★ Consolidation Lesson — Deepening L01–L03. Three students just answered the same exam question about dynamic equilibrium. Only one of them is correct. Before reading on — can you spot who?
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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.
📚 Consolidation Content
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.
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:
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).
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:
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.
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:
Concentration-vs-time 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
| 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 |
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.
Quick-reference: four common equilibrium misconceptions and the correct understanding
✏️ Worked Examples
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) 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.
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.
🧪 Activities
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
❓ 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. 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)
5 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. 8 MARKS
Return to your Think First analysis of the three students. Now that you've worked through the full consolidation:
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. For most reactions Keq ≠ 1 — the equilibrium point occurs before or after the crossing point. 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. A and B are still reacting to form products, but at the same rate as products are reverting to A and B. The net change in concentration is zero, but the forward reaction has not stopped."
Scenario 1 (adding reactant): Escalator analogy: add 100 people to Floor 1 → temporarily more people ride up than down → Floor 2 numbers increase → new balance established at higher counts on both floors. Chemistry captured: the immediate increase in forward rate and eventual re-establishment of a new balance. Not captured: the analogy doesn't show WHY more people ride up — in chemistry, this is because increased [reactant] increases collision frequency. Also, the ratio of people on each floor changes in the analogy, whereas in chemistry Keq (the ratio of concentrations) remains constant at constant T.
Scenario 2 (temperature increase, exothermic equilibrium): Analogy: difficult to model — would require both escalators running faster, but the down-escalator running faster still. The analogy has no mechanism for selectively speeding up one direction more than the other. Chemistry not captured: the Ea argument — why higher T selectively increases the reverse rate more for an exothermic reaction. This is a fundamental limitation of the analogy.
Scenario 3 (catalyst): Analogy: both escalators run faster by the same amount → net flow unchanged → same ratio of people on each floor. Chemistry captured: both rates increase equally → equilibrium position unchanged. Chemistry not captured: the mechanism (alternative pathway, lower Ea) and why the ratio is unchanged (rate constants' ratio = Keq, unchanged by catalyst).
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. Starting with 2 mol HI gives the same equilibrium concentrations as starting with 1 mol H₂ + 1 mol I₂ in the same volume.
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). In general, equilibrium ≠ crossing point.
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, not fixed — the forward rate decreases and the reverse rate increases as equilibrium is approached. The analogy works for the FINAL state (rates equal, concentrations constant) 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 (equal rates, not concentrations; Keq determines ratio). Error 2: "cannot proceed further in either direction" — this describes static equilibrium where reactions have stopped. At dynamic equilibrium, BOTH forward and reverse reactions continue at equal non-zero rates. The minimum free energy statement is correct and not an error.
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 — the number of ions leaving per second = the number returning per second [1]. (b) This demonstrates dynamic (not static) equilibrium because: the total amount of dissolved NaCl is constant (like static equilibrium) BUT molecular activity is ongoing — the ²⁴Na tracer experiment shows radioactive ions appearing in solution while total [NaCl(aq)] is unchanged, proving continuous ion exchange [1]. In static equilibrium, no molecular activity occurs — the system is truly at rest. Here, constant concentration results from equal rates, not stopped reactions [1]. (c) Limitation: this analogy applies only to heterogeneous equilibria involving solids — a solid crystal surface is required for dissolution/recrystallisation to occur. Gas-phase equilibria (e.g. N₂O₄ ⇌ 2NO₂) operate through gas-phase collisions, not surface exchange — the analogy does not apply to these systems [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 (consumed by forward reaction); 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 (reaction irreversible); forward rate curve falls to x-axis on rate-vs-time graph; both rates reach zero; reactant curve falls to zero on concentration-vs-time graph; product curve rises to maximum constant value [2]. Dynamic equilibrium: reversible reaction in closed system (e.g. N₂O₄(g) ⇌ 2NO₂(g)); forward rate = reverse rate ≠ 0; both reactions continue simultaneously; 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 (not necessarily equal) [2]. Contrast: static — true microscopic stillness; dynamic — microscopic activity with macroscopic constancy. Static equilibrium: concentrations constant because reactions finished; dynamic: concentrations constant because forward rate = reverse rate. Static: approached only from reactant side (irreversible); dynamic: same equilibrium reached from either direction [2]. Specific examples and correct use of "rate" vs "concentration" language [2].
Tick when you've finished all activities and checked your answers.