Last updated: 18 August 2026
In kinetics, we studied how quickly reactions occur. Chemical equilibrium asks a different question: what happens when a reaction can proceed in both directions and neither direction eventually produces any overall change?
Products do not always represent the final stopping point of a reaction. In a reversible reaction, the products can react to reform the original reactants. Forward and reverse reactions can therefore take place simultaneously.
Once this microscopic picture is clear, ideas such as Le Chatelier’s Principle, Kc, Kp and ICE-table calculations become much easier to understand. They are all ways of describing the behaviour of the same reversible chemical system.
Diagnostic Quiz
Try these five questions before reading the chapter. They reveal whether some common equilibrium misconceptions are already hiding in your thinking.
- 5 quick questions
- Instant right-or-wrong feedback
- Every option is explained
Reversible reactions proceed in both directions
A reversible reaction is one in which products can react to reform the reactants.
For a general reaction:
we can distinguish two directions:
- A + B → C + D is the forward reaction; and
- C + D → A + B is the reverse reaction.
Both directions can occur within the same reacting mixture.
It does not tell us that equal amounts of reactants and products are present. Nor does it tell us that equilibrium has already been established. It simply shows that both reaction directions are possible.
An example: nitrogen dioxide and dinitrogen tetroxide
A useful gaseous example is:
N₂O₄ molecules can dissociate to produce NO₂ molecules. At the same time, NO₂ molecules can collide and combine to reform N₂O₄.
| Direction | Reaction | What happens? |
|---|---|---|
| Forward | N₂O₄(g) → 2NO₂(g) | N₂O₄ is converted into NO₂. |
| Reverse | 2NO₂(g) → N₂O₄(g) | NO₂ is converted back into N₂O₄. |
What is dynamic equilibrium?
A reversible reaction is not automatically at equilibrium. Equilibrium describes a particular state of the reversible system.
Dynamic equilibrium is established in a closed system when the rate of the forward reaction is equal to the rate of the reverse reaction, so there is no overall change in the concentrations of reactants and products.
Every part of this definition matters:
- the system is closed;
- both forward and reverse reactions continue;
- the forward and reverse reaction rates are equal; and
- the concentrations of reactants and products remain constant with time.
Why is the equilibrium dynamic?
The word dynamic tells us that molecular change continues. Reactant particles are still becoming products, and product particles are still becoming reactants.
If the forward reaction forms a particular amount of product in one second and the reverse reaction converts an equivalent amount back into reactant during the same second, there is no net change in composition.
Incorrect idea: Equilibrium means that the reaction has stopped.
Correct reasoning: Both reactions continue at equal, non-zero rates. The molecular activity continues even though there is no overall observable change.
How dynamic equilibrium becomes established
Consider a simple reversible reaction:
Suppose the system initially contains only A.
- Initially, the forward reaction dominates. A is present but essentially no B has yet formed, so the reverse reaction is initially very slow or zero.
- As A is consumed, the forward reaction generally slows down. The concentration of A decreases.
- As B accumulates, the reverse reaction speeds up. More B particles are available to react back to form A.
- Eventually, the forward and reverse reaction rates become equal.
- The system has now established dynamic equilibrium.
| Stage | Forward reaction | Reverse reaction | Overall composition |
|---|---|---|---|
| Initially | Relatively fast | Very slow or zero | Products form overall |
| As reaction proceeds | Generally slows | Generally speeds up | Composition continues changing |
| At equilibrium | Constant rate | Same constant rate | Composition remains constant |
At equilibrium, the forward reaction has not gradually slowed to zero. Both reaction directions continue at the same non-zero rate.
Equilibrium can be approached from either direction
If the system starts mainly with products instead, the reverse reaction initially dominates. Reactants form, causing the forward reaction rate to increase until the two rates become equal.
Dynamic equilibrium can therefore be approached from either side of a reversible equation.
Recognising equilibrium from graphs
Rate–time graphs
If a system initially contains reactants only, the forward reaction rate begins relatively high. As reactants are consumed, it generally decreases.
The reverse rate initially begins at or near zero because little or no product is available. As products accumulate, the reverse rate increases.
Equilibrium is established when:
From then onwards, both rates remain equal and non-zero while the conditions remain unchanged.
On a rate–time graph: the two rate curves meet and then remain at the same non-zero value.
Do not draw both curves decreasing to zero. That would represent reactions stopping rather than dynamic equilibrium.
Concentration–time graphs
A concentration–time graph tells the same story from a different perspective. Reactant concentrations decrease while product concentrations increase until the system reaches equilibrium.
Once equilibrium is established, each concentration becomes constant. The curves therefore become horizontal.
The concentration remains constant because the species is being formed and consumed at equal overall rates.
Constant does not mean equal
Suppose the equilibrium:
has equilibrium concentrations:
[B] = 0.30 mol dm⁻³
Both concentrations can remain constant with time even though they are clearly not equal.
Constant concentration: the concentration no longer changes with time.
Equal concentrations: two species happen to have the same numerical concentration.
Dynamic equilibrium requires the first condition, not the second.
Why a closed system is required
Dynamic equilibrium is established in a closed system. Matter does not continuously enter or leave the reacting system.
This ensures that reactants and products remain available for both reaction directions.
If a gaseous product continuously escapes from an open container, for example, it is removed from the reaction mixture and cannot readily participate in the reverse reaction.
A closed system does not exchange matter with its surroundings, although energy may still be transferred. An isolated system exchanges neither matter nor energy.
The position of equilibrium
Equilibrium does not mean that half of the reactants have become products. Different reactions establish very different equilibrium compositions.
- If relatively more products are present, equilibrium is described as lying more towards the right.
- If relatively more reactants are present, equilibrium is described as lying more towards the left.
The position of equilibrium can change when the equilibrium conditions are altered. Le Chatelier’s Principle provides a useful qualitative way of predicting the direction of this change.
Le Chatelier’s Principle
When a system at equilibrium is subjected to a change in conditions, the equilibrium position shifts in the direction that tends to reduce the effect of that change.
We use this principle to predict the effect of changes in concentration, pressure and temperature.
Changing concentration
Consider:
If more A is added, the system responds by consuming some of the additional A. The equilibrium position therefore shifts towards the products.
If C is removed, more C tends to form, so the equilibrium also shifts towards the products.
| Change | Predicted shift |
|---|---|
| Add reactant | Towards products |
| Remove reactant | Towards reactants |
| Add product | Towards reactants |
| Remove product | Towards products |
Changing pressure in gaseous equilibria
Pressure changes are important when gases are involved and the two sides of the equation contain different total stoichiometric numbers of moles of gas.
Consider:
There are 4 moles of gas according to the equation on the left and 2 moles of gas on the right.
If the gas mixture is compressed at constant temperature, its pressure increases. The equilibrium shifts towards the side with fewer moles of gas.
- Increasing pressure by decreasing volume favours the side with fewer moles of gas.
- Decreasing pressure by increasing volume favours the side with more moles of gas.
Increasing pressure therefore shifts the Haber equilibrium to the right and increases the equilibrium amount of NH₃.
Pure solids and liquids are not included when deciding which side has fewer moles of gas.
What if both sides contain the same number of moles of gas?
Consider:
There are 2 moles of gas on each side. Compressing the system does not favour either direction, so there is no shift in equilibrium position.
Changing temperature
Temperature is fundamentally different from concentration and pressure because changing temperature also changes the value of the equilibrium constant.
A useful way to apply Le Chatelier’s Principle is to consider heat as being on the appropriate side of the reaction.
Exothermic forward reaction
Increasing temperature favours the endothermic reverse direction, so equilibrium shifts towards the reactants.
- Increase temperature → equilibrium shifts towards reactants.
- Decrease temperature → equilibrium shifts towards products.
- For an exothermic forward reaction, increasing temperature decreases K.
Endothermic forward reaction
Increasing temperature favours the endothermic forward direction.
- Increase temperature → equilibrium shifts towards products.
- Decrease temperature → equilibrium shifts towards reactants.
- For an endothermic forward reaction, increasing temperature increases K.
Changing concentration or pressure may alter the equilibrium composition at a fixed temperature, but the system eventually establishes another composition corresponding to the same equilibrium constant.
What does a catalyst do?
A catalyst provides an alternative reaction pathway with a lower activation energy. It increases both the forward and reverse reaction rates.
A catalyst therefore:
- allows equilibrium to be reached more quickly;
- does not shift the equilibrium position;
- does not alter the equilibrium composition; and
- does not change the equilibrium constant.
| Change | Can equilibrium position change? | Does K change? |
|---|---|---|
| Concentration | Yes | No, at constant temperature |
| Pressure | Yes, where applicable | No, at constant temperature |
| Temperature | Yes | Yes |
| Catalyst | No | No |
The equilibrium constant
Le Chatelier’s Principle gives a qualitative prediction about the direction in which an equilibrium shifts. The equilibrium constant, K, provides a quantitative description of the equilibrium composition at a particular temperature.
A reaction can strongly favour products at equilibrium and still take a long time to reach that equilibrium. Equilibrium and kinetics answer different questions.
Equilibrium constant in terms of concentration, Kc
For the general equilibrium:
the concentration equilibrium constant is written as:
The square brackets represent equilibrium concentrations. The stoichiometric coefficients in the balanced equation become powers in the equilibrium expression.
Example: writing a Kc expression
For:
the expression is:
Which species are included?
Gaseous and aqueous species may appear in a Kc expression. Pure solids and pure liquids are omitted because their effective concentrations are constant.
For example:
gives:
Incorrect: Kc = [CaO][CO₂] / [CaCO₃]
Correct: Kc = [CO₂]
Units of Kc
The units of Kc depend on the equilibrium expression.
For:
the concentration units cancel:
so no units remain in this case.
By contrast, for:
the unit is:
Derive the units from the particular equilibrium expression. In some reactions the units cancel completely; in others they do not.
Using an ICE table for equilibrium calculations
Equilibrium questions often provide some information about the mixture before equilibrium is reached and ask for information about the final equilibrium mixture. An ICE table keeps these stages organised.
- I — Initial: the concentrations before the equilibrium change occurs;
- C — Change: the changes as the reaction moves towards equilibrium; and
- E — Equilibrium: the final concentrations once equilibrium is established.
The concentration changes are linked by the coefficients in the balanced equation. If one species changes by x, the others do not automatically change by the same amount.
Worked example 1: calculating Kc using an ICE table
Consider:
Initially:
- [H₂] = 1.00 mol dm⁻³;
- [I₂] = 1.00 mol dm⁻³; and
- [HI] = 0 mol dm⁻³.
At equilibrium, [HI] = 1.20 mol dm⁻³. Calculate Kc.
Step 1: use the stoichiometric ratio
If H₂ decreases by x mol dm⁻³:
- H₂ changes by −x;
- I₂ changes by −x; and
- HI changes by +2x.
Since [HI] increases from 0 to 1.20 mol dm⁻³:
Step 2: complete the ICE table
| H₂ / mol dm⁻³ | I₂ / mol dm⁻³ | HI / mol dm⁻³ | |
|---|---|---|---|
| Initial, I | 1.00 | 1.00 | 0 |
| Change, C | −0.600 | −0.600 | +1.20 |
| Equilibrium, E | 0.400 | 0.400 | 1.20 |
Step 3: write the Kc expression
Step 4: substitute equilibrium concentrations
Answer: Kc = 9.00
The initial values help you construct the ICE table, but the final E row is what must be substituted into Kc.
Worked example 2: using Kc to find equilibrium concentrations
Consider the same equilibrium:
Initially:
[I₂] = 1.00 mol dm⁻³
[HI] = 0
At this temperature:
Let x mol dm⁻³ of H₂ react.
| H₂ | I₂ | HI | |
|---|---|---|---|
| Initial, I | 1.00 | 1.00 | 0 |
| Change, C | −x | −x | +2x |
| Equilibrium, E | 1.00 − x | 1.00 − x | 2x |
Substitute the equilibrium expressions into Kc:
Taking the positive square root:
Therefore:
Answer:
[H₂] = 0.333 mol dm⁻³
[I₂] = 0.333 mol dm⁻³
[HI] = 1.33 mol dm⁻³
Incorrect ICE-table change: −x, −x, +x
Correct change: −x, −x, +2x because the balanced equation has the stoichiometric ratio 1 : 1 : 2.
Equilibrium constant in terms of partial pressure, Kp
For equilibria involving gases, equilibrium composition may instead be expressed in terms of partial pressures.
For:
the equilibrium constant is:
Only gaseous species appear in a Kp expression.
Finding partial pressure
For a mixture of gases:
where:
Worked example: calculating Kp
Consider:
At equilibrium, a mixture contains:
- 1.00 mol SO₂;
- 0.500 mol O₂; and
- 2.00 mol SO₃.
The total pressure is 350 kPa.
Step 1: find the total number of moles
Step 2: calculate the partial pressures
Step 3: write the Kp expression
Step 4: substitute the equilibrium partial pressures
Answer: Kp = 0.0800 kPa⁻¹
Begin with the balanced equation. Use equilibrium concentrations for Kc and equilibrium partial pressures for Kp, with stoichiometric coefficients becoming powers.
What does the magnitude of K tell us?
The size of an equilibrium constant indicates the relative equilibrium amounts of products and reactants at a particular temperature.
| Magnitude of K | Qualitative interpretation |
|---|---|
| K ≫ 1 | Products are strongly favoured at equilibrium |
| K around 1 | Appreciable amounts of both sides may be present |
| K ≪ 1 | Reactants are strongly favoured at equilibrium |
Large K: tells you that equilibrium lies strongly towards products.
Fast reaction: tells you that products are formed rapidly.
These are different ideas. K describes equilibrium position, not rate.
The Haber process: equilibrium in industry
The Haber process provides an important industrial example in which equilibrium, reaction rate and economic considerations must all be balanced.
The forward reaction is exothermic and changes 4 moles of gas into 2 moles of gas.
Pressure
Increasing pressure shifts equilibrium towards the side with fewer moles of gas.
High pressure therefore increases the equilibrium yield of ammonia.
However, very high pressures require stronger equipment, greater energy input and higher operating costs. Industry therefore uses a high but practical pressure, commonly around 200 atm.
Temperature
Since the forward reaction is exothermic, lower temperature shifts equilibrium towards NH₃ and gives a greater equilibrium yield.
Unfortunately, lowering the temperature also slows the reaction. A very low temperature would therefore give a favourable equilibrium yield but an uneconomically slow production rate.
A compromise temperature of about 450°C is used to provide a reasonable equilibrium yield together with an acceptable reaction rate.
Catalyst
An iron catalyst increases the rates of both the forward and reverse reactions. It allows equilibrium to be established more quickly without changing the equilibrium yield or equilibrium constant.
Removing ammonia
Ammonia is removed from the equilibrium mixture. Removing a product shifts equilibrium towards the products and encourages further ammonia formation.
Unreacted nitrogen and hydrogen can be recycled through the process.
| Industrial condition | Equilibrium consideration | Practical consideration |
|---|---|---|
| High pressure | Favours NH₃ | Very high pressure increases cost and engineering demands |
| Compromise temperature | Lower temperature favours NH₃ | Higher temperature gives a faster rate |
| Iron catalyst | No change in equilibrium position | Equilibrium is reached more quickly |
| Removal of NH₃ | Shifts equilibrium towards NH₃ | Supports continuous production |
An industrial process must also consider reaction rate, energy requirements, equipment cost and safety. The Haber process is therefore a classic chemical compromise.
Common mistakes with chemical equilibria
| Common mistake | Why it is wrong | Correct approach |
|---|---|---|
| Saying both reactions stop at equilibrium | Dynamic equilibrium involves continuing forward and reverse reactions | State that both continue at equal, non-zero rates |
| Saying reactant and product concentrations become equal | Equilibrium requires constant concentrations, not equal concentrations | Use the phrase “remain constant with time” |
| Drawing equilibrium rate curves falling to zero | The reactions continue at equilibrium | Show the two rates becoming equal at a non-zero value |
| Assuming increasing pressure always shifts equilibrium towards products | Pressure favours the side with fewer moles of gas | Count gaseous mole numbers on both sides first |
| Including pure solids in a K expression | The effective concentration of a pure solid is constant | Omit pure solids and pure liquids |
| Forgetting the powers in an equilibrium expression | The powers come from the balanced-equation coefficients | Write the balanced equation before writing K |
| Using initial concentrations directly in Kc | Kc uses equilibrium concentrations | Find the E row first |
| Using the same x-change for every species in an ICE table | The changes must follow the stoichiometric ratio | Use changes such as −x, −x and +2x where required |
| Saying changing concentration changes K | At constant temperature the system readjusts until the same K is satisfied | Remember that temperature changes K |
| Saying a catalyst increases equilibrium yield | A catalyst accelerates both directions | State that the same equilibrium is reached faster |
| Using total pressure directly for every term in Kp | Kp uses the partial pressure of each gas | Calculate the required partial pressures first |
| Saying a large K means a fast reaction | K describes equilibrium position rather than kinetics | Keep equilibrium favourability and reaction rate separate |
Chemical-equilibrium checklist
Dynamic equilibrium
- Recognise a reversible reaction.
- Remember that both reaction directions continue.
- Forward rate = reverse rate at equilibrium.
- Concentrations remain constant, not necessarily equal.
Le Chatelier’s Principle
- Identify exactly which condition changes.
- For pressure, compare gaseous mole numbers.
- For temperature, identify the endothermic direction.
- Remember that a catalyst does not shift equilibrium.
Kc and Kp
- Use equilibrium concentrations for Kc.
- Use equilibrium partial pressures for Kp.
- Use stoichiometric coefficients as powers.
- Exclude pure solids and pure liquids.
ICE calculations
- Write the balanced equation first.
- Set up Initial, Change and Equilibrium rows.
- Make every change follow the mole ratio.
- Substitute equilibrium values into K.
Mastery Quiz
Now apply the chapter to equilibrium shifts, K expressions and calculations. These questions require more than simple definition recall.
- 10 application questions
- Equilibrium calculations and reasoning
- Every answer option is explained
Frequently asked questions
Questions students often ask about chemical equilibrium
These questions connect dynamic equilibrium, Le Chatelier’s Principle and equilibrium calculations so that they become one coherent picture rather than separate rules to memorise.
What is the difference between a reversible reaction and dynamic equilibrium?
A reversible reaction is capable of proceeding in both directions. Dynamic equilibrium is a particular state of that reversible system in which the forward and reverse reactions occur at equal rates in a closed system.
A reaction can therefore be reversible without being at equilibrium at that particular moment.
If concentrations remain constant, how can particles still be reacting?
Each species is being formed and consumed at equal overall rates. Individual particles continue reacting, but the amount being added is balanced by the amount being removed, so there is no net concentration change.
Why does changing concentration not change Kc?
At a fixed temperature, a particular equilibrium equation has a fixed Kc. Changing one concentration disturbs the equilibrium. The system then shifts until new equilibrium concentrations are established that satisfy the same Kc.
Why does temperature change the equilibrium constant?
Changing temperature changes the relative favourability of the endothermic and exothermic directions. The resulting equilibrium composition therefore corresponds to a different value of K.
Why are pure solids and pure liquids omitted from equilibrium expressions?
Their effective concentrations remain constant and are incorporated into the equilibrium constant. They therefore do not appear explicitly in the expression.
Do I always need an ICE table for Kc calculations?
No. If all the required equilibrium concentrations are given directly, you can substitute them immediately into the Kc expression.
An ICE table is especially helpful when initial values are given and you must calculate how the composition changes before equilibrium is reached.
How do I decide whether an ICE-table change is +x or −x?
First decide which reaction direction occurs overall as the initial mixture moves towards equilibrium. Species being consumed receive negative changes, while species being formed receive positive changes.
The coefficients attached to x must then follow the stoichiometric coefficients in the balanced equation.
Does a large K mean that the reaction is fast?
No. K describes the equilibrium composition. A large K means the equilibrium is strongly product-favoured.
Reaction speed is a kinetic property. A strongly product-favoured reaction can still be extremely slow.
Why does increasing pressure not always shift a gaseous equilibrium?
Compression favours one side only when the two sides contain different total stoichiometric numbers of moles of gas. If both sides contain the same number of gaseous moles, neither direction reduces the pressure preferentially, so there is no shift.
Why does the Haber process not use the lowest possible temperature?
A low temperature increases the equilibrium yield of NH₃ because the forward reaction is exothermic, but it also decreases reaction rate.
A compromise temperature gives a sufficiently fast production rate while retaining a useful equilibrium yield.
Does an iron catalyst increase the equilibrium yield of ammonia?
No. It increases both the forward and reverse reaction rates. The same equilibrium composition is reached more quickly, but neither the equilibrium yield nor the value of K changes.
Should Kp be calculated using total pressure or partial pressures?
Kp uses the equilibrium partial pressure of each gaseous species. If only the total pressure is given, first determine each gas mole fraction and use pi = xiPtotal.
Chemical equilibrium brings several ideas together. A reversible system reaches dynamic equilibrium when its forward and reverse reactions occur at equal rates, while Le Chatelier’s Principle predicts how the equilibrium position responds to changing conditions. Kc and Kp describe that equilibrium quantitatively, and ICE tables provide a systematic route from initial conditions to equilibrium quantities. These ideas form the foundation for the acid–base, buffer and solubility equilibria that follow.