Investigate the differences between static and dynamic equilibrium, and reversible and non-reversible reactions, using practical examples

A 4 mark answer needs clear definitions, contrast, and a worked chemical example.

Static equilibrium is a state of balance in which no further change occurs because the system is not reacting. A book resting on a table is the classic physical example. In chemistry, a non-reactive mixture (for example, a sealed flask of argon and helium gas) is in static equilibrium because no forward or reverse process is taking place.

Dynamic equilibrium is the state of a reversible chemical reaction in which the rate of the forward reaction equals the rate of the reverse reaction. Macroscopic properties (concentration, colour, pressure) remain constant, but reactions continue in both directions at the molecular level.

Chemical example. For the reaction $2NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$, a sealed flask of brown $NO_2$ gas gradually pales as colourless $N_2O_4$ forms. The colour eventually stabilises, not because the reaction has stopped, but because $NO_2$ molecules are dimerising into $N_2O_4$ at the same rate as $N_2O_4$ molecules are dissociating back to $NO_2$.

Markers reward (1) the definitions, (2) the explicit "equal forward and reverse rates" criterion, (3) a named chemical example with the equation.

A sealed bottle of soft drink contains dissolved $CO_2$ in liquid and gaseous $CO_2$ above the liquid. The equilibrium is:

At equilibrium, the rate at which $CO_2$ molecules leave the solution equals the rate at which gaseous $CO_2$ molecules dissolve back into the liquid. No net change in $CO_2$ concentration occurs (the drink stays fizzy), but molecules continue to move between phases.

Opening the bottle disturbs the equilibrium by releasing gaseous $CO_2$. The forward rate (dissolved to gas) then exceeds the reverse rate, so the drink slowly loses its dissolved gas and goes flat.

Markers reward (1) the equation, (2) explicit "equal rates" language, (3) a sentence on what happens when the system is disturbed.

(a) Static equilibrium

Argon is a single unreactive gas; there is no forward or reverse chemical process occurring, so nothing is dynamically balanced, only mechanically at rest.

(b) Dynamic equilibrium

The colour is constant because the forward dimerisation ($2NO_2 \\rightarrow N_2O_4$) and reverse dissociation ($N_2O_4 \\rightarrow 2NO_2$) occur at equal rates, even though the system looks unchanging.

(c) Neither

The beaker is open, so water vapour escapes rather than condensing back at an equal rate. Without a closed system, equilibrium (static or dynamic) cannot be reached; the water level simply falls until it is gone.

Marking criteria: 1 mark for (a) and (c) correct, 1 mark for (b) correct with the word "rates" (not just "concentrations") in the justification.

(a) The flask settles at a constant, pale brown colour (a mixture of brown $NO_2$ and colourless $N_2O_4$); the colour stops visibly changing.

(b) At the molecular level, $NO_2$ molecules are continuously colliding and combining into $N_2O_4$ (forward reaction) while $N_2O_4$ molecules are continuously dissociating back into $NO_2$ (reverse reaction), at equal rates.

(c) Evidence of dynamic behaviour: heating (or cooling) the sealed flask causes the colour to change again (darken on heating, pale on cooling), showing the system responds to a disturbance and is not "frozen" as it would be if static.

Marking criteria: 1 mark per part; part (c) must describe an observable change on disturbance, not just restate the definition.

Step 1: convert volume to litres.

Step 2: apply $n = c \\times V$.

Step 3: round to 3 significant figures (matching the given data).

Explanation. The marble chip reaction, $CaCO_{3(s)} + 2HCl_{(aq)} \\rightarrow CaCl_{2(aq)} + H_2O_{(l)} + CO_{2(g)}$, is carried out in an open flask, so the gaseous product $CO_2$ escapes rather than being able to react back with $CaCl_2$ and $H_2O$ at an equal rate. Without a closed system holding all species present, the reverse reaction cannot proceed at a measurable rate, so the forward reaction runs to completion (all the limiting reagent is consumed) instead of settling into a dynamic equilibrium.

Marking criteria: 1 mark for the unit conversion, 1 mark for correct use of $n = cV$, 1 mark for the answer to 3 significant figures with correct units, 1 mark for explicitly linking the open, gas-escaping system to the absence of a measurable reverse rate.

(a) Description. The $[NO_2]$ curve starts high and falls steeply at first, then the rate of decrease slows and the curve flattens from about $t = 30$ s onward. The $[N_2O_4]$ curve starts at zero and rises with a matching, mirrored shape, also flattening from about $t = 30$ s. Both curves become horizontal (constant) beyond $t \\approx 30$ s, which is when dynamic equilibrium is reached.

(b) Explanation. Beyond $t \\approx 30$ s, the forward rate (two $NO_2$ molecules combining to $N_2O_4$) has fallen to exactly equal the reverse rate (an $N_2O_4$ molecule dissociating to two $NO_2$ molecules). Because the rates are equal, $NO_2$ is being consumed exactly as fast as it is being regenerated, so its concentration stops changing net, and the same is true in reverse for $N_2O_4$. Neither concentration reaches zero because the reaction never goes to completion in either direction; both forward and reverse processes continue indefinitely at equal, non-zero rates.

Marking criteria: 1 mark for describing the falling $NO_2$ shape, 1 mark for describing the rising $N_2O_4$ shape (mirrored), 1 mark for correctly reading the approximate time equilibrium is reached from the graph, 1 mark for the "equal, non-zero rates" explanation, 1 mark for explicitly stating why neither concentration reaches zero.

(a) Non-reversible. Combustion of methane proceeds to completion under normal conditions; the single arrow shows the products ($CO_2$ and $H_2O$) do not spontaneously recombine into methane and oxygen at any significant rate, even in a closed vessel, so no reverse rate develops and dynamic equilibrium cannot be established.

(b) Reversible. The Haber process is written with $\\rightleftharpoons$, showing that ammonia can decompose back into nitrogen and hydrogen. In a closed vessel at fixed temperature, the forward (synthesis) and reverse (decomposition) rates become equal over time, so the system reaches dynamic equilibrium, which is exactly why industrial ammonia yield is limited and optimised using Le Chatelier's principle.

Marking criteria: 1 mark per correct classification, 1 mark per justification that explicitly links the arrow type / feasibility of a reverse rate to whether dynamic equilibrium can be reached (max 4).

This is a 6-mark ANALYSE/ASSESS: markers reward a reasoned judgement supported by both systems, not a simple yes/no.

Band 6 plan.

• Thesis: both systems ARE valid dynamic equilibria that satisfy the same three criteria (closed system, reversible process, equal forward/reverse rates), but they are not interchangeable in every HSC context because they demonstrate the concept via different observable evidence (phase change vs colour change), and a well-chosen example should match what the question is testing.
• Soft drink: $CO_{2(aq)} \\rightleftharpoons CO_{2(g)}$ is a physical (phase) equilibrium; the evidence for it being dynamic is indirect, you cannot see gas continuing to dissolve and escape, you infer it from the fact that opening the bottle (making it open, i.e. not closed) causes a one-directional loss of fizz.
• $NO_2$/$N_2O_4$: is a chemical equilibrium with a directly OBSERVABLE, continuous piece of evidence, the colour is intermediate and shifts reversibly with temperature, which is stronger direct evidence of ongoing bidirectional reaction than the soft drink example provides.
• Judgement: for a question specifically asking to "justify with an example that a system is genuinely dynamic" the $NO_2$/$N_2O_4$ (or cobalt chloride) example is more effective because temperature-shift evidence is directly observable; the soft drink example is better suited to a question about the CLOSED SYSTEM requirement, since opening it removes closure and demonstrates why equilibrium collapses when a system stops being closed.

Model answer (excerpt). Both systems meet the formal definition of dynamic equilibrium: each is closed, reversible, and settles to equal forward and reverse rates. However, they are not fully interchangeable as HSC examples because they showcase different evidence. The soft drink equilibrium, $CO_{2(aq)} \\rightleftharpoons CO_{2(g)}$, is best used to explain why a CLOSED system is required, because opening the bottle visibly and irreversibly shifts the system once closure is lost. The $NO_2$/$N_2O_4$ equilibrium is better used to prove that reactions continue at the molecular level even when macroscopic properties look constant, because a further temperature change produces an immediate, reversible colour shift that could not occur if the system were truly static. A student should therefore select whichever example matches the specific criterion the question is targeting, rather than treating the two as generic substitutes.

Marker's note: top-band responses (1) confirm both examples genuinely satisfy all three criteria before critiquing them, (2) identify a concrete point of difference (direct visible evidence vs inferred phase evidence) rather than a vague preference, (3) reach an explicit, justified judgement rather than "it depends", and (4) use correct equations with state symbols for both systems.

This is a 7-mark EVALUATE: markers reward a judgement about industrial practice grounded correctly in the equilibrium definitions.

Band 6 plan.

• Define dynamic equilibrium precisely: forward rate equals reverse rate in a closed, reversible system, with constant (not maximal) product concentration.
• Key industrial insight: once dynamic equilibrium IS reached in a closed vessel, the concentration of $NH_3$ stops increasing net, because ammonia is decomposing back to $N_2$ and $H_2$ at the same rate it forms. A plant that simply waited for equilibrium would get a fixed, sub-100 percent conversion and then produce no MORE ammonia from that batch of gas.
• Continuous, non-equilibrium operation: real Haber plants continuously remove $NH_3$ as it liquefies (it has a much higher boiling point than $N_2$/$H_2$) and recycle unreacted $N_2$/$H_2$. Removing product prevents the reverse rate from ever fully catching up to the forward rate in that reaction vessel, so the system is deliberately kept away from settling at a static equilibrium position, and net forward conversion continues over time.
• Evaluation/judgement: the industrial goal is maximum THROUGHPUT of ammonia per unit time, not reaching the equilibrium state itself; because equilibrium concentration is fixed by $K$ at a given temperature and pressure, allowing the system to truly equilibrate would cap the reaction's usefulness, so continuous removal of product is a rate-and-yield strategy consistent with, not contradictory to, Le Chatelier's principle (removing product shifts the position of equilibrium to the right, favouring more forward reaction before the new equilibrium is ever reached).

Model answer (excerpt). Dynamic equilibrium in a closed vessel means the forward and reverse rates of $N_2 + 3H_2 \\rightleftharpoons 2NH_3$ have become equal, so the concentration of ammonia present is fixed, and no more accumulates from that batch of gas. Industrially, this fixed concentration is a limitation rather than a goal: Haber plants continuously liquefy and remove ammonia as it forms and recycle the unreacted nitrogen and hydrogen, which prevents the reverse (decomposition) rate from ever catching up fully with the forward rate in that vessel. By Le Chatelier's principle, removing a product continuously shifts the reaction further to the right each cycle, so the plant achieves a much higher cumulative conversion of nitrogen and hydrogen into ammonia over time than a single closed batch left to reach a static equilibrium concentration ever could. The system is therefore engineered to approach, but never fully settle into, the equilibrium state, trading a textbook closed-system equilibrium for continuous industrial throughput.

Marker's note: full-mark responses (1) correctly state that equilibrium concentration, not maximum concentration, is what a closed system would reach, (2) explain the specific engineering mechanism (continuous removal of liquefied ammonia and recycling) rather than a vague "they keep it going", (3) connect this explicitly to Le Chatelier's principle on product removal, and (4) reach a clear evaluative statement about why deliberately avoiding equilibrium increases throughput.