Chemistry • Year 12 • Module 5 • Lesson 8
Consolidation — LCP Mastery
Cement the key vocabulary, disturbance rules, and graph signatures for all three Le Chatelier’s Principle stresses before moving on to multi-variable and extended-response work.
1. Term–definition match
Match each term in the right-hand column to its definition. Write the matching term in the blank. 12 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | A system at equilibrium shifts to partially counteract any applied disturbance. | |
| 1.2 | The ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficient. | |
| 1.3 | The only factor that changes the value of the equilibrium constant Keq. | |
| 1.4 | When a reactant is added to an equilibrium mixture, the system shifts in this direction. | |
| 1.5 | When volume is decreased (pressure increased), the equilibrium shifts toward the side with this many moles of gas. | |
| 1.6 | Adding this substance speeds up attainment of equilibrium but does not change the equilibrium position or Keq. | |
| 1.7 | On a concentration–time graph, this disturbance causes ALL gas concentrations to jump simultaneously before re-equilibration. | |
| 1.8 | For an exothermic forward reaction, increasing temperature causes Keq to do this. | |
| 1.9 | The speed at which the system reaches equilibrium (governed by kinetics, not thermodynamics). | |
| 1.10 | The proportion of reactants converted to products at equilibrium; determined by Keq. | |
| 1.11 | For an endothermic forward reaction, increasing temperature shifts equilibrium in this direction. | |
| 1.12 | On a concentration–time graph, adding a catalyst to a system already at equilibrium produces this observable change. |
Term bank: Le Chatelier’s Principle • Keq expression • temperature • forward (right) • fewer • catalyst • volume decrease (pressure increase) • decreases • rate • yield • forward (right) • no observable change
Stuck? Revisit the Key Terms panel and the graph-signatures table in Card 3 of the lesson.2. True or false — with correction
Circle T or F for each statement. If false, write the corrected version on the line below. 10 marks (1 T/F + 1 correction where needed)
2.1 Increasing pressure always shifts a gaseous equilibrium to the right. T / F
2.2 For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ/mol, increasing temperature decreases the value of Keq. T / F
2.3 Adding a catalyst to a system already at equilibrium shifts the equilibrium position to the right. T / F
2.4 Removing a product from an equilibrium mixture causes the system to shift to the right, but Keq is unchanged. T / F
2.5 For N2O4(g) ⇌ 2NO2(g), decreasing the volume shifts the equilibrium to the right because NO2 molecules are smaller. T / F
3. Cloze paragraph — the Haber process
Complete the paragraph by writing one word or short phrase in each blank. Each blank corresponds to one mark. 10 marks
Word bank: decreases • 400–500°C • iron • right • left • fewer • unchanged • exothermic • rate • yield
The Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ/mol, is the industrial synthesis of ammonia. The forward reaction is (3.1) ______________, releasing heat energy. According to Le Chatelier’s Principle, increasing temperature shifts the equilibrium to the (3.2) ______________ because the system counteracts the added heat by favouring the endothermic reverse reaction. This (3.3) ______________ the value of Keq. Increasing the pressure shifts the equilibrium to the right because the product side has (3.4) ______________ moles of gas (4 mol reactants → 2 mol products), favouring the side of (3.5) ______________ gas molecules. Pressure changes leave Keq (3.6) ______________. The industrial catalyst used in the Haber process is (3.7) ______________. This catalyst improves the (3.8) ______________ of ammonia production — how fast equilibrium is reached — but does not affect the equilibrium (3.9) ______________ or Keq. The industrial compromise temperature of (3.10) ______________ balances acceptable rate against acceptable yield.
4. Function recall — what does each factor do?
Answer each in 1–2 sentences using precise lesson terms. 8 marks (2 each)
4.1 What is the function of temperature as a disturbance in an equilibrium system — how does it differ from all other disturbances in its effect on Keq?
4.2 What is the role of a catalyst at equilibrium? Why does it not change the equilibrium position?
4.3 For the equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g), what is the function of increasing pressure on the equilibrium position? Identify which side has fewer gas moles to justify your answer.
4.4 Distinguish between rate of reaction and equilibrium yield. What is an example of a change that improves rate but decreases yield?
5. Build a concept map — Le Chatelier’s Principle
Draw labelled arrows between the six terms below to show how they connect. Each arrow must carry a linking phrase (e.g. “shifts equilibrium toward”, “does not change”, “is the only factor that changes”). Aim for at least 6 labelled arrows. 6 marks
Supplied terms: equilibrium position • Keq value • temperature change • concentration change • pressure change • catalyst
Q1 — Term–definition matches
1.1 Le Chatelier’s Principle • 1.2 Keq expression • 1.3 temperature • 1.4 forward (right) • 1.5 fewer • 1.6 catalyst • 1.7 volume decrease (pressure increase) • 1.8 decreases • 1.9 rate • 1.10 yield • 1.11 forward (right) • 1.12 no observable change.
Q2 — True/false with correction
2.1 False. Increasing pressure shifts the equilibrium toward the side with fewer moles of gas. If moles are equal on both sides, there is no shift.
2.2 True. For an exothermic forward reaction (ΔH < 0), increasing temperature favours the reverse endothermic direction, shifting equilibrium left and decreasing Keq.
2.3 False. A catalyst does not shift the equilibrium position. It lowers the activation energy equally for both forward and reverse reactions, so both rates increase equally — no net change in position or Keq.
2.4 True. Removing a product reduces its concentration → reverse rate decreases → forward rate > reverse rate → shift right. Keq is unchanged because only temperature changes Keq.
2.5 False. Equilibrium shifts toward fewer moles of gas, not toward smaller molecules. Decreasing volume for N2O4(g) ⇌ 2NO2(g) shifts left (toward 1 mol N2O4, away from 2 mol NO2).
Q3 — Cloze paragraph answers
3.1 exothermic • 3.2 left • 3.3 decreases • 3.4 fewer • 3.5 fewer • 3.6 unchanged • 3.7 iron • 3.8 rate • 3.9 yield • 3.10 400–500°C.
Q4.1 — Temperature as a disturbance
Temperature is the only disturbance that changes the value of Keq. All other disturbances (concentration, pressure, catalyst) shift the equilibrium position but leave Keq unchanged. Increasing temperature for an exothermic forward reaction shifts equilibrium left and decreases Keq; for an endothermic forward reaction, it shifts right and increases Keq.
Q4.2 — Role of a catalyst
A catalyst lowers the activation energy for both the forward and reverse reactions equally. At equilibrium, this means both rates increase by the same factor — forward and reverse rates remain equal, so no net shift in equilibrium position occurs and Keq is unchanged. The catalyst only affects how quickly equilibrium is reached (rate), not where it sits (position or Keq).
Q4.3 — Pressure increase for 2SO2 + O2 ⇌ 2SO3
Left side: 2 + 1 = 3 mol gas. Right side: 2 mol gas. Increasing pressure shifts equilibrium toward the side with fewer moles of gas — to the right — producing more SO3. Keq is unchanged (pressure does not change Keq).
Q4.4 — Rate vs yield
Rate = speed at which the system reaches equilibrium (kinetics). Yield = proportion of reactants converted to products at equilibrium (thermodynamics, determined by Keq). For an exothermic forward reaction such as the Haber process, increasing temperature increases rate but decreases yield (Keq decreases). These are independent — a higher rate does not mean a higher yield.
Q5 — Sample concept map
A correct map should include arrows such as:
- temperature change — is the only factor that changes → Keq value
- temperature change — also shifts → equilibrium position
- concentration change — shifts → equilibrium position
- concentration change — does not change → Keq value
- pressure change — shifts (if unequal gas moles) → equilibrium position
- pressure change — does not change → Keq value
- catalyst — changes neither → equilibrium position and Keq value
Award 1 mark per correctly labelled arrow that accurately reflects the chemical relationship (max 6).