Module 5, Chemical Equilibrium

(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).

(a) ICE table (4 marks):

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

(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)

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):

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.

(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).