Chemistry · Year 12 · Module 5 · Lesson 1
HSC Exam Practice
Static vs Dynamic Equilibrium
Short answer
1.Short answer
Define dynamic equilibrium.
Distinguish between static equilibrium and dynamic equilibrium with reference to the forward reaction rate, the reverse reaction rate, and what is happening at the molecular level.
Outline the two conditions that must be satisfied for a chemical system to reach dynamic equilibrium.
Identify whether each of the following systems can reach dynamic equilibrium. Give a reason for each answer.
(a) A sealed aluminium can of sparkling water at room temperature.
(b) An open beaker of ethanol evaporating in a laboratory.
Describe the shape of a rate-vs-time graph for a reversible reaction that begins with pure reactants and reaches dynamic equilibrium in a closed system. In your answer, describe the forward rate curve, the reverse rate curve, and what the graph looks like once equilibrium is established.
Explain why a rusted iron nail sitting on a laboratory bench is at static equilibrium rather than dynamic equilibrium.
Data response
2.Data response — rate-vs-time graph for N2O4 ⇌ 2NO2
A student starts with pure N2O4(g) in a sealed flask at constant temperature. The graph below shows the rate of the forward reaction and the rate of the reverse reaction over time.
(a) Identify the time at which dynamic equilibrium is first established and state the evidence from the graph that supports this. (2 marks)
(b) Explain why the reverse rate starts at zero and increases, using your understanding of the concentrations present in the flask at t = 0. (2 marks)
(c) A student argues that, because the graph shows both rates becoming constant after equilibrium, “the reaction must have stopped.” Evaluate this claim. (2 marks)
Extended response
3.Extended response
Read the following excerpt and answer the question below.
“Equilibrium is just another way of saying the reaction has finished. Whether you call it static or dynamic, at the end of the day nothing is happening — the concentrations are not changing, the colour has stabilised, and the system has come to rest. The difference is that in a closed flask you happen to have both gases still present, whereas in an irreversible reaction only products are left. The molecular behaviour is the same in both cases.”
Identify the central scientific flaw in the claim above and explain the correct chemistry. In your response, refer to what is occurring at the molecular level in each type of equilibrium and explain why the observable similarities (constant colour, constant concentration) do not mean the two types are equivalent.
Evaluate the statement: “A sealed bottle of Australian sparkling water is a better example of chemical equilibrium than a rusted iron nail on a bench, because both appear identical from the outside but are chemically quite different.”
In your response, analyse what is occurring in each system at the molecular level, determine what type of equilibrium (if any) each represents, and assess whether the statement is scientifically valid.
Chemistry · Year 12 · Module 5 · Lesson 1
Answer Key & Marking Guidelines
Section 1 · Short answer · 2 marks · Band 3
Sample response. Dynamic equilibrium is a state in a reversible chemical reaction occurring in a closed system where the forward reaction rate equals the reverse reaction rate, and both rates are non-zero, so the concentration of each species remains constant over time.
Marking notes. 1 mark for equal forward and reverse rates (both non-zero); 1 mark for constant concentrations / closed system / reversible reaction context. Accept any two of these three features if both criteria are clearly captured.
Section 1 · Short answer · 3 marks · Band 3–4
Sample response. In static equilibrium, an irreversible reaction has gone to completion; both the forward and reverse reaction rates are zero, meaning no molecular activity occurs — the system is truly at rest. In dynamic equilibrium, a reversible reaction continues in a closed system; both the forward and reverse rates are non-zero and equal, so reactions are occurring simultaneously in both directions at the molecular level even though no net change in concentration is observed.
Marking notes. 1 mark for static: forward rate = reverse rate = 0, reaction complete, only products remain; 1 mark for dynamic: both rates non-zero and equal, both reactions continuing; 1 mark for contrast of molecular-level activity (truly at rest vs ongoing reactions).
Section 1 · Short answer · 2 marks · Band 3
Sample response. (1) The reaction must be reversible — it must be able to proceed in both the forward and reverse directions. (2) The system must be closed — no matter can enter or leave the system, so concentrations can stabilise.
Marking notes. 1 mark per condition. Must state both “reversible reaction” and “closed system” for full marks. Accept equivalent phrasings.
Section 1 · Short answer · 2 marks · Band 3
Sample response. (a) Yes — the sealed can is a closed system and CO2(g) ⇌ CO2(aq) is a reversible process; dynamic equilibrium can be established. (b) No — the open beaker is an open system; ethanol vapour escapes to the atmosphere and is not replaced, so the vapour concentration cannot stabilise and dynamic equilibrium cannot be established.
Marking notes. 1 mark per system: must identify the correct answer AND give a reason that references closed/open system type or whether concentrations can stabilise.
Section 1 · Short answer · 3 marks · Band 4
Sample response. The forward rate curve starts at its maximum value at t = 0 (high reactant concentration, maximum collision frequency) and decreases steadily as reactants are consumed, eventually levelling off at a constant non-zero value. The reverse rate curve starts at zero (no products present at t = 0) and increases as products accumulate, eventually levelling off at the same constant non-zero value as the forward rate. Once equilibrium is established, both curves are horizontal and equal — they meet at a constant non-zero rate and remain there indefinitely.
Marking notes. 1 mark for forward rate curve (starts high, decreases to plateau); 1 mark for reverse rate curve (starts at zero, increases to same plateau); 1 mark for equilibrium description (both curves equal, horizontal, non-zero).
Section 1 · Short answer · 2 marks · Band 3–4
Sample response. The oxidation of iron (4Fe + 3O2 → 2Fe2O3) is an irreversible reaction: iron oxide does not spontaneously reduce back to iron metal under normal atmospheric conditions. Once all available iron has been oxidised, the forward reaction rate falls to zero. The reverse reaction rate is also zero because the reaction cannot proceed in reverse. Both rates are zero and no molecular activity occurs — this is the defining property of static, not dynamic, equilibrium.
Marking notes. 1 mark for identifying the oxidation of iron as irreversible; 1 mark for stating that both forward and reverse rates are zero at static equilibrium (or that no molecular activity occurs / no reverse reaction is possible).
Section 2 · Data response · 2 marks · Band 4
Sample response. Dynamic equilibrium is first established at approximately t = 27–30 s (the point where the two curves intersect). Evidence: at this point, the forward rate curve and the reverse rate curve meet and both become horizontal (constant) — both rates are equal and non-zero from this point onward.
Marking notes. 1 mark for identifying the intersection/meeting point of the two curves (accept any time consistent with graph); 1 mark for graphical evidence (both curves horizontal/equal/non-zero).
Section 2 · Data response · 2 marks · Band 4
Sample response. At t = 0 the flask contains only N2O4 — there are no NO2 product molecules present. The reverse reaction (2NO2 → N2O4) requires NO2 molecules to collide, so without any NO2, the reverse reaction cannot occur and the reverse rate is zero. As the forward reaction proceeds, NO2 accumulates, increasing the frequency of reverse collisions and causing the reverse rate to rise.
Marking notes. 1 mark for zero NO2 concentration at t = 0 (no product present); 1 mark for explaining that the reverse rate increases as NO2 (product) accumulates.
Section 2 · Data response · 2 marks · Band 4–5
Sample response. The student’s claim is incorrect. After equilibrium, both curves are horizontal at a constant non-zero value — both the forward reaction (N2O4 → 2NO2) and the reverse reaction (2NO2 → N2O4) continue simultaneously at equal rates. The reaction has not stopped; the rates are merely equal, so the net change in concentration is zero. If the reaction had stopped, both curves would reach zero — which would indicate static equilibrium rather than dynamic equilibrium.
Marking notes. 1 mark for stating both rates remain non-zero at equilibrium (reaction has not stopped); 1 mark for linking the horizontal constant rate to equal (not zero) forward and reverse rates, or for contrasting with static equilibrium where rates do reach zero.
Section 3 · Source critique · 5 marks · Band 4–5
Sample response. The central flaw is the claim that “the molecular behaviour is the same in both cases.” It is not. In static equilibrium (e.g. a completed irreversible reaction), all molecular reaction activity has ceased: the forward rate is zero because no reactants remain, and the reverse rate is zero because the reaction cannot run backwards under these conditions. The system is genuinely at rest at the molecular level. In dynamic equilibrium, by contrast, both the forward reaction and the reverse reaction are occurring continuously and simultaneously at equal, non-zero rates. The constant macroscopic concentration and colour are the result of these two processes cancelling each other — not of the reaction having stopped. The observable similarities (stable colour, constant concentration) arise from very different underlying causes: genuine molecular stillness in the static case, and equal and opposing molecular activity in the dynamic case. Concluding that the two types are “equivalent” because the macroscopic observables are similar is an error of reasoning — it ignores the microscopic level. The author also implies that equilibrium means “the reaction has finished,” which is only true of static equilibrium; for dynamic equilibrium this definition is specifically incorrect.
Marking notes. 1 mark — identifies the flaw: the claim that molecular behaviour is the same in both types of equilibrium is incorrect. 1 mark — static equilibrium: both rates are zero, reaction complete, truly at rest. 1 mark — dynamic equilibrium: both rates non-zero and equal, both reactions occurring simultaneously. 1 mark — explains why constant macroscopic properties (colour/concentration) do not imply the same molecular mechanism (two different causes of same observable). 1 mark — evaluative conclusion: the observable similarity does not make the two systems equivalent; the microscopic difference is fundamental.
Section 3 · Extended response · 6 marks · Band 5–6
Sample response. The statement is scientifically valid in its conclusion but requires qualification. From the outside, both systems do indeed appear unchanged — the sealed sparkling water bottle shows no visible bubbling, and the rusted nail shows no visible change. Both exhibit constant macroscopic properties: stable colour (in the N2O4/NO2 analogy), stable pressure, and stable mass. However, the molecular-level reality in each system is fundamentally different, and this is what makes the statement scientifically correct and important. The rusted nail is at static equilibrium. The oxidation of iron (4Fe + 3O2 → 2Fe2O3) is an irreversible reaction that has gone to completion. All available iron has been oxidised; no iron metal remains at the surface to sustain the forward reaction (rate = 0). The reverse reaction (reducing Fe2O3 back to Fe) does not occur spontaneously under ambient conditions (rate = 0). The nail is at rest at the molecular level — nothing is happening. The sealed sparkling water bottle is at dynamic equilibrium. The dissolution of carbon dioxide (CO2(g) ⇌ CO2(aq)) is a reversible process in a closed system (the sealed can). At equilibrium, CO2 molecules continuously leave the solution into the gas headspace at the same rate as CO2 molecules from the gas phase redissolve into the liquid. Both the forward rate (escaping) and the reverse rate (dissolving) are non-zero and equal. The system is “busy” at the molecular level even though no net change in the CO2 concentration is observed. The statement is therefore valid: the sparkling water is genuinely a “better” example of chemical equilibrium in the Module 5 sense, because it demonstrates the defining property of dynamic equilibrium — ongoing molecular activity at equal rates — whereas the rusted nail, while at equilibrium, represents a static case where the system has finished reacting. The two systems are chemically quite different despite identical surface appearances, which is precisely why the lesson uses them as the opening contrast.
Marking notes. 1 mark — identifies rusted nail as static equilibrium with correct reason (irreversible reaction, gone to completion, both rates zero). 1 mark — identifies sealed sparkling water as dynamic equilibrium with correct reason (reversible process, closed system). 1 mark — describes molecular-level behaviour for static (no activity, truly at rest). 1 mark — describes molecular-level behaviour for dynamic (both forward and reverse rates non-zero, equal, ongoing simultaneously). 1 mark — explains why the same macroscopic appearance (no visible change) arises from different molecular causes (genuine stillness vs equal opposing activity). 1 mark — reaches an evidenced evaluative judgement that accepts or qualifies the statement — e.g. agrees the sparkling water is the better example of equilibrium in the Module 5 sense (dynamic, reversible, closed), while acknowledging both are technically at equilibrium.