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Chemistry Y12 · Module 5 · Lesson 9 of 18

IQ3, Keq, ICE Tables & Reaction Quotient

Writing Keq Expressions, The Rules

In 1864, Norwegian chemists Cato Guldberg and Peter Waage at the University of Christiania published the first equilibrium expression, relating the ratio of product to reactant concentrations to a constant they called K. They measured K for the N₂ + O₂ ⇌ 2NO equilibrium and found it was 1 × 10⁻³⁰ at 25°C, explaining why 78% N₂ and 21% O₂ in our atmosphere never spontaneously react.

Understand Apply Interpret

Today's hook: In 1864, Guldberg and Waage at the University of Christiania discovered that a single ratio of concentrations, raised to stoichiometric powers, is always constant at equilibrium. That ratio is Keq, and learning to write it correctly is the gateway to every calculation in this module.

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Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

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

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.

Learning Intentions

Know

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

Understand

Interpret the magnitude of Keq to predict whether a mixture is predominantly products, reactants, or a mixture of both
Justify why only temperature changes Keq, while concentration, pressure, and catalysts do not

Can Do

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

Key Terms

Equilibrium constant (Keq)

The ratio of product to reactant concentrations at equilibrium, each raised to stoichiometric power.

Homogeneous equilibrium

Equilibrium where all species are in the same phase.

Heterogeneous equilibrium

Equilibrium where reactants/products are in different phases; pure solids and liquids omitted from Keq.

Stoichiometric coefficient

The number in front of a species in a balanced equation; used as the exponent in the Keq expression.

Keq expression

Written as Keq = [products]^p / [reactants]^r using molar concentrations in mol L⁻¹.

Inverse equilibrium

If the reaction is reversed, the new Keq = 1/original Keq; if multiplied by n, Keq is raised to the nth power.

Core Concepts

The Keq expression rules

Understand

1. The Keq Expression, Rules and Derivation

The equilibrium constant expression is not an arbitrary formula. Once a reversible reaction in a closed system reaches dynamic equilibrium, the forward and reverse rates are equal, the concentrations stop changing, and one particular ratio of product to reactant concentrations settles on a fixed value at a given temperature. That fixed value is Keq. From Module 4 you already know the same settling point described thermodynamically, as the state where the Gibbs free energy of the mixture (ΔG = ΔH − TΔS) is at its minimum. That is assumed background here; in Module 5 you work directly with the concentration ratio.

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 = [products]^coeff / [reactants]^coeff; powers = stoichiometric coefficients; exclude pure solids (s) and pure liquids/water-as-solvent (l); include gases (g), aqueous species (aq), and H₂O(g) in gas-phase reactions; stoichiometry must match the equation as written.

Pause, copy the highlighted Keq expression rules into your book before moving on.

Migrated Keq expression decision flowchart to list-compare

Which is the correct Keq expression for CaCO₃(s) ⇌ CaO(s) + CO₂(g)?

Homogeneous and heterogeneous examples

Apply

2. Writing Keq Expressions, Homogeneous and Heterogeneous Examples

We just saw the six rules for writing any Keq expression. That raises a question: how do those rules apply to the specific types of reactions we encounter, homogeneous gas-phase, aqueous, and mixed-phase heterogeneous equilibria? This card answers it → four worked examples including the tricky H₂O cases.

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). Forward reference: Ka is simply Keq applied to an acid dissociation. Acid-base equilibrium belongs to Module 6 (previewed in Lesson 14) and is not assessed in this lesson. It appears here only to show the omit-the-solvent rule in action.

Heterogeneous equilibria: check each species' state symbol; exclude (s) and pure liquid solvent; include (g) and (aq); water is the critical edge case, H₂O(l) as solvent is excluded, but H₂O(g) in a gas-phase reaction is included (e.g. Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g) → Keq = [H₂O]⁴/[H₂]⁴).

Add the highlighted heterogeneous Keq rules to your notes before continuing.

True or False: For the equilibrium Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g), the H₂O should be included in the Keq expression.

Interpreting Keq magnitude

Interpret

3. Interpreting Keq Magnitude

We just saw how to write Keq expressions for both homogeneous and heterogeneous equilibria. That raises a question: once we have a Keq value, what does the number itself tell us about the mixture at equilibrium? This card answers it → the three-zone magnitude scale from reactant-favoured to product-favoured.

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
10⁻³ to 10³ (around 1)	Intermediate, appreciable amounts of both	Significant amounts of both reactants and products are present; neither side is exhausted
< 10⁻³	Strongly reactants favoured	Almost entirely reactants; very little product

The three bands are continuous and cover every possible value, there is no unclassified gap. Any Keq between 10⁻³ and 10³ sits in the intermediate band, so a value such as 0.02 or 54 still means appreciable amounts of both reactants and products. The further a value moves past 10³ (or below 10⁻³), the more one-sided the mixture becomes. The boundaries are conventional guides for interpretation, not sharp physical cut-offs, so always phrase your answer comparatively ("much greater than 1", "much less than 1", "close to 1").

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.

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.

Keq magnitude: >10³ → products strongly favoured (mixture almost entirely products); 10⁻³ to 10³ → intermediate, appreciable amounts of both reactants and products; <10⁻³ → reactants strongly favoured (very little product). The three bands are continuous with no gap. Always state the qualitative interpretation in HSC answers alongside the number.

Pause, copy the highlighted Keq magnitude scale into your book before the check below.

A reaction has Keq = 8.5 × 10⁻¹⁰. Which correctly describes the equilibrium mixture?

Reciprocal and multiplier relationships

Apply

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

We just saw how to interpret the magnitude of Keq. That raises a question: if the same reaction can be written in different ways (reversed, scaled), how does Keq change, and why must we always check this before a calculation? This card answers it → the reciprocal and multiplier relationships between equation form and Keq value.

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.

Equation manipulations change Keq: reversed equation → Keq(new) = 1/Keq(original); all coefficients multiplied by n → Keq(new) = Keq(original)ⁿ; always perform two checks before using a Keq value, (1) is the equation reversed? (2) are the coefficients scaled?

Add the highlighted reciprocal and multiplier rules to your notes before the check below.

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

Cross-lesson links: Guldberg and Waage's 1864 equilibrium expression introduced here is the foundation for ICE table calculations in L10, the reaction quotient Q in L12, and Ka/Kb expressions in L14. The coefficient-as-exponent rule in Card 4 is the most common arithmetic error in L10 ICE tables. Temperature dependence of Keq introduced in Card 5 is quantified in L13.

Temperature and Keq

Justify

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

We just saw how to adjust Keq when the equation is reversed or scaled. That raises a question: what is the thermodynamic reason only temperature changes Keq, while concentration, pressure, and catalysts cannot? This card answers it → by treating heat as a reactant or product and applying Le Chatelier's principle, which is the reasoning the HSC expects.

Add a catalyst to a flask at equilibrium, the colour doesn't change. Add more reactant, the colour changes temporarily, then re-establishes. Heat the flask, and the colour shifts permanently to a new equilibrium. Only temperature changes the value of Keq itself; everything else only shifts the equilibrium position.

Keq is fixed by the temperature alone; changing anything else moves the equilibrium position but leaves the value of Keq untouched. The reason is easiest to see by writing heat as a reactant or a product and then applying Le Chatelier's principle. A temperature change is the only disturbance that changes the balance the ratio itself settles to, because it is the only one that changes the relative rates of the forward and reverse reactions by different amounts.

Exothermic forward reaction (heat is a product, A + B ⇌ C + heat): increasing T is like adding a product, so the system shifts left, giving less product and more reactant at the new equilibrium → Keq decreases. Decreasing T shifts right → Keq increases.
Endothermic forward reaction (heat is a reactant, A + B + heat ⇌ C): increasing T is like adding a reactant, so the system shifts right → Keq increases. Decreasing T → Keq decreases.
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

Extension, beyond the syllabus

If you go on to study thermodynamics, the same result falls out of the relationship ΔG° = −RT ln Keq: because ΔG° depends on temperature through the TΔS term, so does Keq. This equation is not part of the NSW Chemistry 2017 course and is never required in an answer. The Le Chatelier reasoning above is the expected justification. (The related Module 4 result ΔG = ΔH − TΔS is assumed knowledge.)

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.

Keq depends ONLY on temperature. Treat heat as a reactant or product and apply Le Chatelier: exothermic forward (ΔH < 0) → increasing T shifts left → Keq decreases; endothermic forward (ΔH > 0) → increasing T shifts right → Keq increases. Concentration, pressure, and catalyst changes shift the equilibrium position but leave Keq unchanged.

Pause, copy the highlighted temperature-Keq relationship into your book before the check below.

A student adds more H₂ gas to the equilibrium N₂ + 3H₂ ⇌ 2NH₃. After the system reaches its new equilibrium, what is the value of Keq compared to before?

✓

Worked examples and activities

Activities

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

Worked Example 1, Writing Keq Expressions

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.)

Worked Example 2, Analyse/Evaluate, 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.

Interactive Tool, Keq Calculations Open fullscreen ↗

Use the Keq Calculator on the heterogeneous equilibrium CaCO₃(s) ⇌ CaO(s) + CO₂(g). Which expression is the correct Keq, and what does its magnitude tell you?

Fill the blanks+4 XP

Write the K_eq expression for: 2SO&sub2;(g) + O&sub2;(g) ⇌ 2SO&sub3;(g)

K_eq = [SO&sub3;] ÷ ([SO&sub2;] × [O&sub2;])

Pure solids and liquids are from K_eq expressions.

A K_eq value much greater than 1 indicates the equilibrium position favours the .

✓

Practice questions

Multiple Choice

+5 XP

A fresh set drawn from this lesson's question bank, with feedback shown immediately. +5 XP per correct · +25 XP all correct

Short Answer

4 marks

Q6: 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

Q7: 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

Q8: 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.

✓

Model Answers

Show Model Answers

Q6 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.

Q7 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.

Q8 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.

✓

Retrieve and reflect

Check what actually stuck

Extended Response

A student is given the following information: For the reaction 2HI(g) ⇌ H₂(g) + I₂(g), Keq = 0.020 at 400°C. Using this information: (a) Write the Keq expression for 2HI(g) ⇌ H₂(g) + I₂(g). (b) Write the Keq expression and calculate Keq for the reverse reaction H₂(g) + I₂(g) ⇌ 2HI(g) at 400°C. (c) Interpret both Keq values qualitatively. (d) The student claims that adding more HI gas to the system at 400°C will increase Keq. Evaluate this claim with full explanation. (e) State what change would be needed to increase Keq for the reverse reaction H₂(g) + I₂(g) ⇌ 2HI(g), which is exothermic.

How did your thinking change?

Look back at your Think First predictions for the three reactions. Recall Guldberg and Waage's 1864 discovery at the University of Christiania, they defined K as the ratio of product to reactant concentrations raised to stoichiometric powers. Were your predictions about which reaction favours products correct? How does the magnitude of Keq relate to what you predicted about the equilibrium mixture? What does the N₂ + O₂ ⇌ 2NO example (Keq = 1 × 10⁻³⁰) tell us about our atmosphere?

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