← Unit 3: Equilibrium, acids and redox reactions
Topic 1: Chemical equilibrium systems
Explain dynamic equilibrium in terms of rates of forward and reverse reactions, and recognise that equilibrium can only be established in a closed system
A focused answer to the QCE Chemistry Unit 3 dot point on dynamic equilibrium. Defines reversible reactions, contrasts dynamic with static equilibrium, explains why equilibrium requires a closed system, and works through a sample concentration-vs-time graph with the kind of stimulus QCAA uses in IA1.
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What this dot point is asking
QCAA wants you to explain chemical equilibrium as a dynamic, rate-based concept and recognise that equilibrium can only be reached when reactants and products are confined to a closed system. Stimulus questions in IA1 typically show a concentration-vs-time graph, ask you to identify when equilibrium is reached, and require you to justify the answer in terms of forward and reverse reaction rates.
The answer
A reversible chemical reaction reaches dynamic equilibrium in a closed system when the forward and reverse reactions occur at equal non-zero rates, so the macroscopic concentrations of all species become constant.
Reversible reactions
Some reactions go essentially to completion (combustion of hydrocarbons in excess oxygen). Others are reversible: products can recombine to reform reactants. Reversible reactions are written with a double arrow.
For a reversible reaction conducted in a sealed system at constant temperature, the forward and reverse processes both proceed throughout the experiment.
Why equilibrium develops
Consider starting with pure N2O4 in a sealed flask.
- At t = 0. Forward rate is at its maximum (concentration of N2O4 is highest). Reverse rate is zero (no NO2 yet).
- As the reaction proceeds. [N2O4] falls, so forward rate decreases. [NO2] rises, so reverse rate increases.
- At equilibrium. Forward and reverse rates become equal. From this point, every N2O4 molecule that decomposes is replaced by an N2O4 reformed from two NO2 molecules. The bulk concentrations stop changing.
This is dynamic, not static: the reactions never stop. Equilibrium is the condition of equal opposing rates, not the condition of "no reaction".
Macroscopic vs microscopic behaviour at equilibrium
| Property | At equilibrium |
|---|---|
| Forward rate | Equal to reverse rate, both non-zero |
| Reverse rate | Equal to forward rate, both non-zero |
| Concentration of each species | Constant |
| Colour, pressure, mass | Constant |
| Particles reacting | Yes, continuously |
A common exam trap is to claim that "the reaction has stopped". It has not. Only the net change has stopped.
Why the system must be closed
A closed system is one in which matter cannot enter or leave. Equilibrium cannot be established in an open system because:
- Reactant or product loss removes a participant. If NO2 escapes through an open container, the reverse rate cannot rise to meet the forward rate. The reaction proceeds to completion of N2O4 decomposition instead of reaching equilibrium.
- External addition keeps the system perturbed. Continuous reactant input pushes the forward rate above the reverse rate indefinitely.
Closed system does not require constant temperature, but most QCAA equilibrium problems also assume constant temperature so that Kc remains fixed. The system being closed concerns matter (no gain or loss); thermal conditions are separately specified.
Recognising equilibrium on a concentration-vs-time graph
Equilibrium is the time region where every curve has zero slope. The curves do not have to be at the same value: reactant concentrations and product concentrations typically differ at equilibrium, but each one is individually constant.
Key features QCAA expects you to identify on a graph:
- The starting concentrations at t = 0.
- The non-zero gradient region while the reaction is proceeding net-forward (or net-reverse).
- The flat region where equilibrium has been established.
- The stoichiometric ratio between species changes (for N2O4 to 2NO2, NO2 changes by twice as much as N2O4).
If a graph shows a sudden perturbation (a vertical step or a sharp change in slope mid-curve), that signals a disturbance to equilibrium, which connects to the Le Chatelier dot point.
Worked example: the iron(III) thiocyanate equilibrium
A classic IA2 system, also examinable in IA1 stimulus form:
(pale yellow + colourless leftrightarrow deep red)
Starting with separated solutions of FeCl3 and KSCN, when mixed and sealed:
- The forward rate is initially high (Fe3+ and SCN- both abundant), the reverse rate is zero.
- The red colour develops, then steadies. Spectrophotometric absorbance increases, then plateaus.
- At the plateau, the forward and reverse rates are equal. FeSCN2+ is continuously forming and dissociating.
Open this system to the air and nothing dramatic happens because none of the species are gases; the closed-system requirement is satisfied for aqueous-only systems by simply preventing evaporation. For gas-phase equilibria like N2O4 / NO2, sealing is essential.
Common traps
Saying the reaction has stopped. It has not. Net change is zero, but molecules are reacting continuously.
Claiming concentrations are equal at equilibrium. They are constant, not equal. Forward and reverse rates are equal, which is what makes the concentrations constant; the actual values depend on Kc.
Forgetting the closed-system requirement. If matter can leave, equilibrium cannot be reached. Always check whether the question describes a sealed system.
Confusing equilibrium with completion. A reaction that goes essentially to completion has Kc very large, but if it is reversible it still has a tiny reverse rate at equilibrium. Equilibrium and completion are different concepts.
In one sentence
Dynamic equilibrium is the state of a closed reversible system in which the forward and reverse reactions occur at equal, non-zero rates, so the bulk concentrations of all species remain constant while particles continue to react.
Past exam questions, worked
Real questions from past QCAA papers on this dot point, with our answer explainer.
2023 QCAA-style5 marksA student seals 1.00 mol of N2O4 gas in a 1.0 L flask at 100 degrees C. The reaction N2O4(g) <-> 2NO2(g) is allowed to proceed. The concentration of NO2 rises rapidly at first, then more slowly, and is constant at 0.62 mol/L after 60 seconds. (a) Sketch the concentration-vs-time graph for both species from t=0 to t=120 s. (b) Explain in terms of forward and reverse reaction rates why both concentrations stop changing after 60 seconds. (c) State why this experiment must be conducted in a sealed flask rather than an open beaker.Show worked answer →
A 5-mark answer needs the graph, the rate-equality explanation, and the closed-system justification.
(a) Graph. Plot concentration (mol/L) against time (s). Two curves:
- N2O4 starts at 1.00 mol/L, falls along a curve that flattens, and levels off at 1.00 - (0.62/2) = 0.69 mol/L (since 2 NO2 are made per N2O4 consumed).
- NO2 starts at 0 mol/L, rises along a curve that flattens, and levels off at 0.62 mol/L.
Both curves flatten at t = 60 s and remain horizontal after that. Mark the equilibrium region.
(b) Rate-equality explanation. At t = 0 the forward rate is at its maximum (high [N2O4], zero [NO2]) and the reverse rate is zero. As the reaction proceeds the forward rate slows (less N2O4) and the reverse rate rises (more NO2). When the two rates become equal, every N2O4 that decomposes is replaced by an N2O4 reformed from two NO2, so net concentration change is zero. The system is at dynamic equilibrium: the reactions continue at the particle level, but bulk concentrations do not change.
(c) Closed system. NO2 is a brown gas. In an open beaker NO2 would escape and the reverse reaction could never reach the same rate as the forward reaction. With reactant continuously removed (or with no barrier to gas exchange), the system never reaches equilibrium. A sealed flask keeps all species in the system so forward and reverse rates can equalise.
Markers reward (1) a correctly shaped graph with labelled equilibrium concentrations, (2) explicit reference to rates equalising rather than reactions stopping, (3) a justification that ties closed-system to non-loss of NO2.
2022 QCAA-style3 marksDistinguish between static equilibrium and dynamic equilibrium. Use one example of each.Show worked answer →
A 3-mark answer needs both definitions and one example for each, with the contrast made explicit.
Static equilibrium. A state in which no change occurs at any level. Forward and reverse processes are both zero. Example: a book resting on a table; the forces are balanced and nothing is moving.
Dynamic equilibrium. A state in which forward and reverse processes both occur at the same non-zero rate, so the macroscopic properties (concentration, pressure, colour) are constant but particles continue to react. Example: N2O4(g) and NO2(g) in a sealed flask at constant temperature, where N2O4 continually decomposes and NO2 continually recombines at equal rates.
Contrast. In static equilibrium nothing happens; in dynamic equilibrium reactions continue but with opposing rates equal. Chemical equilibrium is always dynamic.
Markers reward both definitions, one valid example of each, and the explicit point that chemical systems are dynamic.
Related dot points
- Predict, using Le Chatelier's principle, the qualitative effects of changes in concentration, temperature, pressure and volume on the equilibrium position of homogeneous reactions
A focused answer to the QCE Chemistry Unit 3 dot point on Le Chatelier's principle. Defines the principle, works through how concentration, temperature, pressure and volume changes shift equilibrium position, explains why catalysts do not shift equilibrium, and applies the reasoning to the Haber process and the iron(III) thiocyanate system used in IA1 and IA2.
- Derive and apply the equilibrium law expression (Kc) for homogeneous reactions, including calculating Kc from equilibrium concentrations and predicting the position of equilibrium from the value of Kc
A focused answer to the QCE Chemistry Unit 3 dot point on the equilibrium constant. Sets out the equilibrium law expression, works through Kc calculation from a stimulus ICE table (the dominant IA1 question type), interprets the value of Kc in terms of extent, and addresses why Kc is temperature-dependent but pressure-independent.
- Describe acids and bases using the Bronsted-Lowry model, including the identification of conjugate acid-base pairs, amphiprotic species, and the distinction between strong and weak acids and bases
A focused answer to the QCE Chemistry Unit 3 dot point on Bronsted-Lowry acids and bases. Defines proton donors and acceptors, walks through conjugate acid-base pairs, identifies amphiprotic species (water, hydrogencarbonate, dihydrogenphosphate), and contrasts strong with weak acids and bases using Ka and Kb in equilibrium terms.