Chemical equilibria and Le Chatelier's principle

In a nutshell

Equilibrium is about how far a reversible reaction goes, not how fast (that is kinetics).

Many reactions are reversible and settle into a dynamic equilibrium: both the forward and reverse reactions keep going, but nothing measurable seems to change.

This note covers how an equilibrium responds to changes in temperature, pressure and concentration (Le Chatelier's principle), why industry uses compromise conditions, and how to write and use the equilibrium constant Kc.

Assumed knowledge: Amount of substance, Energetics, Kinetics.

Core content

Reversible reactions and dynamic equilibrium

A reversible reaction can go both ways, shown by the sign ⇌:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

In a closed system (nothing added or removed), a reversible reaction reaches dynamic equilibrium. There are two features you must be able to state:

• the rate of the forward reaction equals the rate of the reverse reaction
• the concentrations of reactants and products remain constant

"Dynamic" means both reactions are still happening; they have not stopped. The forward and reverse changes simply cancel out, so the amounts stay fixed.

As a reaction approaches equilibrium, the forward rate falls (reactants run low) while the reverse rate rises (products build up), until the two rates are equal:

Once the rates are equal, the concentrations stop changing. Notice they become constant, not equal to each other:

We describe the make-up of the mixture as the position of equilibrium. If it lies to the right, the mixture is mostly products; to the left, mostly reactants.

Still don't get it? · how can concentrations be constant if the reaction is still going?

Picture a sink with the tap running and the plug out. Water pours in from the tap and drains out of the plughole at the same time. After a while the water level stops rising or falling, even though water is still flowing in and out constantly. The level is steady not because everything has stopped, but because the two flows are matched.

Now the chemistry. The tap is the forward reaction making products; the drain is the reverse reaction turning products back into reactants. At the start the tap is fast (lots of reactant) and the drain is slow (hardly any product yet). As product builds up the drain speeds up and the tap slows down, until the two are going at exactly the same rate. From that moment the amount of each substance stays fixed.

So the exam points are: the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant (constant, not equal to each other). "Dynamic" is the word that captures the tap and drain still running.

Le Chatelier's principle

Le Chatelier's principle: if a system at equilibrium is subjected to a change in conditions, the position of equilibrium shifts in the direction that opposes the change.

Two habits make the difference between a mark and no mark:

1. Say "the position of equilibrium shifts", and refer to the equilibrium, not to "the system".
2. Always name the specific change being opposed (for example "to oppose the increase in pressure"), never just "to oppose the change".

Effect of temperature

Changing temperature shifts the equilibrium in whichever direction cancels the temperature change:

• Increasing the temperature shifts the equilibrium in the endothermic direction, because that direction takes in heat and lowers the temperature again.
• Decreasing the temperature shifts it in the exothermic direction, which gives out heat.

So for the exothermic forward reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (ΔH = −92 kJ mol^-1), raising the temperature shifts the equilibrium to the left (the endothermic, reverse direction), lowering the yield of ammonia.

Effect of pressure

Pressure only matters for gases. Increasing the pressure shifts the equilibrium towards the side with fewer moles of gas, which reduces the pressure again.

• In N₂(g) + 3H₂(g) ⇌ 2NH₃(g) there are 4 moles of gas on the left and 2 on the right, so increasing the pressure shifts the equilibrium to the right, raising the yield of ammonia.
• If the number of moles of gas is equal on both sides, pressure has no effect. For example H₂(g) + Cl₂(g) ⇌ 2HCl(g) has 2 moles each side.

Effect of concentration

Increasing the concentration of a species shifts the equilibrium in the direction that removes that species.

• Increasing a reactant concentration shifts the equilibrium towards the products (raising the yield).
• Increasing a product concentration, or removing a product, shifts it towards the reactants, or towards making more product respectively.

Coloured equilibria make this visible. Adding concentrated hydrochloric acid to a solution of aqueous copper(II) ions shifts a coloured equilibrium and the colour changes; adding water shifts it back again (skills opportunity PS 1.1).

Effect of a catalyst

A catalyst has no effect on the position of equilibrium and no effect on the value of Kc.

It speeds up the forward and reverse reactions equally, so equilibrium is simply reached faster; the final composition is unchanged.

Still don't get it? · why a catalyst does not change the yield

Imagine two towns with people walking in both directions along a slow mountain path. At any moment, say 30 people per hour cross east and 30 cross west, so the population of each town holds steady. Now someone builds a fast tunnel through the mountain. It helps everyone, whichever way they are going, so now 90 cross each way per hour. The crossings are much faster, but because both directions sped up by the same amount, the population of each town is exactly what it was before.

A catalyst is that tunnel. It provides an easier route (a lower activation energy) for the reaction, and that route is open to both the forward and the reverse reaction. Both rates increase by the same factor.

The exam point: a catalyst increases the rate of the forward and reverse reactions equally, so it has no effect on the position of equilibrium (or on Kc); it only reduces the time taken to reach equilibrium. Saying "a catalyst does not change the equilibrium amounts" on its own just repeats the question and earns nothing; you must give the reason.

Compromise conditions in industrial processes

The conditions that give the best yield are often not the conditions that give a workable rate or an affordable plant, so industry uses a compromise. The Haber process makes this concrete:

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

Typical conditions are about 450 °C, about 200 atm (roughly 20 MPa) and an iron catalyst.

Temperature is a compromise. The forward reaction is exothermic, so a low temperature would give a higher yield. But a low temperature gives a slow rate. A moderate temperature (about 450 °C) is chosen to give an acceptable yield at an acceptable rate.

Pressure is a compromise. There are fewer moles of gas on the right, so a high pressure gives a higher yield and a faster rate. But very high pressures are expensive:

• the energy cost of the pumps needed to produce the high pressure is large
• the plant and pipework must be built to withstand the high pressure, which is costly.

A moderate high pressure is chosen as the balance.

The catalyst does not change the yield; it speeds the reaction, which lets a lower temperature be used (saving energy) while still reaching equilibrium quickly.

The equilibrium constant Kc

For a general homogeneous equilibrium

mA + nB ⇌ pC + qD

the equilibrium constant is

$K_c = \frac{[\text{C}]^{p}\,[\text{D}]^{q}}{[\text{A}]^{m}\,[\text{B}]^{n}}$

where [X] is the equilibrium concentration of X in mol dm^-3, and each concentration is raised to the power of its balancing number from the equation. Homogeneous means every species is in the same physical state (for example all gases, or all in solution).

The units of Kc change from reaction to reaction, so you must work them out each time by putting mol dm^-3 into the expression and cancelling (see Worked examples).

A large Kc means the equilibrium lies towards the products; a small Kc means it lies towards the reactants.

What changes the value of Kc

This is the point examiners catch students on: only a change in temperature changes the value of Kc.

• Concentration and pressure: the position of equilibrium can move, but Kc does not change. The system simply re-adjusts so that the expression gives the same value again.
• A catalyst: no effect on Kc (or on the position).
• Temperature: this is the only thing that changes Kc. For the exothermic forward reaction N₂ + 3H₂ ⇌ 2NH₃, increasing the temperature decreases Kc (the equilibrium moves towards reactants, so there are fewer products). For an endothermic forward reaction, increasing the temperature increases Kc.

Determining Kc for an esterification (skills opportunity PS 2.3)

A classic way to measure Kc is the esterification of ethanol with ethanoic acid, using a strong acid as the catalyst:

CH₃COOH(l) + CH₃CH₂OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

Outline method:

1. Mix known amounts of ethanoic acid, ethanol and a little dilute sulfuric acid in a stoppered tube (work out the initial moles from the volumes and densities).
2. Leave the mixture for about a week to reach equilibrium.
3. Titrate the whole mixture against standard sodium hydroxide using phenolphthalein, to find the total acid present.
4. Subtract the acid due to the sulfuric acid catalyst (found separately) to get the moles of ethanoic acid at equilibrium.
5. Deduce the equilibrium moles of ethanol, ester and water from the mole ratios, then substitute concentrations into:

$K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5]\,[\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}]\,[\text{CH}_3\text{CH}_2\text{OH}]}$

Because there are equal numbers of moles on each side, the volume cancels and Kc has no units here.

Worked examples

Example 1: construct the Kc expression and find its units.

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g):

$K_c = \frac{[\text{SO}_3]^{2}}{[\text{SO}_2]^{2}\,[\text{O}_2]}$

Put in the unit mol dm^-3 and cancel:

$\text{units} = \frac{(\text{mol dm}^{-3})^{2}}{(\text{mol dm}^{-3})^{2}(\text{mol dm}^{-3})} = \frac{1}{\text{mol dm}^{-3}}$

$\text{units} = \text{mol}^{-1}\,\text{dm}^{3}$

Watch the balancing numbers become powers, and always derive the units rather than guessing them.

Example 2: calculate a value for Kc.

In a 2.00 dm³ vessel, 1.60 mol of N₂ and 4.00 mol of H₂ are allowed to reach equilibrium. At equilibrium 25.0% of the nitrogen has reacted. Find Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).

Moles of N₂ reacted:

$n(\text{N}_2)_{\text{reacted}} = 0.250 \times 1.60 = 0.400 \text{ mol}$

Use the 1 : 3 : 2 ratio:

$n(\text{H}_2)_{\text{reacted}} = 3 \times 0.400 = 1.20 \text{ mol}$

$n(\text{NH}_3)_{\text{formed}} = 2 \times 0.400 = 0.800 \text{ mol}$

Each concentration is the equilibrium moles divided by the volume, 2.00 dm³. This division is the step most students forget.

$K_c = \frac{[\text{NH}_3]^{2}}{[\text{N}_2]\,[\text{H}_2]^{3}} = \frac{(0.400)^{2}}{(0.600)(1.40)^{3}}$

$K_c = \frac{0.160}{1.6464} = 0.0972 \text{ mol}^{-2}\,\text{dm}^{6}$

The data are given to 3 significant figures, so the answer is quoted to 3 significant figures, with its derived units mol^-2 dm⁶.

Example 3: perform a calculation using Kc.

For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 49.0 (no units) at the temperature used. 1.00 mol of H₂ and 1.00 mol of I₂ are mixed in a vessel of volume V. Find the equilibrium amount of each substance.

Let x = moles of H₂ that react. Then at equilibrium there are (1.00 − x) mol of H₂, (1.00 − x) mol of I₂ and 2x mol of HI.

Because the numbers of moles of gas are equal on both sides, the volume V cancels throughout, so we can use moles directly:

$K_c = \frac{(2x)^{2}}{(1.00 - x)(1.00 - x)} = \frac{(2x)^{2}}{(1.00 - x)^{2}} = 49.0$

Take the square root of both sides (this avoids a quadratic):

$\frac{2x}{1.00 - x} = 7.00$

$2x = 7.00(1.00 - x) = 7.00 - 7.00x$

$9.00x = 7.00 \quad\Rightarrow\quad x = 0.778 \text{ mol}$

So at equilibrium:

• n(H₂) = 1.00 − 0.778 = 0.222 mol
• n(I₂) = 1.00 − 0.778 = 0.222 mol
• n(HI) = 2 × 0.778 = 1.56 mol

Common exam mistakes

• Defining equilibrium without the word rate: writing "the forward reaction equals the backward reaction". The mark needs the rate of the forward reaction equals the rate of the reverse reaction.
• Saying the concentrations of reactants and products are equal at equilibrium. They are constant, not equal to each other.
• In a Le Chatelier answer, writing only "to oppose the change". You must name the specific change opposed, for example "to oppose the increase in temperature" or "to lower the temperature by taking in heat".
• Referring to "the system" shifting. AQA wants "the equilibrium" or "the position of equilibrium".
• Explaining a catalyst by saying what it does or defining it, instead of the required point: it increases the rate of the forward and reverse reactions equally, so the position of equilibrium is unchanged.
• In a Kc calculation, putting the number of moles straight into the expression instead of the concentration. You must divide the moles by the volume in dm³ first. This is the single most common Kc error.
• Forgetting to raise a concentration to its power (for example not squaring [NH₃] or cubing [H₂]).
• Leaving off the units of Kc, or copying them from another reaction. The units change every time and must be derived; a common slip is wrongly adding s^-1.
• Rounding too early, or giving the wrong number of significant figures. Quote the final answer to the least precise data (often 2 significant figures).
• Thinking pressure, concentration or a catalyst changes the value of Kc. Only a change in temperature changes Kc.
• On compromise conditions, writing "energy is expensive" or "equipment is expensive" with no reason. Link it to the energy needed to drive the pumps that create the pressure, and the cost of plant built to withstand high pressure.
• Discussing only the effect on yield and ignoring the effect on rate (or the reverse), and confusing the equilibrium yield with the amount made in a given time.

Key definitions

• Dynamic equilibrium (features): the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant, in a closed system.
• Le Chatelier's principle - if a system at equilibrium is subjected to a change in conditions, the position of equilibrium shifts in the direction that opposes the change.
• Position of equilibrium - a description of the composition of the equilibrium mixture, that is, whether it lies towards the reactants (left) or the products (right).
• Equilibrium constant Kc - for a homogeneous equilibrium, the value of the equilibrium concentrations (in mol dm^-3) of the products, each raised to the power of its balancing number, divided by those of the reactants, each raised to the power of its balancing number.
• Effect of a catalyst - it increases the rate of the forward and reverse reactions equally, so it has no effect on the position of equilibrium or on the value of Kc; it only reduces the time taken to reach equilibrium.
• Compromise conditions - conditions chosen to give an acceptable yield at an acceptable rate (and cost), rather than the conditions giving the maximum possible yield.

Specification

• I can state that, in a reversible reaction at equilibrium, the forward and reverse reactions proceed at equal rates and the concentrations of reactants and products remain constant.
• I can state Le Chatelier's principle and use it to predict the effect of changes in temperature, pressure and concentration on the position of equilibrium.
• I can state that a catalyst does not affect the position of equilibrium, and explain why (it speeds the forward and reverse reactions equally).
• I can explain why a compromise temperature and pressure may be used in an industrial process.
• I can construct an expression for Kc for a homogeneous system in equilibrium.
• I can calculate a value for Kc from equilibrium concentrations, including working out its units.
• I can perform calculations involving Kc (for example finding an equilibrium amount from a given Kc).
• I can predict the qualitative effect of a change of temperature on the value of Kc.

Related notes

• Amount of substance
• Energetics
• Kinetics
• Equilibrium constant Kp for homogeneous systems (A-level only, builds on this note)