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Direction of Reversible Reactions

Use simultaneous forward/reverse events and stoichiometry to explain net direction.

AP Chemistry Topic 7.2 · CED printed p. 129 / PDF p. 136. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.

Before you start

A reversible reaction can have a clear net direction while both directions are happening. If ten A particles become B while four B particles become A during the same interval, there are six more B particles afterward. The forward rate exceeds the reverse rate, so the mixture moves forward overall. Switch those event counts and the net direction reverses. Equal counts per interval give equilibrium. For a reaction such as 2A ⇌ B, tie the rate definition to stoichiometry: every net B formed consumes two A. The rate difference determines the net progress, while the coefficients determine each species’ change. Do not infer the event rates solely from which species has more particles; the rate constants and mechanism matter too.

Let's work through it

  1. In a model interval, 12 A → B events and 5 B → A events occur for A ⇌ B.
  2. Cancel five events in each direction when calculating the net change. Seven A become seven B overall.
  3. A decreases and B increases, even though the reverse reaction is active.
  4. When both event counts are eight in a later equal-length interval, the average composition is unchanged. Both rates are eight events per interval, not zero.

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)

Q, K, and mass balance schematicInitial and equilibrium concentration bars for A and B use the same scale, with exact rounded concentrations printed above each bar. Q = 0.25, K = 3.98: net forward: form B. Equilibrium [A] = 0.201 and [B] = 0.799 mol L⁻¹.Q = 0.25 · K = 3.98 · net forward: form BInitial mixtureAB0.80.2EquilibriumAB0.2010.799Bar heights: concentration (mol L⁻¹); common scale in both mixtures.
Initial and equilibrium concentration bars for A and B use the same scale, with exact rounded concentrations printed above each bar. Q = 0.25, K = 3.98: net forward: form B. Equilibrium [A] = 0.201 and [B] = 0.799 mol L⁻¹.

On a small screen, scroll inside the graph or diagram to read its labels. The values and explanation also appear below.

Initial Q0.25
Selected K3.98
Final [B]/[A]3.98

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
Q, K, and mass balance: model values and comparisons (numbers rounded for display)
ICE row[A] (mol L⁻¹)[B] (mol L⁻¹)
Initial0.80.2
Change−0.599+0.599
Equilibrium0.2010.799

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 of 20 self-assessed

Question 1

Which way does the mixture move if the reverse rate is larger?

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

Which way does the mixture move if the reverse rate is larger?

Worked answer and reason

Products are consumed overall and reactants form.

Net change equals the difference between competing progress rates.

Retrieval 2

If the rates become equal, must the molecule counts become equal?

Worked answer and reason

No. Equal event rates can occur with unequal populations.

Population affects a rate together with the rate constant; it is not itself a rate.

Catch the tempting mistake

The more abundant side always reacts faster.
Rates depend on both populations and the reaction pathway; use measured rates or a valid rate law.

Put it to work

Give the chemical reason as well as the calculation. Worked solutions are on a separate page.

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For the teacher
  • Require students to describe simultaneous events before using the word shift.

Unit planning and formative guidance →

Source objectives and scope

Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →

7.2.A
Explain the relationship between the direction in which a reversible reaction proceeds and the relative rates of the forward and reverse reactions.
7.2.A.1
If the rate of the forward reaction is greater than the reverse reaction, then there is a net conversion of reactants to products. If the rate of the reverse reaction is greater than that of the forward reaction, then there is a net conversion of products to reactants. An equilibrium state is reached when these rates are equal.
Science references for this teaching
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.

  • Stoichiometry

    Stoichiometry uses a balanced chemical equation to relate amounts of reactants and products. Equation coefficients provide mole ratios rather than direct mass ratios. For example, forming water from hydrogen and oxygen consumes two moles of hydrogen molecules per mole of oxygen molecules.

    Read more: Stoichiometry →

  • Reversible reaction

    A reversible reaction can proceed in both forward and reverse directions under the modeled conditions. Its net direction depends on the current composition and temperature. For example, a product-rich starting mixture can react backward toward equilibrium.

  • Mixture

    A mixture contains two or more substances whose proportions can vary. Its components retain their chemical identities until a chemical reaction changes them. For example, water and ethanol can form mixtures with many different compositions.

    Read more: Composition of Mixtures →

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