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Chemistry Y12 · Module 5 · Lesson 9

IQ3 — Keq, ICE Tables & Reaction Quotient

Writing Keq Expressions — The Rules

The atmosphere is 78% nitrogen and 21% oxygen — yet they don't spontaneously combust to form nitrogen monoxide. The reason is a single number: Keq = 1 × 10⁻³⁰ for N₂ + O₂ ⇌ 2NO at 25°C. Understanding what that number means is the whole point of this lesson.

Understand Apply Interpret

Misconceptions to Fix

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Wrong: A large Keq means the reaction happens quickly.

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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

Write correct Keq expressions for homogeneous and heterogeneous equilibria, excluding pure solids and pure liquids

Explain why pure solids and pure liquids are excluded from Keq expressions

Interpret the magnitude of Keq to predict whether a mixture is predominantly products, reactants, or a mixture of both

Apply the reciprocal and multiplier relationships to convert Keq values when equations are reversed or scaled

Justify why only temperature changes Keq, while concentration, pressure, and catalysts do not

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🤔 Think First — Predicting from Keq Values

Three reactions are written below with their Keq values at 25°C. Before reading anything about Keq expressions or magnitude — predict for each reaction whether you would expect to find mostly reactants, mostly products, or significant amounts of both in an equilibrium mixture.

(1) N₂(g) + O₂(g) ⇌ 2NO(g), Keq = 1 × 10⁻³⁰
(2) H₂(g) + I₂(g) ⇌ 2HI(g), Keq = 54
(3) 2NO(g) + O₂(g) ⇌ 2NO₂(g), Keq = 1 × 10¹²

Write your three predictions and the reasoning behind each before reading on. You will revisit this after Card 3.

Your predictions:

Keq Expression Rules

General form: aA + bB ⇌ cC + dD → Keq = [C]^c[D]^d / ([A]^a[B]^b)

Include: gases (g), aqueous species (aq)

Exclude: pure solids (s), pure liquids (l) including liquid water as solvent

Magnitude guide: Keq >> 1 → products favoured | Keq ≈ 1 → significant both | Keq << 1 → reactants favoured

Reciprocal: reverse equation → Keq(new) = 1/Keq

Multiplier: multiply all coefficients by n → Keq(new) = Keqⁿ

⚠ Keq depends ONLY on temperature

Understand

1. The Keq Expression — Rules and Derivation

The equilibrium constant expression is not an arbitrary formula — it is derived from the thermodynamic condition that at equilibrium, the free energy of the system is at its minimum, and the ratio of products to reactants at that minimum has a fixed value at a given temperature.

For the general reversible reaction aA + bB ⇌ cC + dD:

Keq = [C]^c × [D]^d / ([A]^a × [B]^b)

The rules for writing Keq expressions:

Products appear in the numerator; reactants in the denominator
Each concentration is raised to the power of its stoichiometric coefficient in the balanced equation as written
Pure solids are excluded (activity = 1 by convention)
Pure liquids are excluded, including water when it acts as the solvent
Aqueous species (aq) are included
Gases (g) are included

Example — Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Keq = [NH₃]² / ([N₂][H₂]³)

Powers match stoichiometric coefficients: NH₃ coefficient = 2 → squared; N₂ coefficient = 1; H₂ coefficient = 3 → cubed.

Species Type	Included in Keq?	Reason
Gas (g)	Yes	Concentration affects equilibrium
Aqueous (aq)	Yes	Concentration affects equilibrium
Pure solid (s)	No	Activity = 1; fixed composition
Pure liquid (l) / H₂O(l) as solvent	No	Activity = 1; fixed composition
Water as reactant/product in gas phase	Yes	H₂O(g) — concentration varies

Must do: The stoichiometric coefficients in the Keq expression must match the balanced equation AS WRITTEN. If the Haber process is written as ½N₂ + 3/2H₂ ⇌ NH₃, then Keq = [NH₃]/([N₂]^½[H₂]^(3/2)) — a different numerical value. Always check which version of the equation is given.

Most common error in IQ3: Including pure solids in the Keq expression. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), the correct Keq = [CO₂]. Students write Keq = [CaO][CO₂]/[CaCO₃] — this is wrong. Never include a solid in a Keq expression.

Keq expression writing flowchart — the two questions: (1) write products/reactants^coeff, (2) exclude solids and pure liquid solvents

Apply

2. Writing Keq Expressions — Homogeneous and Heterogeneous Examples

The rule is always the same — products over reactants, stoichiometric powers, exclude solids and pure liquids — but applying it to heterogeneous equilibria (mixed phases) requires careful identification of which species to include.

Homogeneous equilibria (same phase):

Example 1: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) — all gases

Keq = [SO₃]² / ([SO₂]²[O₂])

Example 2: CH₃COOH(aq) + C₂H₅OH(aq) ⇌ CH₃COOC₂H₅(aq) + H₂O(l)

Water is the solvent (pure liquid) → excluded.

Keq = [CH₃COOC₂H₅] / ([CH₃COOH][C₂H₅OH])

Heterogeneous equilibria (mixed phases):

Example 3: CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Both CaCO₃ and CaO are pure solids → excluded. Only CO₂(g) remains.

Keq = [CO₂]

Example 4: Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g)

Fe₃O₄ and Fe are pure solids → excluded. H₂ and H₂O are gases → included. Note: H₂O is a gas here, NOT the solvent — include it.

Keq = [H₂O]⁴ / [H₂]⁴

Water requires special attention: When H₂O(l) is the solvent in an aqueous reaction, exclude it. When H₂O(g) is a product of a gas-phase reaction (e.g. combustion, steam reactions), include it. The state symbol tells you which case applies.

Common error: Including H₂O in Keq expressions for reactions in aqueous solution. For CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq), the Ka expression is [CH₃COO⁻][H⁺]/[CH₃COOH] — water is excluded (it is the solvent).

Interpret

3. Interpreting Keq Magnitude

The numerical value of Keq encodes everything about where an equilibrium lies — and learning to read it at a glance is one of the most powerful interpretive skills in Year 12 Chemistry.

Keq Value	Equilibrium Position	What the Mixture Looks Like
> 10³	Strongly products favoured	Almost entirely products; reactants nearly exhausted
10⁻¹ to 10³ (≈ 1)	Intermediate	Significant amounts of both reactants and products
< 10⁻³	Strongly reactants favoured	Almost entirely reactants; very little product

Examples revisited from Think First:

N₂ + O₂ ⇌ 2NO, Keq = 1 × 10⁻³⁰ — reactants strongly favoured → almost no NO at room temperature → the atmosphere is stable
H₂ + I₂ ⇌ 2HI, Keq = 54 — intermediate → significant HI AND measurable H₂ and I₂ coexist
2NO + O₂ ⇌ 2NO₂, Keq = 1 × 10¹² — products strongly favoured → almost entirely NO₂; virtually no NO and O₂ remain

HSC answer format: "Keq = 3.5 × 10⁻⁸ is much less than 1, so the equilibrium strongly favours reactants — very little product is present at equilibrium." Always interpret the magnitude qualitatively, not just numerically.

Keq magnitude scale — from 10⁻³⁰ (essentially no product) to 10³⁰ (essentially complete reaction)

Insight: The fact that Keq for N₂ + O₂ ⇌ 2NO is 10⁻³⁰ at room temperature is why life exists. If this equilibrium were even Keq = 10⁻¹⁰, significant NO would form in the atmosphere — NO is a toxic pollutant that would have prevented evolution of aerobic life. At high temperatures in lightning and engine combustion, enough thermal energy forces the reaction forward — this is why smog contains NOₓ.

Apply

4. Keq and the Balanced Equation — Reciprocal and Multiplier Relationships

Keq is not an absolute property of a reaction — it is a property of the equation as written. Change the equation and you change the Keq value, even though the underlying chemistry is identical.

Relationship 1 — Reversed equation:

If A + B ⇌ C + D has Keq = K, then C + D ⇌ A + B has Keq = 1/K.

Example

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Keq = 0.013 at 500°C

Reverse: 2NH₃(g) ⇌ N₂(g) + 3H₂(g), Keq = 1/0.013 = 77

The reverse equilibrium strongly favours decomposition of NH₃ back to N₂ and H₂ at 500°C.

Relationship 2 — Multiplied equation:

If all stoichiometric coefficients are multiplied by n, the new Keq = (original Keq)ⁿ.

Example

If N₂ + 3H₂ ⇌ 2NH₃ has Keq = K, then ½N₂ + 3/2H₂ ⇌ NH₃ has Keq = K^(½) = √K

Dividing all coefficients by 2 takes the square root of Keq.

Systematic check before any Keq calculation: (1) Is the equation reversed? → take the reciprocal. (2) Are the coefficients multiplied or divided by a factor? → raise Keq to that power. Do both checks before starting any calculation.

Common error: Using the wrong Keq value because the equation is written differently from the one the Keq was given for. If Keq = 54 is given for H₂ + I₂ ⇌ 2HI, but the question asks about 2HI ⇌ H₂ + I₂, the Keq for the reversed equation is 1/54 = 0.019. Using 54 instead gives the completely wrong answer.

Justify

5. Temperature Dependence of Keq — The Only Variable That Matters

Keq is a thermodynamic constant — it is fixed by the temperature alone, and changing anything else changes the equilibrium position but not the value of Keq.

This is because Keq is related to the Gibbs free energy change: ΔG° = −RT ln Keq. Since ΔG° depends on temperature (through the TΔS term), Keq also depends on temperature.

Exothermic forward reaction: increasing T → ΔG° less negative → Keq decreases; decreasing T → ΔG° more negative → Keq increases
Endothermic forward reaction: opposite — increasing T → ΔG° more negative → Keq increases
Concentration changes: shift equilibrium position but Keq value unchanged
Pressure changes: shift equilibrium position for gas reactions with unequal moles — Keq unchanged
Catalyst: no shift in position, no change in Keq

Common HSC question: "A student claims that adding more reactant to an equilibrium mixture increases Keq. Evaluate this claim." Answer: incorrect. Adding reactant shifts equilibrium position to the right (more products form) but the ratio of equilibrium concentrations at the new equilibrium still satisfies the same Keq expression — Keq is unchanged. Only temperature changes Keq.

Common error: "Increasing pressure increases Keq because more products are formed." Wrong — more products are formed at the new equilibrium (position shifts right) but the Keq expression value at the new equilibrium is the same as before. Position shifted; Keq did not.

Interactive — Keq Expression Builder

Revisit Your Thinking

Keq is written as products over reactants, each raised to the power of their stoichiometric coefficient. Pure solids and pure liquids are omitted because their activities are constant. A large Keq (> 10³) means products are strongly favoured; a small Keq (< 10⁻³) means reactants dominate. Keq is temperature-dependent and has no units in the HSC treatment.

Key Terms — scan these before reading

ExcludeFor CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq), the Ka expression is [CH₃COO⁻][H⁺]/[CH₃COOH] — water is excluded (it is the solvent).

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.

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 — Writing Keq Expressions for Various Equilibria

Problem: Write the Keq expression for each equilibrium and identify any excluded species. (a) 2NO(g) + O₂(g) ⇌ 2NO₂(g). (b) AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). (c) CO₂(g) + C(s) ⇌ 2CO(g). (d) CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq).

(a): All species are gases — all included. Products: NO₂ (coefficient 2) → [NO₂]². Reactants: NO (coefficient 2) → [NO]²; O₂ (coefficient 1) → [O₂]. Keq = [NO₂]² / ([NO]²[O₂])

(b): AgCl is a pure solid → excluded. Ag⁺ and Cl⁻ are aqueous → included. Keq = [Ag⁺][Cl⁻] (This is the Ksp expression for AgCl — studied in L15–L16.)

(c): C is a pure solid → excluded. CO₂ and CO are gases → included. Products: CO (coefficient 2) → [CO]². Reactants: CO₂ → [CO₂]. Keq = [CO]² / [CO₂]

(d): CH₃COOH, CH₃COO⁻, H⁺ are aqueous → included. Water is the solvent → excluded. Keq = [CH₃COO⁻][H⁺] / [CH₃COOH] (This is the Ka expression — studied in L14.)

Band 5–6

Example 2 — Interpreting Keq Magnitude and Reciprocal Relationships

Problem: The equilibrium constant for N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is Keq = 977 at 300°C and Keq = 0.013 at 500°C. (a) Interpret both Keq values in terms of equilibrium position. (b) Write the Keq expression for the reverse reaction 2NH₃(g) ⇌ N₂(g) + 3H₂(g) and calculate its value at 500°C. (c) A student claims Keq = 0.013 means less than 1.3% of N₂ and H₂ has converted to NH₃ at equilibrium. Evaluate this claim.

(a): Keq = 977 at 300°C — much greater than 1 → products strongly favoured → at equilibrium, the mixture is mostly NH₃ with relatively little N₂ and H₂ remaining. Higher yield at lower temperature is consistent with exothermic forward reaction. Keq = 0.013 at 500°C — much less than 1 → reactants strongly favoured → mostly N₂ and H₂ at equilibrium; little NH₃. This is why the industrial process at 450°C gives only ~15–25% conversion per pass.

(b): Reversing the equation → Keq(reverse) = 1/Keq = 1/0.013 = 76.9 at 500°C. Keq(reverse) = [N₂][H₂]³ / [NH₃]² = 76.9 This large value confirms that at 500°C, NH₃ decomposition back to N₂ and H₂ is strongly favoured.

(c): The student's claim is incorrect. Keq = 0.013 tells you the ratio [NH₃]²/([N₂][H₂]³) = 0.013 at equilibrium — it does not directly give percentage conversion without knowing initial concentrations and pressure. At 200 atm, the actual conversion is approximately 15–25% per pass. The percentage conversion depends on BOTH Keq AND the initial conditions. Keq is not a direct percentage.

Answer: (a) Keq = 977: products strongly favoured, mostly NH₃; Keq = 0.013: reactants strongly favoured, mostly N₂ and H₂. (b) Keq(reverse) = 76.9; expression [N₂][H₂]³/[NH₃]². (c) Incorrect — Keq is not a direct percentage; actual conversion depends on initial conditions and pressure.

Checkpoint Questions

1 mark

Q1: Which is the correct Keq expression for: 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)?

A Keq = [Fe₃O₄][H₂]⁴ / ([Fe]³[H₂O]⁴)

B Keq = [H₂]⁴ / [H₂O]⁴

C Keq = [Fe₃O₄][H₂]⁴ / [H₂O]⁴

D Keq = [H₂O]⁴ / [H₂]⁴

1 mark

Q2: The Keq for 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) is 1.8 × 10⁻⁶ at 600°C. What is the Keq for SO₃(g) ⇌ SO₂(g) + ½O₂(g) at the same temperature?

A 3.6 × 10⁻⁶

B 1.8 × 10⁻⁶

C 1.34 × 10⁻³

D 3.24 × 10⁻¹²

Q2: The Keq for 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) is 1.8 × 10⁻⁶ at 600°C. Identify the Keq for SO₃(g) ⇌ SO₂(g) + ½O₂(g) at the same temperature?

A3.6 × 10⁻⁶

B1.8 × 10⁻⁶

C1.34 × 10⁻³

D3.24 × 10⁻¹²

1 mark

Q3: A student sets up the equilibrium H₂(g) + Cl₂(g) ⇌ 2HCl(g) in a sealed flask, then adds more HCl gas. Which correctly describes the effect on Keq and equilibrium position?

A Keq increases and equilibrium shifts left

B Keq decreases and equilibrium shifts left

C Keq is unchanged and equilibrium shifts left

D Keq increases and equilibrium shifts right

Short Answer Practice

4 marks

Q4: Write the Keq expression for each of the following equilibria and explain any exclusions: (a) C(s) + CO₂(g) ⇌ 2CO(g); (b) Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g); (c) Pb²⁺(aq) + 2Cl⁻(aq) ⇌ PbCl₂(s).

3 marks

Q5: The equilibrium constant for 2NO(g) + O₂(g) ⇌ 2NO₂(g) is Keq = 1 × 10¹² at 25°C. (a) Interpret this value — what does it tell you about the equilibrium mixture? (b) Calculate Keq for the reverse reaction NO₂(g) ⇌ NO(g) + ½O₂(g) at 25°C.

3 marks

Q6: A student adds a platinum catalyst to a sealed flask containing SO₂(g), O₂(g), and SO₃(g) at equilibrium. The student claims that since conditions changed, Keq must have changed. Evaluate this claim and write the correct statement about what effect the catalyst has on Keq and equilibrium position.

Q4 Model Answers:

(a) C(s) + CO₂(g) ⇌ 2CO(g): C is a pure solid → excluded. CO₂ and CO are gases → included. Keq = [CO]² / [CO₂]

(b) Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g): Fe₃O₄ and Fe are pure solids → excluded. H₂ and H₂O are gases → included. Keq = [H₂O]⁴ / [H₂]⁴

(c) Pb²⁺(aq) + 2Cl⁻(aq) ⇌ PbCl₂(s): PbCl₂ is a pure solid → excluded. Pb²⁺ and Cl⁻ are aqueous → included. Keq = 1 / ([Pb²⁺][Cl⁻]²). The solid is a product, so it goes in the numerator — but since it is excluded (activity = 1), the numerator = 1. This is the inverse of the Ksp expression: Ksp = [Pb²⁺][Cl⁻]².

Q5 Model Answer:

(a) Keq = 1 × 10¹² is much greater than 1 → products are strongly favoured → at equilibrium, the mixture is almost entirely NO₂; virtually no NO and O₂ remain. The reaction essentially goes to completion in the forward direction at 25°C.

(b) The given reaction has coefficients doubled (2NO + O₂ ⇌ 2NO₂). The reverse of this, written with halved coefficients (NO₂ ⇌ NO + ½O₂), requires two operations: reverse (take reciprocal) AND halve coefficients (take square root). Keq = (1/(1 × 10¹²))^(½) = (10⁻¹²)^(½) = 10⁻⁶. So Keq for NO₂ ⇌ NO + ½O₂ is 1 × 10⁻⁶ at 25°C — reactants (NO₂) strongly favoured, confirming NO₂ does not spontaneously decompose at room temperature.

Q6 Model Answer:

The student's claim is incorrect. Adding a catalyst does NOT change Keq. Keq is a thermodynamic constant that depends only on temperature — it is unchanged by catalysts, concentration changes, or pressure changes. The catalyst lowers the activation energy equally for both the forward and reverse reactions. Both reaction rates increase by the same factor. Since the system was already at equilibrium (forward rate = reverse rate), both rates increase equally and they remain equal. No shift in equilibrium position occurs. The concentrations of SO₂, O₂, and SO₃ are unchanged. Keq is unchanged. The only change is that the system can re-establish equilibrium more quickly if disturbed, but the equilibrium position and Keq are identical with or without the catalyst.