Checkpoint 2 — IQ2: Le Chatelier's Principle | Chem Y12 M5 | HSC Chemistry Year 12 Module 5

What's Covered

L05

LCP — Concentration

Adding/removing species
Shift direction prediction
Rate explanation for shift
Solids & pure liquids excluded

L06

LCP — Temperature

Exo vs endo shifts
Temperature changes Keq
Colour change predictions
Qualitative Keq direction

L07

LCP — Pressure & Catalysts

Moles of gas counting
Pressure/volume effects
Catalysts: rate not position
Inert gas no effect

L08

★ LCP Consolidation

Haber process conditions
Contact process SO₃
Multi-variable predictions
Industrial compromise logic

Section A — Multiple Choice (10 questions)

Question 1

For the reaction $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$, what is the effect of adding extra Cl₂(g) at constant temperature and volume?

A The equilibrium shifts right; Keq increases

B The equilibrium shifts right; Keq remains constant

C The equilibrium shifts left; Keq remains constant

D The equilibrium shifts left; Keq decreases

Correct: C. Adding Cl₂ (a product) increases [Cl₂], which stress the equilibrium. By LCP, the system shifts to oppose the increase — it shifts LEFT to consume the added Cl₂ and produce more PCl₅. Keq is unaffected by concentration changes; only temperature changes Keq. Option A is wrong (direction). Options B is wrong (direction). Option D is wrong (Keq doesn't change with concentration).

Question 2

Consider the exothermic reaction: $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ $\Delta H = -197 \text{ kJ mol}^{-1}$. If the temperature is increased, which prediction is correct?

A The equilibrium shifts left; Keq decreases; [SO₃] decreases

B The equilibrium shifts right; Keq increases; [SO₃] increases

C The equilibrium shifts left; Keq remains constant; [SO₃] decreases

D There is no shift because the reaction is already at equilibrium

Correct: A. This is an exothermic reaction — heat is a product. Increasing temperature adds heat, so LCP predicts a shift LEFT (toward reactants, to absorb the extra heat). When the shift is left, [SO₃] decreases. Importantly, increasing temperature changes Keq — for an exothermic reaction, Keq decreases with increasing temperature. Option B is the opposite direction. Option C incorrectly states Keq stays constant.

Question 3

The reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ is at equilibrium. The volume of the container is halved at constant temperature. Which correctly describes what happens?

A The equilibrium shifts left, producing more N₂ and H₂

B The equilibrium shifts right, producing more NH₃; the side with fewer moles of gas is favoured

C There is no shift; pressure changes do not affect gaseous equilibria

D The equilibrium shifts right, and Keq increases

Correct: B. Halving the volume doubles the pressure. By LCP, the system shifts to reduce pressure — it favours the side with fewer moles of gas. Reactants: 1 + 3 = 4 mol gas. Products: 2 mol gas. The system shifts RIGHT (toward 2 mol, the smaller number), producing more NH₃. Keq is unchanged — only temperature changes Keq (eliminates D). Option A has the wrong direction. Option C is wrong — pressure does affect gaseous equilibria when Δn(gas) ≠ 0.

Question 4

A catalyst is added to an equilibrium system. Which statement is correct?

A The catalyst shifts the equilibrium to the right, increasing product yield

B The catalyst increases Keq, favouring products

C The catalyst increases the activation energy of the forward reaction only

D The catalyst lowers the activation energy of both the forward and reverse reactions equally, so equilibrium is reached faster but the equilibrium position and Keq are unchanged

Correct: D. A catalyst provides an alternative reaction pathway with lower Ea for BOTH forward and reverse reactions — equally. This means both rates increase by the same factor, equilibrium is reached faster, but the equilibrium position (ratio of products to reactants) does not change. Keq is unchanged. Options A and B are wrong — no shift in position or change in Keq. Option C is wrong — catalysts lower (not increase) Ea and affect both directions.

Question 5

In the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (ΔH = −92 kJ mol⁻¹), industrial conditions use approximately 450°C and 200 atm. Why is a temperature of 450°C used rather than a lower temperature like 200°C?

A Lower temperatures would shift the equilibrium left, reducing NH₃ yield

B 450°C maximises the Keq value and gives the highest possible yield

C Lower temperatures give a higher equilibrium yield but an unacceptably slow rate; 450°C is a compromise between yield and a commercially viable reaction rate

D 450°C is needed to activate the N₂ molecules; lower temperatures cannot break the N≡N triple bond

Correct: C. This is the industrial compromise argument. The reaction is exothermic, so lower temperature → higher Keq → higher equilibrium yield. However, lower temperature also means lower reaction rate (kinetics). If the rate is too slow, NH₃ production becomes commercially unviable. 450°C is a compromise: acceptable yield combined with a rate fast enough to be economical (using an iron catalyst). Option B is wrong — lower temperature gives higher Keq for an exothermic reaction. Option D confuses thermodynamics with kinetics.

Question 6

For the equilibrium $\text{CoCl}_4^{2-}(aq) + 6\text{H}_2\text{O}(l) \rightleftharpoons \text{Co(H}_2\text{O)}_6^{2+}(aq) + 4\text{Cl}^-(aq)$, the left side is blue and the right side is pink. The solution appears blue. Adding water shifts the equilibrium to produce a pink colour. What does this tell you about the enthalpy of this reaction?

A Nothing — colour change from adding water is a concentration effect (dilution of Cl⁻), not a temperature effect, so enthalpy cannot be determined from this observation

B The reaction is endothermic because it shifts right when heated

C The reaction is exothermic because the right side (pink) is favoured at lower energy

D The forward reaction is endothermic; adding water provides heat energy

Correct: A. Adding water is a concentration change (diluting Cl⁻ and shifting the equilibrium right), NOT a temperature change. To determine enthalpy, you need to observe the effect of changing temperature. A shift right upon heating would indicate an endothermic forward reaction. The colour change from adding water tells us about the equilibrium position shift due to concentration, not enthalpy. Options B, C, and D incorrectly draw thermodynamic conclusions from a concentration change.

Question 7

An equilibrium mixture of $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ is placed in a piston. The piston is compressed to half the original volume. Which graph correctly shows [NO₂] over time?

A [NO₂] drops instantly to half, then returns exactly to the original concentration

B [NO₂] doubles instantly (due to halved volume), then decreases to a new equilibrium value that is higher than the original concentration but lower than twice the original

C [NO₂] remains constant throughout — pressure has no effect on this equilibrium

D [NO₂] doubles instantly, then stays doubled

Correct: B. Compressing to half the volume instantly doubles all concentrations (including [NO₂] — so it jumps to 2×original). The equilibrium then shifts LEFT (1 mol N₂O₄ vs 2 mol NO₂ — higher pressure favours fewer moles of gas). As the system shifts left, some NO₂ is consumed, so [NO₂] decreases from 2×original to a new equilibrium value. This new value is higher than the original (because compression overall increases concentrations) but lower than 2×original (because some NO₂ was consumed). Option A is wrong — [NO₂] cannot return to exactly the original value under higher total pressure. Option C is wrong — there IS a shift (2 mol gas → 1 mol gas). Option D ignores the shift.

Question 8

The Contact process for sulfuric acid production uses the reaction: $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ $\Delta H = -197 \text{ kJ mol}^{-1}$, with a vanadium(V) oxide (V₂O₅) catalyst at ~450°C. Why is higher pressure NOT used industrially, even though it would increase SO₃ yield?

A Higher pressure would shift the equilibrium left for this reaction

B The vanadium catalyst only works at atmospheric pressure

C Higher pressure decreases Keq and reduces yield

D The yield is already very high (~98%) at moderate pressure, so the engineering cost of very high-pressure equipment cannot be economically justified

Correct: D. The reaction goes from 3 mol gas to 2 mol gas (Δn = −1), so higher pressure does shift the equilibrium right and increases SO₃ yield. However, at moderate pressures the yield is already ~98%, which is commercially acceptable. Building and maintaining equipment for very high pressures is extremely expensive. Since the yield gain from further pressure increase is minimal (you're already near 100%), it's not economically justified. Option A is wrong — higher pressure shifts RIGHT for this reaction (fewer product moles). Option C is wrong — pressure doesn't change Keq.

Question 9

Adding an inert gas (such as argon) to an equilibrium mixture at constant volume has what effect?

A The equilibrium shifts to the side with fewer moles of gas

B The equilibrium shifts to the side with more moles of gas

C No shift occurs; the partial pressures (and concentrations) of the reacting species are unchanged at constant volume

D Keq increases because total pressure increases

Correct: C. Adding an inert gas at constant volume does NOT change the concentrations (or partial pressures) of any reacting species — only the total pressure increases. Since equilibrium depends on the concentrations/partial pressures of reacting species, there is no shift. If the volume were allowed to expand (constant pressure), then concentrations would decrease and the equilibrium would shift toward more moles of gas. But at constant volume, no shift occurs. Option D is wrong — Keq is unchanged by pressure or inert gas addition.

Question 10

At equilibrium, some solid CaCO₃(s) is present in a flask with CaO(s) and CO₂(g): $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$. More CaCO₃(s) is added. What happens?

A The equilibrium shifts right; [CO₂] increases

B No shift occurs; solids do not appear in the equilibrium expression, so adding more solid does not affect the equilibrium position

C The equilibrium shifts left; [CO₂] decreases

D Keq increases because more CaCO₃ is available

Correct: B. Solids and pure liquids are NOT included in the equilibrium expression (Keq) because their "concentrations" are constant (fixed by density). Adding more solid CaCO₃ does not change [CO₂] or Q, so there is no shift. The only thing that matters is whether any solid is present (to maintain the heterogeneous equilibrium) — the amount of solid doesn't matter. Options A and C both predict a shift, which is wrong. Option D is wrong — Keq is unaffected.

Section B — Short Answer

Question 11

The reaction $2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)$ $\Delta H = -57 \text{ kJ mol}^{-1}$ is at equilibrium in a sealed flask. NO₂ is brown; N₂O₄ is colourless. Predict and explain the colour change observed when (a) the flask is cooled and (b) the flask is compressed to a smaller volume. (4 marks)

4 marks

Model Answer (4 marks):

(a) Cooling the flask (2 marks): The reaction is exothermic (ΔH = −57 kJ mol⁻¹), so heat is a product. Cooling removes heat, creating a stress. By LCP, the equilibrium shifts to oppose this — it shifts RIGHT (toward N₂O₄) to produce more heat (1 mark). As more N₂O₄ forms and [NO₂] decreases, the brown colour fades / the mixture becomes lighter (less brown, more colourless) (1 mark).

(b) Compressing to smaller volume (2 marks): Compression increases pressure. By LCP, the system shifts to reduce pressure by favouring the side with fewer moles of gas. Reactants: 2 mol gas; products: 1 mol gas → the equilibrium shifts RIGHT (toward N₂O₄) (1 mark). As more N₂O₄ forms and [NO₂] decreases, the brown colour fades. Note: initially the colour intensifies (due to the immediate concentration increase from compression) before fading as the shift occurs (1 mark).

Question 12

In the Haber process, $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ $\Delta H = -92 \text{ kJ mol}^{-1}$, the operating conditions are 400–500°C, 150–300 atm, with an iron catalyst. For each of the three conditions, explain: (i) why it was chosen, and (ii) the trade-off it involves. (6 marks)

6 marks

Model Answer (6 marks):

Temperature 400–500°C (2 marks): Why chosen: The reaction is exothermic, so lower temperature → higher Keq → higher equilibrium yield. However, lower temperature also means lower reaction rate. 400–500°C is the compromise temperature that balances acceptable yield with a commercially viable reaction rate (1 mark). Trade-off: at this temperature, the equilibrium yield is only ~15–25%, which is low — but the rate is fast enough to produce NH₃ quickly. Unreacted N₂/H₂ are recycled (1 mark).

Pressure 150–300 atm (2 marks): Why chosen: The reaction has 4 mol gas on the left and 2 mol on the right (Δn = −2). Higher pressure shifts equilibrium right (toward fewer gas moles), increasing NH₃ yield. Higher pressure also increases rate (more frequent collisions) (1 mark). Trade-off: very high pressures require expensive, specialised equipment and significant energy to maintain. 150–300 atm is the economic compromise between yield improvement and equipment/energy cost (1 mark).

Iron catalyst (2 marks): Why chosen: The iron catalyst (with K₂O/Al₂O₃ promoters) lowers the activation energy for both forward and reverse reactions, increasing the rate at which equilibrium is reached (1 mark). Trade-off: the catalyst does NOT change the equilibrium position or Keq — it only makes equilibrium reached faster. The yield remains the same as without catalyst, but the process becomes commercially viable because it achieves the equilibrium yield in a much shorter time (1 mark).

Question 13

A student applies LCP to the following reaction at equilibrium and makes a prediction error:

$\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)$

The student claims: "Adding more Fe³⁺ will shift the equilibrium right and increase [SCN⁻]." Identify the error and provide the correct prediction with explanation. (3 marks)

3 marks

Model Answer (3 marks):

Error identified (1 mark): The student correctly predicted the shift direction (right) but incorrectly predicted that [SCN⁻] increases. When the equilibrium shifts right, SCN⁻ is consumed (it is a reactant in the forward reaction), so [SCN⁻] decreases, not increases.

Correct prediction (1 mark): Adding more Fe³⁺ increases [Fe³⁺], creating a stress. By LCP, the equilibrium shifts right to consume the excess Fe³⁺ and re-establish equilibrium. This means more FeSCN²⁺ is produced and SCN⁻ is consumed.

Correct outcome for all species (1 mark): [Fe³⁺] increases overall (even after the shift, it is higher than the original equilibrium value because only some of the added Fe³⁺ is consumed); [SCN⁻] decreases (consumed by the rightward shift); [FeSCN²⁺] increases (product of the rightward shift). The red colour of the solution intensifies due to increased [FeSCN²⁺].

Score Tracker

Self-Assessment

Section A — MC (Q1–10) /10

Q11 — NO₂/N₂O₄ shifts /4

Q12 — Haber process conditions /6

Q13 — LCP prediction error /3

Total /23—

Checkpoint 2 complete — IQ2 Le Chatelier's Principle