unchanged.

• The yield at equilibrium is the same with or without a catalyst.
• The catalyst makes the system reach equilibrium faster.
• That is still very valuable in industry: the same yield is produced in less time.

Practical: investigating changes to the position of equilibrium

You need to know the techniques for showing, qualitatively, how concentration and temperature shift an equilibrium. The trick is to pick an equilibrium where the two sides are different colours, so you can see the shift.

Changing concentration: the chromate(VI) / dichromate(VI) equilibrium

2CrO₄^2-(aq) + 2H⁺(aq) ⇌ Cr₂O₇^2-(aq) + H₂O(l)

yellow ⇌ orange

1. Put the same volume of potassium chromate(VI) solution into several test tubes. Keep one as a control for colour comparison.
2. Add dilute acid dropwise to one tube, shaking after each drop, and compare it with the control against a white background.
3. Add aqueous sodium hydroxide dropwise to the acidified tube and compare again.

Chromium(VI) compounds are toxic: wear gloves and eye protection.

Changing temperature: the nitrogen dioxide / dinitrogen tetroxide equilibrium

2NO₂(g) ⇌ N₂O₄(g) ΔH = -57 kJ mol^-1

brown ⇌ colourless

1. Use identical sealed tubes of the gas mixture (NO₂ is toxic, so it is never opened). Keep one at room temperature as the control.
2. Put one tube in an ice-water bath and one in a hot water bath.
3. Leave each tube long enough to reach the bath temperature, then compare the colours with the control.
4. Return the tubes to room temperature: the colours change back, showing the shift is reversible.

In an exam you may be given an unfamiliar coloured equilibrium. Use exactly the same reasoning: decide the shift, then say which colour you would see.

Le Chatelier's principle: compromise conditions in industry

Industry wants a high yield and a fast rate, made safely and cheaply. Equilibrium and rate often pull in opposite directions, so the operating conditions are a compromise.

The Haber process is the classic example:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ mol^-1

Two more points that earn marks:

• Low temperature: gives a higher equilibrium yield but a slow rate, so the product is made too slowly to be economic.
• Unreacted gases are recycled back into the reactor, so the overall conversion is high even though each pass gives a modest yield.

You do not need to memorise exact conditions: a question will give you the reaction and ΔH. You must be able to explain the trade-off for any industrial equilibrium.

The equilibrium constant, K_c: writing the expression

For a homogeneous equilibrium

aA + bB ⇌ cC + dD

the equilibrium constant is:

$K_c = \frac{[\mathrm{C}]^c\,[\mathrm{D}]^d}{[\mathrm{A}]^a\,[\mathrm{B}]^b}$

• Square brackets [ ] mean equilibrium concentration in mol dm^-3.
• Products on top, reactants on the bottom.
• The balancing numbers become powers, not multipliers.
• Concentrations are multiplied together, never added.

OCR does not ask you to work out the units of K_c in this subtopic, so give the number only.

The equilibrium constant, K_c: calculating K_c

At AS you calculate K_c from equilibrium concentrations that you are given:

1. Write the K_c expression from the balanced equation.
2. Substitute the equilibrium concentrations.
3. Raise each concentration to its power before multiplying.
4. Give the answer to the same number of significant figures as the least precise data.

Worked examples 1 and 2 below show two full calculations.

The equilibrium constant, K_c: what its size tells you

The size of K_c gives a qualitative estimate of the position of equilibrium:

This works because products are on the top of the expression: a big top and a small bottom give a big K_c.

Worked examples

1. Calculating K_c (one squared term)

Hydrogen and iodine reach equilibrium in a sealed container:

H₂(g) + I₂(g) ⇌ 2HI(g)

The equilibrium concentrations are [H₂] = 0.0050 mol dm^-3, [I₂] = 0.0030 mol dm^-3 and [HI] = 0.028 mol dm^-3. Calculate K_c.

Step 1: write the expression. The 2 in front of HI becomes a power.

$K_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$

Step 2: substitute the equilibrium concentrations.

$K_c = \frac{(0.028\ \mathrm{mol\ dm^{-3}})^2}{(0.0050\ \mathrm{mol\ dm^{-3}}) \times (0.0030\ \mathrm{mol\ dm^{-3}})}$

Step 3: work out the top and the bottom.

$\text{top} = 0.028^2 = 7.84 \times 10^{-4}$

$\text{bottom} = 0.0050 \times 0.0030 = 1.5 \times 10^{-5}$

Step 4: divide and round.

$K_c = \frac{7.84 \times 10^{-4}}{1.5 \times 10^{-5}} = 52.27$

$K_c = 52 \ \text{(2 s.f., because the data are 2 s.f.)}$

Where marks go:

• Forgetting to square [HI] gives 1867 (about 1900): wrong.
• Writing the expression upside down gives 0.019: wrong.
• No units are needed for K_c in this subtopic.

2. Calculating K_c (a cubed term and standard form)

For the Haber process at a particular temperature:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

The equilibrium concentrations are [N₂] = 0.400 mol dm^-3, [H₂] = 1.20 mol dm^-3 and [NH₃] = 0.0350 mol dm^-3. Calculate K_c.

Step 1: write the expression.

$K_c = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}$

Step 2: substitute.

$K_c = \frac{(0.0350\ \mathrm{mol\ dm^{-3}})^2}{(0.400\ \mathrm{mol\ dm^{-3}}) \times (1.20\ \mathrm{mol\ dm^{-3}})^3}$

Step 3: apply the powers first.

$0.0350^2 = 1.225 \times 10^{-3}$

$1.20^3 = 1.728$

Step 4: multiply the bottom, then divide.

$\text{bottom} = 0.400 \times 1.728 = 0.6912$

$K_c = \frac{1.225 \times 10^{-3}}{0.6912} = 1.7723 \times 10^{-3}$

$K_c = 1.77 \times 10^{-3} \ \text{(3 s.f.)}$

Where marks go:

• Forgetting the cube on [H₂] gives 2.55 × 10^-3: wrong.
• Using the balancing numbers as multipliers, (2 × 0.0350) ÷ (0.400 × 3 × 1.20), gives 0.0486: wrong.
• Rounding to 1.8 × 10^-3 loses precision: the data are 3 s.f., so give 3 s.f.

3. Estimating the position of equilibrium from K_c

Use the answers above.

• Example 1: K_c = 52, which is greater than 1, so the position of equilibrium lies to the right: the mixture contains more HI than H₂ and I₂.
• Example 2: K_c = 1.77 × 10^-3, which is much less than 1, so the position of equilibrium lies to the left: under these conditions the mixture is mostly N₂ and H₂.

This is only a qualitative estimate. You will not be asked for anything more precise in this subtopic.

4. Model answer: explaining a shift in equilibrium

Methanol can be made from carbon monoxide and hydrogen:

CO(g) + 2H₂(g) ⇌ CH₃OH(g) ΔH = -91 kJ mol^-1

Explain the effect on the equilibrium yield of methanol of (a) increasing the pressure and (b) increasing the temperature.

(a) Increasing the pressure:

1. There are 3 moles of gas on the left and 1 mole of gas on the right.
2. The position of equilibrium shifts to the right, the side with fewer moles of gas, to oppose the increase in pressure.
3. So the yield of methanol increases.

(b) Increasing the temperature:

1. The forward reaction is exothermic (ΔH is negative), so the reverse reaction is endothermic.
2. The position of equilibrium shifts to the left, in the endothermic direction, to oppose the increase in temperature by absorbing heat.
3. So the yield of methanol decreases.

Each part has the same three-step shape: reason from the equation, direction of shift, effect on yield.

Common exam mistakes

• Writing that at equilibrium "the reaction has stopped". Both reactions are still happening; the mark needs the rates of the forward and reverse reactions are equal.
• Writing that at equilibrium "the concentrations of reactants and products are equal". They are constant (do not change), not equal.
• Leaving out closed system from the definition of dynamic equilibrium.
• Giving only the direction ("the equilibrium shifts to the right") with no reason. You need the why: fewer moles of gas, the endothermic direction, or using up the added substance.
• Writing "the equilibrium shifts to oppose the change" and stopping there. Say which way it shifts and what happens to the yield.
• Counting molecules of every substance, or comparing atoms, when deciding the pressure effect. Count moles of gas from the balancing numbers in the equation.
• Forgetting that the quoted ΔH is for the forward reaction. If ΔH is negative, raising the temperature favours the reverse (endothermic) reaction.
• Saying a catalyst "shifts the equilibrium to the right" or "increases the yield". It increases the rates of the forward and reverse reactions by the same amount, so the position of equilibrium is unchanged; equilibrium is just reached faster.
• For industrial conditions, saying high pressure is "expensive" or "dangerous" with no reason. Give the reason: energy to compress the gases and strong, expensive equipment.
• Saying a low temperature is used because it gives a high yield. In practice a compromise temperature is used, because a low temperature makes the rate too slow.
• In K_c expressions: putting reactants on top, adding concentrations instead of multiplying, using balancing numbers as multipliers instead of powers, or using round brackets instead of square brackets.
• Substituting initial concentrations or amounts into K_c. Only equilibrium concentrations go in.
• Rounding K_c to the wrong number of significant figures. Match the least precise data given.
• In the practical, describing the shift but not the observation. Say what you would see (for example, "yellow to orange", "darker brown").

Key definitions

• Dynamic equilibrium: exists in a closed system when the rate of the forward reaction is equal to the rate of the reverse reaction, and the concentrations of reactants and products do not change.
• Closed system: a system in which no substances are added or removed.
• Position of equilibrium: the relative amounts of reactants and products in the equilibrium mixture.
• Homogeneous equilibrium: an equilibrium in which all the species are in the same phase.
• Le Chatelier's principle (formal definition not required by OCR; this is the working statement to apply): if a closed system at equilibrium is changed, the position of equilibrium shifts to oppose (minimise) the change.
• Effect of a catalyst on an equilibrium: a catalyst increases the rate of both the forward and reverse reactions by the same amount, so the position of equilibrium is unchanged.
• K_c for aA + bB ⇌ cC + dD: $K_c = \dfrac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$, where [ ] is the equilibrium concentration in mol dm^-3.

Specification

• I can explain that a dynamic equilibrium exists in a closed system when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products do not change. (3.2.3 a)
• I can use le Chatelier's principle to deduce, qualitatively, how changing temperature, pressure or concentration shifts the position of a homogeneous equilibrium. (3.2.3 b)
• I can explain that a catalyst increases the rates of the forward and reverse reactions by the same amount, so the position of equilibrium is unchanged. (3.2.3 c)
• I can describe techniques to investigate, qualitatively, how changes in concentration and temperature shift the position of equilibrium. (3.2.3 d)
• I can explain why industry uses a compromise between equilibrium yield and reaction rate (with safety and cost) when choosing operating conditions, for example in the Haber process. (3.2.3 e)
• I can write the K_c expression for a homogeneous equilibrium. (3.2.3 f)
• I can calculate K_c from given equilibrium concentrations, to an appropriate number of significant figures. (3.2.3 f)
• I can estimate the position of equilibrium from the size of K_c. (3.2.3 g)

Related notes

• Enthalpy changes: exothermic and endothermic reactions and the sign of ΔH
• Reaction rates: collision theory, catalysts and the Boltzmann distribution
• Amount of substance: concentrations in mol dm^-3
• Alkenes: hydration of alkenes with steam and an acid catalyst
• How far? (A level): K_c and K_p calculations, units, and how temperature changes equilibrium constants
• Acids, bases and buffers (A level): K_a, an equilibrium constant for weak acids