Equilibrium in Context: Analogies & Misconception Deep-Dive

Q1(a) — Errors in Student A’s response (4 marks)

Error 1: “concentrations of SO₂, O₂, and SO₃ are all equal to each other.” Wrong: at dynamic equilibrium, it is the rates that are equal, not the concentrations. The equilibrium concentrations of each species depend on K_eq and the stoichiometry; they will almost certainly be different from each other. [1]

Error 2: “the reaction has reached a point where it has stopped.” Wrong: this describes static equilibrium, not dynamic equilibrium. At dynamic equilibrium, both the forward reaction (2SO₂ + O₂ → 2SO₃) and the reverse reaction (2SO₃ → 2SO₂ + O₂) continue simultaneously at equal, non-zero rates. Molecular activity is continuous. [1]

Error 3: “adding V₂O₅ catalyst will shift the equilibrium to the right and produce more SO₃ (higher yield).” Wrong: a catalyst does not shift the equilibrium position or change K_eq. It lowers E_a equally for forward and reverse reactions — both rates increase by the same factor, so K_eq and the equilibrium position are unchanged. The catalyst only speeds up the rate of reaching equilibrium, not the yield at equilibrium. [1]

Error 4: “molecules don’t have enough energy to react anymore.” Wrong: at dynamic equilibrium, both reactions are still occurring — molecules continuously have sufficient energy and correct orientation for both forward and reverse reactions. Concentrations are constant because the rate of formation of each species exactly equals its rate of consumption, not because reactions have stopped. [1]

Q1(b) — Sample Band 5–6 rewrite (4 marks)

At dynamic equilibrium in the sealed vessel, both the forward reaction (2SO₂(g) + O₂(g) → 2SO₃(g)) and the reverse reaction (2SO₃(g) → 2SO₂(g) + O₂(g)) continue to occur simultaneously [1]. The rate of the forward reaction equals the rate of the reverse reaction — both rates are non-zero. At the molecular level, SO₂ and O₂ molecules are continuously colliding and reacting to form SO₃, while SO₃ molecules are simultaneously decomposing back to SO₂ and O₂. The concentrations of each species are constant (not equal to each other) because the rate of production of each species exactly equals its rate of consumption; their actual values at equilibrium are determined by K_eq at the given temperature [1]. Adding V₂O₅ provides an alternative reaction pathway with lower activation energy for both the forward and reverse reactions by the same amount. Both rates therefore increase by the same factor, maintaining their equality — the equilibrium position does not shift and K_eq is unchanged. The catalyst only shortens the time taken to reach the same equilibrium, not the yield of SO₃ [1]. Full marks require: forward = reverse rate (non-zero), constant concentrations are not equal, catalyst explanation, collision theory reference [1].

Marking criteria:

• 1 mark — both forward and reverse reactions continue simultaneously at equal, non-zero rates at the molecular level.
• 1 mark — concentrations are constant but not necessarily equal; values determined by K_eq.
• 1 mark — catalyst lowers E_a equally for both reactions; does not shift equilibrium position or change K_eq; speeds up reaching equilibrium only.
• 1 mark — uses collision theory: rates are constant because rate of production = rate of consumption for each species; not because reactions have stopped.

Q2(a) — Errors in Student B’s interpretation (4 marks)

Error 1: “equilibrium is established at the point where [A] and [B] are equal (crossing point).” The graph shows [A] and [B] curves never cross — [A]_eq = 0.018 mol L⁻¹ and [B]_eq = 0.008 mol L⁻¹. Equilibrium is correctly identified where both curves simultaneously become horizontal (flat), not where they are equal. Equal concentrations at equilibrium is a misconception; only when K_eq ≈ 1 would concentrations happen to be equal. [1]

Error 2: “K_eq = 1.” K_eq = [B]²/[A] = (0.008)²/0.018 = 0.0000355 mol L⁻¹ ≈ 3.6 × 10⁻³ mol L⁻¹. K_eq ≠ 1 and the student’s claim is incorrect. [1]

Error 3: “there is no equilibrium in this system and the reaction has just stopped.” The graph clearly shows equilibrium: both curves flatten simultaneously at the dashed line. Dynamic equilibrium has been reached. Concentrations being different from each other does not mean equilibrium is absent — only that K_eq ≠ 1. [1]

Error 4: “[A] ran out of energy to convert.” and “the reaction has completely stopped.” This conflates dynamic equilibrium with static equilibrium. At equilibrium, both A → 2B and 2B → A continue at equal non-zero rates; A molecules still react but are simultaneously being regenerated from B. Concentrations are constant because rates cancel, not because reactions have stopped. [1]

Q2(b) — Calculate K_eq (3 marks)

From the graph: [A]_eq = 0.018 mol L⁻¹; [B]_eq = 0.008 mol L⁻¹. [1]

K_eq = [B]² / [A] = (0.008)² / 0.018 = 6.4 × 10⁻⁵ / 0.018 = 3.6 × 10⁻³ mol L⁻¹. [1]

Student B’s claim that K_eq = 1 is not justified. K_eq = 1 would only occur if [B]² = [A] at equilibrium; the graph shows [B]_eq < [A]_eq, so K_eq < 1, indicating the equilibrium strongly favours reactants (A). K_eq = 1 is not a necessary consequence of dynamic equilibrium — it is simply one possible value that K_eq could take. [1]

Q3 — Sample Band 6 response (8 marks), annotated

Escalator analogy — what it captures well (2 features):

(1) The constant flow of people in both directions while the count on each floor remains constant maps directly onto the equal forward and reverse reaction rates producing constant concentrations. This correctly captures the “dynamic” aspect — apparent stillness emerging from continuous, balanced activity. [1]

(2) Adding people to Floor 1 temporarily causes more upward flow until a new balance is established — correctly models Le Chatelier’s Principle for adding reactant: a temporary increase in forward rate followed by re-establishment of equilibrium at new concentrations. [1]

Escalator analogy — significant limitation: Real escalators have fixed, constant speeds. In chemistry, reaction rates depend on concentrations, which change continuously as equilibrium is approached; the “escalator speed” (rate) is not constant but decreases (forward) or increases (reverse) dynamically until they meet. The analogy cannot accurately represent the approach to equilibrium, only the final state. [1]

Saturated NaCl analogy — what it captures well (2 features):

(1) The ²⁴Na tracer experiment proves that ion exchange continues even though [NaCl(aq)] appears constant — directly demonstrating molecular-level activity in an apparently static system, the essence of dynamic equilibrium. [1]

(2) The analogy is based on a real, observable, everyday system (a saturated salt solution) with measurable evidence, making the concept concrete and testable rather than abstract. [1]

Saturated NaCl analogy — significant limitation: This is a heterogeneous equilibrium (solid-liquid interface). The mechanism (surface dissolution/recrystallisation) is specific to solid-liquid equilibria and does not apply to gas-phase equilibria (e.g. N₂O₄(g) ⇄ 2NO₂(g)), where equilibrium is maintained through gas-phase molecular collisions, not surface exchange. [1]

Direction-of-approach modelling: Neither analogy directly models the concept that the same equilibrium is reached from either direction. The escalator analogy implicitly starts from “Floor 1 full” (all reactants); there is no inherent version that starts from “Floor 2 full.” The NaCl analogy is also implicitly unidirectional (we add solid to water, not ions to a crystal). A student wishing to understand approach-from-products would need a separate modification of either analogy (e.g. “now imagine starting with all people on Floor 2 and the down-escalator only running”). [1]

Evidence-based judgement: The NaCl analogy is more valuable for students who need to understand the experimental evidence for dynamic equilibrium (the tracer experiment is a classic HSC exam reference). The escalator analogy is more valuable for understanding Le Chatelier’s Principle and how disturbances shift equilibrium. Neither is universally superior — the most effective teaching uses both in sequence: NaCl first to establish that something is happening at the molecular level, escalator to build intuition about disturbances. [1]

Marking criteria:

• 1 mark per correct, explained feature of escalator analogy (max 2).
• 1 mark for identified and explained limitation of escalator analogy.
• 1 mark per correct, explained feature of NaCl analogy (max 2).
• 1 mark for identified and explained limitation of NaCl analogy.
• 1 mark for direction-of-approach discussion (either correct or honest acknowledgement of limitation).
• 1 mark for evidence-based judgement with context-specific recommendation.