Full chapter
Reaction Kinetics
Infer a rate law from evidence and explain how concentration, temperature and catalysts change reaction speed.
A-Level 8873, revised syllabus (2026-2027)
Rate measures change per time
Name the quantity being followed before writing an equation.
Reaction rate measures how quickly a reactant is consumed or a product appears. A concentration-based rate is commonly in mol dm-3 s-1. Reactant concentration falls, so its graph has a negative slope; report its disappearance rate as the positive magnitude. If the equation has coefficients other than one, rates for different species need their stoichiometric factors.
For 2H2O2 → 2H2O + O2, hydrogen peroxide disappears twice as fast in molar terms as oxygen forms. If peroxide disappearance is 6.0 × 10-4 mol dm-3 s-1, oxygen formation is 3.0 × 10-4 mol dm-3 s-1 for the same volume convention.
| Term | Meaning |
|---|---|
| rate = k[A]m[B]n | A relationship between the stated rate and the concentrations at fixed conditions. |
| Order with respect to A: m | The exponent showing how changing [A] affects rate while other conditions remain fixed. |
| Overall order: m + n | The sum of the concentration exponents. |
| Rate constant, k | The proportionality constant for that equation at a specified temperature and catalytic conditions. |
| Half-life | Time for the amount or concentration being followed to fall to half its current value. |
| Activation energy | The minimum energy barrier for the reacting pathway. |
| Catalysis | Acceleration through an alternative pathway; the catalyst is regenerated overall. |
Worked example
An initial rate from a tangent
A tangent at t = 0 on a reactant concentration-time graph passes through (0 s, 0.160 mol dm-3) and (20.0 s, 0.0860 mol dm-3). Find the initial disappearance rate.
- Tangent slope = (0.0860 - 0.160)/(20.0 - 0) = -0.00370 mol dm-3 s-1.
- Take the positive magnitude for disappearance.
- These points are on the tangent. A chord joining distant measured points would give an average rate, not the initial rate.
Initial disappearance rate = 3.70 × 10-3 mol dm-3 s-1.
Change one concentration at a time
The rate factor reveals the exponent.
| Rate factor when [A] doubles | Order in A | Reason |
|---|---|---|
| 1 | 0 | 20 = 1 |
| 2 | 1 | 21 = 2 |
| 4 | 2 | 22 = 4 |
Use experiments with all other reactant concentrations, temperature and catalyst conditions held constant. The H1 scope uses m and n equal to 0, 1 or 2 in simple single-step cases. Rate orders are read from the stated evidence; do not copy coefficients from an overall equation without justification.
| Run | [A] / mol dm-3 | [B] / mol dm-3 | Rate / mol dm-3 s-1 |
|---|---|---|---|
| 1 | 0.0200 | 0.100 | 2.00 × 10-5 |
| 2 | 0.0400 | 0.100 | 4.00 × 10-5 |
| 3 | 0.0400 | 0.200 | 1.60 × 10-4 |
Worked example
Derive, calculate and predict
Use the dataset to obtain a rate equation and predict the rate at [A] = 0.0300 and [B] = 0.200 mol dm-3.
- Runs 1 and 2: [A] doubles at fixed [B]; rate doubles, so m = 1.
- Runs 2 and 3: [B] doubles at fixed [A]; rate quadruples, so n = 2.
- rate = k[A][B]2. From run 1, k = (2.00 × 10-5)/(0.0200 × 0.1002) = 0.100.
- Units of k = (mol dm-3 s-1)/(mol dm-3)3 = dm6 mol-2 s-1.
- New rate = 0.100 × 0.0300 × 0.2002 = 1.20 × 10-4 mol dm-3 s-1.
rate = k[A][B]2; k = 0.100 dm6 mol-2 s-1; predicted rate = 1.20 × 10-4 mol dm-3 s-1.
Check your understandingDoes zero order in A mean no A is consumed?Think it through, then reveal the answer
Recognise zero and first order from the whole curve
Constant slope differs from constant halving time.
Two concentration-time patterns
Both illustrative curves start at 0.160 mol dm-3 at constant temperature. Zero-order concentration falls linearly to zero at 120 seconds. The first-order curve halves every 30 seconds and progressively flattens.
A zero-order concentration-time graph is a straight line while reactant remains: the same amount is lost per second. Its successive halving intervals become shorter. A first-order graph curves because rate falls with concentration; its half-life is constant even when measured from a later starting concentration.
| Time / s | 0 | 30 | 60 | 90 |
|---|---|---|---|---|
| [A] / mol dm-3 | 0.160 | 0.0800 | 0.0400 | 0.0200 |
Worked example
Half-life is independent of the chosen starting concentration
Show how the table supports first order.
- 0.160 to 0.0800 takes 30 s.
- 0.0800 to 0.0400 also takes 30 s, despite the lower starting concentration.
- 0.0400 to 0.0200 again takes 30 s.
The constant 30 s halving interval supports first order. A falling curve alone is insufficient evidence: many rate laws produce falling curves.
In real data, plot a sensible curve through measurements and read several halving intervals. Keep temperature controlled; warming during the run changes k and can distort the shape. An anomalous measurement needs a reasoned treatment, not automatic deletion to make the curve look ideal.
Separate more collisions from a larger successful fraction
Concentration and temperature affect different parts of the explanation.
For a reaction on the collision model, particles must collide with adequate energy and a suitable orientation. Increasing a reactant concentration gives more particles per unit volume, so collisions between reactants occur more frequently. At unchanged temperature, the energy distribution and k do not change; the rate changes according to the concentration dependence in the rate equation.
Temperature changes the energy distribution
For the same total number of particles, the hotter distribution is broader and has a lower peak. Its area beyond the fixed activation energy is larger. The curves share the same total area over the full energy range.
A Boltzmann distribution shows that particles have a range of energies. The area to the right of Ea represents the fraction with enough energy for the pathway. Raising temperature broadens the distribution and substantially increases that fraction; faster particle motion also raises collision frequency. More effective collisions per second mean a larger k and a faster reaction at the same concentrations.
Check your understandingWhy is "particles collide more often" incomplete for a temperature question?Think it through, then reveal the answer
A catalyst changes the route, not the starting energy distribution
Connect a lowered barrier with a surface process.
A lower threshold at the same temperature
There is one distribution because temperature is unchanged. The catalytic threshold lies to the left of the original threshold, so more particles have sufficient energy. The catalyst does not supply a hotter population.
A catalyst offers a different pathway with lower activation energy, increasing the rate constant. It takes part in steps but is regenerated overall. It does not change the overall enthalpy change, the equilibrium constant or the final equilibrium composition; it helps the system reach that composition faster.
- Adsorb reactants
Gaseous reactants attach to active surface sites. Adsorption is attachment to the surface, not absorption into the bulk.
- React by an easier route
Surface interactions weaken relevant bonds and bring reactants into suitable positions. The pathway has lower activation energy.
- Desorb products
Products leave, freeing sites for another catalytic cycle.
A car catalytic converter uses a solid catalyst while exhaust reactants are gases: the catalyst is in a different phase, so the catalysis is heterogeneous. A representative nitrogen-oxide removal reaction is 2NO(g) + 2CO(g) → N2(g) + 2CO2(g). Adsorption, surface reaction and desorption convert NO to nitrogen while oxidising CO. CO2 is less acutely toxic than CO but remains a greenhouse gas; "all products are harmless" would be misleading.
Check your understandingWhy can a contaminant that binds strongly to the catalyst reduce its activity?Think it through, then reveal the answer
An enzyme has a specific active site and suitable operating conditions
Use the lock-and-key model without confusing cooling with denaturation.
Enzymes in this course are protein molecules acting as biological catalysts. Their folded structures create active sites whose shape and chemical environment suit particular substrates. In the lock-and-key model, a matching substrate binds to the active site; the enzyme catalyses a particular reaction and releases the products to be used again.
- Matching substrate
Its shape and interactions suit the active site.
- Enzyme-substrate complex
Binding enables a lower-activation-energy pathway.
- Products released
The enzyme is available for another suitable substrate.
At lower temperatures, less kinetic energy and fewer effective collisions make activity slower. Raising temperature initially increases rate. At sufficiently high temperatures, disruption of interactions holding the protein's shape changes the active site: denaturation causes activity to fall. Cooling a still-folded enzyme usually just slows it; heating beyond its tolerance can damage function.
Changing pH can alter charged groups and interactions in the protein and at its active site. An enzyme therefore has a limited useful pH range; extreme pH can denature it. Specificity includes both the choice of substrate and the type of reaction catalysed. The same enzyme does not automatically catalyse every reaction its substrate could undergo.
Check your understandingAn enzyme reaction slows in a refrigerator, then recovers on warming gently. Is denaturation the best explanation?Think it through, then reveal the answer
Quick revision
Revisit the essentials, then return to an explanation when you need it.
| Change | Essential explanation |
|---|---|
| Concentration | More particles per volume; collision frequency changes; use the measured order. |
| Temperature | Larger fraction above Ea plus more frequent collisions; k increases. |
| Catalyst | Lower-Ea pathway; distribution unchanged at the same temperature. |
| First-order half-life | Equal times for successive halvings, independent of initial concentration. |
| Enzyme extremes | Active-site shape/interactions change; extremes may denature the protein. |
Scope and references
Learning outcomes and sources
7. Reaction Kinetics (8873, 2026 revision; examinations 2026 and 2027). Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
7(a) Use kinetic vocabulary.
- Rate, rate equation, order, rate constant, half-life, activation energy and catalysis.
Rate measures change per timeChange one concentration at a time
7(b) Construct and use simple rate equations.
- m,n = 0,1,2 for the stated H1 single-step cases.
- (i) Initial-rate order deduction.
- (ii) Zero/first-order concentration-time graphs.
- (iii) Initial rate from concentration data; no integrated laws.
Rate measures change per timeChange one concentration at a timeRecognise zero and first order from the whole curve
7(c) Explain constant first-order half-life.
- Independence from starting concentration.
7(d) Explain concentration effects.
- Collision frequency at fixed temperature.
7(e) Explain activation energy using a distribution.
- Boltzmann distribution and fraction above threshold.
7(f) Explain temperature effects on k and rate.
- Both Boltzmann distribution and collision frequency.
7(g) Explain the catalytic effect.
- (i) Alternative lower-Ea pathway and increased k.
- (ii) Same-temperature Boltzmann interpretation.
A catalyst changes the route, not the starting energy distribution
7(h) Explain heterogeneous catalytic action.
- Adsorption/reaction/desorption; removal of nitrogen oxides from car exhaust.
A catalyst changes the route, not the starting energy distribution
7(i) Explain enzyme properties.
- Protein catalysts; substrate and reaction specificity; lock-and-key; temperature and pH sensitivity; link to 9(m), no named protein levels.
An enzyme has a specific active site and suitable operating conditions
- SEAB H1 Chemistry 8873, 2026 revision
Official scope: section 7, printed pages 17-18. All lettered outcomes and nested requirements checked.
- SEAB H1 Chemistry 8873, 2027
Section 7: same substantive outcomes as the revised 2026 course.
- Holy Grail: NJC 7 Kinetics H1 (Notes), note 9558
Older school notes consulted for background and common question patterns. Current 8873 revised-2026 outcomes determine coverage; wording, graphs and worked examples here are original. Older notes are not authority for current exclusions or enzyme coverage.