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Chemistry Y12 · Module 5 · Lesson 10 of 18

IQ3 — Keq, ICE Tables & Reaction Quotient

Calculating Keq — Substitution & ICE Tables

In 1927, physical chemist Farrington Daniels at the University of Wisconsin used a spectrophotometer to measure [NO₂] in a sealed N₂O₄/NO₂ flask at 55°C and calculated Keq = 0.498 mol/L — the first precise experimental determination of Keq for this equilibrium. His ICE table method, setting up I-C-E rows from initial concentrations and the measured equilibrium value, became the standard technique taught today.

Calculate Apply Analyse

Today's hook: In 1927, Farrington Daniels at the University of Wisconsin measured just one equilibrium concentration — [NO₂] by spectrophotometry — and used an ICE table to calculate every other concentration and Keq. That systematic method is what you master today.

0/5TASKS

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Think First

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.

Learning Intentions

Know

Calculate Keq from given equilibrium concentrations by direct substitution
Set up an ICE table with correct stoichiometric ratios in the Change row

Understand

Use an ICE table to find equilibrium concentrations when one equilibrium concentration is given
Verify ICE table answers by substituting equilibrium concentrations back into the Keq expression

Can Do

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

Key Terms

ICE table

Initial, Change, Equilibrium — a tabular method for tracking concentration changes as equilibrium is established.

Initial concentration

The starting molar concentration of each species before any reaction occurs.

Change row

The stoichiometric change in concentration as the reaction proceeds towards equilibrium (uses variable x).

Equilibrium concentration

The final molar concentration of each species at equilibrium; substituted into Keq expression.

Keq calculation

Dividing the equilibrium concentrations of products by reactants, each raised to stoichiometric powers.

Assumption (small-x)

If Keq is very small, x is negligible compared to initial concentrations — simplifying the algebra.

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

Core Concepts

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.

Direct substitution: write Keq expression first (products/reactants, stoichiometric powers); substitute equilibrium concentrations with correct powers; evaluate numerator and denominator separately; for 2SO₂ + O₂ ⇌ 2SO₃ with [SO₂]=0.200, [O₂]=0.100, [SO₃]=0.350 → Keq = (0.350)²/((0.200)²(0.100)) = 30.6.

Pause — copy the highlighted direct substitution procedure into your book before moving on.

For the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), at equilibrium [SO₂] = 0.200, [O₂] = 0.100, [SO₃] = 0.350 mol/L. What is Keq?

Apply

2. Type 2 — Introducing the ICE Table

We just saw how to calculate Keq by direct substitution when all equilibrium concentrations are given. That raises a question: what do we do when only some equilibrium concentrations are given — how do we find the rest? This card answers it → the ICE table framework using stoichiometric ratios.

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.

ICE table: three rows — Initial (given starting concentrations), Change (stoichiometric ratios, reactants −, products +), Equilibrium (Initial + Change); for N₂O₄ ⇌ 2NO₂ with [NO₂]eq = 0.300 → Δ[N₂O₄] = −0.150 (ratio 1:2); Keq = (0.300)²/0.350 = 0.257.

Add the highlighted ICE table structure and worked example to your notes before continuing.

True or False: In an ICE table for N₂O₄ ⇌ 2NO₂, if [NO₂] increases by 0.300 mol/L, then [N₂O₄] decreases by 0.300 mol/L.

Apply

3. ICE Table — Worked Framework with Stoichiometric Ratios

We just saw the ICE table framework with the N₂O₄/NO₂ example. That raises a question: how do we apply the full eight-step ICE procedure to a different reaction — especially getting the stoichiometric ratios right in the Change row? This card answers it → the 2SO₃ ⇌ 2SO₂ + O₂ worked example showing the 2:2:1 ratio trap.

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.

Eight-step ICE procedure: balanced equation → Keq expression → set up table → fill Initial row → use given equilibrium concentration to find x → Change row using stoichiometric ratios → Equilibrium row (I + C) → substitute into Keq → verify by back-substitution; for 2SO₃ ⇌ 2SO₂ + O₂: ratios 2:2:1, so if Δ[SO₃] = −0.220, Δ[O₂] = +0.110.

Pause — copy the highlighted eight-step procedure into your book before the check below.

For 2SO₃(g) ⇌ 2SO₂(g) + O₂(g), if SO₃ decreases by 0.220 mol/L, what is the change in O₂?

Calculate

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

We just saw how to find Keq from one known equilibrium concentration. That raises a question: how do we work backwards — given Keq and initial concentrations, how do we find all the equilibrium concentrations? This card answers it → the forward ICE calculation with the H₂/I₂/HI square-root shortcut.

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.

Forward ICE (given Keq): write Change row as ±nx (n = stoichiometric coefficient) → substitute Equilibrium row into Keq → solve algebraically; for H₂ + I₂ ⇌ 2HI (Keq = 54.3), Keq = 4x²/(0.100−x)² → take √ → x = 0.0787; [H₂]=[I₂]=0.0213, [HI]=0.157 mol/L.

Add the highlighted forward ICE procedure and shortcut to your notes before continuing.

For H₂ + I₂ ⇌ 2HI with Keq = 54.3, why can we take the square root of both sides of 4x²/(0.100−x)² = 54.3?

Cross-lesson links: Daniels' 1927 colourimetry method introduced here reappears in L13 (temperature and Keq — colourimetry investigation). The given-Keq ICE table introduced in Card 4 is extended in L11 (consolidation — quadratic and simplifying assumption). The N₂O₄/NO₂ system used throughout this lesson is a standard NESA-listed example also featured in L03 (collision theory) and L06 (pressure effects).

Analyse

5. Measuring Keq by Colourimetry — NO₂/N₂O₄ System

We just saw the forward ICE calculation for solving equilibrium concentrations algebraically. That raises a question: how is Keq measured experimentally in the lab — what is the NESA-specified colourimetry method? This card answers it → the five-step colourimetry procedure using Beer-Lambert law and an ICE table.

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.

Colourimetry Keq method: (1) calibration curve from known [NO₂] vs absorbance; (2) measure absorbance of equilibrium mixture; (3) read [NO₂]eq from curve; (4) use ICE table to find [N₂O₄]eq; (5) calculate Keq = [NO₂]²/[N₂O₄]; works because only NO₂ is brown; N₂O₄ is colourless.

Pause — copy the highlighted colourimetry procedure into your book before the check below.

Why is the N₂O₄ ⇌ 2NO₂ equilibrium ideal for colourimetry experiments to determine Keq?

Activities

Worked 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 ✓

Worked Example 2 — Band 5–6 — 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).

Interactive Tool — Keq Calculations Open fullscreen ↗

In the Keq Calculator, if the equilibrium concentration of products is much greater than reactants, K is…

Sort the steps+7 XP

Put the ICE table method for finding K_eq in the correct order.

Fill in the Change row using stoichiometric ratios (express as ±x or known values).
Write the balanced equilibrium equation.
Substitute Equilibrium concentrations into the K_eq expression and solve.
Fill in the Initial row with known starting concentrations.
Calculate the Equilibrium row (I + C for each species).

Complete the Learn phase to unlock practice questions.

✓

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

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

1 mark

Q4: For the equilibrium SO₂(g) + ½O₂(g) ⇌ SO₃(g), Keq = 0.240 at 700°C. What is Keq for the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) at 700°C?

A) 0.480 (doubled)

B) 0.120 (halved)

C) 0.0576 (squared)

D) 0.0576 — coefficients doubled means Keq(new) = Keq(original)² = (0.240)² = 0.0576

1 mark

Q5: 0.800 mol of PCl₅ is placed in a 2.00 L flask. At equilibrium, [PCl₅] = 0.250 mol/L. For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), what are the equilibrium concentrations of PCl₃ and Cl₂?

A) [PCl₃] = [Cl₂] = 0.400 mol/L (initial − equilibrium)

B) [PCl₃] = [Cl₂] = 0.150 mol/L (ICE Change = initial − equilibrium)

C) [PCl₃] = [Cl₂] = 0.250 mol/L (same as PCl₅ equilibrium)

D) [PCl₃] = [Cl₂] = 0.550 mol/L

Short Answer

4 marks

Q6: 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

Q7: 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.)

✓

Model Answers

Show Model Answers

Q6 Model Answer:

(a) ICE table: Initial [H₂] = [I₂] = 0.500; [HI] = 0. At equilibrium, [H₂] = 0.107, so change in H₂ = 0.107 − 0.500 = −0.393. Stoichiometric ratio 1:1:2: Change in I₂ = −0.393; Change in HI = +0.786. Equilibrium: [H₂] = 0.107; [I₂] = 0.107; [HI] = 0.786 mol/L.

(b) Keq = [HI]² / ([H₂][I₂]) = (0.786)² / (0.107)(0.107) = 0.6178 / 0.01145 = 53.9 ≈ 54

(c) Verification: (0.786)² / (0.107)² = 0.6178 / 0.01145 = 53.9 ≈ 54 ✓ (consistent with Keq ≈ 54.3 for H₂ + I₂ ⇌ 2HI at 430°C).

Q7 Model Answer:

ICE: N₂O₄: Initial = 0.400, Change = −x, Equilibrium = 0.400−x. NO₂: Initial = 0, Change = +2x, Equilibrium = 2x.

Keq = [NO₂]² / [N₂O₄] = (2x)² / (0.400−x) = 4x² / (0.400−x) = 0.500

Check simplifying assumption: Keq/[initial] = 0.500/0.400 = 1.25 = 125% >> 5% → simplifying assumption is invalid → must solve quadratic.

4x² = 0.500(0.400 − x) = 0.200 − 0.500x

4x² + 0.500x − 0.200 = 0

Quadratic formula: x = (−0.500 ± √(0.250 + 3.200)) / 8 = (−0.500 ± √3.450) / 8 = (−0.500 ± 1.8574) / 8

Positive root: x = (−0.500 + 1.8574) / 8 = 1.3574 / 8 = 0.1697 mol/L

Equilibrium: [N₂O₄] = 0.400 − 0.170 = 0.230 mol/L; [NO₂] = 2(0.170) = 0.339 mol/L

Verification: (0.339)² / (0.230) = 0.1149 / 0.230 = 0.499 ≈ 0.500 ✓

Extended Response

A sealed 2.00 L flask initially contains 1.20 mol of H₂ and 1.20 mol of I₂. After reaching equilibrium at 430°C for H₂(g) + I₂(g) ⇌ 2HI(g), the flask is analysed and [HI] = 0.480 mol/L is measured. (a) Calculate the initial concentrations of H₂ and I₂. (b) Set up and complete the ICE table, finding x from the given [HI]. (c) Calculate Keq and compare it to the literature value of 54.3. (d) The temperature is now increased to 600°C and Keq decreases to 38.0. Explain what this tells you about whether the forward reaction is exothermic or endothermic. (e) Without re-doing the full ICE calculation, describe qualitatively how the equilibrium concentrations of H₂, I₂, and HI change when temperature increases from 430°C to 600°C.

How did your thinking change?

Look back at your Think First calculation for the N₂O₄/NO₂ scenario. Recall that Farrington Daniels used this exact system in 1927 at the University of Wisconsin — measuring just one equilibrium concentration by spectrophotometry, then using the ICE method to find everything else. Did you get N₂O₄ consumed = 0.150 mol/L and [N₂O₄]eq = 0.350 mol/L, matching Daniels' systematic approach?

I have completed this lesson

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