Year 12 Chemistry Module 5 ⏱ ~35 min Lesson 4 of 18

Equilibrium in Context, Analogies and Misconception Deep-Dive

In 1888 Jacobus van 't Hoff told the chemical world that a system at equilibrium is not at rest, that the forward and reverse reactions are proceeding simultaneously at equal rates. More than a century later, the two errors his statement was written to kill are still the two most common mistakes in HSC equilibrium answers, so this lesson hunts them down one at a time.

Optional consolidation. This lesson introduces no new syllabus dot points, it consolidates material taught earlier in the module. It is valuable practice, and it is not required to complete the core route.

Today's hook, drop a crystal of radioactive ²⁴NaCl into a saturated salt solution that looks completely unchanging. Within minutes the radioactivity is in the liquid, even though not one extra gram of salt has dissolved. Nothing is moving, and yet everything is.

0/5QUESTS

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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.

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

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★ Consolidation Lesson: This lesson contains no new dot points. Its purpose is to deepen your understanding of IQ1 material (L01–L03) through analogies, harder examples, and explicit misconception resolution. If you are confident with L01–L03, use this lesson to push to Band 5–6 exam responses.

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Think First, Who Is Correct?

Before You Read: Spot the Error

In 1888, Jacobus van 't Hoff at the University of Amsterdam published his landmark paper on chemical equilibrium, stating: "at equilibrium, the system is not at rest, the forward and reverse reactions are simultaneously proceeding at equal rates." His examiner class gave three different responses when tested on what this meant.

Three students were asked: "Explain what is happening at the molecular level when a reversible reaction has reached dynamic equilibrium."

Student A: "At equilibrium, the forward and reverse reactions are both happening at the same rate. Molecules are constantly converting between reactants and products, but the overall concentrations stay the same."

Student B: "At equilibrium, the concentrations of reactants and products are equal. The reaction has balanced out so both sides are the same."

Student C: "At equilibrium, the reaction has stopped. There is no more energy available for particles to collide and react, so everything is frozen in place."

Before reading on, which student is correct? What specific errors has each incorrect student made? Write your analysis now. You will return to this at the end of the lesson.

FORMULAS

Key Relationships, Consolidation of L01–L03

Dynamic equilibrium: forward rate = reverse rate ≠ 0   (closed system, reversible reaction)

Static equilibrium: forward rate = 0, reverse rate = 0   (irreversible, completed)

No new formulas, this lesson deepens understanding through analogies, harder examples, and misconception resolution.

LEARNING INTENTIONS

By the end of this lesson

Know

The two most common equilibrium misconceptions and why they are wrong
Two analogies that correctly model dynamic equilibrium
That equilibrium can be approached from either direction

Understand

Why "equal concentrations" and "reaction stopped" are fundamentally wrong descriptions of dynamic equilibrium
Where each analogy breaks down and what it cannot model
How to identify equilibrium on a concentration-vs-time graph

Can Do

Write a Band 6 extended response describing dynamic equilibrium using correct chemical language
Correct flawed student descriptions of equilibrium precisely and completely
Interpret concentration-vs-time graphs including graphs with disturbances

KEY TERMS

Scan these before reading

Dynamic equilibrium

Forward and reverse reaction rates are equal; concentrations remain constant.

Le Chatelier's Principle

A system at equilibrium shifts to counteract any applied disturbance (stress).

Equilibrium constant (K_eq)

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

Reaction quotient (Q)

The same ratio as K_eq calculated at any point; compared to K_eq to predict shift direction.

Entropy (S)

A measure of disorder; drives many spontaneous reactions alongside enthalpy.

Gibbs free energy (ΔG°)

Predicts reaction spontaneity; ΔG° = −RT ln K_eq links thermodynamics to equilibrium.

Fill the blanks+4 XP

Before you read, try these four key equilibrium statements. Whatever you cannot fill in yet is exactly what this lesson repairs.

At equilibrium, the of the forward and reverse reactions are equal, not the concentrations.

A higher K_eq value means the equilibrium position favours the .

Changing concentration shifts the position of equilibrium but does NOT change the value of .

Dynamic equilibrium requires a system.

Who is correct?

Consolidation Content

01 Analyse Analyse

Who Is Correct?, The Three Students Analysed

The fastest path to correct understanding is to precisely identify what went wrong

The three students in the Think First represent the three most common ways Year 12 students misunderstand dynamic equilibrium, and identifying exactly what each one got wrong is the fastest path to getting it right yourself.

Student A is correct. The forward and reverse reactions are both occurring simultaneously at equal, non-zero rates. Concentrations are constant because the rate of production of each species exactly equals its rate of consumption. Molecules are constantly converting, the system is dynamic, not static.

Student B is incorrect and this is the most common error in Module 5. Student B has confused "rates are equal" with "concentrations are equal." At dynamic equilibrium, it is the RATES that are equal, the concentrations can be any values and will almost always be different from each other. A reaction with K_eq = 10⁶ is at dynamic equilibrium with almost entirely products present. Equal concentrations would only occur if K_eq ≈ 1, which is a special case, not the definition.

Student C is incorrect the second most common error. Student C has confused dynamic equilibrium with static equilibrium. At dynamic equilibrium, the reaction has NOT stopped. Individual molecules are constantly being converted between reactants and products, billions of reactions per second at the molecular level. The reason macroscopic properties appear constant is not because nothing is happening, it is because the rates of change cancel each other out.

Self-check rule

Every time you write an HSC answer about dynamic equilibrium, check against these two errors: (1) Did I accidentally say "concentrations are equal"? Change it to "rates are equal." (2) Did I say or imply the reaction has stopped? Add "both reactions continue at equal non-zero rates."

Third misconception

Not shown by the three students but equally common: "equilibrium can only be approached from the reactant side." This is wrong. The same equilibrium position is reached regardless of whether you start with all reactants, all products, or any mixture, as long as the total atomic composition is the same. Equilibrium is a property of the system's composition and temperature, not of the direction you approached from.

At dynamic equilibrium the forward and reverse rates are equal and non-zero, concentrations are constant but not necessarily equal; the most common errors are confusing "equal rates" with "equal concentrations," or assuming the reaction has stopped.

Pause, copy the highlighted definition into your book before moving on.

+5 XP Quick Check

At dynamic equilibrium, which of the following is TRUE?

Analogy 1, the escalator crowd

Interactive, Equilibrium Analogy Evaluator

Use the evaluator. A burning candle stays the same size for a while. Why is that not a system at equilibrium?

02 Understand Apply

Analogy 1, The Escalator Crowd (Traffic Flow Model)

A busy shopping centre with up and down escalators, constant flow, stable numbers on each floor

We just saw the three most common misconceptions about dynamic equilibrium. That raises a question: how do we make the idea of "equal rates but non-equal concentrations" intuitive? This card answers it → the escalator analogy maps traffic flow directly onto forward/reverse reaction rates.

Imagine a shopping centre with an up-escalator and a down-escalator between Floor 1 and Floor 2. During peak hour: the number of people on Floor 1 stays roughly constant at 200; the number on Floor 2 stays constant at 150. Does this mean nobody is moving? Absolutely not. Every minute, 50 people ride up and 50 people ride down. Net change: zero. Activity: constant and enormous.

The chemical mapping:

Floor 1 = reactants; Floor 2 = products
Up-escalator rate = forward reaction rate
Down-escalator rate = reverse reaction rate
Equal flow rates = equal forward and reverse reaction rates
Constant number of people on each floor = constant concentration of each species

If you add 100 more people to Floor 1 (adding reactant), temporarily more people ride up than down, the numbers on Floor 2 increase, until a new balance is established. This maps directly onto Le Chatelier's Principle (IQ2).

Where this analogy breaks down: Real escalators have fixed speeds, you cannot change the "rate" independently. In chemistry, reaction rates depend on concentrations and temperature. Also, escalator capacity is limited; chemical reactions have no such upper limit. Use this analogy to understand the CONCEPT of equal rates with constant concentrations, not the MECHANICS of rate dependence.

The escalator analogy: Floor 1 = reactants, Floor 2 = products; equal people-flow in both directions = equal forward/reverse rates; constant numbers on each floor = constant (not equal) concentrations of each species.

Add the highlighted analogy mapping to your notes before the check below.

+5 XP True or False

In the escalator analogy, the fact that the number of people on each floor stays constant proves that no one is moving between floors.

Analogy 2, the dissolving crystal

03 Understand Apply

Analogy 2, The Dissolving Crystal (Salt in Saturated Solution)

A solid crystal in a saturated salt solution, the ultimate everyday example of dynamic equilibrium

We just saw the escalator model for understanding equal rates with constant concentrations. That raises a question: is there a real chemical system, not an analogy, that shows ongoing activity in an apparently static system? This card answers it → NaCl dissolving in a saturated solution, proven by radioactive tracer evidence.

When you add NaCl to water and stir until no more dissolves, you have a saturated solution. A solid crystal sits in solution with no apparent change. Students often assume "nothing is happening." In reality:

At every instant, Na⁺ and Cl⁻ ions are leaving the crystal surface and entering solution (dissolution, forward reaction)
Simultaneously, Na⁺ and Cl⁻ ions from solution are striking the crystal surface and being incorporated back into the lattice (recrystallisation, reverse reaction)
The two rates are equal, the number of ions leaving per second equals the number returning per second

The formal equilibrium: NaCl(s) ⇌ Na⁺(aq) + Cl⁻(aq)

Evidence that activity is occurring: If you add a crystal of NaCl labelled with a radioactive ²⁴Na isotope to a saturated normal NaCl solution, radioactive Na⁺ ions gradually appear in solution, even though the total dissolved NaCl does not change. The radioactive ions are exchanging with the solution, proof of molecular-level activity in an apparently static system.

Where this analogy breaks down: This is a heterogeneous equilibrium (solid and aqueous phases). Not all equilibria involve solids. The same rates principle applies universally, but the specific mechanism (surface dissolution/recrystallisation) only applies to solid-liquid equilibria. Gas-phase equilibria have no "surface", the same rates principle applies through collision frequency in the gas phase.

NaCl(s) ⇌ Na⁺(aq) + Cl⁻(aq) in a saturated solution is dynamic equilibrium: dissolution rate equals recrystallisation rate; ²⁴Na radioactive tracer appearing in solution while total [NaCl(aq)] stays constant proves ongoing molecular exchange.

Pause, write the highlighted equation and tracer evidence into your book before moving on.

+5 XP Fill the Gap

Complete: In the ²⁴Na tracer experiment, radioactive ions appear in solution even though total dissolved NaCl is __________, which proves that __________ is occurring at the crystal surface.

Show answer

constant / ongoing ion exchange (dissolution and recrystallisation at equal rates). The constancy of total NaCl proves equilibrium; the appearance of ²⁴Na proves the system is dynamic, not static.

Concentration-vs-time graphs

04 Analyse Apply

Concentration-vs-Time Graphs, Reading Equilibrium

The other half of the graphical story, what you can actually measure

We just saw the ²⁴Na tracer experiment proving ongoing molecular exchange in a saturated NaCl solution. That raises a question: how do we show equilibrium and disturbances on a concentration-vs-time graph, what does it actually look like? This card answers it → by explaining how to read c-vs-t graphs and recognise three types of disturbance.

Concentration-vs-time graphs show the same equilibrium approach from the perspective of what you can measure in a flask, they are the experimental partner of the rate-vs-time graphs from L01 and L03.

A concentration-vs-time graph for a system approaching equilibrium from pure reactants shows:

Reactant concentration curves starting high and decreasing, then levelling off at a constant non-zero value
Product concentration curves starting at zero and increasing, then levelling off at a constant non-zero value
All curves becoming horizontal at the same moment when dynamic equilibrium is established

Reactant falls and product rises until both level off, the system reaches dynamic equilibrium.

Reactant concentration decreases and levels off; product concentration increases and levels off, NOT at equal values unless K_eq ≈ 1

Three types of disturbance on a concentration-vs-time graph

Disturbance	What the graph shows	Why
Adding more reactant	Reactant concentration jumps suddenly, then decreases as system re-establishes equilibrium; product concentration increases to new equilibrium value	More reactant → forward rate spikes → net forward reaction until new equilibrium
Removing a product	Product concentration drops suddenly, then increases back toward a new value; reactant concentration decreases as more product is made	Less product → reverse rate drops → net forward reaction to replace product
Temperature increase (exothermic reaction)	All equilibrium concentrations shift, reactant concentrations increase, product concentrations decrease to new horizontal values	Higher T favours reverse → new K_eq at higher T → new equilibrium position

Must know

On a concentration-vs-time graph, equilibrium is established when ALL concentration curves become horizontal simultaneously. A common HSC question asks you to identify this point, it is NOT when the curves cross (they may not cross at all if K_eq is very large or very small).

Most common graph error

Students mark equilibrium on a concentration-vs-time graph at the point where reactant and product concentrations are equal (the crossing point). This is only correct if K_eq ≈ 1. In general, equilibrium is where the curves flatten out, not where they cross.

On a concentration-vs-time graph, equilibrium is established when ALL curves become horizontal simultaneously, not at the crossing point; disturbances cause sudden jumps (adding reactant → spike up; removing product → sudden drop) followed by re-equilibration at new constant horizontal values.

Add the highlighted graph interpretation rules to your notes before the check below.

+5 XP Quick Check

On a concentration-vs-time graph, the equilibrium point is best identified as:

Cross-lesson links: The concentration-vs-time graph interpretation in Card 4 is extended in L12 (reaction quotient Q) and L13 (temperature changes Keq). The three common misconceptions about dynamic equilibrium in Card 5 are the exact errors tested in HSC questions about L05–L07 LCP shifts. Van 't Hoff's framework introduced here underpins the Keq expression you will write in L09.

Hard misconceptions, resolve these now

05 Evaluate Band 5–6

Hard Misconceptions, Resolve These Now

Four persistent misconceptions that appear in HSC extended responses every year

We just saw how to read concentration-vs-time graphs and identify equilibrium from flattening curves. That raises a question: which four persistent misconceptions about dynamic equilibrium appear most often in HSC responses, and exactly how do we correct them? This card answers it → with four wrong/correct pairs to recognise and memorise.

Dynamic equilibrium misconceptions vs the correct understanding: concentrations are constant (equal rates) not equal; both reactions continue (net change zero, not zero activity); a catalyst only speeds reaching equilibrium (Keq and yield unchanged); the same equilibrium is reached from either direction.

Quick-reference: four common equilibrium misconceptions and the correct understanding

✗ WRONG: "At dynamic equilibrium, equal amounts of reactants and products are present."

✓ CORRECT: At dynamic equilibrium, the forward and reverse RATES are equal. Concentrations are constant, not equal to each other.

Test: for N₂ + 3H₂ ⇌ 2NH₃ at 25°C, K_eq ≈ 6 × 10⁵, enormously products-favoured. The concentrations of NH₃ far exceed N₂ and H₂ at equilibrium, yet both rates are equal.

✗ WRONG: "The reaction stops at equilibrium, no more chemical activity occurs."

✓ CORRECT: At dynamic equilibrium, BOTH the forward and reverse reactions continue to occur at equal non-zero rates.

Test: radioactive ²⁴Na added to a saturated NaCl solution gradually appears in solution even though total dissolved NaCl doesn't change, proof of ongoing ion exchange at the surface.

✗ WRONG: "Adding a catalyst increases the yield of the reaction at equilibrium."

✓ CORRECT: A catalyst increases the RATE of reaching equilibrium, it does NOT change the equilibrium position or K_eq.

Test: K_eq depends only on temperature. Adding a catalyst at constant temperature cannot change K_eq, therefore it cannot change the equilibrium position or yield.

✗ WRONG: "The equilibrium can only be reached from the reactant side, you must start with reactants."

✓ CORRECT: The same equilibrium position is reached regardless of the starting direction, as long as the total atomic composition is identical.

This is the fundamental meaning of "equilibrium", the system finds the minimum free energy state regardless of its history.

Four critical corrections: "equal concentrations" → equal RATES (concentrations constant, not equal); "reaction stopped" → both reactions continue at equal non-zero rates; "catalyst increases yield" → catalyst leaves Keq unchanged; "only from reactant side" → same equilibrium reached from either direction.

Pause, write the highlighted four corrections into your book before the check below.

+5 XP True or False

Adding a catalyst to an equilibrium mixture will increase the yield of products because it increases the forward reaction rate.

✓

Worked examples and activity

Worked Examples

Worked Example 1 (Straightforward), Correcting a student's equilibrium description

A student writes: "When N₂(g) + 3H₂(g) ⇌ 2NH₃(g) reaches equilibrium in a sealed flask, the concentrations of N₂, H₂, and NH₃ are all equal, and the reaction has stopped." Identify two errors and write a corrected version.

Error 1: "Concentrations of N₂, H₂, and NH₃ are all equal." This is wrong. At dynamic equilibrium, it is the RATES of the forward and reverse reactions that are equal, not the concentrations. The concentrations at equilibrium depend on the value of K_eq at the given temperature. Concentrations are constant at equilibrium, not equal to each other.

Error 2: "The reaction has stopped." This is wrong. At dynamic equilibrium, both forward and reverse reactions continue to occur simultaneously at equal, non-zero rates. N₂ and H₂ molecules are continuously combining to form NH₃ (forward), and NH₃ molecules are continuously decomposing back to N₂ and H₂ (reverse). The net change in concentration is zero, but molecular activity is constant.

Corrected version: "When the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) reaches equilibrium in a sealed flask, the concentrations of N₂, H₂, and NH₃ are each constant but not necessarily equal to each other, their specific values depend on the value of K_eq at that temperature. Both the forward and reverse reactions continue to occur at equal, non-zero rates, the system is dynamic at the molecular level, even though macroscopic properties appear unchanged."

Worked Example 2 (Intermediate), Interpreting a concentration-vs-time graph with a disturbance

A concentration-vs-time graph for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) shows SO₂ and O₂ concentrations decreasing then levelling off; SO₃ concentration increasing then levelling off. At time t₁, SO₂ and O₂ concentrations suddenly spike upward, then decrease and level off at new values higher than the original equilibrium, while SO₃ increases to a new, higher equilibrium value.

(a) What disturbance most likely occurred at t₁? (b) Explain the spike in SO₂ and O₂ concentrations using collision theory. (c) Why does SO₃ end up higher than its original equilibrium value?

SO₂ and O₂ concentrations suddenly spike upward at t₁ → this is consistent with more SO₂ or O₂ being added to the system (addition of reactant). This immediately increases their concentrations.

Adding more SO₂ or O₂ immediately increases the concentration of reactants → frequency of effective forward collisions increases immediately → forward rate > reverse rate → net forward reaction begins. As the reaction proceeds, SO₂ and O₂ concentrations decrease (reactants consumed) and SO₃ concentration increases (more product formed) until forward rate = reverse rate again at a new equilibrium.

More reactant was added to the system, so more product is produced at the new equilibrium. The system has more total material, the equilibrium concentrations of all species are higher than the original equilibrium. K_eq is unchanged (temperature is unchanged), the same ratio of products to reactants is maintained, but at higher absolute concentrations.

Summary: (a) Addition of SO₂ or O₂ reactant. (b) More reactant → more forward collisions → forward rate spikes; as SO₃ forms, reverse rate catches up; new equilibrium established. (c) More total material in system → all equilibrium concentrations higher; K_eq unchanged.

Worked Example 3 (Hard), Band 6 Extended Response

(7 marks) A chemist studies CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g), ΔH = −206 kJ/mol. The system is at equilibrium in a sealed vessel. Describe and explain, using collision theory and Le Chatelier's Principle, the effect on equilibrium position and on K_eq when: (a) temperature is increased; (b) a nickel catalyst is added; (c) volume of vessel is halved.

The forward reaction is exothermic (ΔH = −206 kJ/mol) → E_a(forward) < E_a(reverse). Increasing temperature increases average kinetic energy of all particles, both forward and reverse rates increase. However, the rate increases proportionally more for the reaction with the higher E_a. Since E_a(reverse) > E_a(forward), the reverse rate increases more than the forward rate.

Reverse rate > forward rate → net reverse reaction → equilibrium shifts LEFT → concentrations of CO and H₂ increase; concentrations of CH₄ and H₂O decrease. Le Chatelier: system opposes the increase in temperature by favouring the endothermic reverse reaction (absorbs heat). K_eq decreases temperature is the only factor that changes K_eq.

The catalyst provides an alternative lower-energy pathway. E_a is lowered by the same amount for BOTH forward and reverse reactions. Both rates increase by the same factor. Since the system is already at equilibrium, both rates were already equal, increasing them by the same factor maintains equality. No shift in equilibrium position. K_eq unchanged it depends only on temperature, which has not changed.

Halving volume doubles concentration of all gas-phase species simultaneously. Both forward and reverse collision frequencies increase. The forward reaction involves 4 mol of gas (1 CO + 3 H₂); the reverse involves 2 mol of gas (1 CH₄ + 1 H₂O). Doubling concentration increases forward collision frequency more (more gas molecules involved in forward collisions).

Forward rate > reverse rate → equilibrium shifts RIGHT → more CH₄ and H₂O produced; CO and H₂ concentrations decrease. Le Chatelier: system opposes the pressure increase by reducing total moles of gas (4 mol → 2 mol). K_eq unchanged pressure does not change K_eq.

Summary: (a) Shift LEFT; K_eq decreases, reverse rate increases more (higher E_a reverse); system opposes added heat. (b) No shift; K_eq unchanged, catalyst lowers E_a equally for both directions; both rates increase equally. (c) Shift RIGHT; K_eq unchanged, pressure increase raises forward collision frequency more (4 mol gas → 2 mol gas); system reduces gas moles.

Activity

Activity 1

Activity 1, Correcting Equilibrium Descriptions

Each statement contains one or more errors. Identify every error precisely, then write a corrected version.

Student A says: "When I heat the equilibrium mixture A + B ⇌ C (ΔH = −50 kJ/mol), the reaction speeds up and more C is produced."
Student B says: "The concentration-vs-time graph shows equilibrium was reached at the point where the [A] and [C] curves crossed, because that's when they became equal."
Student C says: "I know the reaction has reached equilibrium because the rate of the forward reaction has become zero, no more of A and B are reacting."

✓

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

Understand Apply (5 marks)

6. Using the saturated NaCl solution analogy, explain the concept of dynamic equilibrium. Your answer must: (a) describe what is happening at the molecular level in the saturated solution; (b) explain how this demonstrates that the system is at dynamic equilibrium (not static equilibrium); (c) state one limitation of this analogy.

Apply Apply (5 marks)

7. The reaction Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) is blood-red when FeSCN²⁺ is present and colourless when SCN⁻ is in excess. At equilibrium, a deep red colour is observed. More Fe(NO₃)₃ solution is added to the equilibrium mixture.
(a) Predict the colour change and explain using collision theory. (3 marks)
(b) Sketch a description of the concentration-vs-time graph showing what happens to [FeSCN²⁺] after the addition. (2 marks)

Evaluate Evaluate (8 marks)

8. Band 6 Response: Compare and contrast static equilibrium and dynamic equilibrium. In your response, discuss: the types of reactions that can lead to each; the molecular-level activity at each type of equilibrium; and how each would appear on (i) a rate-vs-time graph and (ii) a concentration-vs-time graph. Use specific chemical examples for each type.

Model Answers

Activity 1, Error Corrections

Student A: Two errors. (1) "The reaction speeds up", partially correct (both rates increase) but incomplete. (2) "more C is produced", WRONG for an exothermic reaction. Heating an exothermic equilibrium shifts it LEFT (toward reactants), less C, not more. Corrected: "Heating speeds up both the forward and reverse reactions. However, for this exothermic forward reaction (ΔH = −50 kJ/mol), the reverse rate increases more (higher E_a). Reverse rate > forward rate → equilibrium shifts LEFT → less C is produced at the new equilibrium." K_eq decreases.

Student B: Error, confusing the crossing point with the equilibrium point. Equilibrium is where curves flatten out (become horizontal), not where they cross. The curves cross at equal concentrations, which only happens when K_eq ≈ 1. Corrected: "Equilibrium was reached at the point where both curves become horizontal simultaneously, where concentrations stop changing. This may occur before, at, or after the crossing point depending on K_eq."

Student C: Error, "the rate of the forward reaction has become zero" describes STATIC equilibrium, not dynamic. At dynamic equilibrium, the forward rate equals the reverse rate but both are non-zero. Corrected: "Dynamic equilibrium is reached when the forward rate equals the reverse rate, both are still non-zero."

Multiple Choice, worked explanations for this lesson's question bank

Same equilibrium from either direction. The same equilibrium position is always reached regardless of direction of approach (given same total atomic composition). Equilibrium is a thermodynamic minimum free energy state, it is direction-independent.

Reading the rate-vs-time graph. Equilibrium is where both curves become horizontal simultaneously, concentrations stop changing. The crossing point only equals the equilibrium point when K_eq ≈ 1 (equal concentrations at equilibrium).

The ²⁴Na tracer evidence. The ²⁴Na tracer experiment proves molecular-level activity: radioactive ions appear in solution even though total [NaCl(aq)] is constant. This is impossible if the reaction had stopped (static equilibrium). Ion exchange at the crystal surface continues at equal rates (dissolution = recrystallisation), dynamic equilibrium.

Limits of the escalator analogy. The fixed escalator speeds. In chemistry, reaction rates depend on concentrations (which change as the reaction proceeds) and temperature. The "escalator speed" in chemistry is continuously changing, the analogy works for the FINAL state but not for the approach to equilibrium.

Constant, not equal, concentrations. Two errors. Error 1: "concentrations of reactants and products become equal", wrong; concentrations are CONSTANT but not equal. Error 2: "cannot proceed further in either direction", this describes static equilibrium. At dynamic equilibrium, BOTH forward and reverse reactions continue at equal non-zero rates.

Short Answer Model Answers

Q6 (5 marks): (a) At the surface of the NaCl crystal, Na⁺ and Cl⁻ ions are simultaneously leaving the lattice and entering solution (dissolution, forward reaction) AND ions from solution are returning to the lattice surface (recrystallisation, reverse reaction) [1]. The rate of dissolution equals the rate of recrystallisation [1]. (b) This demonstrates dynamic (not static) equilibrium because: the total amount of dissolved NaCl is constant BUT molecular activity is ongoing, the ²⁴Na tracer experiment shows radioactive ions appearing in solution while total [NaCl(aq)] is unchanged [1]. In static equilibrium, no molecular activity occurs, constant concentration results from equal rates, not stopped reactions [1]. (c) Limitation: this analogy applies only to heterogeneous equilibria involving solids, gas-phase equilibria operate through gas-phase collisions, not surface exchange [1].

Q7 (5 marks): (a) Colour becomes deeper red [1]. Adding Fe³⁺ increases [Fe³⁺] → increases frequency of effective forward collisions between Fe³⁺ and SCN⁻ → forward rate increases immediately [1]. Reverse rate is initially unchanged → forward rate > reverse rate → net forward reaction → more FeSCN²⁺ forms → deeper red until new equilibrium established [1]. (b) [FeSCN²⁺] concentration-vs-time: starts at original equilibrium value; then increases gradually (as new equilibrium is established); levels off at a new, higher equilibrium value [1]. [Fe³⁺] concentration: spikes up suddenly at the moment of addition; then decreases gradually; levels off at a value higher than the original equilibrium [1].

Q8 (8 marks), Key points for Band 6: Static equilibrium: irreversible reaction (e.g. 2Mg + O₂ → 2MgO); forward rate → 0 as reactants exhausted; reverse rate = 0; forward rate curve falls to x-axis on rate-vs-time graph; reactant curve falls to zero on concentration-vs-time graph [2]. Dynamic equilibrium: reversible reaction in closed system (e.g. N₂O₄(g) ⇌ 2NO₂(g)); forward rate = reverse rate ≠ 0; rate-vs-time shows both curves converging at the same non-zero plateau; concentration-vs-time shows reactants decreasing to non-zero constant and products increasing to non-zero constant [2]. Contrast: static, true microscopic stillness; dynamic, microscopic activity with macroscopic constancy [2]. Specific examples and correct use of "rate" vs "concentration" language [2].

✓

Retrieve and reflect

Check what actually stuck

⚔️

Boss Challenge

Consolidation Deep-Dive

Put your understanding of dynamic equilibrium misconceptions and analogies to the ultimate test. Answer correctly to deal damage, get it wrong and the boss hits back.

REVISIT YOUR THINKING

Return to your Think First analysis

Look back at what you wrote about the three students. Recall van 't Hoff's 1888 statement: "the forward and reverse reactions are simultaneously proceeding at equal rates." Now that you've worked through the full consolidation:

Did you correctly identify Student A as the one closest to van 't Hoff's definition? Could you articulate precisely why Student B and Student C were wrong?
Can you now write a Band 5–6 description of dynamic equilibrium that avoids all four misconceptions from this lesson?

✓

Mark Lesson 4 as Complete Tick when you've finished all activities and checked your answers.

← L03: Collision Theory Applied to Equilibrium L05: Le Chatelier's Principle →