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Educerie · IB Diploma · Chemistry
Reactivity 2 How much, how fast and how far? · R2.2 How fast? The rate of chemical change
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Define rate, and determine it from data, including by drawing a tangent | SL, HL | Paper 1B: "Determine the initial rate of reaction from the graph" (2 marks) |
| Describe methods of measuring rate, and when time is the dependent or independent variable | SL, HL | "Outline how the rate of this reaction could be monitored" (2 marks) |
| Explain reaction in terms of collisions with enough energy and the right orientation; link kinetic energy to temperature in kelvin | SL, HL | "State two conditions for a collision to be successful" (2 marks) |
| Predict and explain the effect of concentration, pressure, surface area, temperature and a catalyst on rate | SL, HL | "Explain why increasing the temperature increases the rate" (3 marks) |
| Construct Maxwell–Boltzmann curves to show the effect of temperature and of a lower Eₐ | SL, HL | "Sketch the distribution at a higher temperature and use it to explain…" (3 marks) |
| Sketch and explain energy profiles with and without a catalyst, exothermic and endothermic | SL, HL | "Sketch an energy profile showing the effect of a catalyst" (3 marks) |
| Evaluate mechanisms; identify intermediates, transition states, the rate-determining step and molecularity | HL only | "Deduce which step is rate-determining and explain your choice" (2 marks) |
| Deduce a rate equation from data, and sketch and recognise zero, first and second order graphs | HL only | Initial-rate table: "Deduce the order with respect to each reactant" (3 marks) |
| Solve problems with the rate equation, including the units of k | HL only | "Calculate the value of k and state its units" (2 marks) |
| Use the Arrhenius equation and its linear form to find Eₐ and A | HL only | Paper 1B or 2: "Determine the activation energy from the graph" (3 marks) |
Before you start
You need the kinetic molecular theory from Structure 1.1: particles are always moving, and temperature measures their average kinetic energy. You need energy profiles and the meaning of ΔH from Reactivity 1.1, and concentration and gas volumes from Structure 1.4 and Reactivity 2.1. HL students need logarithms: ln x, and that ln(eˣ) = x.
1The idea in one paragraph
A reaction happens only when particles collide, and only collisions with at least a minimum energy, the activation energy, and the right orientation succeed. So there are two ways to speed a reaction up: make collisions more frequent, or make a bigger fraction of them successful. Concentration, pressure and surface area do the first; temperature does a little of the first and a lot of the second; a catalyst does the second by offering a route with a lower activation energy. At HL, a reaction is a series of simple steps, the slowest sets the pace, and a measured rate equation is the evidence for which steps are plausible.
2What rate means, and how it is measured
The rate of reaction is the change in concentration of a reactant or product per unit time.
rate = change in concentration ÷ time taken, units mol dm⁻³ s⁻¹
Rate is quoted as a positive number, and it depends on which species you follow: in 2H₂O₂ → 2H₂O + O₂, oxygen forms at half the rate at which hydrogen peroxide is used up. Concentration is rarely measured directly; you measure a property that follows it:
| What you measure | When it works |
|---|---|
| Volume of gas, in a gas syringe | a gas is produced, e.g. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ |
| Loss of mass, on a balance | a dense gas escapes from an open flask |
| Colour intensity, with a colorimeter | a reactant or product is coloured, e.g. brown iodine |
| Electrical conductivity or pH | the ions, or H⁺, are used up or made |
| Time for a precipitate to hide a mark | a solid forms, e.g. sodium thiosulfate with acid |
Usually time is the independent variable: you choose the times and record the volume or mass at each, giving a curve like Figure 1. In a "clock" experiment it is the dependent variable: you record the time taken to reach a fixed point, such as a precipitate hiding a cross, so 1/time serves as a measure of rate.
3Reading rate from a graph
On a graph of amount against time, the rate is the gradient. The steeper the curve, the faster the reaction. Figure 1 shows carbon dioxide collected from marble chips and acid.
The curve is steepest at the start, when the reactants are most concentrated, and flat once one has run out. The rate at a particular moment is the gradient of the tangent there: a straight line that touches the curve at that point. Lay a ruler so the gaps either side look equal, then read two points on it far apart.
Worked example 1. Determine the rate at 40 s from Figure 1.
The initial rate, from the tangent at t = 0, is 2.25 cm³ s⁻¹; by 40 s the acid is less concentrated and the rate lower.
To express it as a change in concentration, convert. With gas volumes at STP and 50.0 cm³ of acid:
4Collision theory
Collision theory says particles react only when they collide, and a collision leads to reaction only if two conditions are both met.
Enough energy. The colliding particles must bring at least the activation energy, Eₐ, between them: the minimum energy needed for a collision to lead to reaction. Below it, bonds are not stretched far enough to break and the particles bounce apart.
The right orientation. The particles must meet so that the atoms that need to bond actually touch. Figure 2 shows carbon monoxide and nitrogen dioxide. For an oxygen atom to move from NO₂ to CO, the carbon of CO has to hit an oxygen of NO₂. Hitting it the other way round achieves nothing, however hard.
This is the collision geometry. The energy comes from motion: the average kinetic energy of the particles is proportional to the absolute temperature in kelvin. So going from 20 °C to 40 °C does not double it; it rises from 293 K to 313 K, about 7%.
5Activation energy and the spread of energies
At any temperature, collisions constantly pass energy around, so a few particles are slow, most are near the middle and a few are very fast. The Maxwell–Boltzmann distribution, Figure 3, shows how many particles have each energy.
Features you must draw correctly:
- It starts at the origin: no particle has zero energy.
- It rises to a peak, the most probable energy, and has a long tail to the right.
- The tail never touches the energy axis: there is no maximum energy.
- The area under the curve is the total number of particles.
Mark Eₐ on the energy axis. Only the particles in the shaded area beyond it have enough energy to react when they collide.
Now raise the temperature. Figure 4 draws the new curve.
At the higher temperature the curve is lower and broader, with its peak further right, and the area under it is the same, because the number of particles has not changed. The area beyond Eₐ is much bigger. So temperature raises the rate in two unequal ways:
- Particles move faster, so they collide more often. This is a small effect.
- A much larger fraction of collisions have E ≥ Eₐ. This is the main effect.
For an activation energy of about 50 kJ mol⁻¹, warming by 10 K near room temperature nearly doubles the fraction with enough energy, while collision frequency rises by under 2%.
6The factors that change the rate
Every factor works through the two parts of collision theory. Say which part, every time.
| Change | What happens to particles | Frequency of collisions | Fraction with E ≥ Eₐ |
|---|---|---|---|
| Higher concentration of a solution | more particles in the same volume | increases | unchanged |
| Higher pressure of a gas | the same particles squeezed into a smaller volume: a higher concentration | increases | unchanged |
| Larger surface area of a solid | more particles exposed where collisions can happen | increases | unchanged |
| Higher temperature | particles move faster | increases a little | increases a lot |
| Adding a catalyst | a different route becomes available | unchanged | increases, because Eₐ is lower |
Pressure matters only when gases react. Surface area matters only for a solid, which can be hit only at its surface, so a powder reacts faster than lumps. Figure 5 compares runs of the marble chip experiment, with the acid in excess so that calcium carbonate is limiting.
The powder curve is steeper at the start but levels off at the same final volume, because the amount of CO₂ is fixed by the limiting reactant (Reactivity 2.1), not by speed. Halve the mass of chips and the final volume halves. Rate is the slope; amount is the height the curve levels off at.
To study one factor, change only that one; the rest are controlled variables: here the mass and size of chips, the volume and concentration of acid, and the temperature. On the graph, points scattered either side of a smooth curve show random error; a whole curve shifted, say one that misses the origin because gas escaped before the bung went on, shows systematic error. Repeats reduce the first; only a better method removes the second.
7Catalysts
A catalyst increases the rate of a reaction without being used up. It works by providing an alternative reaction pathway with a lower activation energy. Figure 6 draws the two routes for an exothermic and an endothermic reaction.
The reactants, the products and ΔH stay the same; only the height of the barrier changes. Figure 7 shows why that matters.
The distribution is unchanged; only the Eₐ line moves left, so a much larger fraction of collisions have enough energy.
Enzymes are biological catalysts. A catalyst in the same phase as the reactants is homogeneous, one in a different phase heterogeneous; how they work is not assessed. The catalysed route lowers the barrier for the forward and backward reactions equally, so a catalyst speeds the arrival at equilibrium without moving it (Reactivity 2.3).
8HLReaction mechanisms
SL students can skip to section 12.
Most reactions do not happen in one collision. They happen in a series of simple steps called elementary steps, and the list of steps is the reaction mechanism. The steps must add up to the overall equation.
Take the reaction of nitrogen dioxide with carbon monoxide at low temperature, NO₂(g) + CO(g) → NO(g) + CO₂(g). A proposed mechanism is:
An intermediate, here NO₃, is made in one step and used up in a later one, so it is absent from the overall equation. It really exists, briefly, and sits in a dip on the energy profile. A transition state is the highest-energy arrangement within one step, bonds half broken and half formed. It sits on a peak and cannot be isolated; each step has one.
The rate-determining step is the slowest step, the one with the largest activation energy, and it limits the rate of the whole reaction. Figure 8 draws two cases; in (b) it is the second step.
Molecularity is the number of particles reacting in one elementary step: unimolecular (one particle breaks up), bimolecular (two collide) or termolecular (three collide at once, which is rare). Both steps above are bimolecular. Molecularity belongs to a step; order belongs to a rate equation.
9HLRate equations and orders
A rate equation links the rate to the concentrations of the reactants:
rate = k[A]ᵐ[B]ⁿ
The order with respect to a reactant is the power to which its concentration is raised: m for A, n for B. The overall order is the sum, m + n. k is the rate constant.
The orders cannot be read from the balanced equation; they depend on the mechanism and are found only by experiment. Zero order: changing [A] has no effect on the rate. First order: doubling [A] doubles the rate. Second order: doubling [A] multiplies it by 2² = 4. Figure 9 shows the graphs you must sketch and recognise.
Concentration against time: zero order is a straight line; first order a curve with a constant half-life (as long to fall from 0.8 to 0.4 as from 0.4 to 0.2); second order steeper at first with a longer tail. Rate against concentration: horizontal, a line through the origin, and an upward curve. Only whole-number orders are assessed.
Deducing orders from initial rates. Change one concentration at a time and measure the initial rate. The data below are invented, for 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) at a fixed temperature.
| Experiment | [NO] / mol dm⁻³ | [H₂] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.0100 | 0.0100 | 1.20 × 10⁻⁴ |
| 2 | 0.0200 | 0.0100 | 4.80 × 10⁻⁴ |
| 3 | 0.0100 | 0.0200 | 2.40 × 10⁻⁴ |
| 4 | 0.0300 | 0.0200 | ? |
Worked example 2.
From the rate equation to the mechanism. The orders show which particles take part in the rate-determining step or in fast steps before it; a zero-order reactant enters only after the slow step. For NO₂ + CO, experiment gives rate = k[NO₂]², zero order in CO. The mechanism in section 8 fits: its slow step is two NO₂ colliding. A one-step mechanism would predict rate = k[NO₂][CO], contradicting the data, so it is rejected. A mechanism must pass two tests: its steps add up to the overall equation, and its slow step matches the rate equation.
For the NO and H₂ reaction the slow step is not the first. A mechanism consistent with rate = k[NO]²[H₂] is:
The slow step uses N₂O₂, an intermediate made from two NO, and H₂, so the rate depends on [NO]²[H₂]. When the slow step contains an intermediate, replace it by the reactants that made it.
The evidence has a limit: a single termolecular step, 2NO + H₂ → …, would give the same rate equation. Kinetic data can rule a mechanism out but never prove one, so a mechanism that fits is only a possible mechanism. Reactivity 3.4 uses the same logic: rate = k[halogenoalkane] for a tertiary halogenoalkane with OH⁻ points to a slow step involving the halogenoalkane alone, and rate = k[halogenoalkane][OH⁻] for a primary one points to a single bimolecular step.
10HLThe rate constant and its units
k is constant for a reaction at a given temperature: it changes with temperature (section 11) but not with concentration. Its units depend on the overall order, because rate is always in mol dm⁻³ s⁻¹:
| Overall order | Rate equation (example) | Units of k |
|---|---|---|
| 0 | rate = k | mol dm⁻³ s⁻¹ |
| 1 | rate = k[A] | s⁻¹ |
| 2 | rate = k[A]² or k[A][B] | dm³ mol⁻¹ s⁻¹ |
| 3 | rate = k[A]²[B] | dm⁶ mol⁻² s⁻¹ |
If you forget the table, work it out: for order 3, k = rate ÷ [ ]³ = (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ = mol⁻² dm⁶ s⁻¹.
11HLThe Arrhenius equation
Raising the temperature increases k exponentially. The Arrhenius equation captures this:
k = A e^(−Eₐ/RT)
R = 8.31 J K⁻¹ mol⁻¹, T is in kelvin and Eₐ in J mol⁻¹. The term e^(−Eₐ/RT) is the fraction of collisions with at least Eₐ, the tail of Figure 4 as a number: it grows as T rises and shrinks as Eₐ grows. A, the Arrhenius factor, accounts for the frequency of collisions with the right orientation, and has the same units as k. Taking natural logarithms gives the linear form; both are in the data booklet:
ln k = −(Eₐ/R) × (1/T) + ln A
Plot ln k against 1/T and you get a straight line, y = mx + c, with gradient −Eₐ/R and intercept ln A. A steeper line means a larger Eₐ and a rate more sensitive to temperature.
Worked example 3. Rate constants for a first-order reaction were measured at five temperatures and plotted in Figure 10.
| T / K | 300 | 310 | 320 | 330 | 340 |
|---|---|---|---|---|---|
| k / s⁻¹ | 1.72 × 10⁻³ | 4.54 × 10⁻³ | 1.13 × 10⁻² | 2.65 × 10⁻² | 5.93 × 10⁻² |
| 1/T / 10⁻³ K⁻¹ | 3.333 | 3.226 | 3.125 | 3.030 | 2.941 |
| ln k | −6.365 | −5.395 | −4.483 | −3.631 | −2.825 |
Take the gradient from the best-fit line, using two points far apart on it. The intercept at 1/T = 0 is far off the graph, so ln A is calculated rather than read off. With only two temperatures, subtract the linear form at T₁ from the form at T₂ and ln A cancels:
Your data booklet may print this form too; either way it follows from the linear one.
12Where marks are lost
"Particles collide more" as the whole explanation for temperature. The main reason is that a greater fraction of collisions have E ≥ Eₐ. Give both, and say which matters more.
"Particles have more energy" for concentration. Concentration changes collision frequency only.
Maxwell–Boltzmann curves that touch the axis or change area. Start at the origin, never meet the energy axis, keep the same area.
A catalyst that moves the Maxwell–Boltzmann curve, changes ΔH or is "used up". The curve stays put and the Eₐ line moves left; products and ΔH are unchanged; the catalyst is regenerated.
Confusing rate with amount. A faster reaction reaches the same final amount sooner.
HL: taking orders from the coefficients of the overall equation. Orders come only from experiment. In 2NO + 2H₂, the order in H₂ is 1, not 2.
HL: calling a transition state an intermediate. An intermediate sits in a dip and exists; a transition state sits on a peak and cannot be isolated.
13Draw it right
- Rate graphs: time on the x-axis with units; the curve steepest at the start, then levelling off. A tangent touches the curve and is drawn long, with two widely spaced points read from it.
- Two runs compared: the faster run steeper at the start; the same final level unless the limiting reactant changed.
- Maxwell–Boltzmann: axes "kinetic energy" and "fraction of particles"; from the origin, asymmetric, the tail never touching the axis; Eₐ marked and the area beyond it shaded.
- Higher temperature: peak lower and to the right, same area, above the old curve in the tail; label T₁ and T₂.
- Energy profiles: axes "potential energy" and "reaction coordinate"; Eₐ from the reactants' level up to the peak; ΔH from reactants to products. A catalysed route has the same ends and a lower peak, labelled Eₐ(cat).
- HL: one peak per step in a multistep profile, dips labelled as intermediates; order graphs and Arrhenius plots on exactly the right axes ([A] against time, rate against [A], ln k against 1/T).
14Try it
Marks in brackets. Answers and marker's notes are at the end. Use R = 8.31 J K⁻¹ mol⁻¹ and the data booklet where you need it.
Q1. Excess hydrochloric acid is added to marble chips in an open flask on a balance. The loss of mass is recorded.
| Time / s | 0 | 30 | 60 | 90 | 120 | 180 | 240 | 300 |
|---|---|---|---|---|---|---|---|---|
| Loss of mass / g | 0.00 | 0.62 | 1.04 | 1.34 | 1.54 | 1.77 | 1.88 | 1.93 |
(a) Calculate the average rate of mass loss over the first 60 s, with units. 1 mark
(b) Explain, in terms of particles, why the rate decreases as the reaction goes on. 2 marks
(c) The experiment is repeated with the same mass of marble as a powder. State how the final loss of mass would compare, and explain why. 2 marks
Q2. Using a Maxwell–Boltzmann distribution, explain why a small increase in temperature causes a large increase in the rate of a reaction. 4 marks
Q3. Sketch an energy profile for an endothermic reaction, with and without a catalyst. Label ΔH, Eₐ and Eₐ(cat). 3 marks
Q4 (HL). The acid-catalysed reaction of propanone with iodine was studied at 298 K. The data are invented.
CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + HI(aq)
| Experiment | [CH₃COCH₃] / mol dm⁻³ | [I₂] / mol dm⁻³ | [H⁺] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|---|
| 1 | 0.40 | 0.0020 | 0.20 | 4.8 × 10⁻⁶ |
| 2 | 0.80 | 0.0020 | 0.20 | 9.6 × 10⁻⁶ |
| 3 | 0.40 | 0.0040 | 0.20 | 4.8 × 10⁻⁶ |
| 4 | 0.40 | 0.0020 | 0.40 | 9.6 × 10⁻⁶ |
(a) Deduce the order with respect to each of the three species, and write the rate equation. 3 marks
(b) Calculate the value of k, with its units. 2 marks
(c) Suggest what the order with respect to iodine tells you about the mechanism. 1 mark
Q5 (HL). Nitryl chloride decomposes: 2NO₂Cl(g) → 2NO₂(g) + Cl₂(g). The rate equation is rate = k[NO₂Cl]. A proposed mechanism is: Step 1: NO₂Cl → NO₂ + Cl Step 2: NO₂Cl + Cl → NO₂ + Cl₂
(a) Identify the intermediate. 1 mark
(b) State the molecularity of each step. 1 mark
(c) Deduce which step is rate-determining, and show that the mechanism is consistent with both the equation and the rate equation. 2 marks
Q6 (HL). For a first-order reaction, a graph of ln k against 1/T has a gradient of −6.65 × 10³ K. At 300 K, k = 3.0 × 10⁻⁴ s⁻¹.
(a) Determine the activation energy, in kJ mol⁻¹. 2 marks
(b) Determine the Arrhenius factor, A, with its units. 2 marks
15In one breath
Rate is the change in concentration of a reactant or product per unit time, in mol dm⁻³ s⁻¹; on a graph it is the gradient, found at an instant from a tangent. Particles react only when they collide with at least Eₐ and the right orientation. The Maxwell–Boltzmann curve starts at the origin, never touches the axis and keeps its area; at a higher temperature it flattens and shifts right, so far more particles pass Eₐ, which matters much more than the small rise in collision frequency. Concentration, pressure and surface area raise collision frequency; a catalyst gives a route with a lower Eₐ, leaving the curve, the products and ΔH unchanged. Rate is the slope; the final amount is set by the limiting reactant. HL: reactions go in elementary steps; the slowest sets the rate; intermediates sit in dips and transition states on peaks; molecularity counts the particles in a step. Rate = k[A]ᵐ[B]ⁿ comes only from experiment, the orders show what is in or before the slow step, and a mechanism that fits is only possible. The units of k follow from the overall order. For k = Ae^(−Eₐ/RT), plot ln k against 1/T: the gradient is −Eₐ/R, and A, the frequency of correctly oriented collisions, comes from ln A.
Answers
Q1. (a) 1.04 ÷ 60 = 0.0173 g s⁻¹. (b) The acid is used up, so its concentration falls: fewer acid particles per unit volume, so collisions with the marble surface become less frequent and the rate falls. (c) The same final loss of mass. The acid is in excess, so the marble is the limiting reactant, and the same mass of marble produces the same mass of CO₂; the powder only makes it happen faster. A1 for the rate with units; for (b), 1 for concentration of acid decreasing, 1 for less frequent collisions; for (c), 1 for "same", 1 for the limiting reactant being unchanged. "The particles get tired" or "the reaction slows down" scores 0 in (b).
Q2. A sketch with the energy axis and fraction-of-particles axis labelled, two curves with T₂ > T₁, the T₂ peak lower and to the right, and Eₐ marked. At the higher temperature the area beyond Eₐ is much larger, so a much larger fraction of particles have E ≥ Eₐ. A higher proportion of collisions are therefore successful, and the rate rises sharply. The frequency of collisions also increases, but only slightly, so the main effect is the larger fraction with enough energy. 1 for correct axes and two curves with the T₂ peak lower and to the right, 1 for Eₐ marked with the areas beyond it compared, 1 for more particles or collisions having E ≥ Eₐ, 1 for identifying this as the main reason, rather than collision frequency. Curves that cross the axis or are drawn with clearly different areas lose the first mark.
Q3. Products drawn higher than reactants, with a hump between. ΔH shown as an upward arrow from the reactants' level to the products' level. Eₐ as an arrow from the reactants' level to the top of the uncatalysed hump. A second, lower hump, starting and ending at the same levels, with Eₐ(cat) from the reactants to its top. 1 for an endothermic profile with labelled axes, 1 for ΔH and Eₐ correctly shown, 1 for a lower catalysed hump with Eₐ(cat), starting and ending at the same levels. Arrows for Eₐ drawn from the products' level score 0 for that mark.
Q4. (a) Experiments 1 and 2: [CH₃COCH₃] doubles and the rate doubles, so first order in propanone. Experiments 1 and 3: [I₂] doubles and the rate is unchanged, so zero order in iodine. Experiments 1 and 4: [H⁺] doubles and the rate doubles, so first order in H⁺. Rate = k[CH₃COCH₃][H⁺]. (b) k = 4.8 × 10⁻⁶ ÷ (0.40 × 0.20) = 6.0 × 10⁻⁵ dm³ mol⁻¹ s⁻¹. (c) Iodine is not involved in the rate-determining step or any step before it, so it reacts in a fast step after the slow step. 1 for each order with the pair of experiments shown, the rate equation carried in the third mark only if all three are right; A1 for the value, A1 for the units; 1 for iodine reacting after the rate-determining step. Writing rate = k[CH₃COCH₃[I₂] from the equation scores 0 in (a).]
Q5. (a) Cl, the chlorine atom: made in step 1 and used in step 2. (b) Step 1 is unimolecular; step 2 is bimolecular. (c) Step 1 is rate-determining: the rate equation is first order in NO₂Cl and contains nothing else, which matches a slow step involving one NO₂Cl. Adding the steps: 2NO₂Cl + Cl → 2NO₂ + Cl + Cl₂, and cancelling the Cl gives 2NO₂Cl → 2NO₂ + Cl₂, the overall equation. 1 for Cl; 1 for both molecularities; 1 for step 1 with the reason from the rate equation; 1 for showing that the steps sum to the equation. Answering "step 2 because it is last" scores 0 for the third mark.
Q6. (a) Gradient = −Eₐ/R, so Eₐ = 6.65 × 10³ × 8.31 = 5.53 × 10⁴ J mol⁻¹ = 55.3 kJ mol⁻¹. (b) ln A = ln k + Eₐ/RT = ln(3.0 × 10⁻⁴) + 6.65 × 10³ ÷ 300 = −8.11 + 22.17 = 14.05, so A = e14.05 = 1.3 × 10⁶ s⁻¹. M1 for Eₐ = −gradient × R, A1 for 55.3 kJ mol⁻¹; M1 for the linear form rearranged with a correct substitution, A1 for 1.3 × 10⁶ with the unit s⁻¹. An answer of 55 300 kJ mol⁻¹ loses the A1 for a unit error.
Educerie · written from the published IB Diploma Programme Chemistry guide, first assessment 2025, section R2.2 How fast? The rate of chemical change. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.