Loav · VCE Chemistry · Unit 3 / Area of Study 2
U3A2-02 · 2026 curriculum · Expanded original teaching guide · Study guide
1. Constant does not mean stopped
A reversible reaction can proceed in both directions under the conditions considered. Reversibility alone does not establish equilibrium: a freshly mixed system can be reversible while its composition is still changing. Dynamic equilibrium is reached in a closed system under constant conditions when forward and reverse reaction rates are equal. Both reactions continue. Measurable concentrations become constant because each species is formed and consumed at equal rates, not because all particles have stopped reacting.
Equal rates do not imply equal concentrations of reactants and products. Some equilibria contain mostly products; others mostly reactants. “Homogeneous” means all reacting species occupy the same phase, such as a gas mixture or one aqueous solution. A sealed vessel can exchange heat, so “closed” is not the same as thermally insulated. A constant-temperature condition must be stated separately.
Rate asks how fast composition changes; extent asks how far the reaction has proceeded. A very slow reaction may eventually favour products strongly. A fast reaction may establish an equilibrium containing mostly reactants. If an irreversible approximation is suitable, the reverse reaction is negligible under those conditions; do not infer that every reaction can never reverse under any different conditions.
2. A disturbance has two stages
Le Chatelier’s principle predicts that a system disturbed from equilibrium undergoes a net reaction that tends to oppose the imposed change. “Oppose” does not mean completely undo. The new equilibrium normally differs from the old one. First identify the instantaneous physical effect, then the subsequent chemical adjustment. A balanced equation is essential because its coefficients govern how concentration changes relate.
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), adding N₂ at fixed volume makes [N₂] jump immediately. The other two concentrations do not jump merely because N₂ was added. The subsequent forward net reaction consumes N₂ and H₂ and produces NH₃: [N₂] falls from its post-addition value, [H₂] falls and [NH₃] rises smoothly. N₂ need not return to its original concentration. During adjustment, rates are unequal; at the new plateaus they are equal again.
Removing NH₃ at fixed volume gives an immediate downward jump in its concentration, followed by partial recovery as the net forward reaction produces more. N₂ and H₂ then decline gradually. Never draw a product jump caused by chemical production at the exact instant another species is added: chemical adjustment takes time.
3. Volume, partial pressure and mole counts
Compressing a gas mixture at constant temperature raises every gas concentration instantly: halving volume doubles them. The subsequent net reaction favours the side with fewer total moles of gas. In the ammonia equation, four gaseous reactant moles become two gaseous product moles, so compression favours products. After the jump, reactants decline and product increases further. The relevant comparison is the sum of gaseous coefficients, not the number of different gas species.
If gas mole totals are equal, such as H₂(g) + I₂(g) ⇌ 2HI(g), uniform compression does not shift equilibrium in the ideal-gas concentration model, although all concentrations jump. “Increase pressure” is incomplete unless the mechanism is specified. Adding an inert gas at constant volume leaves reacting-gas partial pressures unchanged and causes no shift. Changes at constant total pressure can instead change volume and reacting-gas concentrations; inspect the stated conditions.
Diluting an aqueous homogeneous equilibrium raises volume and immediately lowers dissolved-species concentrations. Subsequent direction follows the equilibrium expression and coefficients of the dissolved species. Do not transfer a gas-pressure rule to a solution without examining the chemical system.
4. Temperature changes the equilibrium itself
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH < 0 for the forward reaction. Heating favours the endothermic reverse direction, reducing equilibrium ammonia yield. Cooling favours the exothermic forward direction. At fixed volume, changing temperature does not instantly change the number of molecules or their concentration, so concentrations start continuous and then change through reaction. Gas pressure can change immediately because temperature has changed.
Temperature changes the equilibrium constant for a fixed equation; concentration, volume and a catalyst do not. Heating an exothermic forward system decreases Kc; heating an endothermic forward system increases Kc. A catalyst accelerates both directions and helps a disturbed system reach its new equilibrium faster. It cannot shift an undisturbed equilibrium at fixed temperature.
Worked example 1: reconstruct a disturbance
Original synthetic data at fixed temperature and volume: equilibrium concentrations are [N₂] = 0.200, [H₂] = 0.600 and [NH₃] = 0.100 mol L⁻¹. N₂ is added to give an immediate 0.300 mol L⁻¹. At the new equilibrium, N₂ is 0.29276 mol L⁻¹. Thus x = 0.00724 mol L⁻¹ is consumed subsequently. Hydrogen becomes 0.600 − 3x = 0.57828; ammonia becomes 0.100 + 2x = 0.11448 mol L⁻¹. The rounded concentrations preserve the original Kc within rounding. The final nitrogen concentration is still higher than before addition.
Worked example 2: choose industrial conditions
A reversible exothermic synthesis has an unacceptably slow rate at low temperature. Cooling improves equilibrium yield but usually reduces rate. A defensible choice is a moderate operating temperature with a suitable catalyst, alongside separation of product and recycling of unreacted feed where feasible. Explain each benefit separately: the catalyst speeds approach to equilibrium, lower operating temperature can reduce heating energy, and product removal promotes further formation in the remaining reacting mixture. Compression may favour yield if fewer gas moles form, but pressure equipment and compression consume resources.
5. Diagnose a graph by checking what must happen first
Read a concentration–time graph in three passes. First find intervals where all reacting-species concentrations are stable. Then inspect the exact disturbance time: one species jumping suggests selective addition/removal at fixed volume, whereas every gas concentration changing by the same factor suggests a volume change. Finally inspect the smooth adjustment and check that its changes have the ratios required by the balanced equation. These are clues under the specified model, not unique diagnoses when multiple disturbances occur together.
A temperature change at fixed volume causes no instantaneous concentration jump. It can immediately alter forward and reverse rates, which then produces curved concentration changes. A catalyst added to a system already at equilibrium causes no concentration change: both opposing rates increase while their equality is maintained. A catalyst added before equilibrium can steepen concentration curves toward the same endpoint. A flat concentration trace therefore cannot reveal whether the microscopic turnover is fast or slow.
Rate–time graphs require different reading. At equilibrium, forward and reverse reaction rates meet at a common nonzero value, whereas a concentration–time graph has horizontal plateaus at potentially different values. Do not transfer the “equal curves” rule between these representations. Where species formation/disappearance rates are graphed instead of forward/reverse reaction rates, stoichiometric coefficients may mean their numerical magnitudes differ.
Worked example 3: more product but lower concentration
Transfer, original ideal-gas dataset. At a fixed temperature, N₂O₄(g) ⇌ 2NO₂(g) has Kc = 0.400 mol L⁻¹. A 1.00 L equilibrium mixture contains [N₂O₄] = 0.100 and [NO₂] = 0.200 mol L⁻¹. The vessel expands to 2.00 L without matter loss, with temperature controlled. Immediately the amounts are unchanged but concentrations halve to 0.0500 and 0.100 mol L⁻¹. Q = 0.100²/0.0500 = 0.200 mol L⁻¹, below Kc, so net forward dissociation occurs.
If x mol L⁻¹ of N₂O₄ dissociates at the new volume, concentrations become 0.0500 − x and 0.100 + 2x. Substitution gives (0.100 + 2x)²/(0.0500 − x) = 0.400. Rearrangement gives 4x² + 0.800x − 0.0100 = 0. The physically valid root is x = 0.0118034; the negative root is inconsistent with the predicted forward adjustment and gives a negative NO₂ concentration. Final [NO₂] = 0.123607 mol L⁻¹ and [N₂O₄] = 0.0381966 mol L⁻¹.
Final n(NO₂) = 0.123607 × 2.00 = 0.247 mol, greater than its original 0.200 mol. Yet final [NO₂] = 0.124 mol L⁻¹ is lower than its original 0.200 mol L⁻¹. There is no contradiction: amount and volume have both changed. Draw NO₂ jumping down to 0.100, then rising smoothly to 0.124; N₂O₄ jumps down to 0.0500, then falls to 0.0382. Chemical adjustment magnitudes are in the ratio 2:1, and Kc is unchanged.
The quadratic above is an optional algebraic extension showing why the graph is physically consistent; the central course skill is the distinction between immediate dilution, subsequent direction and final amount versus concentration. Even when a question supplies the final composition instead of asking you to solve it, use cV to answer an amount question.
Temperature inference needs a controlled comparison. If the same balanced equation has a larger measured equilibrium Kc at a higher temperature, the forward reaction is endothermic. A higher product concentration alone is insufficient evidence: changed starting quantities or volume can also change composition. If rounded graph readings give slightly different Kc estimates, consider reading precision before asserting an undisclosed temperature change.
For industrial decisions, separate single-pass conversion from overall material recovery and production per hour. Recycling unreacted feed can improve overall use of material even if one pass has moderate conversion. Product removal changes the reacting mixture and enables further reaction; it does not change the fixed-temperature Kc. Heat recovery may reduce external energy demand. A well-justified recommendation weighs actual production rate, separation energy, pressure costs, waste and safety, instead of treating maximum equilibrium yield as the sole objective.
Practice
- For 2NO₂(g) ⇌ N₂O₄(g), predict immediate and later changes after halving volume at fixed temperature.
- A mixture has constant but unequal concentrations. Is that inconsistent with equilibrium?
- For an endothermic forward reaction, predict the effects of heating at fixed volume on concentration jumps, direction and Kc.
- Correct “Adding a catalyst shifts the equilibrium right because it speeds up the forward reaction.”
- For the expansion in worked example 3, give the immediate and final amounts of NO₂. Explain why the immediate amount does not halve.
- A fixed-volume equilibrium shows no instantaneous concentration jumps, followed by increasing products. Give two possible disturbances and explain what additional observation could distinguish them.
- An operator reduces temperature in an exothermic synthesis and adds a catalyst. Explain separately what happens to Kc, equilibrium yield and rate.
Answers with reasoning
- Both concentrations double immediately. Two gaseous moles become one, so a subsequent forward shift decreases NO₂ from its jumped value and increases N₂O₄ further. Kc is unchanged.
- No. Equal forward and reverse rates require constant concentrations, not equal concentrations. Constant readings alone need context: an inactive or exhausted system can also appear constant.
- No instantaneous concentration jump at fixed volume. The endothermic forward direction is favoured; products increase through reaction and Kc increases.
- The catalyst lowers the barriers of both directions. At equilibrium both rates rise while remaining equal; composition and Kc do not change at fixed temperature.
- Immediately n = 0.100 mol L⁻¹ × 2.00 L = 0.200 mol; expansion alone moves no matter out. Finally n = 0.123607 × 2.00 = 0.247 mol. Only the subsequent chemical reaction increases the amount.
- Heating an endothermic forward system or cooling an exothermic forward system can both cause a smooth forward adjustment. Measure temperature and know the sign of ΔH; the graph shape alone cannot choose between them. A catalyst at an already established equilibrium cannot produce that shift.
- Lower temperature increases Kc and favours products. The catalyst itself changes neither Kc nor fixed-temperature equilibrium yield. Cooling tends to slow reactions, while catalysis accelerates them relative to the uncatalysed reaction at that same temperature. Without kinetic data, the combined operating rate cannot be ranked against the original hot uncatalysed run.
