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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.
Key Relationships — Consolidation of L01–L03
No new formulas — this lesson deepens understanding through analogies, harder examples, and misconception resolution.
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
Scan these before reading
Consolidation Content
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 Keq = 106 is at dynamic equilibrium with almost entirely products present. Equal concentrations would only occur if Keq ≈ 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.
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.
At dynamic equilibrium, which of the following is TRUE?
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).
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.
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 (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.
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.
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 — 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 concentration decreases and levels off; product concentration increases and levels off — NOT at equal values unless Keq ≈ 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 Keq at higher T → new equilibrium position |
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.
On a concentration-vs-time graph, the equilibrium point is best identified as:
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.
Quick-reference: four common equilibrium misconceptions and the correct understanding
Test: for N2 + 3H2 ⇌ 2NH3 at 25°C, Keq ≈ 6 × 105 — enormously products-favoured. The concentrations of NH3 far exceed N2 and H2 at equilibrium — yet both rates are equal.
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.
Test: Keq depends only on temperature. Adding a catalyst at constant temperature cannot change Keq — therefore it cannot change the equilibrium position or yield.
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.
Adding a catalyst to an equilibrium mixture will increase the yield of products because it increases the forward reaction rate.
Worked Examples
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 Keq 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 Keq 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."
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. Keq 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; Keq unchanged.
(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 Keq 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) → Ea(forward) < Ea(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 Ea. Since Ea(reverse) > Ea(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). Keq decreases — temperature is the only factor that changes Keq.
The catalyst provides an alternative lower-energy pathway. Ea 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. Keq 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). Keq unchanged — pressure does not change Keq.
Summary: (a) Shift LEFT; Keq decreases — reverse rate increases more (higher Ea reverse); system opposes added heat. (b) No shift; Keq unchanged — catalyst lowers Ea equally for both directions; both rates increase equally. (c) Shift RIGHT; Keq unchanged — pressure increase raises forward collision frequency more (4 mol gas → 2 mol gas); system reduces gas moles.
Activity
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."
Complete these key equilibrium statements — the most common misconceptions fixed.
A higher Keq 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.
✏️ Multiple Choice
1. A student states: "For the reaction 2HI(g) ⇌ H2(g) + I2(g), starting with pure HI will give a different equilibrium position than starting with an equal mixture of H2 and I2 (same total moles of atoms)." Evaluate this statement.
2. A concentration-vs-time graph for A ⇌ B shows [A] and [B] both becoming horizontal. The point of equilibrium is:
3. The radioactive 24Na tracer experiment — adding a labelled NaCl crystal to a saturated solution — is evidence for:
4. In the escalator analogy for dynamic equilibrium, 50 people per minute ride up and 50 ride down, with 200 on Floor 1 and 150 on Floor 2. Which aspect of this analogy CANNOT be directly mapped to chemical equilibrium?
5. A student writes an HSC extended response: "At dynamic equilibrium, the forward and reverse reactions proceed at equal rates. The concentrations of reactants and products become equal and constant. This is because the reaction has reached minimum free energy and cannot proceed further in either direction." Identify all errors in this response.
Short Answer
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.
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)
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 Ea). Reverse rate > forward rate → equilibrium shifts LEFT → less C is produced at the new equilibrium." Keq 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 Keq ≈ 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 Keq."
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
1. A — 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.
2. C — Equilibrium is where both curves become horizontal simultaneously — concentrations stop changing. The crossing point only equals the equilibrium point when Keq ≈ 1 (equal concentrations at equilibrium).
3. B — 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.
4. D — 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.
5. C — 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].
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.
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?