Chemistry • Year 12 • Module 5 • Lesson 1
Static vs Dynamic Equilibrium
Apply the static/dynamic distinction to real concentration-time data, compare open and closed systems, and reason about the N2O4 ⇌ 2NO2 system under experimental manipulation.
1. Interpret a concentration–time graph
The graph below shows the concentration of N2O4(g) and NO2(g) over time in a sealed flask at constant temperature. The reaction is: N2O4(g) ⇌ 2NO2(g). 8 marks
Figure 1.1 Concentration vs time for N2O4(g) ⇌ 2NO2(g) in a sealed flask at constant temperature. (Hypothetical trial data.)
1.1 Describe the trend in concentration of N2O4 and NO2 from t = 0 to equilibrium. 2 marks
1.2 At what time does dynamic equilibrium first appear to be established? Identify one piece of graphical evidence to support your answer. 2 marks
1.3 Explain why the concentrations of N2O4 and NO2 are not equal at equilibrium. Use lesson content to justify your answer. 2 marks
1.4 If the flask were opened at t = 60 s, allowing NO2 gas to escape, predict and justify what would happen to the equilibrium position. 2 marks
2. Compare open and closed systems
Complete the table below by filling in the missing cells. Use precise lesson vocabulary. 6 marks (1 per cell)
| Feature | Open system | Closed system |
|---|---|---|
| Can matter enter or leave? | Yes | |
| Can energy be exchanged? | Yes | |
| Can dynamic equilibrium be established? | ||
| One named Australian example | ||
| What happens to concentrations over time? | Stabilise (become constant) | |
| Forward rate vs reverse rate at equilibrium | Cannot equalise (no equilibrium) |
3. Cause-and-effect chain — approaching dynamic equilibrium
The five cause boxes below describe the sequence of events as a reversible reaction approaches equilibrium in a sealed flask. Fill in the matching effect boxes. Each effect should name the mechanistic change that results from the cause. The final box should state the overall outcome. 5 marks
Cause 1. Only reactant molecules are present at t = 0 in a sealed flask.
Effect 1: What is the forward rate? What is the reverse rate?
Cause 2. Reactant concentration decreases as the forward reaction proceeds.
Effect 2: What happens to the forward rate?
Cause 3. Product concentration increases as the forward reaction proceeds.
Effect 3: What happens to the reverse rate?
Cause 4. Forward rate continues to decrease; reverse rate continues to increase.
Effect 4: What condition is eventually reached?
Cause 5. Forward rate = reverse rate (both non-zero).
Overall outcome (so…):
4. Case study — Australian wine in a sealed bottle
Read the scenario and answer the question below. 4 marks
Scenario. Australian wine contains a small amount of acetic acid (vinegar, CH3COOH) and ethanol (CH3CH2OH) that react reversibly to form the ester ethyl acetate (CH3COOCH2CH3) and water: CH3COOH + CH3CH2OH ⇌ CH3COOCH2CH3 + H2O. In a sealed bottle at cellar temperature, the ester concentration remains constant for months. A wine writer notes: “The reaction has stopped — nothing is happening inside the sealed bottle.”
4.1 Evaluate the wine writer’s claim. Is the sealed bottle at static equilibrium or dynamic equilibrium? Justify your answer with reference to (i) the type of reaction, (ii) the system type, and (iii) what is happening at the molecular level. 4 marks
5. Predict and justify
A sealed flask contains an equilibrium mixture of H2(g), I2(g), and HI(g) at 450°C: H2(g) + I2(g) ⇌ 2HI(g). The flask is opened briefly and some HI gas is removed, then resealed. 3 marks
5.1 Predict what will happen immediately after the flask is resealed and explain the molecular-level reason. What will eventually happen to the concentrations of all three species? 3 marks
Q1.1 — Trend description
[N2O4] decreases from its initial value and levels off at a lower constant value at equilibrium. [NO2] increases from zero and levels off at a higher constant value at equilibrium. Both concentrations become constant at approximately t = 40 s. (1 mark for N2O4 trend; 1 mark for NO2 trend.)
Q1.2 — Time of equilibrium
Equilibrium is first established at approximately t = 40 s. Graphical evidence: both concentration curves become horizontal (flat) and remain constant from this point onward — indicating that the rate of production of each species now equals the rate of its consumption. (1 mark for time; 1 mark for evidence.)
Q1.3 — Concentrations not equal at equilibrium
At dynamic equilibrium the rates of the forward and reverse reactions are equal — not the concentrations. The equilibrium concentrations depend on the equilibrium constant (Keq) for the reaction. For N2O4 ⇌ 2NO2, Keq favours the products at this temperature, so [NO2] > [N2O4] at equilibrium. (1 mark for rates equal not concentrations; 1 mark for correct link to Keq or the reaction-specific equilibrium position.)
Q1.4 — Opening the flask at t = 60 s
Opening the flask makes the system open — NO2 escapes to the atmosphere. The reverse rate (2NO2 → N2O4) decreases because [NO2] falls. The forward rate now exceeds the reverse rate, so more N2O4 decomposes. If the flask remains open, NO2 keeps escaping and dynamic equilibrium cannot be re-established — the system is no longer closed. (1 mark for open system / loss of NO2; 1 mark for explaining the forward > reverse rate imbalance.)
Q2 — Open vs closed table
| Feature | Open system | Closed system |
|---|---|---|
| Can matter enter or leave? | Yes | No |
| Can energy be exchanged? | Yes | Yes |
| Can dynamic equilibrium be established? | No | Yes |
| One named Australian example | Beaker of water evaporating; open beverage container; camp fire | Sealed wine bottle; sealed sparkling water can; industrial reaction vessel |
| What happens to concentrations over time? | Continue to change (never stabilise) | Stabilise (become constant) |
| Forward rate vs reverse rate at equilibrium | Cannot equalise (no equilibrium) | Equal (and both non-zero) |
Q3 — Cause-and-effect chain
Effect 1: Forward rate is at its maximum (high reactant concentration); reverse rate is zero (no products yet).
Effect 2: The forward rate decreases because reactant concentration (and therefore collision frequency) decreases.
Effect 3: The reverse rate increases from zero because product concentration rises, increasing the frequency of reverse collisions.
Effect 4: The forward rate and reverse rate become equal; dynamic equilibrium is established.
Overall outcome: Concentrations of all species remain constant at their equilibrium values; macroscopic properties are stable but molecular activity (both reactions) continues at equal non-zero rates.
Q4.1 — Wine in sealed bottle
The wine writer’s claim is incorrect. The sealed bottle is at dynamic equilibrium, not static equilibrium, for three reasons. (i) The esterification reaction (acid + alcohol ⇌ ester + water) is reversible — written with ⇌ — so the reverse reaction (hydrolysis of the ester) can and does occur. (ii) The sealed bottle is a closed system — no matter can escape. (iii) At the molecular level, ester molecules are continuously forming and hydrolyzing at equal rates — both the forward and reverse reactions are occurring at the same non-zero rate. The constant ester concentration is the result of these two processes cancelling each other out, not of the reaction having stopped. (1 mark per criterion; 1 mark for correctly labelling it dynamic, not static.)
Q5.1 — Removing HI from sealed flask
Immediately after resealing, [HI] has decreased, so the reverse rate (2HI → H2 + I2) decreases. The forward rate (H2 + I2 → 2HI) now exceeds the reverse rate — the system is no longer at equilibrium. Over time, [H2] and [I2] decrease while [HI] increases as the forward reaction predominates. Eventually the forward and reverse rates equalise again at a new equilibrium where all three concentrations are constant, but at different values from the original equilibrium. (1 mark for immediate rate imbalance; 1 mark for correct direction of change; 1 mark for eventual re-establishment at new equilibrium values.)