Module 5 Quiz — Chemical Equilibrium | Chem Y12 | HSC Chemistry Year 12 Module 5

Module Coverage

IQ1 — L01–04

Dynamic Equilibrium

Static vs dynamic equilibrium
Reversibility and entropy
Collision theory rates
Conditions for equilibrium

IQ2 — L05–08

Le Chatelier's Principle

Concentration disturbances
Temperature and Keq
Pressure, volume, catalysts
Haber & Contact processes

IQ3 — L09–14

Keq, ICE & Gibbs

Writing Keq expressions
ICE table calculations
Reaction quotient Q
Ka/Kb, ΔG° = −RT ln K

IQ4 — L15–18

Dissolution & Ksp

NAGSAG solubility rules
Ksp expressions & calculations
Qsp precipitate prediction
Common ion effect

A Multiple Choice — 20 questions (20 marks)

IQ1 — Dynamic Equilibrium

Question 1

Which of the following is the best definition of a dynamic equilibrium?

A A state where all chemical reactions have stopped and concentrations are constant

B A state where the forward reaction is faster than the reverse reaction

C A state where reactant concentrations equal product concentrations

D A state in a closed system where the forward and reverse reactions occur at equal rates, so macroscopic properties remain constant while molecular-level interconversion continues

Correct: D. Dynamic equilibrium requires (1) equal forward and reverse rates, (2) constant macroscopic properties, (3) ongoing molecular-level reactions, and (4) a closed system. Option A describes a static state — wrong (reactions continue). Option B is wrong (equal rates, not unequal). Option C is a common misconception — concentrations at equilibrium depend on Keq and are NOT necessarily equal.

Question 2

Starting from pure reactants, which correctly describes the shape of the concentration vs time graph before equilibrium is reached?

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

B Reactant concentration decreases and levels off; product concentration increases from zero and levels off — both at constant, non-zero equilibrium values

C Both concentrations remain constant throughout because equilibrium exists from the start

D Reactant concentration remains constant; product concentration increases until they are equal

Correct: B. For a reversible reaction from pure reactants: [reactants] decreases (they are consumed) but reaches a non-zero equilibrium value — not zero. [products] starts at zero and increases, also reaching a non-zero equilibrium value. Both level off at constant values. Option A describes an irreversible reaction. Options C and D are wrong — there is an initial change before equilibrium.

Question 3

Collision theory explains why the forward reaction rate decreases as a reaction proceeds toward equilibrium from pure reactants. Which is the best explanation?

A The activation energy increases as reactants are consumed

B Reactant molecules become less energetic over time

C As [reactants] decreases, molecules are further apart on average, so the frequency of effective collisions decreases, reducing the reaction rate

D The temperature drops as the reaction proceeds, slowing the rate

Correct: C. Collision theory: reaction rate depends on frequency of successful collisions (those with energy ≥ Ea). As reactants are consumed, [reactants] decreases → molecules are less densely packed → fewer collisions per second → lower reaction rate. Ea and molecular energy are unchanged (options A, B wrong). Temperature is constant in a closed system (option D wrong).

Question 4

A reaction in a sealed flask is described as having Keq >> 1. Which correctly describes this system?

A At equilibrium, [products] >> [reactants]; the equilibrium position lies far to the right

B The forward reaction is much faster than the reverse reaction at all times

C The system has reached complete conversion to products

D Keq > 1 means the reaction is exothermic and will release heat

Correct: A. Keq = [products]/[reactants] at equilibrium. Keq >> 1 means the numerator is much larger than the denominator — products dominate at equilibrium. Option B confuses equilibrium position (Keq) with kinetics (rate). Option C is wrong — equilibrium still exists (unless Keq → ∞); it is just heavily product-favoured. Option D is wrong — Keq says nothing directly about ΔH sign.

Question 5

Why is an open container of water evaporating NOT an example of dynamic equilibrium?

A Water evaporation is irreversible

B The temperature is too low for equilibrium

C The system is open — water vapour escapes into the atmosphere and cannot return to establish an equal rate of condensation; no equilibrium is reached

D There is no reverse reaction — water vapour does not condense

Correct: C. Dynamic equilibrium requires a CLOSED system. In an open container, water vapour escapes and disperses — the concentration of vapour above the liquid is too low for condensation to balance evaporation. In a closed container (sealed bottle), equilibrium between liquid water and vapour IS established. Option A is wrong — evaporation/condensation IS reversible. Option D is wrong — condensation can and does occur.

IQ2 — Le Chatelier's Principle

Question 6

For the endothermic reaction $\text{A}(g) \rightleftharpoons \text{B}(g) + \text{C}(g)$, which combination of changes will both increase the yield of B and increase the rate at which equilibrium is reached?

A Decrease temperature; add a catalyst

B Increase temperature; add a catalyst

C Increase pressure; add a catalyst

D Decrease pressure; decrease temperature

Correct: B. Increasing temperature: for an endothermic reaction (absorbs heat as a "reactant"), adding heat shifts the equilibrium right → more B and C → higher yield of B. Keq increases. Adding a catalyst: lowers Ea for both directions equally → equilibrium reached faster → increased rate. The catalyst doesn't change equilibrium position, but the higher temperature both increases yield (for endothermic) AND increases rate. Option A is wrong — lower temperature shifts endothermic reaction left (lower yield). Option C: pressure effect depends on Δn(gas) = 2 − 1 = +1, so higher pressure shifts LEFT (lower yield). Option D: lower T shifts left; lower P shifts right (slight yield improvement) but rate decreases drastically.

Question 7

$\text{SO}_2(g) + \frac{1}{2}\text{O}_2(g) \rightleftharpoons \text{SO}_3(g)$ $\Delta H = -99 \text{ kJ mol}^{-1}$. What happens to Keq when temperature is increased?

A Keq increases because higher temperature increases the reaction rate

B Keq is unchanged — it only depends on concentration, not temperature

C Keq increases because more SO₃ is produced

D Keq decreases because the exothermic forward reaction is disfavoured at higher temperature; equilibrium shifts left, reducing [SO₃] relative to [SO₂][O₂]½

Correct: D. This is an exothermic reaction. Increasing temperature shifts the equilibrium LEFT (toward reactants) to absorb the added heat. At the new equilibrium, [SO₃] is lower and [SO₂][O₂]½ is higher → Keq = [SO₃]/([SO₂][O₂]½) decreases. Temperature is the ONLY factor that changes Keq. Options A and B are wrong — Keq does change with temperature, and it is not the same as rate. Option C gets the direction wrong for an exothermic reaction.

Question 8

For $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$, what is the effect of doubling the pressure at constant temperature?

A No shift in equilibrium position; both sides have the same number of moles of gas (2 mol each), so there is no net pressure change from a shift

B Equilibrium shifts right; the reaction produces more moles of gas

C Equilibrium shifts left; higher pressure always favours reactants

D Keq doubles because pressure doubles

Correct: A. Pressure effects only cause a shift when Δn(gas) ≠ 0. Here: reactants = 1 + 1 = 2 mol gas; products = 2 mol gas. Δn = 0. There is no "preferred" side with fewer gas molecules, so changing pressure does not shift the equilibrium. All concentrations increase by the same factor, Q = Keq remains true, and no shift occurs. Option D is wrong — Keq is unchanged by pressure (only temperature changes Keq).

Question 9

In the Haber process $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ at 450°C, what is the role of the iron catalyst?

A It shifts the equilibrium right to produce more NH₃

B It increases Keq so that the equilibrium yield of NH₃ is higher

C It provides an alternative reaction pathway with lower activation energy for both forward and reverse reactions, so equilibrium is reached faster at the same equilibrium position and Keq

D It absorbs excess heat from the exothermic reaction, maintaining temperature

Correct: C. A catalyst lowers Ea for BOTH forward and reverse reactions equally. This means both rates increase by the same factor, so equilibrium is reached more quickly — but the equilibrium position (ratio of concentrations) and Keq are unchanged. The commercial value of the catalyst is time — it makes the 15–25% yield achievable rapidly enough to be economical. Options A and B incorrectly claim the catalyst improves yield. Option D is wrong — catalysts don't absorb heat.

Question 10

Removing a product from an equilibrium mixture causes a shift. Which correctly explains the mechanism using rates, not just LCP?

A Removing product increases the reverse reaction rate, causing a right shift

B Removing product decreases [product], which decreases the reverse reaction rate (fewer product molecules collide). The forward rate now exceeds the reverse rate → net forward reaction occurs, shifting right

C Removing product increases [reactants], which increases the forward rate

D Removing product has no effect if the temperature is constant

Correct: B. Removing product decreases [product] → fewer product molecules per unit volume → fewer product-product collisions → reverse rate decreases. The forward rate is momentarily unchanged. Now: forward rate > reverse rate → net reaction proceeds in the forward direction → more product is formed → system shifts right until forward rate = reverse rate again (new equilibrium). Option A incorrectly says reverse rate increases (it decreases). Option C confuses product removal with reactant addition.

IQ3 — Keq, ICE Tables & Gibbs

Question 11

Write the correct Keq expression for $2\text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}_2(g)$

A $K_{eq} = \dfrac{[\text{NO}]^2[\text{O}_2]}{[\text{NO}_2]^2}$

B $K_{eq} = \dfrac{[\text{NO}_2]}{[\text{NO}][\text{O}_2]}$

C $K_{eq} = \dfrac{[\text{NO}_2]^2}{[\text{NO}]^2[\text{O}_2]}$

D $K_{eq} = \dfrac{2[\text{NO}_2]}{2[\text{NO}][\text{O}_2]}$

Correct: C. Keq = [products]^coefficients / [reactants]^coefficients = [NO₂]²/([NO]²[O₂]¹). The coefficients from the balanced equation (2, 2, 1) become the powers. Option A is the expression for the reverse reaction. Option B omits the powers. Option D incorrectly uses the coefficients as multipliers (not exponents).

Question 12

For $2\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)$, at equilibrium: [HI] = 0.580 mol/L, [H₂] = [I₂] = 0.105 mol/L. Calculate Keq.

A $K_{eq} = \dfrac{(0.105)(0.105)}{(0.580)^2} = \dfrac{0.011025}{0.3364} \approx 0.0328$

B $K_{eq} = \dfrac{(0.580)^2}{(0.105)(0.105)} \approx 30.5$

C $K_{eq} = \dfrac{0.105 + 0.105}{0.580} \approx 0.362$

D $K_{eq} = \dfrac{0.105 \times 0.105}{0.580} \approx 0.019$

Correct: A. For 2HI ⇌ H₂ + I₂: Keq = [H₂][I₂]/[HI]² = (0.105)(0.105)/(0.580)² = 0.011025/0.3364 = 0.0328. Note: [HI]² not [HI] because the coefficient of HI is 2. Option B is Keq for the reverse reaction (HI formation). Option C adds concentrations (wrong). Option D omits the square on [HI].

Question 13

For the reaction $\text{X}(g) \rightleftharpoons 2\text{Y}(g)$ with Keq = 16, a mixture has [X] = 0.50 mol/L and [Y] = 4.0 mol/L. What is Q and which direction does the reaction shift?

A Q = 8.0; Q < Keq; shifts right

B Q = 32; Q > Keq; shifts left

C Q = 16; Q = Keq; no shift — system is at equilibrium

D Q = (4.0)²/0.50 = 32; Q > Keq; shifts left to reduce [Y] and increase [X]

Correct: D. Q = [Y]²/[X] = (4.0)²/0.50 = 16/0.50 = 32. Q = 32 > Keq = 16. The product-to-reactant ratio is too high — too much Y relative to equilibrium. To reach equilibrium, [Y] must decrease and [X] must increase → the reaction shifts LEFT. Option A calculates Q without squaring [Y]. Option C gets Q = 16 by coincidence — the calculation is wrong (it would require [Y]²/[X] = 16 but here it's 32). Option B has the right value but is less complete than D.

Question 14

For a conjugate acid-base pair (weak acid HA and its conjugate base A⁻), which relationship is always correct at 25°C?

A Ka × Kb = 1

B Ka × Kb = Kw = 1.0 × 10⁻¹⁴

C Ka = Kb for any conjugate pair

D Ka + Kb = Kw

Correct: B. Adding the ionisation of HA and the hydrolysis of A⁻ gives the autoionisation of water. By Hess's Law applied to equilibrium: Ka(HA) × Kb(A⁻) = Kw = 1.0 × 10⁻¹⁴ at 25°C. This allows calculation of Kb from Ka and vice versa. Option A is wrong (Kw ≠ 1). Option C is wrong — Ka = Kb only when the acid is exactly half-way between strong and the most extreme weak, which is a coincidence. Option D uses addition, not multiplication.

Question 15

A reaction has ΔG° = +20 kJ mol⁻¹ at 298 K. Using $\Delta G° = -RT\ln K_{eq}$ (R = 8.314 J mol⁻¹K⁻¹), which best describes Keq and the spontaneity of the reaction under standard conditions?

A Keq > 1; reaction is spontaneous under standard conditions

B Keq = 1; reaction is at equilibrium under standard conditions

C Keq < 1; equilibrium lies to the left; reaction is non-spontaneous under standard conditions (ΔG° > 0)

D Keq is undefined; only reactions with ΔG° < 0 have a Keq

Correct: C. From ΔG° = −RT ln Keq: ln Keq = −ΔG°/(RT) = −20,000/(8.314 × 298) = −8.07. Keq = e^(−8.07) ≈ 3.1 × 10⁻⁴ < 1. Since ΔG° > 0, the reaction is non-spontaneous under standard conditions and the equilibrium lies to the left (reactants favoured). All reactions have a Keq (even very unfavourable ones), so option D is wrong. Option A gets the direction backwards. Option B would require ΔG° = 0.

IQ4 — Dissolution, Ksp & Qsp

Question 16

Which of the following compounds is predicted to be SOLUBLE using NAGSAG rules?

A BaSO₄

B AgCl

C PbI₂

D Ca(NO₃)₂

Correct: D. NAGSAG: Nitrates Always soluble. Ca(NO₃)₂ contains NO₃⁻, so it is soluble. BaSO₄: sulfates are generally soluble BUT Ba²⁺ is an exception — BaSO₄ is insoluble. AgCl: Ag⁺ precipitates halides — insoluble. PbI₂: Pb²⁺ precipitates halides — insoluble (forms bright yellow precipitate).

Question 17

For $\text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+}(aq) + 2\text{PO}_4^{3-}(aq)$, if molar solubility = s, which gives the correct Ksp?

A $K_{sp} = (3s)^3(2s)^2 = 108s^5$

B $K_{sp} = (3s)(2s) = 6s^2$

C $K_{sp} = s^3 \cdot s^2 = s^5$

D $K_{sp} = (3s)^2(2s)^3 = 72s^5$

Correct: A. From s mol/L dissolving: [Ca²⁺] = 3s (3 ions per formula unit), [PO₄³⁻] = 2s (2 ions per formula unit). Ksp = [Ca²⁺]³[PO₄³⁻]² = (3s)³(2s)² = 27s³ × 4s² = 108s⁵. Option B doesn't apply the powers and just multiplies stoichiometric coefficients. Option C uses s for each ion instead of 3s and 2s. Option D reverses the powers (uses 3 on Ca²⁺ → should give (3s)³; here it applied power 2 to Ca²⁺).

Question 18

20.0 mL of 0.050 mol/L BaCl₂ is mixed with 30.0 mL of 0.040 mol/L Na₂SO₄. Ksp(BaSO₄) = 1.1 × 10⁻¹⁰. Does a precipitate form?

A No; Qsp = (0.050)(0.040) = 2.0 × 10⁻³ and this must be compared to Ksp — insufficient data

B No; Qsp = (0.050)(0.040) = 2.0 × 10⁻³ < Ksp (wait, 2.0 × 10⁻³ > Ksp — yes precipitate forms)

C Yes; after mixing (total volume = 50.0 mL), [Ba²⁺] = 0.020 mol/L, [SO₄²⁻] = 0.024 mol/L, so Qsp = (0.020)(0.024) = 4.8 × 10⁻⁴ >> Ksp = 1.1 × 10⁻¹⁰

D No; BaSO₄ is soluble according to the sulfate rule

Correct: C. Always dilute first! Total volume = 20.0 + 30.0 = 50.0 mL. [Ba²⁺] = 0.050 × 20.0/50.0 = 0.020 mol/L. [SO₄²⁻] = 0.040 × 30.0/50.0 = 0.024 mol/L. Qsp = (0.020)(0.024) = 4.8 × 10⁻⁴. Since 4.8 × 10⁻⁴ >> Ksp = 1.1 × 10⁻¹⁰, a precipitate of BaSO₄ forms. Option A/B forget to dilute. Option D is wrong — Ba²⁺ is a major exception to the "sulfates are soluble" rule.

Question 19

Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq), Ksp = 5.6 × 10⁻¹². What is the molar solubility of Mg(OH)₂ in pure water?

A $s = \sqrt{5.6 \times 10^{-12}} = 2.4 \times 10^{-6}$ mol/L

B $s = \sqrt[3]{\dfrac{5.6 \times 10^{-12}}{4}} = 1.12 \times 10^{-4}$ mol/L

C $s = \dfrac{5.6 \times 10^{-12}}{2} = 2.8 \times 10^{-12}$ mol/L

D $s = \sqrt[3]{5.6 \times 10^{-12}} = 1.78 \times 10^{-4}$ mol/L

Correct: B. For Mg(OH)₂ (1:2 type): [Mg²⁺] = s, [OH⁻] = 2s. Ksp = s(2s)² = 4s³. s = ∛(Ksp/4) = ∛(5.6×10⁻¹²/4) = ∛(1.4×10⁻¹²) = 1.12 × 10⁻⁴ mol/L. Option A uses the wrong formula (s² instead of 4s³). Option D forgets the 4 (uses ∛Ksp instead of ∛(Ksp/4)). Option C divides by 2 (wrong approach entirely).

Question 20

A saturated solution of AgBr (Ksp = 5.0 × 10⁻¹³) has solid present. NaBr is added until [Br⁻] = 0.100 mol/L. What is the new equilibrium [Ag⁺]?

A [Ag⁺] remains at √(5.0 × 10⁻¹³) = 7.1 × 10⁻⁷ mol/L; Ksp is unchanged

B [Ag⁺] = 5.0 × 10⁻¹² mol/L because Ksp decreases due to the common ion

C [Ag⁺] = 0 mol/L; all Ag⁺ precipitates when [Br⁻] = 0.100 mol/L

D $[\text{Ag}^+] = \dfrac{K_{sp}}{[\text{Br}^-]} = \dfrac{5.0 \times 10^{-13}}{0.100} = 5.0 \times 10^{-12}$ mol/L

Correct: D. Ksp = [Ag⁺][Br⁻] is always satisfied at equilibrium. With Ksp constant and [Br⁻] = 0.100 mol/L (the common ion, dominated by NaBr): [Ag⁺] = Ksp/[Br⁻] = 5.0 × 10⁻¹³/0.100 = 5.0 × 10⁻¹² mol/L. This is ~140× less than in pure water. Ksp is UNCHANGED (only temperature changes Ksp) — so option B is correct in [Ag⁺] but wrong about Ksp decreasing. Option A is wrong — [Ag⁺] must decrease when [Br⁻] is raised. Option C is wrong — equilibrium still exists; [Ag⁺] → 0 only if Ksp → 0.

B Extended Response — 4 questions (45 marks)

IQ1 + IQ2

Question 21 — Dynamic Equilibrium and Le Chatelier's Principle (12 marks)

The reaction $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ $\Delta H = -197 \text{ kJ mol}^{-1}$ is used in the Contact process for producing sulfuric acid.

(a) Explain what is meant by "dynamic equilibrium" using this reaction as an example. Your answer must include evidence from both macroscopic and molecular levels. (3 marks)

(b) State and explain the effect of each of the following changes on the equilibrium position and on Keq: (i) increasing temperature; (ii) increasing pressure; (iii) adding a V₂O₅ catalyst. (6 marks)

(c) Explain why the industrial Contact process uses ~450°C rather than a lower temperature, even though a lower temperature would give a higher Keq. (3 marks)

12 marks total

Model Answer (12 marks):

(a) Dynamic equilibrium (3 marks): At dynamic equilibrium, the rate of the forward reaction (2SO₂ + O₂ → 2SO₃) equals the rate of the reverse reaction (2SO₃ → 2SO₂ + O₂) (1 mark). Macroscopic level: observable properties — [SO₂], [O₂], [SO₃], pressure, colour — remain constant over time (1 mark). Molecular level: SO₂ and O₂ molecules continue to collide and form SO₃, while SO₃ molecules simultaneously decompose back to SO₂ and O₂ — the rates of these opposing processes are equal, so the macroscopic properties appear static (1 mark).

(b) Effects of changes (6 marks):

(i) Increasing temperature (2 marks): Effect on position: The reaction is exothermic; heat is a product. Adding heat stresses the equilibrium → by LCP, the system shifts LEFT (toward SO₂ and O₂) to absorb the heat (1 mark). Effect on Keq: Keq DECREASES — at higher temperature, the new equilibrium has more SO₂/O₂ and less SO₃ relative to the original equilibrium → [SO₃]²/([SO₂]²[O₂]) ratio is smaller → Keq is smaller (1 mark).

(ii) Increasing pressure (2 marks): Effect on position: Reactants = 2 + 1 = 3 mol gas; products = 2 mol gas. Higher pressure → LCP shifts toward fewer moles of gas → shifts RIGHT (toward 2 mol SO₃), increasing [SO₃] (1 mark). Effect on Keq: Keq is UNCHANGED — only temperature changes Keq. Pressure changes the equilibrium position (concentration ratio at a new balance point) without changing the value of Keq (1 mark).

(iii) V₂O₅ catalyst (2 marks): Effect on position: NO shift in equilibrium position — the catalyst lowers Ea for BOTH forward and reverse reactions equally, so the ratio of rates at any concentration is unchanged → the equilibrium position is the same (1 mark). Effect on Keq: Keq is UNCHANGED. The catalyst only allows the system to reach the same equilibrium faster — it does not alter the thermodynamics or the equilibrium concentrations (1 mark).

(c) Temperature compromise (3 marks): The reaction is exothermic, so lower temperature gives higher Keq → theoretically higher equilibrium yield of SO₃ (1 mark). However, at lower temperatures the reaction rate is unacceptably slow — even with the V₂O₅ catalyst, the rate of approach to equilibrium is too slow for commercial viability (1 mark). ~450°C is the industrial compromise: it gives a yield of ~98% (which is high due to the pressure also being moderate but sufficient) at a rate that is commercially viable with the catalyst. The small yield lost at this temperature is offset by the dramatically improved production rate (1 mark).

IQ3

Question 22 — ICE Table Calculation (10 marks)

The equilibrium reaction $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ has Keq = 0.0211 at 250°C.

(a) Initially, a 5.00 L flask contains 0.200 mol PCl₅, and no PCl₃ or Cl₂. Convert to concentrations and complete a full ICE table showing all steps. (4 marks)

(b) Set up and solve the Keq expression for x. (4 marks)

(c) State all equilibrium concentrations and verify your answer by substituting back into the Keq expression. (2 marks)

10 marks total

Model Answer (10 marks):

(a) ICE table (4 marks):

[PCl₅]₀ = 0.200 mol / 5.00 L = 0.0400 mol/L (1 mark)

	PCl₅	PCl₃	Cl₂
Initial	0.0400	0	0
Change	−x	+x	+x
Equil.	0.0400 − x	x	x

(3 marks for correct ICE: 1 for initial row, 1 for change row with correct stoichiometry, 1 for equilibrium expression row)

(b) Solve for x (4 marks): $K_{eq} = \dfrac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \dfrac{x \cdot x}{0.0400 - x} = \dfrac{x^2}{0.0400 - x} = 0.0211$ (1 mark)

Cross-multiply: $x^2 = 0.0211(0.0400 - x) = 8.44 \times 10^{-4} - 0.0211x$ (1 mark)

Rearrange: $x^2 + 0.0211x - 8.44 \times 10^{-4} = 0$ (1 mark)

Quadratic formula: $x = \dfrac{-0.0211 \pm \sqrt{(0.0211)^2 + 4(8.44 \times 10^{-4})}}{2} = \dfrac{-0.0211 \pm \sqrt{4.455 \times 10^{-4} + 3.376 \times 10^{-3}}}{2}$

$= \dfrac{-0.0211 \pm \sqrt{3.821 \times 10^{-3}}}{2} = \dfrac{-0.0211 \pm 0.06182}{2}$

Taking positive root: $x = \dfrac{-0.0211 + 0.06182}{2} = \dfrac{0.04072}{2} = 0.02036 \approx 0.0204$ mol/L (1 mark)

(c) Equilibrium concentrations and verification (2 marks):

[PCl₅] = 0.0400 − 0.0204 = 0.0196 mol/L; [PCl₃] = [Cl₂] = 0.0204 mol/L (1 mark)

Verify: $K_{eq} = \dfrac{(0.0204)(0.0204)}{0.0196} = \dfrac{4.16 \times 10^{-4}}{0.0196} = 0.0212 \approx 0.0211 ✓$ (1 mark)

IQ3 + IQ4

Question 23 — Gibbs Free Energy, Ksp & Qsp (12 marks)

Part A — Gibbs Free Energy (5 marks)

For the reaction $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$, ΔG° = −2.60 kJ mol⁻¹ at 703 K and ΔG° = +2.63 kJ mol⁻¹ at 298 K. (R = 8.314 J mol⁻¹K⁻¹)

(a) Calculate Keq at each temperature. (2 marks)

(b) Is this reaction exothermic or endothermic? Justify using the Keq values. (2 marks)

Part B — Ksp and Qsp (7 marks)

The molar solubility of silver chromate, Ag₂CrO₄, is 6.5 × 10⁻⁵ mol/L in pure water.

(d) Write the dissolution equation and Ksp expression for Ag₂CrO₄. (1 mark)

(e) Calculate Ksp. (2 marks)

(f) 50.0 mL of 2.0 × 10⁻³ mol/L AgNO₃ is mixed with 50.0 mL of 2.0 × 10⁻⁴ mol/L K₂CrO₄. Determine whether a precipitate of Ag₂CrO₄ will form. Show all steps. (4 marks)

12 marks total

Model Answer (12 marks):

Part A:

(a) Keq at each temperature (2 marks):

At 703 K: $\ln K_{eq} = \dfrac{-(-2600)}{8.314 \times 703} = \dfrac{2600}{5844.8} = 0.445$. $K_{eq} = e^{0.445} = 1.56$ (1 mark)

At 298 K: $\ln K_{eq} = \dfrac{-(+2630)}{8.314 \times 298} = \dfrac{-2630}{2477.6} = -1.061$. $K_{eq} = e^{-1.061} = 0.346$ (1 mark)

(b) Exothermic or endothermic (2 marks): Keq = 1.56 at 703 K and Keq = 0.346 at 298 K. As temperature increases (298 K → 703 K), Keq increases. For an endothermic reaction, higher temperature shifts equilibrium right → Keq increases (1 mark). Since Keq increases with temperature, the reaction is endothermic (ΔH > 0) — consistent with the reaction absorbing heat from the surroundings (1 mark).

(c) ΔG° = 0, Keq = 1 (1 mark): When ΔG° = 0, the equilibrium position lies at equal concentrations of reactants and products — at equilibrium under standard conditions (1 mol/L each), Keq = 1, meaning [HI]² = [H₂][I₂]. This means standard-state conditions are approximately at the equilibrium point for this reaction.

Part B:

(d) Dissolution equation and Ksp (1 mark): $\text{Ag}_2\text{CrO}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{CrO}_4^{2-}(aq)$. $K_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}]$

(e) Calculate Ksp (2 marks): s = 6.5 × 10⁻⁵ mol/L. [Ag⁺] = 2s = 1.30 × 10⁻⁴ mol/L (1 mark). [CrO₄²⁻] = s = 6.5 × 10⁻⁵ mol/L. $K_{sp} = (1.30 \times 10^{-4})^2(6.5 \times 10^{-5}) = 1.69 \times 10^{-8} \times 6.5 \times 10^{-5} = 1.10 \times 10^{-12}$ (1 mark)

(f) Precipitate prediction (4 marks):

⚠️ Critical step: Apply dilution formula before calculating Qsp

Total volume = 50.0 + 50.0 = 100.0 mL (1 mark for this step)

[Ag⁺] = 2.0 × 10⁻³ × 50.0/100.0 = 1.0 × 10⁻³ mol/L

[CrO₄²⁻] = 2.0 × 10⁻⁴ × 50.0/100.0 = 1.0 × 10⁻⁴ mol/L (1 mark for both dilutions)

$Q_{sp} = (1.0 \times 10^{-3})^2(1.0 \times 10^{-4}) = 1.0 \times 10^{-6} \times 1.0 \times 10^{-4} = 1.0 \times 10^{-10}$ (1 mark)

Compare: Qsp = 1.0 × 10⁻¹⁰ > Ksp = 1.10 × 10⁻¹² (1 mark). Since Qsp > Ksp, the solution is supersaturated — a brick-red precipitate of Ag₂CrO₄ WILL form.

IQ2 + IQ4

Question 24 — Dissolution Equilibria, LCP & ATSI Knowledge (11 marks)

(a) Dissolution of ionic compounds involves a balance of lattice energy and hydration energy. (i) Define lattice energy and hydration energy and explain how their relative magnitudes determine whether dissolution is exothermic or endothermic. (3 marks)

(b) For the dissolution of ammonium nitrate: $\text{NH}_4\text{NO}_3(s) \rightarrow \text{NH}_4^+(aq) + \text{NO}_3^-(aq)$, ΔH = +25.7 kJ mol⁻¹. Explain, using Le Chatelier's Principle, why (i) adding water increases dissolution, and (ii) heating the solution increases the concentration of dissolved NH₄NO₃. (4 marks)

(c) Aboriginal and Torres Strait Islander peoples traditionally detoxified cycad seeds by leaching in running water over several days. Apply equilibrium principles to explain: (i) why running water is more effective than still water, and (ii) why slicing the seeds increases the rate of detoxification. (4 marks)

11 marks total

Model Answer (11 marks):

(a) Lattice energy and hydration energy (3 marks): Lattice energy is the energy required to separate an ionic solid into its gaseous ions — it measures the strength of ionic bonds in the crystal lattice (breaking bonds requires energy input) (1 mark). Hydration energy is the energy released when ions are surrounded and stabilised by water molecules (ion-dipole interactions) — it is always negative (exothermic) (1 mark). If hydration energy > lattice energy (in magnitude), dissolution is exothermic (ΔH < 0). If lattice energy > hydration energy (in magnitude), dissolution is endothermic (ΔH > 0) — this is the case for NH₄NO₃ where the lattice energy exceeds the hydration energy (1 mark).

(b) LCP applied to NH₄NO₃ dissolution (4 marks):

(i) Adding water (2 marks): Adding water dilutes the solution — [NH₄⁺] and [NO₃⁻] decrease. This disturbs the dissolution equilibrium: the ion concentrations fall below the equilibrium (saturated) values → Qsp < Ksp → the system is no longer at equilibrium (1 mark). By LCP, the system shifts in the direction that increases [NH₄⁺] and [NO₃⁻] — more NH₄NO₃ dissolves (shifts right) until a new saturation equilibrium is established (1 mark).

(ii) Heating the solution (2 marks): Dissolution of NH₄NO₃ is endothermic (ΔH = +25.7 kJ mol⁻¹). Increasing temperature provides thermal energy. By LCP, the system shifts to absorb the added heat — it shifts in the endothermic direction (forward/dissolution direction) (1 mark). At higher temperature, the equilibrium position shifts right: more NH₄NO₃ dissolves, the solubility increases, and the equilibrium [NH₄NO₃]dissolved increases. Keq also increases (higher temperature favours endothermic reactions) (1 mark).

(c) Cycad detoxification — equilibrium principles (4 marks):

(i) Running water vs still water (2 marks): The toxins (cycasin, macrozamin) in cycad seeds dissolve into the surrounding water according to a dissolution equilibrium: toxin(seed) ⇌ toxin(aq). In still water, the [toxin(aq)] gradually increases until equilibrium is reached — at that point, the concentration in the water matches the concentration remaining in the seed and net dissolution stops (1 mark). Running water continuously removes the dissolved toxins, maintaining [toxin(aq)] near zero. By Le Chatelier's Principle, this low product concentration shifts the equilibrium right — continuously driving more toxin out of the seed. The concentration gradient between seed and water is maintained, keeping dissolution ongoing until most toxin is removed (1 mark).

(ii) Slicing the seeds (2 marks): Slicing increases the surface area of the seed exposed to water (1 mark). Greater surface area means more toxin molecules are in contact with the surrounding water at any given time — more frequent interactions between water molecules and seed surface → increased rate of dissolution and faster approach to equilibrium. The toxin is removed more quickly because the rate at which the dissolution equilibrium is disturbed (toxin constantly carried away by running water) and re-established is faster (1 mark).

Score Tracker

Self-Assessment — Full Module 5

Section A — Multiple Choice

Q1–5 — IQ1 Dynamic Equilibrium /5

Q6–10 — IQ2 Le Chatelier's Principle /5

Q11–15 — IQ3 Keq, ICE & Gibbs /5

Q16–20 — IQ4 Dissolution & Ksp /5

Section B — Extended Response

Q21 — Dynamic equilibrium & Contact process /12

Q22 — ICE table: PCl₅ ⇌ PCl₃ + Cl₂ /10

Q23 — Gibbs, Ksp, Qsp (Ag₂CrO₄) /12

Q24 — Dissolution, LCP & ATSI /11

Total /65—

Grade guide: 59–65 (91%+) Band 6 | 52–58 (80–90%) Band 5 | 39–51 (60–79%) Band 4 | Below 39 — revisit checkpoints

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Module 5 Complete

Chemical Equilibrium — all inquiry questions covered.