Chemistry • Year 12 • Module 5 • Lesson 4

Equilibrium in Context: Analogies & Misconception Deep-Dive

Apply equilibrium concepts to real data, interpret concentration-vs-time graphs, and reason from collision theory to correct student misconceptions.

Apply • Band 4–5

1. Interpret a concentration-vs-time graph — Fe³⊃(aq) + SCN⊃−(aq) ⇄ FeSCN²⊃(aq)

The graph below shows changes in concentration over time for a sealed reaction vessel initially containing Fe³⁺(aq) and SCN⁻(aq) only. At time t₁, additional SCN⁻ solution is injected into the vessel. The red colour is due to [FeSCN²⁺]. 8 marks

Figure 1.1. Stylised concentration-vs-time graph for Fe³⁺(aq) + SCN⁻(aq) ⇄ FeSCN²⁺(aq) in a sealed vessel.

1.1 Identify the moment at which the system first reaches dynamic equilibrium. Justify your answer using features of the graph. 2 marks

1.2 Describe and explain the change in [FeSCN²⁺] after t₁, using collision theory. 3 marks

1.3 A student claims: “After t₁, the new equilibrium concentrations of Fe³⁺ and SCN⁻ must be equal because the reaction has re-balanced.” Identify the error and write the correct statement. 3 marks

Stuck? Equilibrium is established when curves flatten. For 1.2, trace the collision frequency argument from [SCN⁻] increasing. For 1.3, the lesson’s Misconception 1 applies directly.

2. Interpret experimental data — catalyst and equilibrium

A Year 12 class investigated the effect of a vanadium(V) oxide catalyst on the equilibrium 2SO₂(g) + O₂(g) ⇄ 2SO₃(g) at 450 °C. They measured the time to reach equilibrium and the % SO₃ at equilibrium. Results are shown below. 7 marks

Trial	Catalyst present?	Initial [SO₂] (mol L⁻¹)	Time to equilibrium (min)	% SO₃ at equilibrium
1	No	0.20	185	72
2	Yes (V₂O₅)	0.20	28	72
3	No	0.40	190	72
4	Yes (V₂O₅)	0.40	29	72

2.1 Compare the time to equilibrium in Trials 1 and 2. What does this difference tell you about the catalyst? 2 marks

2.2 The % SO₃ at equilibrium is identical in Trials 1 and 2. What does this confirm about the effect of a catalyst on K_eq? 2 marks

2.3 Using collision theory, explain why the catalyst speeds up reaching equilibrium without shifting it. 3 marks

Stuck? Focus on what the catalyst does to the activation energy of both reactions. If E_a drops by the same amount for forward and reverse, what happens to the ratio of rates?

3. Cause-and-effect chain — misconception correction cascade

A student says: “I heated the equilibrium mixture N₂(g) + 3H₂(g) ⇄ 2NH₃(g), ΔH = −92 kJ mol⁻¹. The reaction sped up so more NH₃ must be produced.” Complete the correction chain below. Each box leads to the next consequence. 6 marks (1 each)

Student’s claim:
Heating speeds up the reaction → more NH₃ produced.

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Error 1 — what the student got partially right:

Error 2 — what the student got wrong about direction:

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Why? (Activation energy argument):

Net shift direction:

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Effect on [NH₃] and K_eq:

Stuck? The reaction is exothermic. Le Chatelier’s Principle says heating an exothermic equilibrium shifts it in the endothermic (reverse) direction. Worked Example 3(a) in the lesson gives the full argument.

4. Predict and justify — approaching equilibrium from the product side

A chemist places 2.00 mol of HI(g) in a 1.0 L sealed flask at 425 °C. No H₂ or I₂ is present at the start. The equilibrium is: 2HI(g) ⇄ H₂(g) + I₂(g). A separate experiment starts with 1.00 mol H₂(g) + 1.00 mol I₂(g) in the same 1.0 L flask at 425 °C, with no HI present. 6 marks

4.1 Predict whether the two experiments will reach the same equilibrium concentrations. Justify your prediction using the concept of thermodynamic equilibrium and total atomic composition. 3 marks

4.2 In Experiment 1, at the start Q = 0 (no products). In Experiment 2, at the start Q = ∞ (no reactant HI). Describe the direction of the net reaction for each experiment as it moves toward equilibrium, and explain why this result is consistent with “the same equilibrium is reached regardless of direction of approach.” 3 marks

Stuck? Recall Misconception 4 from the lesson: equilibrium is a property of the system’s total composition and temperature, not the direction of approach. Both experiments have the same atoms: 2 mol H and 2 mol I in 1.0 L at 425 °C.

5. Apply the escalator analogy to a disturbance — Australian industry context

The Haber process in an Australian ammonia plant operates at equilibrium: N₂(g) + 3H₂(g) ⇄ 2NH₃(g). Engineers inject additional N₂ into the system to temporarily boost ammonia output. Using the escalator analogy from the lesson, explain what happens immediately after the N₂ is added, and predict the final outcome for [NH₃]. Identify one feature of the analogy that does not accurately represent this disturbance. 5 marks

Stuck? In the escalator analogy: N₂ is people on Floor 1. Adding N₂ adds people to Floor 1. What temporarily happens to the escalator flow? Then trace through to the new equilibrium count on Floor 2. For the limitation, think about how the up-escalator “speeds up” in chemistry vs. in the shopping centre.

Answers — Do not peek before attempting

Q1.1 — Identifying equilibrium on the graph (2 marks)

The system first reaches dynamic equilibrium at the point labelled “Equil. reached” on the graph (before t₁), where all three concentration curves simultaneously become horizontal [1]. This is the moment at which the forward rate (Fe³⁺ + SCN⁻ → FeSCN²⁺) equals the reverse rate (FeSCN²⁺ → Fe³⁺ + SCN⁻); concentrations cease to change because net reaction is zero [1].

Q1.2 — [FeSCN²⁺] after t₁ (3 marks)

After t₁, [SCN⁻] spikes upward (SCN⁻ added) [1]. Higher [SCN⁻] increases the collision frequency between Fe³⁺ and SCN⁻ ions, so the forward reaction rate immediately exceeds the reverse rate [1]. Net forward reaction proceeds: more FeSCN²⁺ forms and [FeSCN²⁺] rises to a new, higher equilibrium value; the solution becomes a deeper red [1].

Q1.3 — Error in student claim (3 marks)

The student is confusing “equal rates” (the definition of equilibrium) with “equal concentrations” (a misconception) [1]. At dynamic equilibrium, the forward rate equals the reverse rate, but the concentrations of different species are determined by K_eq and will almost always differ [1]. Corrected statement: “After t₁, the system reaches a new equilibrium in which the forward rate again equals the reverse rate. The equilibrium concentrations of Fe³⁺, SCN⁻, and FeSCN²⁺ are constant but not necessarily equal to each other — their values are determined by K_eq at 450 °C” [1].

Q2.1 — Catalyst and time to equilibrium (2 marks)

The catalyst reduces the time to reach equilibrium from 185 minutes to 28 minutes (Trial 1 vs Trial 2) [1]. This demonstrates that V₂O₅ increases the rate of approaching equilibrium — it does not change the equilibrium position, only the time taken to get there [1].

Q2.2 — Catalyst and K_eq (2 marks)

The identical % SO₃ (72%) in Trials 1 and 2 confirms that the catalyst does not change K_eq [1]. The equilibrium position (the ratio of product to reactant concentrations at equilibrium) is unchanged by the catalyst; K_eq depends only on temperature, which was the same in both trials [1].

Q2.3 — Why catalyst speeds equilibrium without shifting it (3 marks)

The catalyst (V₂O₅) provides an alternative reaction pathway with a lower activation energy [1]. Crucially, it lowers E_a by the same amount for both the forward and reverse reactions [1]. Because both rates increase by the same factor, the ratio of forward to reverse rate (which determines K_eq) is unchanged — so the equilibrium position is not shifted, only the time to reach it is shortened [1].

Q3 — Cause-and-effect correction chain (6 marks)

Error 1 — partially right: Heating does increase both the forward and reverse reaction rates (both reactions speed up when temperature rises) [1].

Error 2 — wrong about direction: For an exothermic forward reaction (ΔH = −92 kJ mol⁻¹), heating does not produce more NH₃; it shifts equilibrium in the reverse (endothermic) direction [1].

Activation energy argument: The reverse reaction has a higher activation energy than the forward reaction (since the forward is exothermic). Increasing temperature increases the rate of the reaction with higher E_a proportionally more — reverse rate increases more than forward rate [1].

Net shift direction: Reverse rate > forward rate → net reverse reaction → equilibrium shifts LEFT [1].

Effect on [NH₃] and K_eq: [NH₃] decreases (consumed by reverse reaction); K_eq decreases (temperature change is the only factor that changes K_eq) [1].

Corrected student statement: “Heating the exothermic equilibrium N₂ + 3H₂ ⇄ 2NH₃ speeds up both reactions but increases the reverse rate more. Equilibrium shifts left — [NH₃] decreases, K_eq decreases” [1].

Q4.1 — Same equilibrium? (3 marks)

Yes, both experiments will reach the same equilibrium concentrations [1]. Both flasks contain the same total atomic composition: 2 mol H and 2 mol I in 1.0 L at 425 °C [1]. Equilibrium is a thermodynamic minimum free energy state determined solely by temperature and the total composition of the system — it is independent of the direction from which it is approached [1].

Q4.2 — Direction of net reaction (3 marks)

Experiment 1 (Q = 0, starts with HI only): net forward reaction (2HI → H₂ + I₂) until Q = K_eq [1]. Experiment 2 (Q = ∞, starts with H₂ + I₂ only): net reverse reaction (H₂ + I₂ → 2HI) until Q = K_eq [1]. Both converge on the same equilibrium because K_eq is a fixed ratio at 425 °C — the system always adjusts from its starting Q toward K_eq, regardless of which direction that requires; the two experiments approach from opposite sides and meet at the same equilibrium concentrations [1].

Q5 — Escalator analogy + Haber process (5 marks)

In the escalator analogy: N₂ molecules are “people on Floor 1.” Adding N₂ suddenly increases the number of people on Floor 1 — more than the normal equilibrium count [1]. Temporarily, more people ride the up-escalator than ride down (forward rate > reverse rate) — this represents the net forward reaction: more NH₃ forms [1]. As more people accumulate on Floor 2 (product floor), the down-escalator flow increases too, until a new balance is established — a new equilibrium with more people on both floors (higher [NH₃] at the new equilibrium) [1]. K_eq is unchanged (temperature unchanged) so the equilibrium ratio of concentrations is the same, just at higher absolute levels [1]. Limitation of the analogy: in the shopping centre, escalator speeds are fixed; in chemistry, the forward rate depends on [N₂], [H₂] and is continuously changing as concentrations change during the approach to equilibrium — the “escalator speed” is not constant [1].