Checkpoint 1 — IQ1: Dynamic Equilibrium | Chem Y12 M5 | HSC Chemistry Year 12 Module 5

What's Covered

L01

Static vs Dynamic Equilibrium

Closed vs open systems
Macroscopic vs molecular level
Evidence for dynamic equilibrium
Conditions for establishment

L02

Reversibility & Entropy

ΔG and equilibrium position
Entropy and spontaneity
Why equilibrium is reached
Free energy minimum

L03

Collision Theory

Forward and reverse rates
Rate vs time graphs
Equilibrium as equal rates
Concentration-rate relationship

L04

★ Consolidation

Static vs dynamic distinction
Graph interpretation
Real-world applications
Misconception correction

Section A — Multiple Choice (10 questions)

Question 1

A sealed flask contains N₂O₄(g) ⇌ 2NO₂(g) at equilibrium. The flask is opaque and cannot be observed. Which of the following is evidence that this is a dynamic equilibrium rather than a static one?

A The total mass of gas in the flask remains constant

B If radioactively labelled ¹⁵N₂O₄ is added, ¹⁵N appears in NO₂ over time even though total NO₂ concentration stays constant

C The pressure inside the flask remains constant over time

D The temperature inside the flask remains constant

Correct: B. Dynamic equilibrium requires that both forward and reverse reactions continue. The radioactive label appearing in NO₂ while total [NO₂] stays constant proves both reactions are occurring — molecules are being converted both ways, but at equal rates. Options A, C, and D are consistent with both static and dynamic equilibrium.

Question 2

Which statement correctly describes the equilibrium position of a reaction with a very large Keq (> 10⁶)?

A The reaction is fast and reaches equilibrium quickly

B The forward and reverse reactions occur at equal rates, but the forward rate is faster

C At equilibrium, product concentrations are much larger than reactant concentrations; the equilibrium position lies far to the right

D The reaction will eventually reach complete conversion to products

Correct: C. Keq is the ratio of products to reactants at equilibrium. Keq > 10⁶ means [products] >> [reactants] at equilibrium — the equilibrium position lies far to the right. Option A confuses Keq (position) with rate (kinetics). Option B is wrong — at equilibrium, forward rate = reverse rate. Option D is wrong — equilibrium exists; it never becomes truly complete unless Keq → ∞.

Question 3

A student graph shows forward reaction rate decreasing and reverse reaction rate increasing after a pure-reactant start. At what point on the graph is equilibrium established?

A When the two rate curves intersect (forward rate = reverse rate)

B When the forward rate reaches zero

C When the forward rate is exactly twice the reverse rate

D When the concentration of products equals the concentration of reactants

Correct: A. Equilibrium is established when the forward rate equals the reverse rate — the intersection point on the rate vs time graph. Option B is wrong — both rates remain non-zero at equilibrium (dynamic). Option C is wrong — equilibrium requires equal rates, not a 2:1 ratio. Option D is wrong — equilibrium concentrations are NOT necessarily equal (that would require Keq = 1).

Question 5

For the reversible reaction $\text{A}(g) + \text{B}(g) \rightleftharpoons \text{C}(g) + \text{D}(g)$, starting from pure A and B, which correctly describes the rate vs time graph before equilibrium is reached?

A Forward rate remains constant; reverse rate increases from zero

B Forward rate decreases (as A and B are consumed); reverse rate increases from zero (as C and D accumulate)

C Both forward and reverse rates decrease over time

D Forward rate increases as temperature rises; reverse rate stays constant

Correct: B. Starting from pure A and B: forward rate is highest initially (high [A] and [B]) and decreases as they are consumed. Reverse rate starts at zero (no C or D) and increases as C and D accumulate. They meet at equilibrium. Option A is wrong — forward rate decreases as reactants are consumed. Option C is wrong — reverse rate increases (not decreases) as products form.

Question 6

According to collision theory, why does the forward reaction rate decrease as reactants are consumed?

A The activation energy increases as reactants are consumed

B The temperature of the system decreases as reactants are converted to products

C Lower reactant concentrations mean fewer collisions per unit time between reactant molecules, so fewer successful collisions occur per second

D Reactant molecules become less reactive over time

Correct: C. Collision theory: reaction rate depends on the frequency of successful collisions. As [reactants] decreases, molecules are further apart on average → fewer collisions per unit time → fewer successful (energy ≥ Ea) collisions → lower reaction rate. Activation energy and reactivity don't change (options A and D). Temperature doesn't typically change in a closed system at constant conditions (option B).

Question 7

A reaction reaches equilibrium in a closed container. The system is described as "macroscopically static but microscopically dynamic." Which statement correctly explains this description?

A Observable properties (colour, concentration, pressure) appear constant, but at the molecular level, reactants continuously convert to products and products continuously revert to reactants at equal rates

B All molecular motion stops at equilibrium, but observable properties continue to change slowly

C Observable properties are constant because the forward reaction is much faster than the reverse reaction

D Macroscopic and microscopic descriptions of equilibrium are contradictory — only one can be correct

Correct: A. "Macroscopically static" means observable properties (concentration, colour, pressure) don't change. "Microscopically dynamic" means at the molecular level, both reactions continue at equal rates — individual molecules are still reacting. This is the defining characteristic of dynamic equilibrium. Option B is wrong — molecular motion never stops above absolute zero. Option C confuses rate with equilibrium position.

Question 8

A student connects Gibbs free energy to equilibrium: "At equilibrium, ΔG = 0 but ΔG° is not necessarily zero." Which interpretation is correct?

A ΔG = 0 at equilibrium means the reaction has no tendency to proceed; ΔG° = 0 means Keq = 1

B ΔG° = 0 always at equilibrium because the system has reached minimum energy

C ΔG and ΔG° are the same quantity under all conditions

D ΔG = 0 at equilibrium for any system (minimum free energy); ΔG° is the standard-state free energy change and equals zero only when Keq = 1 (from ΔG° = −RT ln Keq)

Correct: D. ΔG = 0 at equilibrium for any system — it is at the free energy minimum, so no further spontaneous change occurs. ΔG° is the free energy change under standard conditions; from ΔG° = −RT ln Keq, ΔG° = 0 only when ln Keq = 0, i.e. Keq = 1. Option A partially correct about ΔG = 0 but incorrectly implies "no tendency to proceed" — at equilibrium, both reactions proceed at equal rates. Option B is wrong about ΔG°.

Question 9

Which of the following systems is capable of reaching dynamic equilibrium?

A A burning candle in an open room

B A sealed flask containing NO₂(g) and N₂O₄(g) at constant temperature

C A glass of water left open on a bench until it evaporates

D Combustion of petrol in a car engine

Correct: B. Dynamic equilibrium requires: (1) reversible reaction, (2) closed system, (3) constant conditions. A sealed flask with NO₂/N₂O₄ satisfies all three — the interconversion is reversible, the system is closed, and temperature can be kept constant. Options A and D are open systems with essentially irreversible combustion. Option C is open to the atmosphere — water molecules escape and cannot return, so equilibrium cannot be established.

Question 10

At equilibrium, which of the following concentration graphs is correct for a reaction starting from pure reactants only?

A Reactant concentration decreases to zero; product concentration increases to a maximum

B Reactant and product concentrations both remain constant from the start

C Reactant concentration decreases and levels off; product concentration increases and levels off — both reaching constant non-zero values

D Reactant concentration decreases to zero as the reaction is irreversible

Correct: C. For a reversible reaction starting from pure reactants: reactant concentration decreases (some is consumed) but levels off at a non-zero equilibrium value. Product concentration increases from zero but also levels off at a non-zero equilibrium value. Both reach constant values at equilibrium. Options A and D describe an irreversible reaction. Option B is wrong — there's an initial change before equilibrium is established.

Section B — Short Answer

Question 11

(a) Define dynamic equilibrium. Your definition must include reference to the molecular level and macroscopic level. (2 marks)

2 marks

Model Answer (2 marks):

Dynamic equilibrium is a state in a closed system where the forward reaction and reverse reaction occur at the same rate (1 mark), so the macroscopic properties of the system remain constant (concentrations, colour, pressure unchanged) even though reactants and products are continuously interconverting at the molecular level (1 mark).

Question 12

The reaction $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ reaches equilibrium in a sealed flask. Explain, using collision theory, why the forward reaction rate decreases and the reverse reaction rate increases as the system approaches equilibrium from pure reactants. (4 marks)

4 marks

Model Answer (4 marks):

Forward rate decreasing (2 marks): As H₂ and I₂ are consumed in the forward reaction, their concentrations decrease. Lower [H₂] and [I₂] means fewer molecules per unit volume → fewer collisions per second between H₂ and I₂ molecules → fewer successful collisions (frequency of collisions with energy ≥ Ea decreases) → forward reaction rate decreases (1 mark). The forward rate is proportional to [H₂][I₂], so as these decrease, the rate decreases (1 mark).

Reverse rate increasing (2 marks): As HI accumulates (product of the forward reaction), [HI] increases from zero. Higher [HI] means more HI molecules per unit volume → more collisions between HI molecules per second → more successful HI–HI collisions (energy ≥ Ea for the reverse reaction) → reverse reaction rate increases (1 mark). The reverse rate is proportional to [HI]², so as [HI] increases from zero, the reverse rate increases from zero (1 mark).

Question 13

A student observes a sealed flask containing a brown gas (NO₂) and colourless gas (N₂O₄) at equilibrium: $2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)$. The student claims: "Because the colour stops changing, the reaction has stopped." Identify the error in this reasoning and provide the correct explanation with evidence from the experiment. (3 marks)

3 marks

Model Answer (3 marks):

Error identified (1 mark): The student has confused dynamic equilibrium with a static (stopped) state. The constant colour does not mean the reaction has stopped — it means forward and reverse reactions are occurring at equal rates, so [NO₂] (which causes the brown colour) remains constant.

Correct explanation (1 mark): At dynamic equilibrium, the forward reaction (2NO₂ → N₂O₄) and reverse reaction (N₂O₄ → 2NO₂) both continue at equal rates. The rate of NO₂ consumption equals the rate of NO₂ production, so [NO₂] — and hence the colour intensity — remains constant.

Evidence (1 mark): If radioactively labelled ¹⁵NO₂ were added to the flask, ¹⁵N would appear in N₂O₄ molecules over time (even though total colour stays constant), confirming that molecules are still interconverting at equilibrium.

Score Tracker

Self-Assessment

Section A — MC (Q1–10) /10

Q11 — Define equilibrium /2

Q12 — Collision theory /4

Q13 — NO₂/N₂O₄ misconception /3

Total /19—

Checkpoint 1 complete — IQ1 Dynamic Equilibrium