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Chemistry Y12 · Module 5 · Lesson 11

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

★ Consolidation — Deepening L10

Consolidation — ICE Table Mastery

In the 2022 NSW HSC Chemistry exam, Question 28 (7 marks) required a full ICE table for 2HI(g) ⇌ H₂(g) + I₂(g) at 430°C. Only 12% of students earned full marks — the three most common errors were: forgetting stoichiometric coefficients in the Change row, incorrect simplifying assumption, and not verifying the answer by substituting back into the Keq expression.

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

Today's hook: In the 2022 NSW HSC, only 12% of students earned full marks on the 7-mark ICE table question — the errors were always in the Change row coefficients or the simplifying assumption. Today you eliminate those errors permanently.

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 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?

Learning Intentions

Know

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

Understand

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

Can Do

Solve ICE table problems requiring the quadratic formula, including selecting the physically meaningful root
Verify every ICE table answer by substituting equilibrium concentrations back into the Keq expression

Key Terms

ICE table

A systematic tool for calculating equilibrium concentrations from initial conditions and Keq.

Quadratic equation

Arises when the small-x approximation is not valid; solved using the quadratic formula.

5% rule

If x/[initial] × 100 < 5%, the small-x assumption is valid and simplification is justified.

Keq vs Q comparison

If Q < Keq, reaction proceeds forward; if Q > Keq, reverse; if Q = Keq, at equilibrium.

Multiple equilibria

When a substance participates in two simultaneous equilibria, both must be satisfied simultaneously.

Concentration unit

mol L⁻¹ (molarity); all ICE table values must be in these units for Keq calculations.

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

Content

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.

The three fatal ICE table errors: (1) wrong Change-row ratios — always use stoichiometric coefficients (N₂ + 3H₂ → N₂: −x, H₂: −3x, NH₃: +2x); (2) reporting x instead of E-row values — if Change = +2x, then equilibrium [HI] = 2x; (3) skipping verification — always substitute E-row values back into Keq and confirm within ±2%.

Copy the three error/fix pairs into your notes before the check below.

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), what is the correct Change row if the shift is to the right by x?

Analogy 1

2. The Bank Account — Understanding ICE Table Logic

We just saw the three most common ICE table errors — wrong stoichiometric ratios, reporting x not the E-row, and skipping verification. That raises a question: why does the ICE table work the way it does — what's the underlying logic that makes those errors so easy to commit? This card answers it → by using the bank account analogy to make ICE logic concrete and memorable.

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 logic = bank account: I row = opening balance; C row = transactions at fixed stoichiometric ratios (not independent); E row = closing balance = I + C. Critical check: all E-row values must be positive — a negative equilibrium concentration means an error in sign or stoichiometric ratio.

Write the bank-account analogy and the "all E-row positive" check into your notes before continuing.

If any equilibrium concentration in the E row is negative, the ICE table calculation must have an error.

Diagram

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

Analogy 2

3. The Funnel — Simplifying Assumption Logic

We just saw the ICE table structure as a bank account — opening balance, transactions, closing balance. That raises a question: ICE often involves solving a quadratic; is there a shortcut, and when is it valid to use? This card answers it → by explaining the simplifying assumption (x ≈ 0) using the funnel analogy and the 5% validity threshold.

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.

Simplifying assumption rule: if Keq/[initial] × 100 < 5%, then x is negligible and (initial − x) ≈ initial. Pre-check before solving; post-check by calculating x/[initial] × 100 after solving — if ≥ 5%, discard the answer and solve the full quadratic ax² + bx + c = 0 instead.

Add the 5% pre-check and post-check rules to your notes before the diagram and check below.

Keq = 0.041 and [initial] = 0.500 mol/L. Should the simplifying assumption be used?

Non-Zero Initial Products

4. ICE Table with Non-Zero Initial Product Concentrations

We just saw that the simplifying assumption (x negligible) saves you from a quadratic if Keq/[initial] < 5%. That raises a question: what if the problem gives you non-zero initial concentrations for BOTH reactants and products — which way does the ICE table shift? This card answers it → by showing you must calculate Q first to determine shift direction before writing any Change row.

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.

Non-zero initial products procedure: (1) calculate Q using initial concentrations; (2) compare — Q < Keq → shift right (reactants −, products +); Q > Keq → shift left (products −, reactants +); Q = Keq → no shift; (3) write Change row using x in the direction determined by step 2. Never assume direction — always use Q vs Keq.

Pause — record the Q-first procedure in your notes before the check below.

A mixture has Q = 0.025 and Keq = 0.50. In what direction does the system shift?

Cross-lesson links: The three errors identified in the 2022 NSW HSC exam — stoichiometric coefficients, simplifying assumption, and back-verification — are exactly the skills consolidated here for use in L12 (Q calculations) and L13 (temperature and Keq). The quadratic ICE solution in Card 5 is required for HSC extension problems. Verification by substitution introduced here prevents errors in Ka/Kb calculations in L14.

Quadratic Solutions

5. Quadratic Solutions for ICE Problems

We just saw that when all species have non-zero initial concentrations, you must calculate Q to determine shift direction before writing the Change row. That raises a question: when the 5% assumption fails and you can't simplify, how do you actually solve the quadratic from an ICE table systematically? This card answers it → by walking through the full six-step quadratic procedure with a worked example.

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.

Quadratic ICE procedure: (1) set up E row in terms of x; (2) substitute into Keq expression; (3) rearrange to ax² + bx + c = 0; (4) apply x = (−b ± √(b²−4ac))/2a; (5) choose the root giving all positive E-row values (the physically valid root); (6) verify by substituting back into Keq — must match within ±2%.

Write the six-step quadratic procedure into your notes before the check below.

When the quadratic formula gives two roots for x, you should always choose the larger root.

Activities

Worked Examples

Worked Example 1 — Simplifying Assumption Valid

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.

Worked Example 2 — 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 ✓

Worked Example 3 — 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 ✓

Check Your Understanding

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.)

Show Model Answers

Q4 Model Answer:

(a) [HI]₀ = 1.00 mol/L. Keq/[HI] = 0.0184/1.00 = 1.84% < 5% → simplifying assumption appears valid. However, after solving, verify x/[HI] post-check.

(b) ICE: HI: Initial = 1.00, Change = −2x, E = 1.00−2x. H₂: Initial = 0, Change = +x, E = x. I₂: Initial = 0, Change = +x, E = x.

(c) Keq = [H₂][I₂]/[HI]² = x²/(1.00−2x)² = 0.0184. Apply simplifying assumption: (1.00−2x) ≈ 1.00. x²/1.00² = 0.0184 → x² = 0.0184 → x = √0.0184 = 0.1356 mol/L. Post-check: 2x/[HI] = 2(0.1356)/1.00 = 27.1% > 5% → assumption INVALID! Solve exactly: x²/(1.00−2x)² = 0.0184 → x/(1.00−2x) = √0.0184 = 0.1356 → x = 0.1356(1.00−2x) = 0.1356 − 0.2713x → 1.2713x = 0.1356 → x = 0.1067 mol/L. Equilibrium: [HI] = 1.00 − 2(0.1067) = 0.787 mol/L; [H₂] = [I₂] = 0.1067 mol/L.

(d) Verify: (0.1067)²/(0.787)² = 0.01138/0.6194 = 0.01837 ≈ 0.0184 ✓

Q5 Model Answer:

(a) Keq expression for PCl₃ + Cl₂ ⇌ PCl₅: Keq = [PCl₅]/([PCl₃][Cl₂]). Q = [PCl₅]/([PCl₃][Cl₂]) = 0.050/((0.200)(0.200)) = 0.050/0.040 = 1.25. Q = 1.25 < Keq = 65.0 → system shifts RIGHT → PCl₃ and Cl₂ decrease; PCl₅ increases.

(b) ICE table (shift right): PCl₃: 0.200 − x; Cl₂: 0.200 − x; PCl₅: 0.050 + x. Keq expression: Keq = (0.050 + x)/((0.200 − x)(0.200 − x)) = 65.0.

Interactive Tool — Keq Calculations Open fullscreen ↗

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

Fill the blanks+4 XP

For H&sub2;(g) + I&sub2;(g) ⇌ 2HI(g), initial [H&sub2;] = [I&sub2;] = 0.100 mol L¹, [HI] = 0. At equilibrium [HI] = 0.160 mol L¹. Complete the ICE table.

Change in [HI] = +0.160, so change in [H&sub2;] = mol L¹
[H&sub2;]_eq = 0.100 − 0.080 = mol L¹
K_eq = [HI]² ÷ ([H&sub2;][I&sub2;]) = (0.160)² ÷ (0.020 × 0.020) =

Complete the Learn phase to unlock practice questions.

Multiple Choice

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?

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?

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?

Q4: For A(g) ⇌ B(g) + C(g) with Keq = 0.25, [A]₀ = 0.600 mol/L, what is the correct approach?

Q5: A system has Q = 3.13 × 10⁻³ and Keq = 0.50. Initial [NO₂] = 0.400, [NO] = 0.100, [O₂] = 0.050 mol/L for 2NO₂ ⇌ 2NO + O₂. Which Change row is correct?

Short Answer

4 marks

SAQ 1: Explain the three most common ICE table errors and how to avoid each one. Include a specific example for each.

3 marks

SAQ 2: A student claims that when Keq < 1, the simplifying assumption is always valid. Explain why this claim is incorrect and describe the correct threshold to use.

Show Answers

SAQ 1:

Error 1 — Wrong stoichiometric ratios: students write ±x for all species. Fix: use stoichiometric coefficients from the balanced equation. Example: for N₂ + 3H₂ ⇌ 2NH₃, Change row must be N₂: −x, H₂: −3x, NH₃: +2x, not −x, −x, +x.

Error 2 — Not converting x to equilibrium concentrations: students report x instead of the E row value. Fix: always read equilibrium concentrations from the E row. Example: for HI with Change +2x, [HI]eq = 2x, not x.

Error 3 — Not verifying: students stop after solving. Fix: always substitute E values into Keq and confirm ≈ given Keq within ±2%. A 15% discrepancy signals an error.

SAQ 2:

The claim is incorrect. The validity of the simplifying assumption depends on the ratio Keq/[initial], not on whether Keq < 1. For example, if Keq = 0.4 and [initial] = 0.100, then Keq/[initial] = 400% >> 5% — the assumption is completely invalid despite Keq < 1. The correct threshold is: if Keq/[initial] × 100 < 5%, the assumption may be used; otherwise the full quadratic must be solved.

Extended Response

A 1.00 L flask at 250°C contains 0.600 mol of PCl₅(g). The equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) has Keq = 0.041. Using correct ICE table procedure: (a) check the 5% threshold; (b) solve for equilibrium concentrations using the quadratic formula; (c) verify your answer by substituting back into Keq; (d) identify which of the three common errors you were most careful to avoid and why.

How did your thinking change?

You were asked to identify the error in a Change row where HI was written as +x instead of +2x, and to find the correct [HI]. Recall that in the 2022 NSW HSC exam, this exact error (wrong stoichiometric coefficient in the Change row) was the single most common reason students lost marks on the 7-mark ICE question. Look back at what you wrote. Was your diagnosis correct? What do you now understand about why the Change row error is the most common ICE mistake?

I have completed this lesson

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