Reaction Kinetics
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.)
- 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 concentration–time graph = rate of reaction.
- Go to the time asked (e.g. t1) on the curve.
- Draw a neat tangent to the curve at that point.
- Work out the tangent’s gradient with (y2−y1)/(x2−x1) — that is the rate at t1.
② Trend of the rate over time
Draw tangents at every point and see what the gradient does — exactly like reading gradients in kinematics:
- 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.
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 |
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.
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.
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.
- 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.
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.
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.
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).
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.
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
Think: fast particles ⇒ crowded market ⇒ more collisions.
| Peak | Area beyond Ea | Ea itself | |
|---|---|---|---|
| T ↑ | shifts right & lower | increases | unchanged |
| T ↓ | shifts left & higher | decreases | unchanged |
5. Catalysts
Using a catalyst increases the rate of 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.
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 |
Catalyst on the Boltzmann graph & energy profile
- 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.
| 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.
Which curve is obtained when rate is plotted against time for an autocatalytic reaction (a reaction where one of the products catalyses the 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
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.
Answer: C
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
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?
| 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)
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)?
Y ends with more O2 ⇒ more product ⇒ more moles of reactant were needed. Only extra reactant can make more product.
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
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).
Answer: C
7. Quick Recap — 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)).