Notes · ChemistryUK · A-Levels

Chemical equilibria, Le Chatelier's principle and Kc

Many reactions are reversible and settle into a dynamic equilibrium in which forward and reverse reactions continue at equal rates. This chapter explains what dynamic equilibrium means, uses Le Chatelier's principle to predict how changing conditions shifts the position of equilibrium, defines and calculates the equilibrium constant Kc, and applies all of this to the compromise conditions of industrial processes such as the Haber process.

4 sections~13 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 1

T·0666 / 18

Exam profile

AO1 · Define dynamic equilibrium and state Le Chatelier's principleAO2 · Write and calculate Kc, deduce its units, and predict how changing conditions shifts the equilibriumAO3 · Explain the compromise conditions of an industrial process in terms of rate, yield and cost

Operators:statedefinepredictcalculatededuceexplain

basic level

AS-Level requires dynamic equilibrium, Le Chatelier's principle, the calculation of Kc and equilibrium concentrations, and the application to industrial processes.

higher level

The full A-Level extends this to Kp for gaseous equilibria and to the quantitative treatment of acids, bases and buffers in the thermodynamics topic; the Le Chatelier reasoning here carries throughout.

Depth

Text

Contents · 4 sections

Chemical equilibria, Le Chatelier's principle and Kc

Dynamic equilibrium in a closed system#

FoundationAQA 7405 3.1.6.1DfE GCE Chemistry - equilibria

Forward and reverse rates reaching equilibrium

Fig. 1Equilibrium is reached when the forward and reverse rates become equal; the reaction continues but there is no net change.

A reversible reaction can go in both directions, shown by the equilibrium arrow

\rightleftharpoons

. In a closed system - one that exchanges no matter with the surroundings - such a reaction reaches a state of dynamic equilibrium. 'Dynamic' means the forward and reverse reactions are both still happening; 'equilibrium' means they occur at exactly equal rates, so there is no net change. The concentrations of reactants and products remain constant, but individual molecules are continually reacting in both directions.

At the start of a reaction the reactants are at their highest concentration, so the forward rate is fastest, while there is no product yet, so the reverse rate is zero. As the reaction proceeds the forward rate falls (reactants are used up) and the reverse rate rises (product accumulates), until the two rates become equal. From that moment the concentrations no longer change and equilibrium has been reached - although the reaction has not stopped.

The position of equilibrium describes how far the reaction has gone when equilibrium is reached: it lies 'to the right' if there is a high proportion of products, 'to the left' if reactants predominate. This is different from the rate at which equilibrium is reached - a reaction can reach an equilibrium that lies well to the right either quickly or slowly. Equilibrium can be approached from either direction and, for a given temperature, reaches the same position regardless of the starting side.

Two conditions are essential for the definition. First, the system must be closed, because if a gas or product escaped, the reverse reaction could not keep pace and true equilibrium could never be established. Second, the temperature must be constant, because temperature is the only factor that changes the actual value of the equilibrium constant. State both conditions when you define dynamic equilibrium in an exam answer.

Worked example

Describing the approach to equilibrium

For a reversible reaction starting from pure reactants, describe how the forward and reverse rates change until equilibrium is reached.

At the start
Reactant concentration is highest, so the forward rate is at its maximum; there is no product, so the reverse rate is zero.
During the reaction
As reactants are used up the forward rate falls; as product builds up the reverse rate rises.
At equilibrium
When the two rates become equal, the concentrations stop changing - dynamic equilibrium is established, though both reactions continue.

Result: The forward rate falls and the reverse rate rises until they are equal, at which point the concentrations become constant.

Typical mistakes

Saying the reaction 'stops' at equilibrium - it continues in both directions at equal rates.
Omitting the requirement that the system is closed and at constant temperature.

Active revision

Explain why a reversible reaction can only reach a true dynamic equilibrium in a closed system, using the example of a gas that would otherwise escape.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Chemistry 7405 specification (AQA)

Le Chatelier's principle#

StandardAQA 7405 3.1.6.1DfE GCE Chemistry - Le Chatelier's principle

Le Chatelier responses

Fig. 2The equilibrium shifts to oppose each change; only temperature alters the value of Kc.

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in conditions, the position of equilibrium shifts in the direction that tends to oppose the change. It is a qualitative predictive tool: work out what the change is, then ask which way the equilibrium must move to partly counteract it. The word 'partly' matters - the shift reduces but never fully cancels the change.

Changing concentration shifts the equilibrium away from the added substance. Increasing the concentration of a reactant shifts the equilibrium to the right (more product), because the system responds by using up some of the added reactant; removing a product has the same effect. This is why continually removing a product drives a reaction towards completion. Changing the concentration does not change the value of

K_c

; the system simply moves to a new position that restores the same

K_c

Changing pressure affects only equilibria involving gases, and only when the number of gas molecules differs between the two sides. Increasing the total pressure shifts the equilibrium towards the side with fewer gas molecules, reducing the pressure. In the Haber process

\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3

there are four gas molecules on the left and two on the right, so high pressure favours ammonia. If the number of gas molecules is equal on both sides, pressure has no effect on the position.

Changing temperature is the only change that alters the value of the equilibrium constant. Increasing the temperature shifts the equilibrium in the endothermic direction (which absorbs the added heat), and decreasing it shifts towards the exothermic direction. For an exothermic forward reaction, raising the temperature therefore shifts the equilibrium back to the left and decreases

K_c

. A catalyst changes neither the position of equilibrium nor

K_c

; it speeds up the forward and reverse reactions equally, so equilibrium is simply reached sooner.

Worked example

Predicting a shift in the Contact process

The forward reaction 2SO2 + O2 -> 2SO3 is exothermic. Predict and explain the effect on the equilibrium yield of SO3 of increasing the temperature.

Identify the change
The temperature is increased, adding heat to the system.
Apply Le Chatelier
The equilibrium shifts in the endothermic direction to absorb the added heat - here that is the reverse reaction.
State the effect
The equilibrium shifts to the left, so the yield of SO3 falls and Kc decreases.

Result: Raising the temperature shifts the equilibrium left, lowering the yield of SO3 and decreasing Kc.

Typical mistakes

Applying a pressure change to an equilibrium with equal gas moles on both sides (no effect), or to reactions involving only solids and liquids.
Claiming a catalyst or a concentration change alters Kc - only temperature does.

Active revision

For $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ (exothermic forward reaction), predict the effect on the yield of SO3 of (a) increasing the pressure and (b) increasing the temperature.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

The equilibrium constant Kc#

StandardAQA 7405 3.1.6.2DfE GCE Chemistry - the equilibrium constant Kc

ICE table for an equilibrium calculation

Fig. 3The ICE table gives the equilibrium amounts, from which Kc = [HI]^2 / ([H2][I2]) = 64.

The equilibrium constant

K_c

gives a quantitative measure of the position of equilibrium. For a general reaction

a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}

, it is the product of the equilibrium concentrations of the products, each raised to the power of its balancing number, divided by the same for the reactants:

K_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}

. The concentrations used are the equilibrium concentrations in

\text{mol dm}^{-3}

. A large

K_c

means the equilibrium lies to the right (mostly products); a small

K_c

means it lies to the left.

The units of

K_c

are not fixed - they depend on the equation, and must be worked out each time by substituting

\text{mol dm}^{-3}

for each concentration and cancelling. For

\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}

there are two concentration terms on top and two on the bottom, so the units cancel and

K_c

has no units. For

\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3

the powers are

2

on top and

1 + 3 = 4

on the bottom, giving units of

(\text{mol dm}^{-3})^{-2} = \text{mol}^{-2}\,\text{dm}^{6}

To calculate

K_c

you usually need the equilibrium concentrations, which are found with an ICE table (Initial, Change, Equilibrium). Enter the initial amounts, use the equation's ratio to fill in the change in terms of the amount that reacts, and read off the equilibrium amounts; then divide each by the volume to get concentrations. It is the concentration, not the amount, that goes into the

K_c

expression - so never forget to divide by the volume (although for a reaction with equal moles on both sides, such as

\text{H}_2 + \text{I}_2

, the volume happens to cancel).

Two properties of

K_c

are examinable. First, it is constant at a fixed temperature: adding more reactant, changing the pressure or adding a catalyst all leave

K_c

unchanged (the system moves to a new position that restores the same value). Second, only a change of temperature changes

K_c

, and the direction follows Le Chatelier - for an exothermic forward reaction, raising the temperature lowers

K_c

K_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}

Equilibrium constant

Products over reactants, each raised to its balancing number, using equilibrium concentrations in mol dm^-3.

Worked example

Calculating Kc for the hydrogen-iodine equilibrium

1.00 mol of H2 and 1.00 mol of I2 are mixed in a 1.00 dm^3 flask. At equilibrium 0.20 mol of H2 remains. Calculate Kc for H2 + I2 = 2HI.

Equilibrium amounts
0.80 mol H2 has reacted, so I2 = 0.20 mol and HI = 2 x 0.80 = 1.60 mol; in 1.00 dm^3 these are also the concentrations.
Write the expression
$K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$ .
Substitute
$K_c = \dfrac{(1.60)^2}{0.20 \times 0.20} = \dfrac{2.56}{0.040} = 64$ .
$K_c = \dfrac{(1.60)^2}{0.20 \times 0.20} = 64$
Units
Two concentration terms on top and two on the bottom cancel, so Kc has no units.

Result: Kc = 64 (no units); the large value shows the equilibrium lies well to the right.

Typical mistakes

Using initial concentrations instead of equilibrium concentrations in the Kc expression.
Quoting Kc without its correct units, or forgetting to divide amounts by the volume to get concentrations.

Active revision

For $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ , write the expression for Kc and deduce its units.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Industrial processes and compromise conditions#

AdvancedAQA 7405 3.1.6.1DfE GCE Chemistry - industrial equilibria

Ammonia yield against pressure and temperature

Fig. 4Illustrative trend: a higher pressure and a lower temperature both raise the equilibrium yield of ammonia; industry compromises for rate and cost.

Industrial equilibrium processes must balance three competing demands: a high yield of product, a fast enough rate, and low cost. Le Chatelier's principle predicts the conditions that maximise yield, but those are often not the conditions that give an acceptable rate or that are affordable, so real plants run under compromise conditions - a deliberate trade-off rather than the theoretical optimum for yield alone.

The Haber process,

\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

, with an exothermic forward reaction, illustrates every trade-off. A high pressure favours ammonia (fewer gas molecules on the right) and also speeds the reaction, so a high pressure of around

20\ \text{MPa}

(

200\ \text{atm}

) is used - limited only by the cost and safety of building and running high-pressure plant. A low temperature would maximise yield (the forward reaction is exothermic), but the rate would be far too slow, so a compromise temperature of about

400\text{-}450\ ^{\circ}\text{C}

is used to give an acceptable rate at a modest yield.

An iron catalyst is added because it speeds up the attainment of equilibrium without shifting its position or changing

K_c

; this lets the plant use a lower temperature (saving energy and improving yield) for the same rate. Crucially, the unreacted nitrogen and hydrogen are recycled, and the ammonia is removed by cooling and liquefying it, which continually shifts the equilibrium to the right, so the overall conversion is high even though the single-pass yield is modest.

The same reasoning applies to the Contact process for sulfuric acid,

2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3

(exothermic, fewer moles of gas on the right), which uses a moderate temperature of about

450\ ^{\circ}\text{C}

, a vanadium(V) oxide catalyst and a pressure only slightly above atmospheric (because the yield is already high, so expensive high-pressure plant is not justified). When you answer a compromise-conditions question, always explain the choice of pressure, temperature and catalyst separately, referring to yield, rate and cost.

Worked example

Justifying Haber-process conditions

Explain why the Haber process is run at about 450 C rather than at a much lower temperature, given that the forward reaction is exothermic.

Yield argument
The forward reaction is exothermic, so a lower temperature would shift the equilibrium right and increase the yield of ammonia.
Rate argument
However, at a low temperature the rate would be far too slow, so equilibrium would take too long to reach and the plant would be uneconomic.
Compromise
A temperature of about 450 C, with an iron catalyst, is chosen to give an acceptable rate at a reasonable yield.

Result: 450 C is a compromise: it sacrifices some yield to gain an acceptable rate, with the catalyst allowing this without shifting the equilibrium.

Typical mistakes

Stating the conditions that give the maximum yield without recognising the compromise for rate and cost.
Claiming the catalyst increases the yield - it only increases the rate at which equilibrium is reached.

Active revision

The Haber process uses about 200 atm and 450 C with an iron catalyst. Explain why a higher pressure is not used and why a lower temperature is not used, even though both would increase the yield.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Chemistry 7405 specification

Chemical equilibria, Le Chatelier's principle and Kc

4 sections~13 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 1

Chemical equilibria, Le Chatelier's principle and Kc

Key points

Describing the approach to equilibrium

Key points

Predicting a shift in the Contact process

Key points

Calculating Kc for the hydrogen-iodine equilibrium

Key points

Justifying Haber-process conditions

Sources

Chemical equilibria, Le Chatelier's principle and Kc

Key points

Describing the approach to equilibrium

Key points

Predicting a shift in the Contact process

Key points

Calculating Kc for the hydrogen-iodine equilibrium

Key points

Justifying Haber-process conditions

Sources