Introduction to Equilibrium
Distinguish constant unequal concentrations from equal nonzero forward and reverse rates.
AP Chemistry Topic 7.1 · CED printed p. 128 / PDF p. 135. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Closed systems
- Reading graphs
Let’s start with what equilibrium actually looks like. In a closed container, water can evaporate and its vapor can condense; a salt can dissolve and its ions can rejoin the solid. Gas adsorption and desorption are another pair. Chemical reversibility includes proton transfer in acid–base reactions and electron transfer in redox reactions. The paired arrows tell you that the reverse process is possible. They do not promise that it will be fast enough to observe during a short experiment.
At equilibrium, reactants and products coexist and their concentrations or partial pressures stay constant under fixed conditions. That sounds quiet, but the particles are still busy. Each forward event is balanced, on average, by a reverse event. Equal rates produce zero net change. Equal amounts are not required, and neither rate must be zero. A stable reading from an open or kinetically trapped system alone is not proof of equilibrium; you need the reversible process and the relevant conditions too.
On a concentration–time graph, equilibrium appears as plateaus, usually at different heights. On a forward-rate/reverse-rate graph, the curves meet at the same nonzero height. These two graphs answer different questions. Rate says how quickly particles change identity; concentration says how much of each identity is present. Label time, concentration, or rate and their units before interpreting a curve. In a constant-volume 1:1 conversion, the fall in one concentration equals the rise in the other.
Let's work through it
- For A ⇌ B, begin with 0.90 M A and 0.10 M B in a sealed, constant-volume flask. Measurements eventually settle at 0.60 M A and 0.40 M B.
- A has fallen by 0.30 M and B has risen by 0.30 M, consistent with one A forming one B. The total is still 1.00 M.
- Draw two concentration curves leveling at 0.60 and 0.40 M. Draw two rate curves meeting at the same height, not at those concentration values.
- Explain the plateau: forward and reverse rates are equal. The unequal plateau concentrations do not contradict equilibrium.
Make the picture move
Interactive concept model
Q tells you which way; K tells you the destination ratio.
Predict first. Predict net direction from Q compared with K. Check whether the final concentrations must be equal, and whether the total concentration changes.
A(aq) ⇌ B(aq); Q = [B]/[A]; x = (K[A]₀ − [B]₀)/(1 + K)
On a small screen, scroll inside the graph or diagram to read its labels. The values and explanation also appear below.
Q = 0.25, K = 3.98: net forward: form B. Equilibrium [A] = 0.201 and [B] = 0.799 mol L⁻¹.
Read the data and comparisons
| ICE row | [A] (mol L⁻¹) | [B] (mol L⁻¹) |
|---|---|---|
| Initial | 0.8 | 0.2 |
| Change | −0.599 | +0.599 |
| Equilibrium | 0.201 | 0.799 |
Macroscopic / data. Concentration bars describe initial and final mixtures on one common scale.
Particle view. Net interconversion redistributes A and B while conserving their total; both directions continue at equilibrium.
Symbolic view. [A]eq = [A]₀ − x; [B]eq = [B]₀ + x; x = 0.599 mol L⁻¹.
Why it works. The signed x is the forward change in concentration. A decreases by x while B increases by x, preserving [A] + [B]. Substitute both final concentrations into K. Equilibrium requires equal forward and reverse rates; it does not generally require equal concentrations.
Conditions and limits. Abstract 1:1 dissolved interconversion at fixed temperature (298 K) and volume; ideal dilute activities are approximated by concentration divided by a 1 mol L⁻¹ standard state, which cancels in this ratio. Selecting K chooses a different hypothetical system at that fixed temperature, not a temperature change. This simple linear ICE solution does not cover other stoichiometries, solubility, or coupled equilibria.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
Does an equilibrium mixture need equal amounts of reactants and products?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
Does an equilibrium mixture need equal amounts of reactants and products?
Worked answer and reason
No. Equal forward and reverse rates maintain constant amounts, which may be very unequal.
A rate balance is not an amount balance.
Retrieval 2
A sealed bottle keeps the same liquid level. Has evaporation stopped?
Worked answer and reason
No. At vapor–liquid equilibrium evaporation and condensation continue at equal rates.
The observable level tracks net transfer, not individual molecular events.
Retrieval 3
How do you distinguish equilibrium on a concentration–time graph and on a forward/reverse-rate graph?
Worked answer and reason
Concentrations plateau, possibly at unequal heights; forward and reverse rates meet at the same nonzero height. Label the relevant quantity and units on each vertical axis.
Composition is constant because competing event rates balance, not because concentrations must be equal.
Catch the tempting mistake
- Equilibrium means the reaction stops.
- Individual forward and reverse events continue; only their net effect vanishes.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 7 problem 1 · ap application · data
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Scientific calculator
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For the teacher
- Ask students to draw both kinds of graph. If they draw equal concentration plateaus by default, use an unequal-composition example immediately.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 7.1.A
- Explain the relationship between the occurrence of a reversible chemical or physical process, and the establishment of equilibrium, to experimental observations.
- 7.1.A.1
- Many observable processes are reversible. Examples include evaporation and condensation of water, absorption and desorption of a gas, or dissolution and precipitation of a salt. Some important reversible chemical processes include the transfer of protons in acid-base reactions and the transfer of electrons in redox reactions.
- 7.1.A.2
- When equilibrium is reached, no observable changes occur in the system. Reactants and products are simultaneously present, and the concentrations or partial pressures of all species remain constant.
- 7.1.A.3
- The equilibrium state is dynamic. The forward and reverse processes continue to occur at equal rates, resulting in no net observable change.
- 7.1.A.4
- Graphs of concentration, partial pressure, or rate of reaction versus time for simple chemical reactions can be used to understand the establishment of chemical equilibrium.
Science references for this teaching
- AP Chemistry Course and Exam Description, effective Fall 2024 — Required topic-numbered learning objectives and essential knowledge, all continuations and AP exclusions; supplied PDF SHA256 b5dfe8677ef3d88c613865d2e2a3e8d6125d652e2b24c71ef1e8ce4e011094f0.
- OpenStax Chemistry 2e §13.2, Equilibrium Constants — Pressure/concentration quotients, pure-phase activities, and the distinction between composition and rate.
Glossary and deeper reading · 3 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Evaporation
Evaporation is vaporization occurring at a liquid's surface. It can occur below the boiling point as some particles escape into the gas phase. For example, liquid water can evaporate at room temperature and leave less liquid behind.
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Condensation
Condensation is a phase change from gas to liquid. It releases energy to the surroundings as particles form more closely associated liquid arrangements. For example, water vapor condensing on a cooler surface forms liquid droplets.
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Partial pressure
A gas's partial pressure is the pressure it would exert alone in the mixture's volume at the same temperature. In an ideal mixture, the component partial pressures add to the total pressure. For example, a component with mole fraction 0.20 contributes 0.20 of the total ideal pressure.