Chemistry • Year 12 • Module 5 • Lesson 8
Consolidation — LCP Mastery
Apply Le Chatelier’s Principle to real data, multi-stress graph interpretation, data-table comparisons, and contemporary Australian contexts including the Haber process and the blood buffer system.
1. Multi-stress concentration–time graph interpretation
The graph below shows the concentrations of N2O4(g) (colourless) and NO2(g) (brown) in a sealed syringe for the equilibrium:
N2O4(g) ⇌ 2NO2(g) ΔH = +57 kJ/mol
Two disturbances are applied in sequence. 9 marks
Figure 1.1. Concentration–time graph for N2O4(g) ⇌ 2NO2(g) in a sealed syringe under two sequential disturbances. (Illustrative — data are consistent with known Keq values for this equilibrium.)
1.1 Describe the immediate observation at t1 and identify the disturbance. Explain why both concentrations change simultaneously. 3 marks
1.2 After t1, the N2O4 concentration decreases from its peak while NO2 decreases even more sharply. Using Le Chatelier’s Principle, explain the direction of the equilibrium shift and state whether Keq changes. 2 marks
1.3 Describe the changes observed after t2 and explain this disturbance using Le Chatelier’s Principle, collision theory, and its effect on Keq. 4 marks
2. Data-table interpretation — industrial SO3 production conditions
The table below shows experimental data for the equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −197 kJ/mol. Each row shows different industrial conditions. 7 marks
| Condition | Temperature (°C) | Pressure (atm) | Catalyst | % SO3 at equilibrium | Time to reach equilibrium (min) |
|---|---|---|---|---|---|
| A | 450 | 1 | None | 80 | 420 |
| B | 600 | 1 | None | 54 | 190 |
| C | 450 | 5 | None | 91 | 400 |
| D | 450 | 1 | V2O5 | 80 | 18 |
| E | 600 | 5 | V2O5 | 62 | 8 |
2.1 Compare Conditions A and B. What do these results confirm about the effect of temperature on Keq for this reaction? Explain using ΔH. 2 marks
2.2 Compare Conditions A and C. Using Le Chatelier’s Principle, explain why the SO3 yield increases from 80% to 91% when pressure is raised from 1 to 5 atm. 2 marks
2.3 Compare Conditions A and D. What does this comparison demonstrate about the role of the catalyst? Why does the yield remain 80% in both cases? 2 marks
2.4 Predict and justify what would happen to the % SO3 yield if Condition E were repeated at 450°C instead of 600°C (pressure 5 atm, V2O5 catalyst). 1 mark
3. Case study — the blood buffer system in Australian emergency medicine
In 2023, Westmead Hospital’s intensive care unit managed patients with severe respiratory acidosis — a condition in which elevated CO2 from hypoventilation disturbs blood pH. The key equilibrium governing blood pH is:
CO2(aq) + H2O(l) ⇌ H2CO3(aq) ⇌ H+(aq) + HCO3−(aq)
Blood pH is normally 7.35–7.45. When blood pH drops below 7.35 (acidosis) or rises above 7.45 (alkalosis), life-threatening consequences can occur. The kidneys and lungs are both involved in regulating this equilibrium. 6 marks
3.1 In respiratory acidosis, excess CO2 accumulates in the blood. Using Le Chatelier’s Principle, explain the direction of the equilibrium shift and its effect on blood pH. 2 marks
3.2 A clinician administers intravenous sodium bicarbonate (NaHCO3), which releases HCO3− into the blood. Predict the direction of the equilibrium shift and explain how this counteracts acidosis. 2 marks
3.3 A medical student argues: “Adding bicarbonate permanently increases Keq for the blood buffer reaction, making the blood more resistant to future acid loads.” Evaluate this claim using lesson content. 2 marks
4. Predict and justify — multi-stress methane reforming
An engineer is operating the reaction:
CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g) ΔH = −206 kJ/mol
The system is at equilibrium. The engineer simultaneously decreases the temperature by 100°C AND adds more CO(g) to the reactor. 5 marks
4.1 Use the six-step multi-variable method to predict the overall effect on equilibrium position and Keq. Address each disturbance separately before combining. 4 marks
4.2 If the engineer also adds a nickel catalyst at the same time, how does this change your prediction for equilibrium position and Keq? 1 mark
Q1.1 — t1 identification (3 marks)
The disturbance at t1 is a decrease in volume (increase in pressure) — the syringe was compressed. The immediate observation is that both [N2O4] and [NO2] increase simultaneously and proportionally [1]. This is because compressing to half the volume instantaneously doubles the concentration of all gas species before any equilibrium shift takes place — the shift requires time for collisions to occur [1]. The signature of “all gas concentrations jump up simultaneously” uniquely identifies a pressure increase; adding a single reactant would only increase that one species [1].
Q1.2 — Shift after t1 (2 marks)
LCP: increasing pressure shifts equilibrium toward the side with fewer moles of gas. Left side: 1 mol N2O4. Right side: 2 mol NO2. Shift is to the left — more N2O4 is formed and [NO2] falls [1]. Keq is unchanged because pressure does not alter Keq; only temperature does [1].
Q1.3 — t2 disturbance (4 marks)
After t2, [NO2] rises gradually while [N2O4] falls gradually — no sudden jump, indicating a temperature increase (not a concentration change) [1]. LCP: the forward reaction is endothermic (ΔH = +57 kJ/mol); increasing temperature shifts equilibrium in the endothermic (forward/right) direction to absorb the added heat, producing more NO2 and consuming N2O4 [1]. Collision theory: increasing temperature raises average kinetic energy. The forward (endothermic) reaction has a higher activation energy than the reverse. A proportionally greater fraction of particles now exceed the forward Ea, so the forward rate increases more than the reverse rate → net forward shift [1]. Keq increases because temperature increased for an endothermic forward reaction; a higher Keq means more products at the new equilibrium [1].
Q2.1 — Temperature effect on Keq (2 marks)
Increasing temperature from 450°C (Condition A) to 600°C (Condition B) decreases % SO3 from 80% to 54% [1]. This confirms that Keq decreases with increasing temperature because the forward reaction is exothermic (ΔH = −197 kJ/mol); the system shifts left (endothermic direction) to counteract added heat, reducing the proportion of SO3 at equilibrium [1].
Q2.2 — Pressure effect (2 marks)
Left side: 2 + 1 = 3 mol gas. Right side: 2 mol gas. Increasing pressure from 1 to 5 atm shifts equilibrium toward the side with fewer gas moles (right) [1], producing more SO3 and raising the yield from 80% to 91%. Keq is unchanged (pressure does not change Keq) [1].
Q2.3 — Catalyst role (2 marks)
Comparing A and D: the V2O5 catalyst reduces the time to reach equilibrium from 420 to 18 minutes — it dramatically increases the rate of reaching equilibrium [1]. The yield remains 80% in both cases because the catalyst does not change the equilibrium position or Keq; it lowers activation energy equally for forward and reverse reactions, so both rates increase but the equilibrium ratio is unaffected [1].
Q2.4 — Prediction (1 mark)
At 450°C (lower temperature than 600°C in Condition E), Keq is larger (exothermic forward). With pressure still at 5 atm, the yield would increase above 91% (higher than Condition C at 1 atm), likely approaching approximately 97–99% — both temperature decrease and pressure increase favour more SO3 [1]. Accept any answer identifying that lower T increases yield and the combined effect of low T + high P gives the highest SO3 %.
Q3.1 — Respiratory acidosis (2 marks)
Excess CO2 is added to the equilibrium. LCP: adding a reactant (CO2) shifts the equilibrium to the right, producing more H+ and HCO3− [1]. The increase in [H+] decreases blood pH — the blood becomes more acidic (below 7.35), constituting acidosis [1].
Q3.2 — Bicarbonate treatment (2 marks)
Adding HCO3− increases the concentration of a product. LCP: adding a product shifts the equilibrium to the left, consuming H+ and HCO3− and reducing [H+] in the blood [1]. This raises blood pH back toward normal (7.35–7.45), counteracting acidosis [1].
Q3.3 — Evaluating the claim (2 marks)
The claim is incorrect. Adding bicarbonate is a concentration change, not a temperature change. Concentration changes shift the equilibrium position but never change the value of Keq [1]. Keq for the blood buffer reaction depends only on temperature; adding HCO3− transiently shifts the position left but once equilibrium is re-established at the same temperature, Keq is unchanged. The blood is not permanently more “resistant” — the underlying equilibrium constant is the same [1].
Q4.1 — Multi-variable method (4 marks)
Disturbance 1 — temperature decrease: Forward reaction is exothermic (ΔH = −206 kJ/mol). Decreasing temperature shifts equilibrium in the exothermic direction (forward/right) to release heat. Keq increases [1].
Disturbance 2 — adding CO: Increasing reactant concentration shifts equilibrium to the right. Keq unchanged [1].
Combined effect: Both disturbances shift equilibrium to the right — effects reinforce [1]. Net: equilibrium position shifts right (more CH4 and H2O produced). Keq increases (due to the temperature decrease only) [1].
Q4.2 — Catalyst addition (1 mark)
The nickel catalyst does not change the equilibrium position or Keq; the prediction of a right shift with increased Keq is unchanged. The catalyst merely allows the system to reach the new equilibrium position more quickly [1].