Calculating Equilibrium Concentrations
Choose net progress from Q/K and solve checked equilibrium concentration changes.
AP Chemistry Topic 7.7 · CED printed p. 135 / PDF p. 142. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Q comparison
- Stoichiometry
- Algebra and physical roots
Before building a change table, calculate Q if the initial mixture contains both sides. If Q < K, net forward reaction increases the quotient; if Q > K, net reverse reaction lowers it. If Q = K, the system is already at dynamic equilibrium. “Forward” always means the direction of the written arrow. It does not mean that reverse events disappear.
An initial–change–equilibrium table is bookkeeping with a chemical reason. Choose x as the amount of reaction progress per liter, or a consistent pressure change in a fixed-temperature, fixed-volume ideal-gas problem. Multiply x by each stoichiometric coefficient and put the signs in the chosen net direction. Add the changes to the starting values, substitute the final expressions into K, and solve. Check that every concentration is nonnegative, conserved atoms still balance, and the quotient equals K. Reject mathematically valid roots that describe impossible negative populations.
Sometimes a small-change approximation makes the algebra manageable, but you must test it. “K is small” is not enough: K and the starting concentrations together determine the fractional change. If a proposed x is not small relative to the term you approximated, solve the complete equation. A common classroom 5% check is a useful tolerance test, not a chemical law. The more accurate model may also require activities at high ionic strength. State the ideal/dilute assumptions used in the problem, keep extra digits until the last line, and report concentrations with units.
Let's work through it
- A ⇌ B has Kc = 3.00, initially [A] = 0.200 M and [B] = 0.100 M. Q = 0.500 < K, so choose a forward x.
- Write equilibrium concentrations 0.200 − x and 0.100 + x. Solve 3.00 = (0.100 + x)/(0.200 − x).
- 0.600 − 3x = 0.100 + x, so x = 0.125 M. Final [A] = 0.0750 M and [B] = 0.225 M.
- Check: their sum is 0.300 M and the ratio is 3.00. Here x is large relative to 0.200 M, so an assumption of negligible reactant depletion would fail.
Now use a non-1:1 equation: N2O4(g) ⇌ 2NO2(g), Kc = 0.200, initially 0.300 M N2O4 and zero NO2.
| Stage | N2O4 | NO2 |
|---|---|---|
| Initial | 0.300 | 0 |
| Change | −x | +2x |
| Equilibrium | 0.300 − x | 2x |
- Initially Q = 0, so dissociation is net forward. Each N2O4 consumed makes two NO2; that is the reason for 2x.
- 0.200 = (2x)²/(0.300 − x), so 4x² + 0.200x − 0.0600 = 0. The roots are +0.100 and −0.150 M. The negative root gives negative NO2, so reject it.
- Final [N2O4] = 0.200 M and [NO2] = 0.200 M. Their quotient is 0.200, and nitrogen-unit concentration is conserved: 2(0.300) = 2(0.200) + 0.200.
- A trial neglect of depletion gives x = √(0.200 × 0.300/4) = 0.122 M, about 41% of the initial N2O4. That is not negligible; use the full solution.
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
For A ⇌ 2B, is the product change +x?
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
For A ⇌ 2B, is the product change +x?
Worked answer and reason
It is +2x if the reactant change is −x.
Stoichiometry constrains every change.
Retrieval 2
What checks belong after solving?
Worked answer and reason
Nonnegative concentrations, conserved atoms, Q = K, and any neglected-change tolerance.
Arithmetic alone cannot identify a physically valid equilibrium.
Catch the tempting mistake
- A negative algebraic root is always physically valid.
- Use the resulting species concentrations to select the physical root.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 7 problem 7 · ap application · symbolic
- Unit 7 problem 8 · ap synthesis · multiple
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For the teacher
- If a student can solve a quadratic but cannot explain 2x, return to particle stoichiometry.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 7.7.A
- Identify the concentrations or partial pressures of chemical species at equilibrium based on the initial conditions and the equilibrium constant.
- 7.7.A.1
- The concentrations or partial pressures of species at equilibrium can be predicted given the balanced reaction, initial concentrations, and the appropriate K.
- 7.7.A.2
- When Q < K, the reaction will proceed with a net consumption of reactants and generation of products. When Q > K, the reaction will proceed with a net consumption of products and generation of reactants. When Q = K, the system is at dynamic equilibrium; both forward and reverse reactions proceed at the same rate, and the proportion of reactants and products remains constant.
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 · 5 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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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.
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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.
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Chemical equilibrium
Chemical equilibrium is a dynamic state in which forward and reverse reaction rates are equal. Macroscopic composition remains constant in a closed system under fixed conditions, but reactant and product concentrations need not be equal. For example, a reversible reaction can reach equilibrium with much more reactant than product.
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Ionic strength
Ionic strength summarizes dissolved-ion concentrations with weighting by the square of each ion's charge number. It helps describe deviations between ionic activities and concentrations. For example, at equal concentration a divalent ion contributes more to ionic strength than a monovalent ion.
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Dilution
Dilution lowers solute concentration by adding solvent while conserving the solute amount when no reaction or loss occurs. Concentration calculations use the final solution volume rather than simply the volume of solvent added. For example, increasing a solution's volume from 100 to 200 milliliters halves its concentration if its solute amount is unchanged.