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Chemistry Y12 · Module 5 · Lesson 11

IQ3 — Keq, ICE Tables & Reaction Quotient · ★ Consolidation

★ Consolidation — Deepening L10

ICE Table Mastery — Harder Problems & Simplifying Assumptions

The ICE table is the most failed calculation type in Module 5 HSC exams — not because students don't know the method, but because they make the same three errors every time. This lesson is the cure.

No new dot points. Dedicated ICE table consolidation: given-Keq problems, simplifying assumption, quadratic, non-zero initial products, verification.

Misconceptions to Fix

✗

Wrong: You can ignore the minus x in ICE tables when x is small compared to the initial concentration.

✗

Right: The simplifying assumption [initial] − x ≈ [initial] is only valid when x < 5% of the initial concentration (or more strictly, Ka/c < 0.0025). You must always verify this assumption after solving. If invalid, you must solve the full quadratic equation.

Learning Intentions

Identify and fix the three most common ICE table errors: wrong stoichiometric ratios, missing final conversions, and not verifying

Apply the simplifying assumption correctly, including checking its validity with the 5% threshold

Solve ICE table problems requiring the quadratic formula, including selecting the physically meaningful root

Solve ICE problems where initial product concentrations are non-zero, using Q to determine direction of shift first

Verify every ICE table answer 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 — ICE Table Diagnostic

A student is solving: "0.200 mol of H₂ and 0.200 mol of I₂ are placed in a 1.00 L flask at 430°C. Keq = 54.3 for H₂(g) + I₂(g) ⇌ 2HI(g). Find the equilibrium concentrations." The student sets up the ICE table and writes:

Change row: H₂: −x; I₂: −x; HI: +x

They substitute into Keq and get: 54.3 = x²/((0.200−x)(0.200−x)). They solve and get x = 0.155 mol/L. They report [HI] = 0.155 mol/L.

Before reading on — identify the error in the Change row AND the error in the final answer. How should the Change row for HI read? What should the correct [HI] be?

Your diagnosis:

ICE Method — Harder Problems

Simplifying assumption: if Keq/[initial] < 0.05 (i.e. < 5%), then (initial − x) ≈ initial → avoids quadratic

When to use quadratic: when Keq/[initial] ≥ 5% — assumption invalid; solve ax² + bx + c = 0 using x = (−b ± √(b²−4ac)) / 2a

Non-zero initial products: calculate Q first; if Q < Keq → shift right; if Q > Keq → shift left

Always verify: substitute final E values back into Keq — must equal given Keq within rounding

Error Diagnosis

1. The Three Most Common ICE Table Errors

Before doing harder problems, it is worth being precise about exactly which errors cause ICE table calculations to fail.

Error 1 — Wrong stoichiometric ratios in the Change row (most common)

Students write ±x for every species regardless of stoichiometric coefficients. For N₂ + 3H₂ ⇌ 2NH₃: writing N₂: −x, H₂: −x, NH₃: +x gives wrong equilibrium concentrations and a wrong Keq.

Fix 1 — Write out the mole ratios from the balanced equation first

Correct Change row for N₂ + 3H₂ ⇌ 2NH₃: N₂: −x; H₂: −3x; NH₃: +2x

Error 2 — Not converting x to actual equilibrium concentrations

Students solve for x and report x as the equilibrium concentration, forgetting that [HI] = 2x, or that other species have different relationships to x.

Fix 2 — Always write out ALL equilibrium concentrations from the E row

For H₂ + I₂ ⇌ 2HI with Change row +2x: if x = 0.0787, then [HI] = 2(0.0787) = 0.1574, NOT 0.0787.

Error 3 — Not verifying the answer

Students get a value of x, report concentrations, and stop. Without verification, arithmetic errors in the quadratic go undetected.

Fix 3 — Always substitute back into the Keq expression

The verified Keq must equal the given Keq within rounding (typically ±2%). A 15% discrepancy means there is an error to find.

Think First resolution: The student's error was writing +x for HI instead of +2x (coefficient of HI = 2). The correct Change row is H₂: −x; I₂: −x; HI: +2x. Substituting into Keq: 54.3 = (2x)²/((0.200−x)²) → 2x/(0.200−x) = √54.3 = 7.369 → x = 0.0787 → [HI] = 2(0.0787) = 0.1574 mol/L, not 0.155 mol/L.

Analogy 1

2. The Bank Account — Understanding ICE Table Logic

The Bank Account Analogy

Think of each species in the equilibrium as a separate bank account. The Initial row is the opening balance — what you start with before any chemistry happens. The Change row is the set of transactions: deposits (+, species being formed) and withdrawals (−, species being consumed). For every withdrawal from one account (reactant), there is a corresponding deposit to another account (product) in a fixed ratio (the stoichiometric ratio). The Equilibrium row is the closing balance — opening balance plus all transactions.

Key check: Just as a bank account cannot go negative, the Equilibrium row values must all be positive. If you get a negative equilibrium concentration, you have made an error — either in the sign of the Change row, the stoichiometric ratio, or the direction of the shift.

Where this analogy breaks down: Bank transactions are independent; ICE table changes are coupled — changing one "account" changes all others according to fixed stoichiometric ratios. Also, bank accounts are linear; equilibrium concentrations are related by a nonlinear Keq expression. Use this analogy for the STRUCTURE of ICE (initial → change → final), not for the chemistry of how values are related.

Quick diagnostic: If any equilibrium concentration in your E row is negative, stop and find the error before proceeding. The "closing balance must be positive" check catches direction-of-shift errors before they propagate into the Keq calculation.

ICE table decision flowchart — two paths depending on whether Keq or an equilibrium concentration is given

Key Terms — scan these before reading

Simplifying assumptionInvalid, you need the quadratic formula — and the key is setting up the equation correctly from the ICE table before app.

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.

Analogy 2

3. The Funnel — Simplifying Assumption Logic

The Funnel Analogy

Imagine a large funnel with 1 litre of water (the initial concentration) and a very slow leak at the bottom (the x that drains away). If the leak is tiny — say, 0.1% of the total — the amount remaining is 1.000 − 0.001 ≈ 1.000. Subtracting the tiny x makes no practical difference to the arithmetic. This is the simplifying assumption: when x is very small relative to the initial concentration, (initial − x) ≈ initial.

When is it valid? The rule of thumb: if Keq/[initial] < 0.05 (less than 5%), then x/[initial] will be less than 5% and the simplifying assumption gives less than 5% error — acceptable for HSC.

How to check after solving: calculate x/[initial] as a percentage. If <5% → assumption was valid. If ≥5% → must solve the full quadratic.

Where this analogy breaks down: The funnel leak is continuous; the equilibrium shift is self-limiting (as x increases, the rate of change decreases until equilibrium). The analogy captures only the "x is small" concept, not the dynamic process that determines equilibrium.

Common error: Applying the simplifying assumption without checking validity. If Keq is not much smaller than the initial concentration (Keq/[initial] > 5%), the assumption introduces significant error and the quadratic must be solved. Always check the threshold after using the assumption.

Non-Zero Initial Products

4. ICE Table with Non-Zero Initial Product Concentrations

Most ICE table problems start with pure reactants — products at zero. But if the problem gives non-zero initial concentrations for products, you must first calculate Q to determine which direction the system will shift.

Procedure for Non-Zero Initial Products

Step 1: Calculate Q using the initial concentrations

Step 2: Compare Q to Keq:

Q < Keq → system shifts RIGHT → reactants decrease, products increase (Change row: reactants −, products +)
Q > Keq → system shifts LEFT → products decrease, reactants increase (Change row: products −, reactants +)
Q = Keq → already at equilibrium → no change

Step 3: Write Change row with x in the direction determined by Q vs Keq

Example setup: 2NO₂(g) ⇌ 2NO(g) + O₂(g), Keq = 0.50. Initial: [NO₂] = 0.400, [NO] = 0.100, [O₂] = 0.050 mol/L.

Q = [NO]²[O₂]/[NO₂]² = (0.100)²(0.050)/(0.400)² = (0.010)(0.050)/0.160 = 0.000500/0.160 = 3.13 × 10⁻³

Q = 3.13 × 10⁻³ < Keq = 0.50 → system shifts RIGHT → NO₂ decreases (−2x), NO increases (+2x), O₂ increases (+x).

ALWAYS calculate Q first when non-zero initial concentrations of all species are given. Writing the Change row before knowing the direction of shift is a common error — if you assume the shift is right but Q > Keq (meaning the shift is left), all your signs will be wrong.

Common error: Assuming the reaction always shifts right when products are present. The direction of shift depends entirely on Q vs Keq — not on which species are present initially.

Quadratic Solutions

5. Quadratic Solutions for ICE Problems

When the simplifying assumption is invalid, you need the quadratic formula — and the key is setting up the equation correctly from the ICE table before applying the formula.

General procedure:

Set up ICE table with E row in terms of x
Substitute into Keq expression → equation in x
Rearrange to standard form: ax² + bx + c = 0
Apply quadratic formula: x = (−b ± √(b²−4ac)) / 2a
Choose the physically meaningful root — x must give all positive E row values
Verify by substitution

Example: A(g) ⇌ B(g) + C(g), Keq = 0.25. Initial: [A] = 0.600; [B] = [C] = 0.

Check: Keq/[A] = 0.25/0.600 = 42% >> 5% → must use quadratic.

ICE: A: 0.600−x; B: x; C: x. Keq = x²/(0.600−x) = 0.25

x² = 0.25(0.600−x) = 0.150 − 0.25x → x² + 0.25x − 0.150 = 0

x = (−0.25 ± √(0.0625 + 0.600)) / 2 = (−0.25 ± √0.6625) / 2 = (−0.25 ± 0.8140) / 2

Positive root: x = (−0.25 + 0.8140) / 2 = 0.5640/2 = 0.282 mol/L

Equilibrium: [A] = 0.318; [B] = [C] = 0.282 mol/L. Verify: (0.282)(0.282)/(0.318) = 0.250 ✓

Choose the positive root. The negative root gives a negative x value, which means negative equilibrium concentrations — physically impossible. Always select the root that gives all positive E row values.

Common error: Applying the simplifying assumption without checking the 5% threshold. In this example, Keq/[A] = 42% → using the assumption would give x ≈ √(0.25 × 0.600) = 0.387, giving [A] = 0.213 — significantly different from the correct 0.318. Always check first.

Interactive — ICE Table Builder

Revisit Your Thinking

For harder ICE tables, always define x as the change in the limiting reagent's concentration. Use the stoichiometry to express all other changes in terms of x. When Keq is very small and the initial concentration is relatively large, you can often approximate [reactant]eq ≈ [reactant]initial. Always verify this assumption using the 5% rule: if x / [initial] < 0.05, the approximation is valid.

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

Apply

Example 1 (Straightforward) — Simplifying Assumption Valid: Tiny Keq

Problem: N₂(g) + O₂(g) ⇌ 2NO(g), Keq = 1.0 × 10⁻³⁰ at 25°C. A flask contains [N₂] = 0.780 mol/L and [O₂] = 0.210 mol/L. Calculate [NO] at equilibrium. State and verify the simplifying assumption.

Check assumption: Keq/[N₂] = 1.0 × 10⁻³⁰/0.780 = 1.28 × 10⁻³⁰ << 0.05 ✓ → simplifying assumption valid

ICE table:

Initial

N₂: 0.780

O₂: 0.210

NO: 0

Change

N₂: −x

O₂: −x

NO: +2x

Equilibrium

N₂: ≈0.780

O₂: ≈0.210

NO: 2x

Solve: Keq = (2x)²/([N₂][O₂]) = 4x²/(0.780 × 0.210) = 4x²/0.1638 = 1.0 × 10⁻³⁰. x² = (1.0 × 10⁻³⁰ × 0.1638)/4 = 4.095 × 10⁻³². x = 2.024 × 10⁻¹⁶ mol/L. [NO] = 2x = 4.05 × 10⁻¹⁶ mol/L

Verify assumption: x/[N₂] = 2.024 × 10⁻¹⁶/0.780 = 2.6 × 10⁻¹⁶ << 5% ✓

Interpretation: [NO] = 4.05 × 10⁻¹⁶ mol/L — essentially zero. This confirms why the atmosphere contains no measurable NO at room temperature. At high temperatures (lightning, engines), enough thermal energy forces this forward reaction, producing NOₓ smog.

Band 5

Example 2 (Intermediate) — Quadratic Required

Problem: 0.500 mol of PCl₅(g) is placed in a 1.00 L flask at 250°C. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Keq = 0.041. (a) Check the simplifying assumption. (b) Set up and solve. (c) Verify.

(a) Check: [PCl₅]₀ = 0.500 mol/L. Keq/[PCl₅] = 0.041/0.500 = 8.2% > 5% → simplifying assumption INVALID → must use quadratic

(b) ICE table:

	PCl₅	PCl₃	Cl₂
Initial	0.500	0	0
Change	−x	+x	+x
Equilibrium	0.500−x	x	x

Keq = x²/(0.500−x) = 0.041 → x² = 0.041(0.500−x) = 0.0205 − 0.041x → x² + 0.041x − 0.0205 = 0
x = (−0.041 ± √(0.001681 + 0.0820)) / 2 = (−0.041 ± √0.08368) / 2 = (−0.041 ± 0.2893) / 2
Positive root: x = (−0.041 + 0.2893)/2 = 0.2483/2 = 0.124 mol/L
Equilibrium: [PCl₅] = 0.376; [PCl₃] = [Cl₂] = 0.124 mol/L

(c) Verify: Keq = (0.124)(0.124)/0.376 = 0.01538/0.376 = 0.0409 ≈ 0.041 ✓

Band 6 — 7 marks

Example 3 (Hard) — Non-Zero Initial Products with Q

Problem: A flask at 430°C contains: [H₂] = 0.100, [I₂] = 0.100, [HI] = 0.100 mol/L. H₂(g) + I₂(g) ⇌ 2HI(g), Keq = 54.3. (a) Calculate Q and determine the direction of shift. (b) Set up and solve the ICE table. (c) Verify.

(a) Q: Q = [HI]²/([H₂][I₂]) = (0.100)²/((0.100)(0.100)) = 0.010/0.010 = 1.00. Q = 1.00 < Keq = 54.3 → system shifts RIGHT → H₂ and I₂ decrease; HI increases

(b) ICE (shift right):

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

Keq = (0.100+2x)²/(0.100−x)² = 54.3 → take square root → (0.100+2x)/(0.100−x) = 7.369
0.100 + 2x = 7.369(0.100−x) = 0.7369 − 7.369x → 9.369x = 0.6369 → x = 0.0680 mol/L
Equilibrium: [H₂] = [I₂] = 0.100 − 0.0680 = 0.0320 mol/L; [HI] = 0.100 + 2(0.0680) = 0.2360 mol/L

(c) Verify: Keq = (0.2360)²/((0.0320)(0.0320)) = 0.05570/0.001024 = 54.4 ≈ 54.3 ✓

Checkpoint Questions

1 mark

Q1: For the equilibrium A(g) ⇌ 2B(g), a student writes the Change row as A: −x; B: +x. What is wrong and what is the correct Change row?

A The sign on A is wrong — it should be +x; correct: A: +x; B: +x

B The change for B should use the stoichiometric ratio — correct Change row: A: −x; B: +2x

C The change for A should be −2x to match the coefficient of B; correct: A: −2x; B: +x

D The Change row is correct — both species change by x

Q1: For the equilibrium A(g) ⇌ 2B(g), a student writes the Change row as A: −x; B: +x. What is wrong and Identify the correct Change row?

AThe sign on A is wrong — it should be +x; correct: A: +x; B: +x

BThe change for B should use the stoichiometric ratio — correct Change row: A: −x; B: +2x

CThe change for A should be −2x to match the coefficient of B; correct: A: −2x; B: +x

DThe Change row is correct — both species change by x

1 mark

Q2: An ICE table problem gives Keq = 4.5 × 10⁻³ and initial concentration [A] = 0.200 mol/L. Should the simplifying assumption be applied?

A Yes — Keq is less than 1, so x is always negligible

B No — the simplifying assumption is never valid for equilibrium problems

C Yes — Keq/[A] = 4.5 × 10⁻³/0.200 = 2.25% < 5%, so the assumption is valid

D It cannot be determined without knowing the stoichiometry

1 mark

Q3: After solving an ICE table problem, a student substitutes equilibrium concentrations back into Keq and gets 0.047 instead of the given Keq = 0.041. What is the most likely cause?

A The equilibrium concentrations are correct — small discrepancies always occur in Keq calculations

B The student used incorrect stoichiometric ratios in the Change row, leading to wrong equilibrium concentrations

C The value of Keq changes slightly when concentrations are substituted back — this is normal

D The student should have used a different Keq expression for verification

Q3: After solving an ICE table problem, a student substitutes equilibrium concentrations back into Keq and gets 0.047 instead of the given Keq = 0.041. Identify the most likely cause?

AThe equilibrium concentrations are correct — small discrepancies always occur in Keq calculations

BThe student used incorrect stoichiometric ratios in the Change row, leading to wrong equilibrium concentrations

CThe value of Keq changes slightly when concentrations are substituted back — this is normal

DThe student should have used a different Keq expression for verification

Extended Practice

5 marks

Q4: The equilibrium 2HI(g) ⇌ H₂(g) + I₂(g) has Keq = 0.0184 at 430°C. If 1.00 mol of HI is placed in a 1.00 L flask, (a) check whether the simplifying assumption is valid, (b) set up the ICE table, (c) solve for equilibrium concentrations, and (d) verify your answer.

5 marks

Q5: A flask at 250°C contains: [PCl₃] = 0.200 mol/L, [Cl₂] = 0.200 mol/L, [PCl₅] = 0.050 mol/L. The equilibrium PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) has Keq = 65.0. (a) Calculate Q and determine the direction of shift. (b) Set up the ICE table and write the Keq expression in terms of x. (You do not need to solve for x — just set up the expression.)