What is a reversible reaction in the context of chemical equilibria?

In the context of chemical equilibria, a reversible reaction is a chemical reaction where the reactants form products, which can then react together to give the reactants back. This means the reaction can proceed in both forward and backward directions. At equilibrium, the rate of the forward reaction equals the rate of the backward reaction, resulting in no net change in the concentrations of reactants and products.

How is dynamic equilibrium defined in AS-Level Physical Chemistry?

Dynamic equilibrium in AS-Level Physical Chemistry refers to the state of a reversible reaction where the rates of the forward and backward reactions are equal, leading to constant concentrations of reactants and products. Despite the ongoing reactions, there is no observable change in the system, as the processes are balanced. This concept is crucial for understanding how chemical systems behave under different conditions.

What factors can affect the position of equilibrium in a chemical reaction?

The position of equilibrium in a chemical reaction can be affected by changes in concentration, temperature, and pressure. According to Le Chatelier's Principle, if a system at equilibrium is subjected to a change in these conditions, the system will adjust to partially counteract the change and restore equilibrium. For example, increasing the concentration of reactants typically shifts the equilibrium position towards the products.

Why is the concept of equilibrium important in Cambridge International A Levels Chemistry?

The concept of equilibrium is important in Cambridge International A Levels Chemistry because it helps students understand how chemical reactions behave under various conditions. It provides insights into reaction kinetics, the extent of reactions, and how external factors influence chemical processes. Mastery of this concept is essential for predicting the outcomes of reactions and for applications in industrial and laboratory settings.

How does temperature influence dynamic equilibrium in reversible reactions?

Temperature can significantly influence dynamic equilibrium in reversible reactions. According to Le Chatelier's Principle, increasing the temperature of an exothermic reaction will shift the equilibrium position towards the reactants, while for an endothermic reaction, it will shift towards the products. This is because heat can be considered as a reactant or product, and the system will adjust to absorb or release heat accordingly to maintain equilibrium.

Why is "Saying concentrations are equal at equilibrium." a common mistake in Chemical equilibria: reversible reactions, dynamic equilibrium?

Why it happens: Confusing 'constant' with 'equal'. How to avoid it: Concentrations are constant, not necessarily equal. Source: 9701 Examiner Reports 2022-2024

Why is "Thinking the reaction stops at equilibrium." a common mistake in Chemical equilibria: reversible reactions, dynamic equilibrium?

Why it happens: Seeing constant concentrations. How to avoid it: Both reactions continue at equal rates — it is dynamic, not static.

Why is "Claiming a catalyst increases yield or changes K." a common mistake in Chemical equilibria: reversible reactions, dynamic equilibrium?

Why it happens: Overstating the catalyst's role. How to avoid it: A catalyst only speeds up reaching equilibrium; position and K are unchanged.

Why is "Saying pressure or concentration changes the value of K." a common mistake in Chemical equilibria: reversible reactions, dynamic equilibrium?

Why it happens: Confusing a position shift with K. How to avoid it: Only temperature changes the value of K.

Why is "Assuming Kc always has units of mol dm⁻³ (or none)." a common mistake in Chemical equilibria: reversible reactions, dynamic equilibrium?

Why it happens: Memorising one case. How to avoid it: Work out the units from the powers each time; they can cancel or be mol⁻² dm⁶, etc.

Equilibria(AS-Level Physical Chemistry)Cambridge International A Level Chemistry 9701

Chemical equilibria: reversible reactions, dynamic equilibrium

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Chemical Equilibria: Reversible Reactions and Dynamic Equilibrium — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

Dynamic equilibrium and Le Chatelier's principle, writing and using Kc and Kp, calculating equilibrium quantities, what changes the value of K, and the conditions in the Haber and Contact processes.

What you’ll learn

Mapped to the Cambridge International A Level 9701 syllabus (2025-2027).

7.1.1 — Understand reversible reactions, dynamic equilibrium (equal rates, constant concentrations) and the need for a closed system.
7.1.2 — Define Le Chatelier's principle.
7.1.3 — Use Le Chatelier's principle to deduce the effects of changes in temperature, concentration, pressure or a catalyst.
7.1.4–7.1.6 — Deduce Kc and Kp expressions; use mole fraction and partial pressure.
7.1.7–7.1.8 — Use Kc and Kp expressions in calculations (no quadratics) and calculate equilibrium quantities from data.
7.1.9 — State whether temperature, concentration, pressure or a catalyst affect the value of the equilibrium constant.
7.1.10 — Describe and explain the conditions used in the Haber and Contact processes.

Reversible reactions and dynamic equilibrium

At equilibrium the forward and reverse reactions still happen, at equal rates, in a closed system.

A reversible reaction can go both ways: $\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}$ .

Dynamic equilibrium is reached when:

the rate of the forward reaction equals the rate of the reverse reaction, and
the concentrations of reactants and products remain constant (but are not necessarily equal).

It is dynamic because both reactions continue — molecules are constantly converting both ways — but at the same rate, so nothing appears to change.

A closed system is required: nothing can enter or leave, so no reactant or product escapes (otherwise equilibrium can't be maintained).

Forward rate = reverse rate.
Concentrations constant (not equal).
Both reactions still occur (dynamic); closed system needed.

Le Chatelier's principle

Impose a change and the equilibrium shifts to oppose (minimise) it.

Le Chatelier's principle: if a change is made to a system at dynamic equilibrium, the position of equilibrium shifts to minimise that change.

Change	Equilibrium shifts…	Example
↑ Temperature	in the endothermic direction (absorbs heat)	exothermic forward → shifts back, less product
↓ Temperature	in the exothermic direction	exothermic forward → more product
↑ Pressure	towards the side with fewer moles of gas
↑ Concentration of a species	away from that species	add reactant → more product
Catalyst	no shift (speeds both directions equally)	reaches equilibrium faster only

Worked. For $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ ( $\Delta H$ negative): increasing pressure shifts right (4 moles gas → 2), giving more NH₃; increasing temperature shifts left (endothermic direction), giving less NH₃.

System shifts to oppose the imposed change.
↑T → endothermic direction; ↑P → fewer gas moles.
↑[species] → shifts away; catalyst → no shift.

Writing Kc and Kp expressions

Kc from concentrations, Kp from partial pressures — products over reactants, each to its power.

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

$K_c$ (concentrations): $K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$

$K_p$ (partial pressures of gases): $K_p = \frac{(p_\text{C})^c (p_\text{D})^d}{(p_\text{A})^a (p_\text{B})^b}$

Partial pressure of a gas: $p_x = (\text{mole fraction of } x) \times (\text{total pressure}) = \frac{n_x}{n_{\text{total}}} \times P$

Units of K depend on the expression — work them out by substituting the units of each term (mol dm⁻³ for Kc, Pa or atm for Kp). They may cancel (e.g. when moles of gas are equal on both sides).

$K_c$ = [products]^powers / [reactants]^powers.
$K_p$ uses partial pressures.
$p_x$ = mole fraction × total pressure.
Derive K's units by substitution.

Equilibrium calculations and the value of K

Use an ICE-style table to find equilibrium amounts, then substitute. Only temperature changes K.

Method (no quadratics needed at 9701):

Tabulate initial, change and equilibrium amounts (an "ICE" table).
Convert to concentrations (÷ volume) or partial pressures (mole fraction × P).
Substitute into the $K_c$ or $K_p$ expression.

Worked. For $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ in a 1 dm³ vessel, at equilibrium [H₂] = 0.20, [I₂] = 0.20, [HI] = 1.60 mol dm⁻³: $K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(1.60)^2}{(0.20)(0.20)} = 64\;(\text{no units})$

What changes the value of K?

Only temperature changes $K$ . (For an exothermic reaction, raising T decreases K.)
Concentration, pressure and catalysts do NOT change K — they may shift the position of equilibrium, but the system adjusts so K stays the same.

Use an initial–change–equilibrium table.
Substitute equilibrium concentrations/partial pressures.
Only temperature changes the value of K.

The Haber and Contact processes

Industrial compromises between yield (Le Chatelier) and rate (kinetics).

Both processes choose conditions that compromise between a good equilibrium yield and a fast enough rate.

Haber process — $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ , $\Delta H$ negative.

~450 °C — a compromise: a lower temperature would give more NH₃ (exothermic) but too slowly; higher would be fast but low yield.
~200 atm (high pressure) — favours the side with fewer gas moles (2 vs 4) → more NH₃; limited by cost/safety.
Iron catalyst — speeds up the reaction (no effect on yield).

Contact process — $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$ , $\Delta H$ negative.

~450 °C — same temperature compromise.
~1–2 atm — the equilibrium yield is already very high at low pressure, so high pressure is not needed (saves cost).
Vanadium(V) oxide (V₂O₅) catalyst — speeds up the reaction.

Key exam point: the moderate temperature is a kinetic compromise, and the catalyst affects rate, not yield.

Haber: ~450 °C, ~200 atm, iron catalyst.
Contact: ~450 °C, ~1–2 atm, V₂O₅ catalyst.
Temperature = compromise (yield vs rate); catalyst affects rate not yield.

How it’s examined

Le Chatelier predictions, Kc/Kp expressions and equilibrium calculations are guaranteed Paper 2 marks, with the Haber/Contact conditions a frequent 'explain' question. Examiner reports flag: saying concentrations are 'equal', claiming a catalyst increases yield or changes K, and forgetting that only temperature changes K.

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 7.1; 9701 Examiner Reports 2022-2024; 9701/22 & 9701/42 May/Jun 2024 question papers and mark schemes. Last reviewed 2026-05-20.

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Step-by-step worked examples — Chemical equilibria: reversible reactions, dynamic equilibrium

Step-by-step solutions to past-paper-style questions on chemical equilibria: reversible reactions, dynamic equilibrium, written exactly the way a tutor would explain them at the board.

1Writing a Kc expression (2 marks)
Getting started• Kc
Question
Write the Kc expression for $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ and state its units.
Step-by-step solution
1. Step 1
  Products over reactants, each raised to its balancing number.
  $K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$
2. Step 2
  Units: $\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^4} = \text{mol}^{-2}\,\text{dm}^{6}$ .
Answer
$K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$ , units mol⁻² dm⁶.
Examiner tip
The units depend on the equation — work them out from the powers each time; they are not always mol dm⁻³.
2Le Chatelier: concentration change (2 marks)
Getting started• Le Chatelier
Question
For $\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{ester} + \text{H}_2\text{O}$ , what happens to the ester yield if more ethanol is added?
Step-by-step solution
1. Step 1
  Adding a reactant shifts the position of equilibrium to the right to oppose the increase.
2. Step 2
  More ester (and water) is formed.
Answer
Equilibrium shifts right → ester yield increases.
Examiner tip
Adding a reactant or removing a product both drive the equilibrium toward products.
3Le Chatelier: temperature and pressure (4 marks)
Building confidence• Le Chatelier
Question
For $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ , ΔH = −197 kJ mol⁻¹, predict the effect on the yield of SO₃ of (a) increasing temperature and (b) increasing pressure.
Step-by-step solution
1. Step 1
  (a) ↑T shifts equilibrium in the endothermic (reverse) direction → less SO₃.
2. Step 2
  (b) ↑P shifts towards fewer gas moles: 3 mol (left) → 2 mol (right) → more SO₃.
Answer
↑T decreases SO₃ yield; ↑P increases SO₃ yield.
Examiner tip
Always justify with the direction (endothermic) and the mole count, not just 'shifts right/left'.
4Calculating Kc (3 marks)
Building confidence• Kc
Question
At equilibrium in a 2.0 dm³ flask: 0.40 mol N₂O₄ and 0.80 mol NO₂ for $\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2$ . Calculate Kc.
Step-by-step solution
1. Step 1
  Concentrations (÷ 2.0 dm³).
  $[\text{N}_2\text{O}_4] = 0.20;\quad [\text{NO}_2] = 0.40$
2. Step 2
  Substitute into Kc.
  $K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \frac{(0.40)^2}{0.20}$
3. Step 3
  Evaluate (units mol dm⁻³).
  $K_c = 0.80\,\text{mol dm}^{-3}$
Answer
$K_c = 0.80\,\text{mol dm}^{-3}$ .
Examiner tip
Convert moles → concentrations FIRST. If amounts change during the reaction, use an ICE table (initial–change–equilibrium).
5Partial pressure and Kp (4 marks)
Stretch• Kp
Question
An equilibrium mixture contains 2.0 mol N₂, 4.0 mol H₂ and 4.0 mol NH₃ at a total pressure of 100 kPa, for $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ . Calculate Kp.
Step-by-step solution
1. Step 1
  Total moles = 10; partial pressure = mole fraction × total P.
  $p_{\text{N}_2} = 20,\; p_{\text{H}_2} = 40,\; p_{\text{NH}_3} = 40\,\text{kPa}$
2. Step 2
  Kp expression.
  $K_p = \frac{(p_{\text{NH}_3})^2}{(p_{\text{N}_2})(p_{\text{H}_2})^3}$
3. Step 3
  Substitute.
  $K_p = \frac{40^2}{20 \times 40^3} = 1.25 \times 10^{-3}\,\text{kPa}^{-2}$
Answer
$K_p = 1.25 \times 10^{-3}\,\text{kPa}^{-2}$ .
Examiner tip
Mole fractions must sum to 1; check the partial pressures add back to the total (20 + 40 + 40 = 100).
6Justifying industrial conditions (4 marks)
Stretch• Haber
Question
The Haber process ( $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ , ΔH < 0) uses ~450 °C, ~200 atm and an iron catalyst. Justify each condition. (4)
Step-by-step solution
1. Step 1
  Temperature ~450 °C: low T favours yield (exothermic) but is too slow; high T is fast but low yield → a compromise.
2. Step 2
  High pressure ~200 atm: fewer moles on the right (4 → 2), so high P increases yield AND rate.
3. Step 3
  Pressure is not even higher because of the cost and safety of building/running high-pressure plant.
4. Step 4
  Iron catalyst: speeds up attainment of equilibrium without changing the position or yield.
Answer
450 °C = rate/yield compromise; 200 atm boosts yield (fewer gas moles) within cost limits; Fe catalyst speeds equilibrium only.
Examiner tip
State that the catalyst does NOT change the yield — it only shortens the time to reach equilibrium.

Key Formulae — Chemical equilibria: reversible reactions, dynamic equilibrium

The formulae you need to memorise for chemical equilibria: reversible reactions, dynamic equilibrium on the Cambridge International A Level 9701 paper, with every variable defined in plain English and a note on when to use it.

Equilibrium constant Kc
$K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$
$[X]$
equilibrium concentration / mol dm⁻³
$a, b, c, d$
balancing numbers from the equation
When to use
Equilibria expressed in concentrations; units depend on the powers.
Equilibrium constant Kp
$K_p = \frac{(p_\text{C})^c(p_\text{D})^d}{(p_\text{A})^a(p_\text{B})^b}$
$p_X$
equilibrium partial pressure of gas X
When to use
Gaseous equilibria expressed in partial pressures.
Partial pressure (from mole fraction)
$p_x = \frac{n_x}{n_{\text{total}}} \times P$
$n_x$
moles of gas x
$n_{total}$
total moles of all gases
$P$
total pressure
When to use
Finding each partial pressure before substituting into Kp.

Key Definitions and Keywords — Chemical equilibria: reversible reactions, dynamic equilibrium

Definitions to memorise and the exact keywords mark schemes credit for chemical equilibria: reversible reactions, dynamic equilibrium answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9701 sitting.

Dynamic equilibrium
Examiner keyword
A state in a closed system where the forward and reverse reactions occur at equal rates and the concentrations of reactants and products remain constant.
Le Chatelier's principle
Examiner keyword
If a change is made to a system at dynamic equilibrium, the position of equilibrium shifts to oppose (minimise) that change.
Position of equilibrium
Examiner keyword
How far a reaction has proceeded toward products; it shifts with concentration, pressure or temperature, but only T changes the value of K.
Kc / Kp
Examiner keyword
Equilibrium constants in terms of concentrations (Kc) or partial pressures (Kp); their value changes only with temperature.
Mole fraction
Examiner keyword
The fraction of the total moles that a component contributes (n_x / n_total); mole fractions sum to 1.
Partial pressure
Examiner keyword
The pressure a gas in a mixture would exert alone; equals its mole fraction × the total pressure.

Common Mistakes and Misconceptions — Chemical equilibria: reversible reactions, dynamic equilibrium

The traps other students keep falling into on chemical equilibria: reversible reactions, dynamic equilibrium questions — taken from recent Cambridge International A Level 9701 examiner reports and mark schemes — and how to avoid them.

✕Saying concentrations are equal at equilibrium.
9701 Examiner Reports 2022-2024
Why it happens
Confusing 'constant' with 'equal'.
How to avoid it
Concentrations are constant, not necessarily equal.
✕Thinking the reaction stops at equilibrium.
Why it happens
Seeing constant concentrations.
How to avoid it
Both reactions continue at equal rates — it is dynamic, not static.
✕Claiming a catalyst increases yield or changes K.
Why it happens
Overstating the catalyst's role.
How to avoid it
A catalyst only speeds up reaching equilibrium; position and K are unchanged.
✕Saying pressure or concentration changes the value of K.
Why it happens
Confusing a position shift with K.
How to avoid it
Only temperature changes the value of K.
✕Assuming Kc always has units of mol dm⁻³ (or none).
Why it happens
Memorising one case.
How to avoid it
Work out the units from the powers each time; they can cancel or be mol⁻² dm⁶, etc.

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