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Chemistry Y12 · Module 5 · Lesson 10

IQ3 — Keq, ICE Tables & Reaction Quotient

Calculating Keq — Substitution & ICE Tables

In a sealed flask at 100°C, a pale brown mixture of NO₂ and N₂O₄ reaches equilibrium. By measuring the brown colour intensity with a spectrophotometer, chemists can calculate the exact concentration of NO₂ — and from that, calculate Keq for the reaction.

Calculate Apply Analyse

Misconceptions to Fix

✗

Wrong: A large Keq means the reaction happens quickly.

✗

Right: Keq indicates the equilibrium position — a large Keq means products are favoured at equilibrium. It says nothing about reaction rate. Rate depends on activation energy, temperature, and catalysts, not thermodynamic favourability. A reaction can have a huge Keq but be extremely slow.

Learning Intentions

Calculate Keq from given equilibrium concentrations by direct substitution

Set up an ICE table with correct stoichiometric ratios in the Change row

Use an ICE table to find equilibrium concentrations when one equilibrium concentration is given

Solve forward ICE problems where Keq is given and initial concentrations are known

Verify ICE table answers by substituting equilibrium concentrations back into the Keq expression

Printable worksheet

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Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.

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🤔 Think First — Before You Set Up the Table

A chemist seals 0.500 mol of N₂O₄(g) in a 1.00 L flask at 100°C. After reaching equilibrium, they measure the concentration of NO₂ as 0.300 mol/L. Before reading on: (1) How much N₂O₄ has been consumed? (Think about the mole ratio in N₂O₄ ⇌ 2NO₂.) (2) What is the equilibrium concentration of N₂O₄? Write your reasoning before reading the worked solution.

Your reasoning:

ICE Table Framework

I = Initial concentration (mol/L) | C = Change (mol/L) | E = Equilibrium (mol/L)

Change row rules:

Use stoichiometric ratios from the balanced equation — NOT ±x for every species
For N₂ + 3H₂ ⇌ 2NH₃: Change = −x, −3x, +2x (ratios 1:3:2)
Equilibrium row = Initial + Change (negative changes decrease the value)

Verification: always substitute final E values back into Keq — must equal the given Keq

Calculate

1. Type 1 — Direct Substitution

The simplest Keq calculation requires no algebra — you are given the equilibrium concentrations directly and simply substitute them into the Keq expression.

Procedure:

Write the correct Keq expression first
Substitute values with correct stoichiometric powers
Evaluate numerator and denominator separately before dividing
Report to appropriate significant figures

Example: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) at 600°C. At equilibrium: [SO₂] = 0.200 mol/L, [O₂] = 0.100 mol/L, [SO₃] = 0.350 mol/L.

Keq expression: Keq = [SO₃]² / ([SO₂]²[O₂])

Substitution: Keq = (0.350)² / ((0.200)²(0.100)) = 0.1225 / (0.0400 × 0.100) = 0.1225 / 0.00400 = 30.6

Always write the expression first. Students who jump straight to substituting numbers frequently use the wrong powers or invert the expression. Writing the expression first forces you to apply the L09 rules before any arithmetic.

Common error: Squaring the entire bracket instead of each concentration separately. Apply powers to each concentration individually: [SO₃]² = (0.350)² = 0.1225; [SO₂]² = (0.200)² = 0.0400. Do not group [SO₂]² and [O₂] together before applying the power.

ICE table structure — three rows (Initial, Change, Equilibrium) with stoichiometric ratios in the Change row

Key Terms — scan these before reading

VerificationVerification catches arithmetic mistakes before you lose marks.

Dynamic equilibriumA state where forward and reverse reaction rates are equal.

Equilibrium constant (Keq)The ratio of product to reactant concentrations at equilibrium.

Le Chatelier's PrincipleA system at equilibrium shifts to minimise applied disturbances.

Reaction quotient (Q)The ratio of product to reactant concentrations at any instant.

Solubility product (Ksp)The equilibrium constant for dissolution of a sparingly soluble salt.

Apply

2. Type 2 — Introducing the ICE Table

When you are given initial concentrations and only some equilibrium concentrations, you need the ICE table to find the missing values. The ICE table is the single most important calculation tool in Module 5.

ICE table rows:

Initial (I): starting concentrations before equilibrium is reached. Products start at zero if the reaction begins with pure reactants.
Change (C): how much each concentration changes, using stoichiometric ratios. All changes are expressed in terms of one variable (x or a given equilibrium value).
Equilibrium (E): concentration of each species at equilibrium. Equilibrium = Initial + Change.

Think First scenario — N₂O₄ ⇌ 2NO₂:

Initial: [N₂O₄] = 0.500 mol/L; [NO₂] = 0. At equilibrium, [NO₂] = 0.300 mol/L.

Change in NO₂ = +0.300 mol/L. Stoichiometric ratio: 1 mol N₂O₄ → 2 mol NO₂. So if NO₂ increased by 0.300, N₂O₄ decreased by 0.300/2 = 0.150 mol/L.

N₂O₄

0.500

−0.150

0.350

NO₂

+0.300

0.300

Keq = [NO₂]² / [N₂O₄] = (0.300)² / (0.350) = 0.0900 / 0.350 = 0.257

Must use stoichiometric ratios in the Change row. For N₂ + 3H₂ ⇌ 2NH₃: if N₂ decreases by x mol/L, then H₂ decreases by 3x mol/L and NH₃ increases by 2x mol/L. The 1:3:2 ratio comes from the coefficients. Writing ±x for every species is the single most common ICE table error.

Common error: Dividing the change in NO₂ by 1 instead of 2 when calculating the change in N₂O₄. The ratio is 1 N₂O₄ : 2 NO₂. If 0.300 mol/L of NO₂ forms, then 0.300/2 = 0.150 mol/L of N₂O₄ was consumed — not 0.300.

Apply

3. ICE Table — Worked Framework with Stoichiometric Ratios

The ICE table is most useful when you see it applied step by step. The framework is always the same, and once you have used it three or four times it becomes automatic.

Standard ICE procedure when one equilibrium concentration is given:

Write the balanced equation and Keq expression
Set up ICE table with all species as columns
Fill in Initial row from problem data
Use the given equilibrium concentration to find x directly
Calculate all Change row values using stoichiometric ratios
Write Equilibrium row as Initial + Change
Substitute Equilibrium row into Keq expression
Solve and verify

Extended example — 2SO₃(g) ⇌ 2SO₂(g) + O₂(g):

Initial: [SO₃] = 1.00 mol/L; [SO₂] = [O₂] = 0. At equilibrium, [SO₃] = 0.780 mol/L.

Change in SO₃ = 0.780 − 1.00 = −0.220 mol/L. Stoichiometric ratios — 2 SO₃ : 2 SO₂ : 1 O₂. Change in SO₂ = +0.220 mol/L; Change in O₂ = +0.110 mol/L.

Initial

SO₃: 1.00

SO₂: 0

O₂: 0

Change

SO₃: −0.220

SO₂: +0.220

O₂: +0.110

Equilibrium

SO₃: 0.780

SO₂: 0.220

O₂: 0.110

Keq = [SO₂]²[O₂] / [SO₃]² = (0.220)²(0.110) / (0.780)² = (0.04840)(0.110) / 0.6084 = 0.005324 / 0.6084 = 8.75 × 10⁻³

Always verify: substitute the equilibrium concentrations back into the Keq expression. If the result doesn't equal the Keq you calculated, you have made an error somewhere. Verification catches arithmetic mistakes before you lose marks.

Common error: Applying the stoichiometric ratio to O₂ as if it were the same as SO₂. Coefficient of O₂ = 1, coefficient of SO₃ = 2. The ratio of O₂ change to SO₃ change = 1:2. If SO₃ decreases by 0.220, O₂ increases by 0.220/2 = 0.110, NOT 0.220.

Calculate

4. Forward ICE Calculation — Solving for x When Keq Is Given

The harder version of the ICE table gives you Keq and initial concentrations and asks you to find the equilibrium concentrations — this requires writing the Change row in terms of x and solving algebraically.

Example: H₂(g) + I₂(g) ⇌ 2HI(g), Keq = 54.3 at 430°C.

Initial: [H₂] = [I₂] = 0.100 mol/L; [HI] = 0. Set x = decrease in [H₂] = decrease in [I₂]; increase in [HI] = 2x.

	H₂	I₂	HI
Initial	0.100	0.100	0
Change	−x	−x	+2x
Equilibrium	0.100−x	0.100−x	2x

Keq = [HI]² / ([H₂][I₂]) = (2x)² / (0.100−x)² = 4x² / (0.100−x)² = 54.3

Take square root of both sides (since both sides are perfect squares): 2x / (0.100−x) = √54.3 = 7.369

2x = 7.369(0.100 − x) = 0.7369 − 7.369x

9.369x = 0.7369 → x = 0.0787 mol/L

Equilibrium: [H₂] = [I₂] = 0.100 − 0.0787 = 0.0213 mol/L; [HI] = 2(0.0787) = 0.1574 mol/L

Verify: Keq = (0.1574)² / (0.0213)(0.0213) = 0.02478 / 0.000454 = 54.6 ≈ 54.3 ✓

Square root shortcut: For the special case where [H₂] = [I₂] initially (or any case where the Keq expression simplifies to a perfect square), you can take the square root of both sides directly and avoid the quadratic formula. Always check for this pattern first.

Common error: Writing +x for HI instead of +2x. The coefficient of HI is 2, not 1. If you write +x for HI, the Keq expression becomes x²/(0.100−x)² = 54.3, giving x = 0.155 — but this is wrong because [HI] = x, not 2x. The correct [HI] is 2(0.0787) = 0.157 mol/L.

Analyse

5. Measuring Keq by Colourimetry — NO₂/N₂O₄ System

Colourimetry turns an invisible number — Keq — into something you can measure by shining a light through a brown gas and reading an absorbance value.

The equilibrium N₂O₄(g) ⇌ 2NO₂(g) is ideal because only NO₂ is coloured (brown) while N₂O₄ is colourless. The concentration of NO₂ can be determined directly from absorbance using Beer-Lambert law.

Experimental procedure:

Build a calibration curve: measure absorbance for known NO₂ concentrations
Measure absorbance of the equilibrium mixture
Read off [NO₂] at equilibrium from the calibration curve
Use ICE table to find [N₂O₄] at equilibrium (from initial amount minus consumed)
Calculate Keq = [NO₂]² / [N₂O₄] by direct substitution

Using the Think First scenario: Starting with 0.500 mol/L N₂O₄; measuring [NO₂]eq = 0.300 mol/L; ICE gives [N₂O₄]eq = 0.350 mol/L; Keq = 0.257.

In real experiments, multiple measurements at different temperatures map the temperature-dependence of Keq — confirming that only temperature changes Keq, while concentration changes shift position without changing the constant.

NESA-specified investigation: You need to describe the colourimetry method (measure absorbance → calibration curve → find [NO₂] → ICE table → calculate Keq) and explain why it works (only NO₂ absorbs visible light; absorbance is proportional to concentration). A fuller treatment with data analysis is in L13.

Beer-Lambert Law (conceptual): Absorbance ∝ concentration of coloured species × path length. You do not need the full mathematical equation for HSC Module 5, but understanding that a deeper brown colour = higher [NO₂] is the core idea.

Interactive — ICE Table Builder

Revisit Your Thinking

ICE tables organise the Initial concentrations, the Change (using −x for reactants and +x for products based on stoichiometry), and the Equilibrium concentrations. Substitute the equilibrium row into the Keq expression, solve for x, and then calculate the equilibrium concentrations. If Keq is very small, the −x approximation may be valid (check with the 5% rule).

Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

Worked Examples

Calculate

Example 1 — Direct Substitution Keq

Problem: The equilibrium PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) is established at 250°C. At equilibrium: [PCl₃] = 0.130 mol/L, [Cl₂] = 0.045 mol/L, [PCl₅] = 0.380 mol/L. (a) Write the Keq expression. (b) Calculate Keq. (c) What does the magnitude of Keq indicate?

(a): All species are gases, all included. Stoichiometric coefficients all = 1. Keq = [PCl₅] / ([PCl₃][Cl₂])

(b): Keq = 0.380 / (0.130 × 0.045) = 0.380 / 0.00585 = 65.0

(c): Keq = 65.0 >> 1 → products are strongly favoured → at equilibrium, the mixture contains significantly more PCl₅ than PCl₃ or Cl₂. The forward reaction proceeds substantially to the right at 250°C.

Verification: (0.130)(0.045) = 0.00585; 0.380/0.00585 = 65.0 ✓

Band 5–6

Example 2 — ICE Table: Find Keq from Initial Concentration and One Equilibrium Concentration

Problem: 1.00 mol of SO₃(g) is placed in a 1.00 L flask at 700°C. At equilibrium, [SO₃] = 0.780 mol/L. Equilibrium reaction: 2SO₃(g) ⇌ 2SO₂(g) + O₂(g). (a) Set up and complete the ICE table. (b) Calculate Keq. (c) Verify and interpret the magnitude.

(a) — ICE setup: Initial: [SO₃] = 1.00 mol/L (1.00 mol / 1.00 L); [SO₂] = 0; [O₂] = 0. Change in SO₃ = 0.780 − 1.00 = −0.220 mol/L. Stoichiometric ratios (2 SO₃ : 2 SO₂ : 1 O₂): Change in SO₂ = +0.220 mol/L; Change in O₂ = +0.110 mol/L.

	SO₃	SO₂	O₂
Initial	1.00	0	0
Change	−0.220	+0.220	+0.110
Equilibrium	0.780	0.220	0.110

(b): Keq = [SO₂]²[O₂] / [SO₃]² = (0.220)²(0.110) / (0.780)² = (0.04840)(0.110) / 0.6084 = 0.005324 / 0.6084 = 8.75 × 10⁻³

(c): Verification: all equilibrium values substituted give 8.75 × 10⁻³ ✓. Magnitude: Keq = 8.75 × 10⁻³ < 1 → reactants strongly favoured → most SO₃ remains unreacted at 700°C; little SO₂ and O₂ present (consistent with [SO₃] = 0.780 >> [SO₂] = 0.220 and [O₂] = 0.110).

Checkpoint Questions

1 mark

Q1: The equilibrium 2HI(g) ⇌ H₂(g) + I₂(g) is established with [HI] = 0.400 mol/L, [H₂] = 0.040 mol/L, [I₂] = 0.040 mol/L. What is Keq?

A 0.010

B 0.040

C 0.100

D 0.400

1 mark

Q2: In an ICE table for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), if N₂ decreases by x mol/L, what is the correct Change row for H₂ and NH₃?

A H₂: −x; NH₃: +x

B H₂: −3x; NH₃: +2x

C H₂: −x/3; NH₃: +x/2

D H₂: −2x; NH₃: +3x

Q2: In an ICE table for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), if N₂ decreases by x mol/L, Identify the correct Change row for H₂ and NH₃?

AH₂: −x; NH₃: +x

BH₂: −3x; NH₃: +2x

CH₂: −x/3; NH₃: +x/2

DH₂: −2x; NH₃: +3x

1 mark

Q3: A student calculates Keq = 45.2 for an equilibrium at 400°C. They then add more reactant to the flask. After the system reaches its new equilibrium, what is the value of Keq?

A Greater than 45.2 — more product has formed

B Less than 45.2 — the added reactant dilutes the equilibrium

C 45.2 — Keq is unchanged by concentration changes; only temperature changes Keq

D It depends on how much reactant was added

Q3: A student calculates Keq = 45.2 for an equilibrium at 400°C. They then add more reactant to the flask. After the system reaches its new equilibrium, Identify the value of Keq?

AGreater than 45.2 — more product has formed

BLess than 45.2 — the added reactant dilutes the equilibrium

C45.2 — Keq is unchanged by concentration changes; only temperature changes Keq

DIt depends on how much reactant was added

Short Answer Practice

4 marks

Q4: 0.500 mol of H₂ and 0.500 mol of I₂ are placed in a 1.00 L flask at 430°C. At equilibrium, [H₂] = 0.107 mol/L. The equilibrium is H₂(g) + I₂(g) ⇌ 2HI(g). (a) Set up and complete the ICE table. (b) Calculate Keq. (c) Verify by substitution.

4 marks

Q5: The equilibrium N₂O₄(g) ⇌ 2NO₂(g) has Keq = 0.500 at 55°C. Starting with [N₂O₄] = 0.400 mol/L and [NO₂] = 0, set up the ICE table, write the Keq expression in terms of x, and solve for x. What are the equilibrium concentrations of both species? (Hint: this simplifies to a quadratic — or try the simplifying assumption first and check if it is valid.)

Calculating Keq — Substitution & ICE Tables

Misconceptions to Fix

Download this lesson's worksheet

🤔 Think First — Before You Set Up the Table

ICE Table Framework

1. Type 1 — Direct Substitution

2. Type 2 — Introducing the ICE Table

3. ICE Table — Worked Framework with Stoichiometric Ratios

Initial

Change

Equilibrium

4. Forward ICE Calculation — Solving for x When Keq Is Given

5. Measuring Keq by Colourimetry — NO₂/N₂O₄ System

Worked Examples

Example 1 — Direct Substitution Keq

Example 2 — ICE Table: Find Keq from Initial Concentration and One Equilibrium Concentration

Checkpoint Questions

Short Answer Practice

Calculating Keq & ICE Tables