When you squeeze a syringe full of brown NO₂ gas, it gets paler — not because gas escapes, but because the increased collision frequency drives the equilibrium toward the colourless N₂O₄, and watching that colour change in real time is watching collision theory in action.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A sealed container holds N₂O₄(g) (colourless) and NO₂(g) (brown) at equilibrium at 25°C. The mixture is a pale brown colour — both species are present.
The container is now placed in an ice bath. Before reading any theory — predict what happens to the colour over the next few minutes. Does the brown get darker, lighter, or stay the same? Explain your reasoning using what you know about how particles collide. Write your prediction and reasoning now. You will revisit this at the end of the lesson with a full collision theory explanation.
📚 Core Content
Wrong: A catalyst increases the amount of product formed at equilibrium.
Right: A catalyst speeds up both forward and reverse reactions equally, so equilibrium is reached faster but the equilibrium position and product yield do not change. Catalysts lower activation energy but do not alter ΔH or the thermodynamic favourability of a reaction.
Collision theory is the particle-level language for everything in equilibrium chemistry — it explains not just whether reactions happen, but how fast, and why the rates change as the system evolves.
From Module 3, you learned that for a reaction to occur, particles must collide with correct orientation and sufficient energy (≥ the activation energy Ea). The rate of reaction depends on the frequency of effective collisions.
In Module 5, this same framework explains the approach to dynamic equilibrium. As a reversible reaction proceeds forward, three things change simultaneously:
This is not coincidence — it is a direct consequence of collision theory applied to a reversible system.
The four stages of approaching dynamic equilibrium — collision theory language at each stage
The activation energy diagrams you studied in Module 4 take on new meaning in equilibrium chemistry — the relative heights of the forward and reverse energy barriers determine which direction is naturally preferred.
An energy diagram for a reversible reaction shows two activation energies:
The difference is the enthalpy change: $\Delta H = E_a(\text{forward}) - E_a(\text{reverse})$
For an exothermic forward reaction (ΔH < 0): Products are lower in energy than reactants. The energy barrier going forward (Ea forward) is smaller than the barrier going backward (Ea reverse). At a given temperature, a larger fraction of particles can cross the forward barrier → forward rate is inherently faster at the same concentrations → equilibrium position lies on the products side.
For an endothermic forward reaction (ΔH > 0): Products are higher in energy than reactants. Ea(reverse) is smaller than Ea(forward) → more particles can cross the reverse barrier → equilibrium position lies on the reactants side.
Energy profile for an exothermic forward reaction — products are lower in energy, so Ea(forward) < Ea(reverse)
A catalyst is the great equaliser of equilibrium chemistry — it accelerates both directions of a reversible reaction proportionally, getting you to equilibrium faster but leaving the destination unchanged.
A catalyst works by providing an alternative reaction pathway with a lower activation energy. In a reversible reaction, a catalyst lowers the Ea for BOTH the forward and reverse reactions by the same amount. Because both barriers are lowered equally:
The equilibrium between nitrogen dioxide and dinitrogen tetroxide is chemistry's most photogenic demonstration of collision theory at work — you can watch equilibrium shift in real time just by changing the temperature or pressure.
The equilibrium: 2NO₂(g) ⇌ N₂O₄(g), ΔH = −57 kJ/mol (exothermic)
Temperature effect (collision theory): The forward reaction is exothermic → Ea(forward) < Ea(reverse). At lower temperature, fewer particles exceed either activation energy, but proportionally more can still cross the lower forward barrier. The forward rate decreases less than the reverse rate → forward rate > reverse rate → equilibrium shifts right → more N₂O₄ forms → mixture becomes paler when cooled.
Pressure effect: Increasing pressure by compressing the gas increases the concentration of all species → both forward and reverse collision frequencies increase. But the forward reaction converts 2 moles of NO₂ into 1 mole of N₂O₄ — reducing total gas molecules. The forward rate increases more → equilibrium shifts toward fewer moles of gas (right) → more N₂O₄ → paler at higher pressure.
A rate-vs-time graph tells you the entire story of how a system approaches equilibrium — and knowing how to read and draw one is a core skill that distinguishes Band 4 answers from Band 6 answers in Module 5.
Three types of scenarios you must be able to draw and interpret:
| Scenario | What the graph shows | Why |
|---|---|---|
| Approach from pure reactants | Forward rate starts high, decreases; reverse rate starts at zero, increases; both meet at non-zero plateau | Forward collisions decrease as reactants consumed; reverse collisions increase as products form |
| Approach from pure products | Mirror image: reverse rate starts high, decreases; forward rate starts at zero, increases; both meet at same plateau | Reverse collisions decrease as products consumed; forward collisions increase as reactants form |
| Adding reactant to a system at equilibrium | Forward rate spikes up suddenly; reverse rate unchanged initially; both re-equalise at a new higher plateau | More reactant → more forward collisions immediately; reverse catches up as more product forms |
Key: the same equilibrium rate plateau is reached regardless of direction of approach (as long as total composition is the same). The equilibrium rate value represents the equal forward and reverse rates — both are non-zero.
✏️ Worked Examples
(a) Explain, using collision theory, how the system approaches dynamic equilibrium from this initial condition.
(b) Indicate which activation energy is larger — Ea(forward) or Ea(reverse) — and explain why using the sign of ΔH.
At t = 0, only PCl₅ is present. Concentration of PCl₅ is at maximum → frequency of effective forward collisions is at maximum → forward rate is at its maximum. There are no products (PCl₃ and Cl₂) → reverse rate is zero.
As the reaction proceeds: PCl₅ is consumed → its concentration decreases → frequency of effective forward collisions decreases → forward rate decreases. PCl₃ and Cl₂ are produced → their concentrations increase → frequency of effective reverse collisions increases → reverse rate increases.
When forward rate = reverse rate (both non-zero), dynamic equilibrium is established. Concentrations of all species become constant.
The forward reaction is endothermic (ΔH = +88 kJ/mol) → products are higher in energy than reactants → the energy barrier going forward (from low-energy reactants up to the transition state) is larger than the barrier going backward (from high-energy products up to the transition state).
Using ΔH = Ea(forward) − Ea(reverse) = +88 kJ/mol → confirms Ea(forward) > Ea(reverse).
→ Ea(forward) > Ea(reverse)
Summary: (a) Forward rate starts maximum and decreases as PCl₅ is consumed; reverse rate starts at zero and increases as PCl₃ and Cl₂ accumulate; equilibrium established when rates equalise. (b) Ea(forward) > Ea(reverse) because the endothermic forward reaction means products sit higher in energy — it takes more energy to climb the hill going forward than backward.
(a) Using a reaction energy diagram, explain why the V₂O₅ catalyst does not shift the equilibrium position.
(b) Explain why the catalyst is still valuable industrially.
(c) Decreasing temperature shifts equilibrium to the right for this exothermic reaction. Explain using Ea and collision theory.
The V₂O₅ catalyst provides an alternative lower-energy pathway. On the energy diagram, the transition state peak is lowered by the same amount for both forward and reverse reactions. Both Ea(forward) and Ea(reverse) decrease by the same value x.
Because both barriers are lowered equally, the rate of the forward reaction increases by the same factor as the rate of the reverse reaction. The ratio of forward to reverse rates is unchanged — and it is this ratio that determines Keq and equilibrium position.
→ No shift in equilibrium position; Keq unchanged.
Although the catalyst does not improve yield, it dramatically increases the rate at which equilibrium is reached. At 450°C, the uncatalysed reaction would be too slow to be economically viable. The catalyst allows equilibrium to be reached quickly at a moderate temperature, making the process commercially feasible.
Without the catalyst, a higher temperature would be needed for sufficient rate — but this would decrease yield (exothermic reaction shifts left at higher T). The catalyst allows the compromise of moderate temperature with acceptable rate.
The forward reaction (SO₂ + O₂ → SO₃) is exothermic → Ea(forward) < Ea(reverse). At lower temperature, average kinetic energy of particles decreases — both forward and reverse rates decrease.
However, the rate decreases proportionally more for the reaction with the higher Ea. Since Ea(reverse) > Ea(forward), the reverse rate decreases more than the forward rate. Now forward rate > reverse rate → net forward reaction → equilibrium shifts right → more SO₃ produced.
Summary: (a) Catalyst lowers Ea equally for both directions — ratio of rates unchanged — no shift in equilibrium position or Keq. (b) Catalyst enables acceptable rate at lower temperature, preserving higher yield from the exothermic equilibrium. (c) Decreasing T reduces both rates but reduces the reverse rate more (higher Ea reverse for exothermic reaction) → forward rate temporarily exceeds reverse → shift right to more SO₃.
🧪 Activities
| Change applied | Colour change? | Collision theory explanation | Effect on Keq |
|---|---|---|---|
| Syringe compressed (volume halved) | Your answer | Your answer | Your answer |
| Syringe placed in hot water bath (temperature increases) | Your answer | Your answer | Your answer |
| A catalyst is added to the syringe | Your answer | Your answer | Your answer |
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
❓ Multiple Choice
1. A reversible reaction starts with pure reactants in a sealed flask. Which description correctly explains how dynamic equilibrium is approached?
2. For the exothermic reaction A + B ⇌ C + D, which correctly describes the relative activation energies?
3. A V₂O₅ catalyst is added to the Contact process reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) at equilibrium. Which correctly describes the effect?
4. A sealed syringe contains an equilibrium mixture of NO₂(g) (brown) and N₂O₄(g) (colourless). The syringe is compressed to half its original volume. What happens to the colour of the mixture?
5. An industrial chemist argues: "We should increase the temperature for the Haber process N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92 kJ/mol) to increase the rate of reaching equilibrium, AND add more catalyst to increase the yield of NH₃." Evaluate both parts of this statement.
✍️ Short Answer
6. Explain, using collision theory, why a catalyst does not change the position of equilibrium or the value of Keq for a reversible reaction. 3 MARKS
7. The equilibrium 2NO₂(g) ⇌ N₂O₄(g) is established in a sealed container at 25°C. The container is placed in a hot water bath at 80°C.
(a) Using collision theory and activation energy, explain why the mixture becomes darker brown at 80°C. (3 marks)
(b) State whether Keq increases, decreases, or stays the same when temperature increases. Justify. (1 mark)
4 MARKS
8. Extended Response (7 marks): A chemist studies the equilibrium CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g), ΔH = −206 kJ/mol in a sealed vessel. The system is at equilibrium. Describe and explain, using collision theory and the concept of activation energy, the effect on the equilibrium position and on Keq when: (a) the temperature is increased; (b) a nickel catalyst is added; (c) the volume of the vessel is halved. 7 MARKS
Go back to your Think First prediction about the NO₂/N₂O₄ syringe placed in an ice bath.
Reaction 1 (SO₂ + O₂ ⇌ SO₃, ΔH = −196): Products (SO₃) sit BELOW reactants on the energy axis. Transition state is the peak. Ea(forward) is the height from reactants to the peak — smaller. Ea(reverse) is the height from products to the peak — larger. Ea(forward) < Ea(reverse) because ΔH is negative: ΔH = Ea(fwd) − Ea(rev) = −196 kJ/mol → Ea(rev) is 196 kJ/mol larger than Ea(fwd). ΔH arrow points downward from reactants to products.
Reaction 2 (N₂ + O₂ ⇌ 2NO, ΔH = +180): Products (NO) sit ABOVE reactants on the energy axis. Ea(forward) is larger than Ea(reverse) because ΔH = Ea(fwd) − Ea(rev) = +180 kJ/mol. ΔH arrow points upward from reactants to products.
Catalyst added to Reaction 1: Both peaks drop by the same amount (x kJ/mol). New Ea(fwd,cat) = Ea(fwd) − x; New Ea(rev,cat) = Ea(rev) − x. Reactant and product energy levels are UNCHANGED. ΔH is UNCHANGED. Equilibrium position and Keq are UNCHANGED — the catalyst only lowers the barrier, not the destination.
Compressed (volume halved): Colour becomes briefly darker then paler. Immediately after compression, [NO₂] and [N₂O₄] both double → both rates increase. But the forward reaction (2 mol NO₂ → 1 mol N₂O₄) involves more gas molecules, so its collision frequency increases proportionally more → forward rate > reverse rate → equilibrium shifts right → more N₂O₄ → mixture becomes paler. Keq = unchanged (pressure does not change Keq).
Hot water bath (temperature increases): Colour becomes darker. 2NO₂ → N₂O₄ is exothermic → Ea(fwd) < Ea(rev). Increasing T raises both rates but the reverse (endothermic, higher Ea) rate increases proportionally more → reverse rate > forward rate → equilibrium shifts left → more NO₂ → darker brown. Keq = decreases (temperature is the only factor that changes Keq; for an exothermic reaction, increasing T decreases Keq).
Catalyst added: Colour unchanged. Catalyst lowers Ea equally for both directions → both rates increase by the same factor. System is already at equilibrium; both rates are already equal; increasing them proportionally maintains equality. No net change in concentrations. Keq = unchanged.
1. B — Forward rate starts at maximum and decreases (reactants consumed, fewer collisions). Reverse rate starts at zero and increases (products accumulate, more reverse collisions). Both converge to a non-zero equilibrium rate. Option A is wrong — forward rate starts at maximum and decreases. Option C is wrong — both rates are non-zero at equilibrium.
2. C — Exothermic: products are at lower energy → transition state is closer in energy to products → Ea(reverse) is larger. ΔH = Ea(fwd) − Ea(rev) = negative → Ea(fwd) < Ea(rev). Option B is wrong — same transition state but different energy gaps to it from reactants vs products.
3. B — Catalyst lowers Ea equally for both directions; both rates increase by the same factor; since rates were already equal, they remain equal; no shift. Keq unchanged — it depends only on temperature. Options A, C, D all incorrectly claim the equilibrium position or Keq changes.
4. A — Immediately after compression, [NO₂] doubles → mixture is momentarily darker. But the forward reaction (2 → 1 mol gas) has its collision frequency increased more than the reverse → forward rate > reverse rate → equilibrium shifts right → [NO₂] decreases, [N₂O₄] increases → mixture becomes paler than original.
5. D — Temperature: correctly increases rate (faster collisions), but DECREASES yield for this exothermic reaction (shift left at higher T). Catalyst: does NOT change yield — only increases the rate of reaching equilibrium without shifting the equilibrium position.
Q6 (3 marks): A catalyst provides an alternative lower-energy reaction pathway — it lowers the activation energy [1]. For a reversible reaction, the catalyst lowers Ea for both the forward AND reverse reactions by the same amount [1]. Therefore, both the forward and reverse reaction rates increase by the same factor. Since the ratio of rates is unchanged, the equilibrium position is unchanged and Keq is unchanged — Keq depends only on temperature [1].
Q7 (4 marks): (a) 2NO₂ → N₂O₄ is exothermic (ΔH = −57 kJ/mol) → Ea(forward) < Ea(reverse) [1]. Increasing temperature increases the average kinetic energy of all particles — both forward and reverse reaction rates increase [1]. However, the rate increases proportionally more for the reaction with the higher activation energy. Since Ea(reverse) > Ea(forward), the reverse rate (N₂O₄ → 2NO₂) increases more than the forward rate — reverse rate > forward rate — net reverse reaction — equilibrium shifts left — more NO₂ is produced — mixture becomes darker brown [1]. (b) Keq decreases — for an exothermic forward reaction, increasing temperature favours the reverse (endothermic) reaction, shifting equilibrium left and reducing the ratio of products to reactants — Keq = [products]/[reactants] decreases [1].
Q8 (7 marks): (a) ΔH = −206 kJ/mol → exothermic forward → Ea(fwd) < Ea(rev) [1]. Increasing T → both rates increase → but reverse rate increases more (higher Ea) [1] → reverse > forward → shift LEFT → less CH₄ and H₂O, more CO and H₂ [1]. Keq decreases [½]. (b) Ni catalyst lowers Ea equally for both forward and reverse directions [1]. Both rates increase by the same factor — system already at equilibrium remains at equilibrium; no shift; equilibrium position unchanged; Keq unchanged [1]. (c) Halving volume doubles all concentrations → both rates increase → forward reaction involves 4 mol gas (CO + 3H₂); reverse involves 2 mol gas (CH₄ + H₂O) → forward collision frequency increases more [1] → forward > reverse → shift RIGHT → more CH₄ and H₂O; Keq unchanged — pressure does not change Keq [½].
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