Chemical equilibrium: practice problems with step-by-step solutions

Free Chemical equilibrium practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: Kc from concentrations, ICE table, ICE table with a quadratic.

Free practice problemEasy
N2​0.2mol/L+3H2​0.4mol/L⇌2NH3​0.3mol/LKc​ = ?

For the reaction N2 + 3H2 ⇌ 2NH3, the equilibrium concentrations are [N2] = 0.2 mol/L, [H2] = 0.4 mol/L, [NH3] = 0.3 mol/L. Find the equilibrium constant Kc.

[N2] = 0.2[H2] = 0.4[NH3] = 0.3
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What you practise

  1. EasyKc from concentrations
  2. MediumICE table
  3. HardICE table with a quadratic

How to solve Chemical equilibrium problems

  1. Write Kc with products over reactants, each raised to the power of its coefficient.
  2. Set up an ICE table: Initial, Change (in steps of x times the coefficients), Equilibrium.
  3. Put the equilibrium expressions into Kc and solve for x.
  4. Check whether x is small enough for an approximation; otherwise solve the quadratic.

Common mistakes

  • Forgetting the exponents in Kc.
  • Using initial instead of equilibrium concentrations.
  • Using the small-x approximation when x is not small.

Key formulas

Equilibrium constant
Kc = [C]c[D]d/([A]a[B]b)
ICE table
Initial, Change (×coefficient), Equilibrium
Quadratic
x = (−b + √(b² − 4ac))/(2a)

Learn Chemical equilibrium from scratch

Before this: Molarity & dilution

1What chemical equilibrium is

Many reactions do not run to completion. As soon as some product forms, it starts to react back into the starting materials. The double arrow ⇌ shows this.

H₂ + I₂ ⇌ 2HI

At first the forward reaction is fast, because there is a lot of H₂ and I₂. As they are used up, it slows down, while the backward reaction speeds up.

After a while both reactions run at the same speed. The concentrations then stop changing. This state is called equilibrium.

tc[H₂] = [I₂][HI]0.50.25

The reactions have not stopped. Molecules still react both ways, but every HI that forms is balanced by one that falls apart. Think of an escalator where people walk up and down at the same rate: the number on each floor stays the same.

Self-check What is true at chemical equilibrium?
  1. All reactants have turned into products
  2. Both reactions have stopped
  3. The concentrations of reactants and products are equal
  4. The forward and backward reactions run at the same rate

Answer: The forward and backward reactions run at the same rate. At equilibrium the two rates are equal, so the concentrations stay constant. They do not have to be equal to each other, and the reactions keep running.

2The equilibrium constant Kc

At equilibrium the concentrations always obey one fixed ratio, the equilibrium constant. For a reaction aA + bB ⇌ cC + dD:

Kc = [C]c · [D]d / ([A]a · [B]b)

Square brackets mean concentration in mol/L at equilibrium. Products go on top, reactants below, and each is raised to the power of its coefficient.

For H₂ + I₂ ⇌ 2HI that gives Kc = [HI]² / ([H₂] · [I₂]).

  • A large Kc (much more than 1): at equilibrium there are mostly products.
  • A small Kc (much less than 1): there are mostly reactants.
  • Kc only changes with temperature, not with the starting amounts.

Le Chatelier’s principle: if you disturb an equilibrium, it shifts to undo part of the disturbance. Adding a reactant pushes it towards the products; removing a product does the same.

Self-check For N₂O₄ ⇌ 2NO₂, [N₂O₄] = 0.08 mol/L and [NO₂] = 0.2 mol/L at equilibrium. What is Kc?
  1. 0.4
  2. 0.5
  3. 2.5
  4. 2

Answer: 0.5. Kc = [NO₂]² / [N₂O₄] = 0.04 / 0.08 = 0.5. Forgetting the square gives 0.2 / 0.08 = 2.5, and turning the fraction upside down gives 2.

3The ICE table

Often you know only the starting concentrations and Kc. An ICE table keeps track of how everything changes:

  • Initial: the concentrations at the start.
  • Change: each substance changes by x times its coefficient. Reactants go down (−), products go up (+).
  • Equilibrium: initial plus change.

For H₂ + I₂ ⇌ 2HI starting from 0.5 mol/L of H₂ and of I₂:

  • H₂: 0.5 − x
  • I₂: 0.5 − x
  • HI: 0 + 2x

Then put the equilibrium row into the Kc expression and solve for x.

When both sides are perfect squares, as here, take the square root of both sides. Otherwise you get a quadratic equation ax² + bx + c = 0, and you keep the root that gives positive concentrations:

x = (−b + √(b² − 4ac)) / (2a)
Self-check You start with 1.0 mol/L of N₂O₄ only, and x mol/L of it reacts: N₂O₄ ⇌ 2NO₂. What is [NO₂] at equilibrium?
  1. 1.0 − 2x
  2. 2x
  3. x
  4. x²

Answer: 2x. Each N₂O₄ that reacts gives two NO₂, so the change for NO₂ is +2x, starting from 0. Writing x forgets the coefficient; the square appears only later, inside Kc.

4The method step by step

  1. Write the balanced equation and the Kc expression: products over reactants, each to the power of its coefficient.
  2. If the equilibrium concentrations are given, put them in and calculate Kc. You are done.
  3. Otherwise set up the ICE table with one unknown x, multiplied by the coefficients.
  4. Put the equilibrium expressions into Kc.
  5. Solve for x: with a square root if both sides are squares, otherwise with the quadratic formula.
  6. Find the concentration that is asked, and check that every concentration is positive and gives back Kc.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

0.5 mol/L of H₂ and 0.5 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 4. Find the equilibrium concentrations [HI] and [H₂].

1

Write Kc

Kc = [HI]² / ([H₂] · [I₂]) = 4
2

Set up the ICE table

Initial: H₂ 0.5, I₂ 0.5, HI 0. Change: −x, −x, +2x.

Equilibrium: [H₂] = [I₂] = 0.5 − x and [HI] = 2x.

3

Fill in Kc

4 = (2x)² / (0.5 − x)²

Both the top and the bottom are squares.

4

Take the square root

2 = 2x / (0.5 − x)
1 − 2x = 2x ⇒ x = 0.25 mol/L
5

Find the concentrations

[HI] = 2x = 0.50 mol/L and [H₂] = [I₂] = 0.5 − 0.25 = 0.25 mol/L.

6

Check the answer

0.50² / (0.25 · 0.25) = 0.25 / 0.0625 = 4

This gives back Kc = 4, and all concentrations are positive.

Answer

[HI] = 0.50 mol/L and [H₂] = [I₂] = 0.25 mol/L

6Common mistakes

  • Forgetting the exponents. A coefficient of 2 means the concentration is squared.
  • Putting reactants on top. Kc is always products over reactants.
  • Using initial concentrations in Kc instead of equilibrium concentrations.
  • Forgetting to multiply x by the coefficient in the change row: 2HI changes by 2x, not x.
  • Using the approximation “x is small” when it is not. If x is more than about 5% of the start value, solve the quadratic.
  • Keeping the negative root of the quadratic, which gives a negative concentration.
Self-check For H₂ + I₂ ⇌ 2HI, the equilibrium concentrations are [H₂] = 0.1, [I₂] = 0.1 and [HI] = 0.4 mol/L. What is Kc?
  1. 0.0625
  2. 16
  3. 40
  4. 80

Answer: 16. Kc = 0.4² / (0.1 · 0.1) = 0.16 / 0.01 = 16. Without the square you get 40, and products below reactants give 0.0625.

7How exam questions look

  • Easy: Kc from concentrations. All equilibrium concentrations are given; write Kc and calculate it.
  • Medium: ICE table. H₂ + I₂ ⇌ 2HI from equal starting amounts; take the square root to find x, then a concentration.
  • Hard: ICE table with a quadratic. A substance such as PCl₅ ⇌ PCl₃ + Cl₂ decomposes from a known start; solve x² + Kc·x − Kc·c₀ = 0.

Example of a hard question: 0.5 mol/L of PCl₅ with Kc = 0.2 gives x² + 0.2x − 0.1 = 0, so x = 0.232 mol/L of Cl₂ and [PCl₅] = 0.268 mol/L.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
N2​O4​0.08mol/L⇌2NO2​0.05mol/LKc​ = ?

For the reaction N2O4 ⇌ 2NO2, the equilibrium concentrations are [N2O4] = 0.08 mol/L, [NO2] = 0.05 mol/L. Find the equilibrium constant Kc.

[N2O4] = 0.08[NO2] = 0.05

Find Kc

Show hints
  1. Products go on top, reactants below.
  2. Each concentration is raised to the power of its coefficient.
  3. K = [NO2]2/([N2O4]).
Show the answer and the step-by-step solution

Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.

Expression for Kc

Kc = [NO2]2/([N2O4])

Substitute

Kc = 0.052/(0.08) = 0.03125

Answer

Kc = 0.03125
Example 2Medium
H2​I2​HII0.20.20C−x−x+2xE0.2 − x0.2 − x2x

0.2 mol/L of H₂ and 0.2 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 4. Find the equilibrium concentration [HI].

[H₂]₀ = [I₂]₀ = 0.2 mol/LKc = 4

Find [HI] (mol/L)

Show hints
  1. Set up an ICE table with x mol/L of H₂ reacting.
  2. Kc = (2x)²/((c₀ − x)²). Both sides are squares, so take the square root.
  3. √K = 2 = 2x/(0.2 − x).
Show the answer and the step-by-step solution

Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.

Equilibrium expression

4 = (2x)²/(0.2 − x)²

Take the square root

2 = 2x/(0.2 − x) → x = 0.2·2/(2 + 2) = 0.1 mol/L

Equilibrium concentration

[HI] = 2x = 0.2 mol/L

Answer

[HI] = 0.2 mol/L
Example 3Hard
PCl5​PCl3​Cl2​I0.2500C−x+x+xE0.25 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.04 at the reaction temperature. A container starts with 0.25 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 0.25 mol/LKc = 0.04

Find [Cl2] (mol/L)

Show hints
  1. ICE table: PCl₅ goes down by x, PCl₃ and Cl₂ each go up by x.
  2. K = x²/(c₀ − x), which rearranges to x² + K·x − K·c₀ = 0.
  3. Use the quadratic formula and keep the positive root.
Show the answer and the step-by-step solution

Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.

Equilibrium expression

0.04 = x²/(0.25 − x) → x² + 0.04·x − 0.01 = 0

Solve the quadratic

x = (−0.04 + √(0.04² + 4·0.01))/2 = 0.08198 mol/L

Equilibrium concentration

[Cl₂] = x = 0.08198 mol/L

Answer

[Cl2] = 0.08198 mol/L

Questions and answers

What is Chemical equilibrium?

Many reactions do not run to completion. As soon as some product forms, it starts to react back into the starting materials. The double arrow ⇌ shows this.

At first the forward reaction is fast, because there is a lot of H₂ and I₂. As they are used up, it slows down, while the backward reaction speeds up.

How do I solve Chemical equilibrium problems step by step?
  1. Write Kc with products over reactants, each raised to the power of its coefficient.
  2. Set up an ICE table: Initial, Change (in steps of x times the coefficients), Equilibrium.
  3. Put the equilibrium expressions into Kc and solve for x.
  4. Check whether x is small enough for an approximation; otherwise solve the quadratic.
What are common mistakes in Chemical equilibrium?
  • Forgetting the exponents in Kc.
  • Using initial instead of equilibrium concentrations.
  • Using the small-x approximation when x is not small.
What do the Chemical equilibrium exercises cover?
  • Easy: Kc from concentrations
  • Medium: ICE table
  • Hard: ICE table with a quadratic

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Molarity & dilution

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pH & buffers

Courses with this topic

General ChemistryChemical Reaction EngineeringFE Chemical exam practice

More topics in Natural sciences

StoichiometryMolarity & dilutionCalorimetry & enthalpypH & buffersGeneticsCell biology

Printable worksheet

Chemical equilibrium worksheet with answers
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