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QLDChemistrySyllabus dot point

Topic 1: Chemical equilibrium systems

Predict, using Le Chatelier's principle, the qualitative effects of changes in concentration, temperature, pressure and volume on the equilibrium position of homogeneous reactions

A focused answer to the QCE Chemistry Unit 3 dot point on Le Chatelier's principle. Defines the principle, works through how concentration, temperature, pressure and volume changes shift equilibrium position, explains why catalysts do not shift equilibrium, and applies the reasoning to the Haber process and the iron(III) thiocyanate system used in IA1 and IA2.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
  2. The answer
  3. Worked example: applying Le Chatelier in IA2
  4. Examples in context
  5. Try this

What this dot point is asking

QCAA wants you to apply Le Chatelier's principle to predict the direction of equilibrium shifts under four kinds of disturbance (concentration, temperature, pressure or volume in gaseous systems) and to explain the predictions in terms of how the system opposes the change. The principle reappears in IA1 stimulus, IA2 experimental design (most IA2s involve perturbing equilibrium and observing the response), and the EA.

The answer

Le Chatelier's principle states: when a system at equilibrium is subjected to a change in conditions, the system will shift its equilibrium position in the direction that partially opposes the change.

A shift "to the right" favours product formation; a shift "to the left" favours reactant reformation. The four disturbances examinable in Unit 3 are concentration, temperature, total pressure (or volume) in gaseous systems, and the addition of a catalyst (which is a control case, not a shift).

Concentration changes

For any equilibrium A + B leftrightarrow C + D:

  • Increase [A] or [B]. Forward rate rises (more reactant particles colliding). Shift right; [C] and [D] increase, [A] and [B] partially consumed.
  • Decrease [A] or [B]. Forward rate falls. Shift left; [C] and [D] partially consumed.
  • Increase [C] or [D]. Reverse rate rises. Shift left.
  • Decrease [C] or [D]. Reverse rate falls. Shift right.

Important. The shift partially opposes the change. After re-equilibration the disturbed species does not return to its original value; only a fraction of the disturbance is undone. Kc is unchanged.

Temperature changes

Temperature is the only disturbance that actually changes Kc. Whether the equilibrium shifts forward or back depends on the thermicity of the forward reaction.

  • Exothermic forward reaction (Ξ”H\Delta H negative). Treat heat as a product. Increasing T adds product; shift left. Kc decreases.
  • Endothermic forward reaction (Ξ”H\Delta H positive). Treat heat as a reactant. Increasing T adds reactant; shift right. Kc increases.

The Haber process N2 + 3H2 leftrightarrow 2NH3 has Ξ”H\Delta H = -92 kJ/mol (exothermic forward). Increasing T shifts left and decreases yield. This is why the industrial compromise sets temperature only as high as needed to keep the iron catalyst active.

Pressure and volume changes (gaseous systems only)

These two are linked: at constant temperature, decreasing volume increases pressure and vice versa. The shift direction depends on the relative moles of gas on each side of the equation.

For a gas-phase equilibrium with delta-n moles of gas:

  • Increase pressure (decrease volume). Shift toward the side with fewer moles of gas.
  • Decrease pressure (increase volume). Shift toward the side with more moles of gas.
  • No moles-of-gas difference. No shift in equilibrium position.

For N2 + 3H2 leftrightarrow 2NH3: reactant side has 4 mol gas, product side has 2 mol gas. Increasing pressure favours the product side (fewer mol). This is why industrial Haber reactors operate at high pressure (250 atm).

For H2 + I2 leftrightarrow 2HI: 2 mol gas each side. Pressure change has no effect on position.

Adding an inert gas at constant volume. Total pressure rises but partial pressures of reacting species are unchanged. No shift.

Adding an inert gas at constant pressure. The container expands; partial pressures of reacting species fall. Equivalent to a volume increase; shift toward more moles of gas.

Catalysts

A catalyst speeds the forward and reverse reactions equally. The equilibrium position does not shift and Kc does not change. The system reaches equilibrium faster.

This is examined as a distractor: a stimulus showing equilibrium re-established at the same concentrations after catalyst addition is consistent with a catalyst having no shift effect.

Visualising the shifts

A useful approach is the "what would the system do to undo the change?" test:

  • Added a reactant? The system consumes it -> shift right.
  • Removed a product? The system replaces it -> shift right.
  • Heated an exothermic forward reaction? The system absorbs heat -> shift left (endothermic direction).
  • Compressed a gas system? The system reduces gas pressure -> shift toward fewer moles of gas.

Sample concentration-vs-time graph after a disturbance

A typical IA1 stimulus shows a system at equilibrium, then a sudden disturbance (e.g. extra reactant injected), then re-equilibration.

Key features to identify on such a graph:

  • The first plateau before the disturbance: original equilibrium.
  • The vertical step at the disturbance: a concentration spike (or drop) of the affected species.
  • The adjustment curve following the step: the system responding (forward rate rises if reactant was added, etc.).
  • The second plateau at the new equilibrium: different concentrations than before but consistent with the same Kc (unless T was changed).

Recognising these features under the criteria "interpretation of stimulus" is exactly what IA1 rewards.

Worked example: applying Le Chatelier in IA2

A common IA2 design uses the iron(III) thiocyanate system. The student designs perturbations and predicts then tests the colour change.

Perturbation Le Chatelier prediction Observation
Add KSCN solution Shift right (consumes added SCN-) Solution deepens red
Add FeCl3 solution Shift right (consumes added Fe3+) Solution deepens red
Add NaOH solution Shift left (OH- removes Fe3+ as Fe(OH)3) Solution fades
Heat the tube (assume forward exothermic) Shift left Solution fades
Cool the tube Shift right Solution deepens

In the IA2 report, the strongest students explicitly state the principle, predict before observing, and then justify the observation against the equilibrium expression. The criterion "claim, evidence, reasoning" rewards each component shown clearly.

Examples in context

Example 1. Ocean acidification on the Great Barrier Reef. Dissolved CO2\text{CO}_2 equilibrates with seawater: CO2(aq)+H2Oβ‡ŒH2CO3β‡ŒH++HCO3βˆ’β‡Œ2H++CO32βˆ’\text{CO}_2(aq) + \text{H}_2 \text{O} \rightleftharpoons \text{H}_2 \text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- \rightleftharpoons 2 \text{H}^+ + \text{CO}_3^{2-}. Industrial-era atmospheric CO2\text{CO}_2 rising from 280280 to 420 ppm420 \, \text{ppm} pushes the first equilibrium right (Le Chatelier: added reactant). Increased H+\text{H}^+ pushes the third equilibrium left, lowering [CO32βˆ’][\text{CO}_3^{2-}]. Coral aragonite CaCO3\text{CaCO}_3 relies on Ca2++CO32βˆ’β†’CaCO3\text{Ca}^{2+} + \text{CO}_3^{2-} \rightarrow \text{CaCO}_3, so reduced carbonate ion crashes calcification rates. AIMS field data show a 14%14\% decline in calcification on the southern GBR since 1990, a striking real-world Le Chatelier outcome.

Example 2. Contact process at Mount Isa sulfuric-acid plant. Mount Isa Mines' acid plant runs 2SO2(g)+O2(g)β‡Œ2SO3(g)2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) exothermically (Ξ”H=βˆ’198 kJΒ molβˆ’1\Delta H = -198 \, \text{kJ mol}^{-1}). Le Chatelier favours forward reaction with high pressure (4 mol β†’\rightarrow 2 mol of gas) and low temperature. Reality: kinetics demand 400∘C400^{\circ}\text{C} even with V2O5\text{V}_2 \text{O}_5 catalyst. Compromise temperature plus interstage SO3\text{SO}_3 removal (Haber's trick: remove product to push equilibrium right) delivers 99.5%99.5\% overall conversion, capturing roof-stack SO2\text{SO}_2 from the lead smelter to make 700 t/day700 \, \text{t/day} of H2SO4\text{H}_2 \text{SO}_4.

Try this

Q1. State Le Chatelier's principle. Predict and justify the effect of (a) increasing T\text{T} and (b) increasing pressure on N2+3H2β‡Œ2NH3\text{N}_2 + 3 \text{H}_2 \rightleftharpoons 2 \text{NH}_3, Ξ”H<0\Delta H < 0. [4 marks]

  • Cue. Principle: system shifts to counter disturbance. (a) Forward exothermic β†’\rightarrow shift left, [NH3][\text{NH}_3] decreases. (b) Forward fewer moles gas β†’\rightarrow shift right.

Q2. Adding Ag+\text{Ag}^+ to a saturated AgCl(s)β‡ŒAg++Clβˆ’\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^- solution: predict observation and justify with KspK_{sp}. [3 marks]

  • Cue. Equilibrium shifts left; more AgCl(s)\text{AgCl}(s) precipitates; common ion effect; [Clβˆ’][\text{Cl}^-] falls so product [Ag+][Clβˆ’]=Ksp[\text{Ag}^+][\text{Cl}^-] = K_{sp} restored.

Q3. CoCl42βˆ’(blue)+6H2Oβ‡ŒCo(H2O)62+(pink)+4Clβˆ’\text{CoCl}_4^{2-}(\text{blue}) + 6 \text{H}_2 \text{O} \rightleftharpoons \text{Co(H}_2 \text{O})_6^{2+}(\text{pink}) + 4 \text{Cl}^-, Ξ”H>0\Delta H > 0 forward. (a) Predict observation on heating. (b) Predict observation on adding AgNO3\text{AgNO}_3. (c) Justify each with Le Chatelier. [2+2+3 marks]

  • Cue. (a) Becomes pinker? No: endothermic forward, heating shifts forward (more pink); reverse. Actually forward is endothermic so heating shifts pink-side; check direction. (b) Ag+\text{Ag}^+ removes Clβˆ’\text{Cl}^-, shift right β†’\rightarrow pink. Judgement criterion.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 QCAA-style6 marksThe Haber process produces ammonia by the equilibrium reaction N2(g) + 3H2(g) <-> 2NH3(g), Delta H = -92 kJ/mol. Industrial reactors operate at approximately 450 degrees C and 250 atm with an iron catalyst, achieving around 15 percent conversion per pass. (a) Using Le Chatelier's principle, predict and explain the effect on equilibrium yield of (i) increasing pressure, (ii) increasing temperature. (b) Explain why the industrial conditions are a compromise rather than the ones that maximise equilibrium yield.
Show worked answer β†’

A 6-mark answer needs both shifts explained with stoichiometry/thermodynamics, then the compromise argument.

(a)(i) Increasing pressure
The forward reaction has 4 mol of gas on the reactant side and 2 mol on the product side. Increasing total pressure (decreasing volume) favours the side with fewer moles of gas, so the equilibrium shifts right toward NH3. Yield of NH3 increases.
(a)(ii) Increasing temperature
The forward reaction is exothermic (Delta H = -92 kJ/mol). Increasing temperature adds energy; the system opposes the change by favouring the endothermic direction (reverse). Equilibrium shifts left, and yield of NH3 decreases.
(b) Why the compromise
Maximum equilibrium yield would suggest very high pressure (favours products) and very low temperature (favours products). But low temperature gives a very slow rate, so even though the yield is high, the time to reach equilibrium is impractical. Industrial conditions are chosen to give an economically acceptable rate at acceptable yield, supported by a catalyst (iron) that speeds attainment of equilibrium without shifting position. 250 atm is the upper bound where reactor cost remains viable; 450 degrees C is the lowest temperature at which the iron catalyst is still active. Unreacted gases are recycled.

Markers reward (1) correct shift direction with stoichiometric justification on pressure, (2) correct shift direction with enthalpy justification on temperature, (3) the rate/yield/cost trade-off explicit in the compromise answer.

2022 QCAA-style4 marksThe reaction Fe3+(aq) + SCN-(aq) <-> FeSCN2+(aq) was at equilibrium when 5 mL of 0.10 mol/L KSCN was added. (a) State the direction of the equilibrium shift. (b) Describe the colour change observed in the test tube. (c) Explain why the addition of solid KCl (potassium chloride) would have no effect on this equilibrium.
Show worked answer β†’

A 4-mark answer needs the shift, the colour, and the spectator-ion explanation.

(a) Shift direction
Adding KSCN increases [SCN-]. The system opposes the increase by consuming SCN-, so the equilibrium shifts right toward more FeSCN2+.
(b) Colour change
[FeSCN2+] increases, so the solution becomes a deeper red.
(c) Why KCl has no effect
K+ and Cl- are spectator ions; neither appears in the equilibrium expression. K+ does not react with SCN- or with Fe3+, and Cl- does not react with Fe3+ at appreciable extent in this concentration range. Adding KCl does not change [Fe3+], [SCN-] or [FeSCN2+] and so the position of equilibrium is unchanged.

Markers reward shift direction, the corresponding colour deepening, and explicit spectator-ion reasoning rather than just "no effect".

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