Year 12

AQA A level Chemistry Reaction Kinetics Notes

Reaction Kinetics

— Chemistry · AS Level (7405) —
★ Exam Tip: Almost all of this is O-level content — the new bits are the Boltzmann distribution and catalyst graphs. It is asked almost only as MCQs (never stand-alone long questions), sometimes mixed with equilibrium ideas. Learn the exact wording of the definitions.

1. Rate of Reaction

Definition: Rate of reaction is the concentration change per unit time of a reactant or product. (Basically it just tells us the speed of the reaction — but write the formal wording in the exam.)

rate = change in concentrationchange in time = Δ[conc]Δt
  • Δ means change. Use Δ[reactant] or Δ[product] — both give the rate.
  • Measure whichever is easier: a solid reactant → its mass falls; a gas product → collect it and watch its pressure / volume rise.

① Getting the rate from a concentration–time graph

Gradient of a straight linegradient = ΔyΔx = y2 − y1x2 − x1
Link: y = concentration, x = time ⇒ Δy/Δx = Δ[conc]/Δt = rate. Same thing!

Gradient of a concentration–time graph = rate of reaction.

  1. Go to the time asked (e.g. t1) on the curve.
  2. Draw a neat tangent to the curve at that point.
  3. Work out the tangent’s gradient with (y2−y1)/(x2−x1) — that is the rate at t1.
[reactant]time ΔyΔxt1tangent at t1gradient = rate
draw tangent at the time asked → gradient = Δy/Δx = rate
Note: a reactant curve slopes down so its gradient is negative — the magnitude of the gradient is the rate. Same method for a product graph (tangent slopes up).

② Trend of the rate over time

Draw tangents at every point and see what the gradient does — exactly like reading gradients in kinematics:

[reactant]time steeplesssteepm = 0rate ↓ to zero
reactant → 0 : fully used up ⇒ not reversible
[reactant]time t1 t2 t3 t 4rate ↓ to zeroconst. ⇒ equilibrium
levels off above 0 ⇒ reversible; equilibrium at t4
[product]time steepm = 0rate decreases to zero →
gradient ↓ to 0 — can’t tell if reversible
  • Very steep → less steep → even less → completely horizontal (m = 0). Horizontal line has gradient 0.
  • So the rate is highest at the start (t = 0) and falls to zero at the end — same story whether you plot reactant or product.
Why is rate fastest at the start? The reactant concentration is highest at the start, so collision frequency is high, hence more successful collisions. As reactants get used up (fewer moles left) the concentration falls, so the rate falls — until it reaches zero at the end.

③ Is the reaction reversible? (only from a reactant graph)

Reactant [conc]–time graph… Meaning
falls all the way to 0 reactants fully converted to products ⇒ not a reversible (complete) reaction
falls, then goes constant above 0 reactants not fully converted ⇒ reversible; mixture holds reactants and products ⇒ equilibrium established (at t4 in the sketch)
a product graph can’t tell! Product always rises then levels off, reversible or not
Board slip corrected: the second reactant graph was written “not fully converted → not a reversible reaction”. It should read reversible (as the teacher explained): concentration stops above zero because products turn back into reactants. Assume no reactant was added in silly excess. Examples of reversible reactions: Haber (NH3) and Contact (SO3).

2. Collision Theory & Activation Energy

Collision theory: two species (atoms / molecules / particles) only react if they ① collide with ② a high enough energy — greater than or equal to the activation energy. Colliding weakly, or not colliding at all, ⇒ no reaction.

slow, E < Ea bounce off → no reaction
fast, E ≥ Eabonds break → new bonds form✓

Activation energy (Ea): the minimum energy required to start the reaction.

Reactants must collide with this or a greater energy to break bonds and then form new ones. Remember: bond breaking is endothermic.

Why is Ea always positive (unlike ΔH)? Bonds can’t form unless others are broken first. Like two strong magnets stuck together — you must put energy in to pull them apart. So Ea is always needed, even for exothermic reactions (that’s why every energy profile has a “bump”). It can never be negative. (This is the simple model we use; real mechanisms vary.)
Collision vs successful collision: collisions can be successful or unsuccessful — only a small fraction succeed (those with E ≥ Ea).

3. Boltzmann Distribution

Boltzmann plotted number of particles (y-axis) against energy (x-axis) for the particles in a sample (e.g. a gas). It looks like a normal bell curve, but it is not symmetrical: it starts at the origin and has a long tail to the right.

No. of particlesEnergy most: moderate Earea = total no.of particles(constant, closed sys.)Eaonly these can react(E ≥ Ea )few have too low / too high E
Boltzmann distribution: green = can’t react (E < Ea), red = can react
  • Starts at the origin (0,0) — energy can’t be negative, so the curve stops at zero. Almost no particles have zero energy.
  • Most particles have a moderate energy (the peak ≈ the average). Few have very low or very high energy.
  • Never touches the x-axis on the right — a rare particle may have monstrously high energy.
Area under the graph = total number of particles — and it stays constant for a closed system. You can’t change the area (can’t make particles appear). You are not asked to calculate it (no integration in this course) — just judge it by eye.
Closed vs isolated system: closed = energy can go in/out, but no particles / molecules / moles cross the boundary. Isolated = neither energy nor matter cross (an isolated system is always closed too).

Where is Ea on this graph?

Normally things don’t react at room temperature — you have to heat them / give them energy. So Ea sits on the right-hand side, in the region where few particles have that much energy.

Careful: a “low Ea” = left on the x-axis (small energy values are on the left).

Area beyond the Ea line (red) = particles with E ≥ Ea — these can react if they collide.

Area before it (green) = particles that cannot react — not enough energy.

No. of particlesEnergy low Eaalmost ALLparticles react
if Ea were low (near the peak)…

4. Factors Affecting Rate of Reaction

① Concentration

Increasing the concentration of reactants usually increases the rate (not always!) — the “not always” is A2 stuff. For AS just assume it does. Doubling the concentration does not automatically double the rate (could be ×4, or no change).

Exam wording (learn it): more particles per unit volume, so higher collision frequency → more successful collisions → rate ↑.
Why not “kinetic energy”? Changing concentration does not change the particles’ kinetic energy — it only adds more particles, so more collisions.
Before: 100 collisions100 × 2%= 2 successful
After: 1000 collisions (same 2%)1000 × 2%= 20 successful
Gases — raise concentration by raising pressureconcentration = molesvolume   ⇒ squash the gas (↑ pressure) → smaller V → ↑ concentration

② Surface area

Increasing the surface area (e.g. powdering a solid reactant) increases the rate. The other reactant can attack from more sides — more of the solid is exposed.

increaseS.A.big lump: inner particles can’t be reachedpowdered: reactants exposed from all sides
pink = solid reactant, blue = other reactant particles
S.A. ↑ → collision frequency ↑ → successful collisions ↑ → rate ↑

Try to put the words “collision frequency” into your answers.

③ Temperature

Increasing temperature increases the rate — in two ways:

  • (1) collision frequency ↑ as particles travel faster (more collisions).
  • (2) increases the number of particles with energy ≥ Ea (so each collision is more likely to be successful). ← the bigger effect
Trick (market): Chaand-raat bazaar = packed, people bump into each other constantly. Same market on a quiet holiday = hardly any collisions.

Think: fast particles ⇒ crowded market ⇒ more collisions.

No. of particlesEnergy T ↓T ↑Eaextra area past Ea= more can react
T ↑ : peak shifts right & lower — area under whole curve stays the same
Peak Area beyond Ea Ea itself
T ↑ shifts right & lower increases unchanged
T ↓ shifts left & higher decreases unchanged
How to draw it: keep the total area constant. More area beyond Ea must be balanced by less area elsewhere — so lower the peak and push it to the right. Temperature does NOT change Ea — it only changes the area beyond Ea. (Ea only moves with a catalyst!) Raising just the tail while not lowering the peak would add particles — not allowed.

5. Catalysts

Using a catalyst increases the rate of reaction.

Definition (exam wording): A substance which increases the rate of reaction by lowering the activation energy by providing an alternate pathway / mechanism, and is chemically unchanged at the end of the reaction.
  • It only lowers Ea — nothing else. Lowered “automatically” because the alternative route is easier.
  • It must not be used up: if it is consumed it’s a reactant. If it’s regenerated (same amount at the end) → catalyst.
A + B → C : 200 JA + X → AX(50 J)AX + B → C + X2 easier steps, lower overall Ea Energy
catalyst X is used in step 1 and regenerated in step 2

Example: A + B → C needs 200 J. With catalyst X: first A + X → AX (needs only 50 J), then AX + B → C + X. Same overall reaction, but the energy needed to complete it has dropped.

Homogeneous vs heterogeneous

Homogeneous catalyst: in the same phase (same state) as the reactants.

Heterogeneous catalyst: in a different phase (different state) to the reactants. (homo = same, hetero = different)

Reaction Catalyst Catalyst vs reactants Type
1) Decomposition of H2O2
2H2O2 → 2H2O + O2
MnO2 solid vs liquid Heterogeneous
2) Haber process
N2 + 3H2 ⇌ 2NH3
Fe (iron) solid vs gases Heterogeneous
3) Contact process
2SO2 + O2 ⇌ 2SO3
V2O5
(vanadium(V) oxide / pentoxide)
solid vs gases Heterogeneous
4) Hydrogenation of C=C bond Ni (nickel) — or Pt solid vs gas / liquid alkene Heterogeneous
5) Esterification
alcohol + carboxylic acid ⇌ ester + water
H2SO4 aqueous vs aqueous / liquid Homogeneous
Note: most catalysts are heterogeneous. Nickel is the more commonly used one for hydrogenation. Know these five — more than enough for now; a couple more homogeneous ones appear later in the course, and why the two types work differently is an A2 topic.

Catalyst on the Boltzmann graph & energy profile

No. of particlesEnergy old Eanew Ea catalyst lowers Ea(moves LEFT)more area ≥ Ea
same curve shape — only the Ea line moves back a little
  • The Ea line moves left / back → more area beyond it → more successful collisions.
  • Be reasonable: shift it a little, keep it near the right end — don’t drag it right to the peak.
  • Temperature changes the curve’s shape; a catalyst does NOT — it only moves the Ea line.
Energyprogress of reactionreactantsproductsuncatalysedcatalysedEa(uncat)Ea (cat)ΔHsame
catalysed hump is lower but ends at the SAME product level (ΔH unchanged)
Energy-profile mistakes: for a catalysed reaction ΔH must stay the same (curve ends at the same product level) — only Ea decreases. Common wrong drawings: whole curve shifted down, or curve stopping short of the products.
Change Boltzmann curve shape Ea line Area beyond Ea
Temperature ↑ changes (peak right & lower) same place ↑
Catalyst unchanged moves LEFT ↑

6. Past-Paper Questions (all MCQs)

Six questions covered at the end of the class — this topic comes almost only in MCQs. Tip: if unsure, first cross out options you know are wrong.

Q1MCQ · rate–time graph

Which curve is obtained when rate is plotted against time for an autocatalytic reaction (a reaction where one of the products catalyses the reaction)?

ratetime
A ✗ rate stays high, never falls
ratetime
B ✗ rate stays high, never falls
ratetimeproductcatalyses ↑reactantused up ↓
C ✓ starts slow, peaks, falls to 0
ratetime
D ✗ highest at start = normal reaction

Careful: the y-axis is rate, not concentration. Autocatalytic: at first there is little catalyst so the rate is low; as product (= catalyst) builds up the rate rises; then reactants get used up so it falls to zero → hump shape.

A, B ✗ the class ruled them out first (rate never drops to 0 at the end). D ✗ that is the normal-reaction shape (fastest at start). The teacher paused to double-check how the rate is plotted, so re-read the axes in the exam.

Answer: C

Q3MCQ · Maxwell–Boltzmann, two temperatures

Two Maxwell–Boltzmann curves for a gas at different temperatures. Which letter on the axes is the most probable energy at the lower temperature?

Step 1 — find the lower-T curve: draw a vertical line anywhere on the right; the curve with less area past that line is the lower T (also: taller peak, further left).

Step 2 — most probable energy = the energy at its peak ⇒ read the x-axis (energy), not the y-axis (number of particles). A / B on the y-axis are wrong axes.

number of moleculesenergy AB CDlower T(taller, left)higher T
A, B are heights (no. of molecules); C, D are energies — we need an energy ⇒ C

Answer: C

Q35MCQ · temperature decreased

Which row correctly describes what happens when the temperature of a reaction is decreased?

activation energy (Ea) no. of successful collisions
A decreases decreases
B decreases increases
C ✓ remains the same decreases
D remains the same increases

Temperature never changes Ea (only a catalyst does); lower T ⇒ less area beyond Ea ⇒ fewer successful collisions.

Answer: C

Q6MCQ · two Boltzmann diagrams

Diagram 1: curves P and Q — one is a higher temperature. Diagram 2: lines X and Y — one is Ea with a catalyst, the other Ea uncatalysed. Which row gives the higher temperature and the catalysed line?

no. of moleculesmolecular speed PQQ: flatter, morearea at high E
diagram 1: Q = higher temperature
no. of moleculesmolecular speed XYX = lower Ea= catalysed
diagram 2: X = catalysed
higher temperature presence of catalyst
A P X
B P Y
C ✓ Q X
D Q Y

Curve with more area past Ea (flatter, lower peak, dotted Q) = higher T. A catalyst moves Ea backwards (left) so more area lies beyond it ⇒ X.

Answer: C (Q, X)

Q13MCQ · volume of O2 vs time

Curve X: decomposition of 100 cm³ of 1.0 mol dm⁻³ H2O2 catalysed by MnO2 (2H2O2 → 2H2O + O2). Which change gives curve Y (which levels off higher)?

volume of O2 formedtime XYX: originalY: same start-ish,but MORE O2 at end
extra H2O2 ⇒ more moles of reactant ⇒ higher final volume

Y ends with more O2 ⇒ more product ⇒ more moles of reactant were needed. Only extra reactant can make more product.

A ✗ adding water — same moles of H2O2, so no change in final volume (I don’t think it matters)B ✓ adding some 0.1 mol dm⁻³ H2O2 — increases the amount of reactant ⇒ more O2C ✗ less MnO2 — a catalyst can’t change the final volume (it doesn’t react)D ✗ lower temperature — only changes speed; eventually the reaction completes and the same final volume forms

Extra (not from class): Y starts a bit slower because the added 0.1 mol dm⁻³ solution dilutes the 1.0 mol dm⁻³ mixture.

Answer: B

Q47MCQ · temperature ↑ AND catalyst

Boltzmann distribution of a gas at T1 with Ea(uncat) marked. Which diagram shows the distribution and the new Ea(cat) when temperature is raised to T2 and a catalyst is used?

Two checks together: (1) temperature ↑ ⇒ T2 curve has a lower peak shifted right (more area past Ea, total area same). (2) catalyst ⇒ Ea(cat) is behind / to the left of Ea(uncat).

proportion of moleculesmolecular energy E T1T2 (higher)Ea(cat)Ea(uncat)
answer C: T₂ curve flatter & shifted right, and Ea(cat) is to the LEFT of Ea(uncat)
A, B, D ✗ all wrongD ✗ vs C: D has the labels reversed — Ea(uncat) drawn behind and Ea(cat) in front. Ea(cat) is behind (left); Ea(uncat) is in front (right). This is the easy one to forget.
Board note: on the board D is circled, but the teacher’s working and explanation in the recording give C — D was the tempting wrong answer.

Answer: C

7. Quick Recap — Cheat-Sheet

✓ Full chapter cheat-sheet
  • Rate = concentration change per unit time of a reactant or product = Δ[conc]/Δt. Gradient (tangent) of conc–time graph = rate.
  • Rate highest at start, → 0 at end (gradient → 0). Reactant → 0 : complete reaction; constant above 0 : reversible (equilibrium). Product graph can’t tell.
  • Collision theory: react only if they collide with energy ≥ Ea. Ea = minimum energy to start a reaction; always positive (bond breaking is endothermic).
  • Boltzmann: starts at origin; most particles moderate E; tail never touches axis; area = total particles (constant, closed system); area past Ea = particles that can react.
  • Concentration ↑ (gas: pressure ↑): more particles per unit volume → higher collision frequency → more successful collisions → rate ↑ (not KE!).
  • Surface area ↑: more exposed → higher collision frequency → more successful collisions → rate ↑.
  • Temperature ↑: (1) collision frequency ↑ (faster particles), (2) more particles with E ≥ Ea. Curve: peak right & lower, area constant. Ea unchanged.
  • Catalyst: lowers Ea by an alternate pathway/mechanism, chemically unchanged at the end. Boltzmann: Ea line moves left, curve shape same. Profile: lower hump, same ΔH.
  • Homogeneous = same phase as reactants, heterogeneous = different. MnO2 (H2O2), Fe (Haber), V2O5 (Contact), Ni/Pt (C=C hydrogenation) → hetero; H2SO4 (esterification) → homo.
  • Autocatalytic: a product catalyses the reaction ⇒ rate–time graph is a hump.
  • MCQ traps: read the axes (rate vs conc; energy vs no. of particles); T ↑ & catalyst together ⇒ check both the curve and Ea(cat) position (left of Ea(uncat)).