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.
Assess whether a rusted appearance alone shows that an iron nail on a laboratory bench is at static equilibrium rather than undergoing continuing corrosion.
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 · Recall
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 · Recall–4
Sample response. In the idealised static/completed-process model, the selected effectively irreversible reaction has no ongoing forward or reverse conversion after completion. For example, the forward reaction may stop when its limiting reactant is exhausted, while an excess reactant remains; the reverse reaction is negligible under the stated conditions. These are statements about reaction conversion, not particle motion: particles still undergo thermal motion. In dynamic equilibrium, a reversible reaction continues in a closed system at constant temperature. 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. Constant composition therefore has two different explanations: no continuing conversion in the completed-process model, or continuing conversions whose opposing effects balance in dynamic equilibrium.
Marking notes. 1 mark for completed process: no ongoing conversion of the selected reaction; 1 mark for dynamic: both rates non-zero and equal; 1 mark for distinguishing reaction conversion from particle motion, which continues in both cases. Do not require that only products remain, since an excess reactant may remain.
Section 1 · Short answer · 2 marks · Recall
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 · Recall
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 · Apply
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 · Recall–4
Sample response. No. A rusted appearance does not establish that all available iron has reacted or that corrosion has stopped. In moist air, iron can continue to oxidise; rust does not form a protective barrier. The visible rust is evidence that oxidation has occurred, not that the supply of iron, oxygen or moisture has been exhausted. Although rust does not spontaneously turn back into iron metal under these conditions, a negligible reverse rate does not establish a zero forward rate. Completion must be independently established before using a completed-process model. If corrosion continues, the nail is neither a completed-process example nor a system with equal non-zero forward and reverse rates. Neither a completed reaction nor dynamic equilibrium implies that particles stop moving.
Marking notes. 1 mark for rejecting the inference from appearance to stopped reaction; 1 mark for a valid reason, such as remaining iron continuing to corrode in oxygen and moisture, or the need for evidence that the selected reaction is complete.
Section 2 · Data response · 2 marks · Apply
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 · Apply
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 · Apply–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 · Apply–5
Sample response. The central flaw is treating constant macroscopic properties as proof that a reaction has stopped. The claim that “the molecular behaviour is the same in both cases” confuses an unchanging bulk observation with the processes causing it. In dynamic equilibrium, both the forward reaction and the reverse reaction are occurring continuously and simultaneously at equal, non-zero rates. Reactants are being converted into products while products are being converted back into reactants. For each species, formation and consumption balance, so its concentration remains constant even though individual particles continue to react. The concentrations of different species do not have to be equal. In an idealised completed effectively irreversible process, no continuing conversion occurs for the selected reaction under the stated conditions. For example, the limiting reactant may be exhausted while the reverse reaction is negligible; an excess reactant may remain, so the assertion that “only products are left” is not a universal rule. Thermal motion of particles continues in both cases: a completed chemical process does not mean that atoms or molecules have become motionless. The observable similarities therefore arise from different causes: no continuing selected conversion in the completed case, and equal opposing conversions in the dynamic case. Constant colour or concentration alone cannot distinguish these mechanisms or show that either model applies. Additional evidence of continuing exchange, obtained without materially disturbing the system, would support the dynamic interpretation. The conclusion that the two cases are equivalent is therefore unjustified. In particular, saying that equilibrium means the reaction has finished contradicts the defining equal, non-zero rates of dynamic equilibrium.
Marking notes. 1 mark, identifies the false inference from constant properties to stopped reaction. 1 mark, completed-process model: no continuing conversion for the selected reaction, with thermal motion retained. 1 mark, dynamic: equal non-zero forward and reverse rates. 1 mark, explains why similar bulk observations do not identify the same mechanism. 1 mark, reaches an evaluative conclusion supported by that distinction.
Section 3 · Extended response · 6 marks · Analyse–6
Sample response. The sealed sparkling water is a useful dynamic-equilibrium example if it has equilibrated at constant temperature. For CO2(g) ⇌ CO2(aq), the forward direction as written is dissolution: carbon dioxide moves from the gas headspace into solution. The reverse direction is outgassing: dissolved carbon dioxide returns to the gas phase. At equilibrium these transfers continue at equal non-zero rates, so the amount of dissolved CO2 and the gas pressure remain constant under unchanged conditions. The bottle is a closed system because its seal prevents carbon dioxide escaping to the surroundings; it does not prevent exchange between the liquid and the headspace. No visible bubbling is required for molecular exchange to continue. This dissolution example is a reversible physical process demonstrating dynamic equilibrium, rather than proof that all particles have stopped or that a chemical reaction has gone to completion. A rusted nail, however, cannot be assigned static equilibrium from its appearance alone. Corrosion may continue where iron, oxygen and moisture are present, because rust does not reliably protect the remaining iron. The fact that the reverse conversion of rust to iron is negligible in moist air does not imply that the forward oxidation rate has fallen to zero. A completed-process model would require independent evidence of completion under stated conditions; it must not be assumed that all iron has been consumed. Even in a completed process particles retain thermal motion. Both objects may look unchanged over a short observation, but this does not establish their reaction rates or make their molecular behaviour equivalent. Slow corrosion can escape visual detection, while balanced dissolution and outgassing can produce no net macroscopic change. The statement is therefore only partly defensible: an equilibrated sealed bottle illustrates the defining ongoing exchange of dynamic equilibrium, while the nail is not automatically an equilibrium example. A sound comparison qualifies the bottle's conditions and rejects the unsupported inference that a rusty appearance establishes a completed reaction.
Marking notes. 1 mark, challenges automatic static-equilibrium classification of the nail. 1 mark, explains possible continuing corrosion or need for completion evidence. 1 mark, identifies a sealed equilibrated bottle as reversible and closed. 1 mark, describes equal non-zero dissolution and outgassing rates. 1 mark, explains why unchanged appearance does not establish microscopic rates. 1 mark, reaches a justified qualified judgement. Credit a scientifically reasoned rejection of the nail premise; do not require molecular stillness.