At AS, Kinetics (3.1.5) explained rate qualitatively — collisions, activation energy and the factors that speed a reaction up. Rate equations make that quantitative: they express exactly how rate depends on concentration, and that relationship — found by experiment — is a window into the mechanism. The topic is calculation-heavy, so lay every step out clearly.
Orders of reaction & the rate constant
The rate equation links the rate of a reaction to the concentrations of the reactants:
rate = k[A]m[B]n
The powers m and n are the orders of reaction, restricted to 0, 1 or 2 at A-level, and k is the rate constant. Crucially, the orders are found by experiment — they cannot be read off the stoichiometric equation.
The order of reaction with respect to a reactant is the power to which the concentration of that reactant is raised in the rate equation.
The overall order is the sum of the powers of the concentration terms in the rate equation.
The rate constant, k, is the constant of proportionality in the rate equation at a given temperature.
A reactant that is zero order ([X]0 = 1) does not appear in the rate equation, so changing its concentration has no effect on the rate.
Working out the units of k
The units of k depend on the overall order. Rearrange the rate equation to make k the subject and cancel the units. Rate is always in mol dm−3 s−1; each concentration is mol dm−3.
| Overall order | Rearranged k | Units of k |
|---|---|---|
| 0 | k = rate | mol dm−3 s−1 |
| 1 | k = rate / [A] | s−1 |
| 2 | k = rate / [A]2 | mol−1 dm3 s−1 |
| 3 | k = rate / [A]3 | mol−2 dm6 s−1 |
Interactive — rate equations and the units of k
Exam questions
What are the units of the rate constant for a third-order reaction? Tick (✓) one box.
The rate expression for the reaction between X and Y is rate = k[X]2[Y]. Which statement is correct? Tick (✓) one box.
Define the term overall order of reaction.
Show answer
The sum of the powers (indices) to which the concentrations are raised in the rate equation. 1 mark
The thermal decomposition of but-3-en-1-ol was investigated at several temperatures. In the results table, the rate constant column is headed k / s−1. The overall order of the reaction can be deduced from a piece of information in one of the column headings.
Identify this piece of information and deduce the overall order.
Show answer
The piece of information is the unit of k, which is s−1. 1 mark
s−1 is the unit of a rate constant for a first-order reaction, so the overall order is 1 (first order). 1 mark
The two marks are awarded independently. Allow “s−1”, “k / s−1” or “k in s−1” for the first mark — but not “k” on its own.
An acidified solution of butanone reacts with iodine: CH3CH2COCH3 + I2 → CH3CH2COCH2I + HI. The rate equation is rate = k[CH3CH2COCH3][H+]. The initial concentrations used in one experiment are shown.
| CH3CH2COCH3 | I2 | H+ | |
|---|---|---|---|
| Initial concentration / mol dm−3 | 4.35 | 0.00500 | 0.825 |
The initial rate of reaction is 1.45 × 10−4 mol dm−3 s−1. Calculate the value of the rate constant, k, and give its units.
Show answer
k = rate ÷ ([CH3CH2COCH3][H+]) = 1.45 × 10−4 ÷ (4.35 × 0.825) 1 mark
k = 1.45 × 10−4 ÷ 3.58875 = 4.04 × 10−5 1 mark
Overall order 2, so units = mol dm−3 s−1 ÷ (mol dm−3)2 = mol−1 dm3 s−1 1 mark
The iodine concentration is given but must be ignored — I2 is not in the rate equation. Turning the expression upside down gives 24 752 mol dm−3 s; multiplying instead of dividing gives 5.20 × 10−4 mol3 dm−9 s−1.
For the reaction in Q16, rate = k[CH3CH2COCH3][H+] and the initial rate is 1.45 × 10−4 mol dm−3 s−1. Calculate the initial rate of reaction when all of the initial concentrations are halved.
Show answer
Only butanone and H+ appear in the rate equation, so the rate is multiplied by (½) × (½) = ¼: 1.45 × 10−4 ÷ 4 = 3.63 × 10−5 mol dm−3 s−1 1 mark
The mark scheme gives 3.6(25) × 10−5 and allows 3.59 × 10−5 to 3.63 × 10−5. Halving the iodine has no effect — it is not in the rate equation.
The rate equation for a reaction is rate = k[L][M]2. Deduce the overall effect on the rate of reaction when the concentrations of both L and M are halved.
Show answer
(½) × (½)2 = ⅛, so the rate falls by a factor of 8 (is divided by 8). 1 mark
Also allowed: “multiplied by ⅛”, “halved then quartered”, or “decreases by 23”.
Source: AQA A-Level Chemistry past papers.
Finding orders from initial-rate data
To find an order experimentally, change one concentration at a time and see how the initial rate responds. The rate scales as (factor)order, so:
- double a concentration and the rate stays the same → zero order in that reactant;
- double it and the rate doubles (×2) → first order;
- double it and the rate quadruples (×4) → second order.
Deduce each order in turn, write the rate equation, then substitute one experiment’s data to find k (with units). Rearrange the equation the right way up — k = rate divided by the concentration terms, never the other way round.
Interactive — find the orders from initial-rate data
Exam questions
Substances P and Q react in solution. The initial rate was measured in three experiments:
| Experiment | [P] / mol dm−3 | [Q] / mol dm−3 | Initial rate / mol dm−3 s−1 |
|---|---|---|---|
| 1 | 1.00 × 10−2 | 1.25 × 10−2 | 1.6 × 10−4 |
| 2 | 2.00 × 10−2 | 1.25 × 10−2 | 3.2 × 10−4 |
| 3 | 0.50 × 10−2 | 2.50 × 10−2 | 3.2 × 10−4 |
Determine the order with respect to P and the order with respect to Q.
Show answer
Experiments 1→2: [P] doubles, [Q] constant, rate doubles → order with respect to P = 1. 1 mark
Experiments 1→3: [P] halves (rate ×½ from P), [Q] doubles, yet rate doubles overall; so [Q] must multiply rate by 4 → order with respect to Q = 2. 1 mark
A reaction between R and S is second order with respect to R and second order with respect to S. The initial rate was 1.20 × 10−3 mol dm−3 s−1 when [R] = 1.00 × 10−2 and [S] = 2.45 × 10−2 mol dm−3. Calculate k and give its units.
Show answer
k = rate / ([R]2[S]2) = 1.20 × 10−3 / [(1.00 × 10−2)2(2.45 × 10−2)2] 1 mark
k = 2.00 × 104 1 mark
Units: mol dm−3 s−1 ÷ (mol dm−3)4 = mol−3 dm9 s−1 1 mark
A reaction has rate = k[L][M]2. The rate is 0.0250 mol dm−3 s−1 when [L] = 0.0155 mol dm−3 and k = 21.3 mol−2 dm6 s−1. Calculate [M].
Show answer
[M]2 = rate / (k[L]) = 0.0250 / (21.3 × 0.0155) = 7.57 × 10−2 1 mark
Rearrange and take the square root. 1 mark
[M] = √(7.57 × 10−2) = 0.275 mol dm−3 1 mark
Substances P and Q react in solution at constant temperature. The initial rate was studied in three experiments by measuring the change in concentration of P over the first 5.0 seconds.
| Experiment | [P] at t = 0 / mol dm−3 | [P] at t = 5.0 s / mol dm−3 | Initial rate / mol dm−3 s−1 |
|---|---|---|---|
| 1 | 1.00 × 10−2 | 0.92 × 10−2 | 1.6 × 10−4 |
| 2 | 2.00 × 10−2 | 1.84 × 10−2 | ? |
| 3 | 0.50 × 10−2 | 0.34 × 10−2 | ? |
Calculate the initial rate of reaction of P in Experiments 2 and 3.
Show answer
Initial rate = change in concentration ÷ time.
Experiment 2: (2.00 × 10−2 − 1.84 × 10−2) ÷ 5.0 = 3.2 × 10−4; Experiment 3: (0.50 × 10−2 − 0.34 × 10−2) ÷ 5.0 = 3.2 × 10−4 mol dm−3 s−1 1 mark
Both values are needed for the mark.
Phosphinate ions react with hydroxide ions to produce hydrogen: H2PO2− + OH− → HPO32− + H2. A student reacted different initial concentrations of phosphinate ions with an excess of hydroxide ions and measured the time (t) taken to collect 15 cm3 of hydrogen. Each experiment was at the same temperature.
| Initial [H2PO2−] / mol dm−3 | t / s |
|---|---|
| 0.25 | 64 |
| 0.35 | 32 |
| 0.50 | 16 |
| 1.00 | 4 |
State the relationship between the initial concentration of phosphinate and the time (t), and deduce the order of reaction with respect to phosphinate.
Show answer
Relationship: [H2PO2−]2 is proportional to 1/t — doubling the concentration quarters the time (0.25 → 0.50 mol dm−3 takes t from 64 s to 16 s). 1 mark
Order with respect to phosphinate = 2. 1 mark
A time argument (“if the concentration doubles the time is quartered”) or any wording implying a square or square-root relationship is accepted. A simple description such as “as concentration increases the time decreases” scores nothing. Since 1/t measures rate, rate ∝ [H2PO2−]2.
Propanone reacts with bromine in alkaline conditions: CH3COCH3 + Br2 + OH− → CH3COCH2Br + Br− + H2O, with rate = k[CH3COCH3][OH−]. The table shows initial rates for three mixtures.
| Experiment | [CH3COCH3] / mol dm−3 | [Br2] / mol dm−3 | [OH−] / mol dm−3 | Initial rate / mol dm−3 s−1 |
|---|---|---|---|---|
| 1 | 1.50 × 10−2 | 2.50 × 10−2 | 2.50 × 10−2 | 2.75 × 10−11 |
| 2 | 1.50 × 10−2 | 2.50 × 10−2 | ? | 8.25 × 10−11 |
| 3 | 3.75 × 10−3 | 5.00 × 10−2 | 1.00 × 10−1 | ? |
Calculate the two missing values.
Show answer
Experiment 2: [CH3COCH3] is unchanged and the rate is 3× that of Experiment 1, so [OH−] = 3 × 2.50 × 10−2 = 7.50 × 10−2 mol dm−3 1 mark
Experiment 3: [CH3COCH3] is ×0.25 and [OH−] is ×4, so rate = 2.75 × 10−11 × 0.25 × 4 = 2.75 × 10−11 mol dm−3 s−1 1 mark
Bromine is not in the rate equation, so its changing concentration is irrelevant to both calculations.
A and B react in the presence of an acid catalyst: A(aq) + 2B(aq) → C(aq) + D(aq), with rate = k[B]2[H+]. The table shows how the relative initial rate varies with the concentrations of each reagent at the same temperature.
| Experiment | [A] / mol dm−3 | [B] / mol dm−3 | [H+] / mol dm−3 | Relative initial rate |
|---|---|---|---|---|
| 1 | 0.40 | 0.20 | 0.10 | 1.00 |
| 2 | 0.50 | 0.20 | 0.10 | ? |
| 3 | 0.40 | ? | 0.10 | 0.64 |
| 4 | 0.50 | 0.30 | 0.06 | ? |
Calculate the three missing values.
Show answer
Experiment 2: only [A] has changed, and A is not in the rate equation (zero order), so the relative rate is unchanged: 1.00 1 mark
Experiment 3: relative rate 0.64 = ([B]÷0.20)2, so [B]÷0.20 = √0.64 = 0.80 and [B] = 0.80 × 0.20 = 0.16 mol dm−3 1 mark
Experiment 4: [B] is ×1.5 and [H+] is ×0.6, so relative rate = (1.5)2 × 0.6 = 2.25 × 0.6 = 1.35 1 mark
A is zero order — changing [A] in Experiments 2 and 4 has no effect at all, even though A appears in the balanced equation.
Source: AQA A-Level Chemistry past papers.
Finding orders from graphs & Required Practical 7
When you monitor a single reaction over time, the shape of the graph reveals the order. Two kinds of graph are used.
Concentration–time graphs
- Zero order: a straight line with a constant (negative) gradient — the rate does not change as the reactant is used up. Because rate = k for a zero-order reactant, the magnitude of this gradient is the rate constant k.
- First order: a curve with a constant half-life — the concentration takes the same time to halve, again and again.
- Second order: a curve with an increasing half-life.
The initial rate is the gradient of the tangent drawn at t = 0; rates at other times come from tangents there.
Rate–concentration graphs
- Zero order: a horizontal line (rate independent of concentration).
- First order: a straight line through the origin (rate directly proportional to concentration).
- Second order: an upward curve (rate proportional to concentration squared).
Interactive — read the order off the graph
Two methods you must know. An initial-rate method (such as a clock reaction) measures how long a fixed small amount of product takes to form for different starting concentrations. A continuous monitoring method follows one reaction over time — collecting gas in a syringe, recording mass loss, or measuring colour — and the order is read from the graph. Draw tangents to a concentration–time curve to find rates.
The method marks all come from the same few places:
- Keep the total volume constant in a clock reaction. Water is added to make up the difference so that the only concentration changing is the one you are investigating — say this if asked why water is added.
- Add the last reagent and start the timer together. Any delay is a systematic error that makes every time too short.
- 1/t is a valid measure of the initial rate only because so little reactant is used before the end point appears — the concentrations are still effectively their starting values.
- Draw the tangent at t = 0 to the curve itself, touching it at the origin, and take the gradient over a large triangle. A chord, or a tiny triangle, is the commonest lost mark.
- Gas escapes before the bung is in. In a continuous-monitoring run that loses volume from every reading, so the measured rate comes out too low.
This is Required Practical 7. The full apparatus, method and safety are collected in the required practicals guide.
Interactive — read successive half-lives off the curve
Exam questions
Hydrogen peroxide decomposes in the presence of a manganese(IV) oxide catalyst, monitored by collecting oxygen. Explain why the reaction is fastest at the start.
Show answer
The concentration of hydrogen peroxide is highest at the start. 1 mark
So there are more frequent successful collisions (the rate falls as it is used up). 1 mark
A graph of rate against [H2O2] is a straight line passing through the origin. State the order of reaction with respect to H2O2, and how the graph shows this.
Show answer
First order. 1 mark
The graph is a straight line through the origin, so the rate is directly proportional to the concentration. 1 mark
In the iodination of propanone, a graph of iodine concentration against time is a straight line. Explain how this shows the reaction is zero order with respect to iodine.
Show answer
The graph is a straight line (constant gradient). 1 mark
So the rate does not change as the iodine concentration changes — the iodine is used up at a constant rate. 1 mark
In an experiment the concentration of a reactant is measured at one-minute intervals as it reacts. Explain how graphical methods can be used to process the measured results to confirm that the reaction is first order.
Show answer
Plot concentration (y-axis) against time (x-axis) and draw tangents to the curve, calculating the gradient of each tangent to give the rate at that concentration. 1 mark
Plot these rates (gradients) against concentration. 1 mark
A straight line through the origin (rate directly proportional to concentration) confirms first order. 1 mark
Alternative routes for the last two marks: measure at least two half-lives from the concentration–time curve (tangents not then needed) and show the half-life is constant; or plot log rate against log concentration and show the gradient is 1; or compare gradients at different concentrations and show the rate halves when the concentration halves. A time–concentration graph (axes the wrong way round) is not accepted unless clarified in words or a sketch.
A student followed the reaction H2PO2− + OH− → HPO32− + H2 by measuring the volume of hydrogen collected every 20 seconds at constant temperature, and plotted volume of hydrogen (cm3) on the y-axis against time (s) on the x-axis. The tangent used to find the initial rate passes through the points (0 s, 0 cm3) and (20 s, 110 cm3).
State the point on the curve at which this tangent must be drawn, calculate the initial rate of reaction, and state its units.
Show answer
The tangent must be drawn to the curve at the origin, (0, 0) — that is, at t = 0. 1 mark
Initial rate = gradient = 110 ÷ 20 = 5.5 1 mark
Units: cm3 s−1 1 mark
The mark scheme allows a rate of 5–7 but writes the unit as cm3 s−1 only — not cm3 / s. The units are not mol dm−3 s−1, because the quantity plotted is a gas volume, not a concentration.
Hydrogen peroxide decomposes over a manganese(IV) oxide catalyst: 2H2O2(aq) → 2H2O(l) + O2(g). A student plots [H2O2] against time and draws a tangent to the curve at the point where [H2O2] = 0.05 mol dm−3. The tangent passes through (10 s, 0.092 mol dm−3) and (50 s, 0.036 mol dm−3).
State the two requirements a correctly drawn tangent must meet here, and calculate the gradient of the curve at this point.
Show answer
The tangent must be drawn with a ruler so that it touches the curve at [H2O2] = 0.05 mol dm−3, and it must not cross the curve. 1 mark
Gradient = (0.036 − 0.092) ÷ (50 − 10) = −0.056 ÷ 40 = −1.4 × 10−3 mol dm−3 s−1 1 mark
The mark scheme accepts a gradient of −0.00120 to −0.00155 mol dm−3 s−1 and ignores the sign. If any white space is visible between the ruled line and the curve, the tangent has crossed the curve and the first mark is lost.
In the same experiment, the concentration of hydrogen peroxide at time t can be calculated from
[H2O2]t = [H2O2]initial × (Vmax − Vt) ÷ Vmax
Vmax is the total volume of oxygen collected during the whole experiment and Vt the volume collected at time t. In this experiment [H2O2]initial = 0.083 mol dm−3 and Vmax = 100 cm3. Calculate [H2O2]t when 20 cm3 of oxygen has been collected.
Show answer
(Vmax − Vt) ÷ Vmax = (100 − 20) ÷ 100 = 0.80 1 mark
[H2O2]t = 0.083 × 0.80 = 0.0664 mol dm−3 1 mark
The mark scheme allows 0.0656–0.0672 and requires a minimum of 2 significant figures.
Propanone reacts with bromine in alkaline conditions: CH3COCH3 + Br2 + OH− → CH3COCH2Br + Br− + H2O, with rate = k[CH3COCH3][OH−]. Describe the shape of the graph obtained when the concentration of bromine is plotted against time at constant temperature.
Show answer
A falling curve with a decreasing gradient — steepest at the start and levelling off as the reaction proceeds. 1 mark
Bromine is zero order, so the instinct is a straight line (compare Q9). A zero-order reactant only gives a straight concentration–time line while the rate is genuinely constant — that is, when the other reactants are in large excess. Here propanone and hydroxide are consumed alongside the bromine and are not stated to be in excess, so the rate falls and the bromine curve flattens.
Iodine and propanone react in an acid-catalysed reaction: CH3COCH3(aq) + I2(aq) → CH3COCH2I(aq) + HI(aq). A student mixed 25 cm3 of 1.0 mol dm−3 propanone, 10 cm3 of 1.0 mol dm−3 HCl(aq) and 25 cm3 of 5.0 × 10−3 mol dm−3 I2(aq), then at one-minute intervals removed a 1.0 cm3 sample and added it to a separate beaker containing an excess of NaHCO3(aq) before titrating with sodium thiosulfate.
Suggest why the 1.0 cm3 portions of the reaction mixture are added to an excess of sodium hydrogencarbonate solution.
Show answer
The sodium hydrogencarbonate neutralises the acid catalyst. 1 mark
This stops (quenches) the reaction, so the iodine concentration is fixed at that instant and can be titrated. 1 mark
In the experiment in Q28, the student was determining the order with respect to iodine. Suggest why the order of this reaction with respect to propanone can be ignored in this experiment.
Show answer
The amount of propanone (25 cm3 of 1.0 mol dm−3 = 25 mmol) is about 200 times the amount of iodine (25 cm3 of 5.0 × 10−3 mol dm−3 = 0.125 mmol). 1 mark
So the concentration of propanone stays effectively constant — the change in it is negligible. 1 mark
The volume of sodium thiosulfate used in each titration in Q28 is proportional to the concentration of iodine in that beaker. The results are shown.
| Time / minutes | Volume of sodium thiosulfate / cm3 |
|---|---|
| 1 | 41 |
| 2 | 35 |
| 3 | 24 |
| 4 | 22 |
| 5 | 16 |
| 6 | 10 |
Identify the anomalous result, and state how it should be treated when a line of best fit is drawn.
Show answer
The 3-minute result (24 cm3) is anomalous — the other results fall by about 6 cm3 per minute, so a value near 29 cm3 is expected. The straight line of best fit is drawn ignoring this plot. 1 mark
The mark scheme awards its mark for “a straight line of best fit which avoids the anomalous plot” — the anomaly is not deleted from the table, it is simply not allowed to pull the line.
In a different experiment, the time t taken for a solution of iodine to react completely was measured when the iodine was added to an excess of an acidified solution of butanone. Suggest an observation used to judge when all the iodine had reacted.
Show answer
The brown colour of the iodine disappears — the mixture goes colourless. 1 mark
“(Orange) brown to colourless” and “purple to colourless” are both accepted.
In the reaction H2PO2− + OH− → HPO32− + H2, the hydrogen produced must be collected and its volume measured at intervals. Name one suitable piece of apparatus for doing this, and state one feature the assembled apparatus must have.
Show answer
A gas syringe (with a free-moving plunger), or an inverted measuring cylinder or burette filled with water over a trough; the tubing carrying the gas from the flask must not be closed or sealed. 1 mark
A lack of graduations on the drawn apparatus is ignored, but a sealed delivery tube, or a gas syringe without a plunger, loses the mark.
The rate of reaction between calcium carbonate and hydrochloric acid is investigated by continuous monitoring. A conical flask on a balance is charged with about 20 g of large marble chips and 50 cm3 of 0.4 mol dm−3 hydrochloric acid, a loose cotton wool plug is placed in the neck, the balance is zeroed and the loss in mass is recorded every 30 seconds.
Suggest why a loose cotton wool plug is used, rather than (i) leaving the flask open, and (ii) inserting a bung.
Show answer
Instead of leaving the flask open: to stop acid or solution splashing, spraying or spitting out of the flask (which would add to the recorded mass loss). 1 mark
Instead of inserting a bung: to allow the carbon dioxide to escape. 1 mark
Evaporation, spilling, “to avoid loss of product” and “pressure would build up” are all ignored. Naming the wrong gas loses the second mark.
In the experiment in Q33, 20 g of large marble chips is a large excess of calcium carbonate. Suggest why using a large excess of calcium carbonate means that the rate is only affected by the changing concentration of the hydrochloric acid.
Show answer
The surface area (or mass, or amount) of calcium carbonate stays effectively constant through the reaction, so it is not a changing variable. 1 mark
“The concentration stays constant” and “so HCl is the limiting factor” are both ignored — the point is the constant surface area of the solid.
In the experiment in Q33, the mass of carbon dioxide produced in time t is equal to mt, and the total mass of CO2 produced during the whole reaction is equal to mtotal. Explain why (mtotal − mt) is proportional to the concentration of hydrochloric acid remaining in the flask at time t.
Show answer
mt, the mass of CO2 produced in time t, is proportional to the amount of HCl that has already reacted at time t. 1 mark
mtotal is proportional to the total amount of HCl present initially, so the difference (mtotal − mt) is proportional to the HCl left at time t. 1 mark
Alternative route, equally creditworthy: (mtotal − mt) equals the mass of CO2 still to be produced, and the CO2 still to be produced is proportional to the HCl still to react. “Equal to” or “represents” are allowed in place of “proportional to”.
In the experiment in Q33 the variable measured is mass loss. Suggest two other variables that could be measured instead to follow the rate of this reaction at constant temperature.
Show answer
Any two from: volume of gas (CO2) collected; pH; concentration of HCl (acid / H+); conductivity. 2 marks
Temperature is not accepted. “Volume” or “concentration” unqualified are not accepted — say volume of CO2 and concentration of HCl. “Time for the calcium carbonate to dissolve or disappear” is not accepted, because the calcium carbonate is in excess. Mass loss and “amount of CO2” are ignored.
Source: AQA A-Level Chemistry past papers.
Orders and the rate-determining step
Many reactions happen in several steps. The slowest step — the rate-determining step — controls the overall rate, so the rate equation only contains species involved up to and including that step. This is why the rate equation is a window into the mechanism.
- A species that appears in the rate equation is involved in the rate-determining step; its order tells you how many of that species enter by that step.
- A species that is absent from the rate equation (zero order) is not in the rate-determining step — it reacts in a later, faster step.
- The proposed slow step must be consistent with the observed rate equation.
Exam questions
A reaction A + 2B → C + D has rate = k[B]2[H+]. A suggested mechanism is: Step 1 B + H+ → BH+; Step 2 BH+ + B → B2H+; Step 3 B2H+ + A → C + D. Deduce the rate-determining step and give a reason.
Show answer
Rate-determining step: Step 2. 1 mark
By the end of Step 2, one H+ and two B have been used — matching [B]2[H+] in the rate equation. 1 mark
Propanone reacts with bromine in alkaline conditions with rate = k[CH3COCH3][OH−]. Use the rate equation to explain why the step involving propanone and hydroxide is the rate-determining step.
Show answer
That step contains CH3COCH3 and OH−, which are both in the rate equation, whereas Br2 is absent from the rate equation, so bromine is not in the rate-determining step. 1 mark
Source: AQA A-Level Chemistry past papers.
The Arrhenius equation
The rate constant depends on temperature, and the Arrhenius equation makes that quantitative:
k = Ae−Ea/RT
Here A is the Arrhenius constant, Ea the activation energy, R the gas constant and T the temperature in kelvin. As T rises, the term −Ea/RT becomes less negative, so k increases and the reaction speeds up — the quantitative counterpart of a greater proportion of molecules having energy ≥ Ea in the Maxwell–Boltzmann distribution.
Taking natural logs rearranges the equation into a straight-line form:
ln k = −(Ea/R)(1/T) + ln A
So a plot of ln k (y) against 1/T (x) is a straight line with gradient −Ea/R and intercept ln A. Hence Ea = −gradient × R (then divide by 1000 for kJ mol−1).
Interactive — the Arrhenius equation
R = 8.31 J K−1 mol−1, and every temperature is in kelvin.
Exam questions
For the decomposition of ethanoic anhydride, Ea = 34.5 kJ mol−1, A = 1.00 × 1012 s−1 and k = 2.48 × 108 s−1 at temperature T1. Calculate T1 (R = 8.31 J K−1 mol−1).
Show answer
k / A = e−Ea/RT, so 2.48 × 108 / 1.00 × 1012 = 2.48 × 10−4. 1 mark
ln(2.48 × 10−4) = −8.302 = −34 500 / (8.31 T1). 1 mark
T1 = 34 500 / (8.31 × 8.302) = 500 K. 1 mark
Convert Ea to joules (34 500) to match R in J.
The rate constant is 1.55 × 10−5 s−1 at 303 K and 1.70 × 10−4 s−1 at 333 K. Using ln(k2/k1) = (Ea/R)(1/T1 − 1/T2), calculate Ea in kJ mol−1 (R = 8.31 J K−1 mol−1).
Show answer
ln(k2/k1) = ln(1.70 × 10−4 / 1.55 × 10−5) = 2.395. 1 mark
1/T1 − 1/T2 = 1/303 − 1/333 = 2.973 × 10−4 K−1. 1 mark
Ea = R × ln(k2/k1) / (1/T1 − 1/T2) = 8.31 × 2.395 / (2.973 × 10−4) 1 mark
Ea = 66 900 J mol−1 1 mark
Ea = 66.9 kJ mol−1 1 mark
Carry unrounded values through — rounding ln(k2/k1) to 2.40 gives ≈67 kJ mol−1. Reversing k1 and k2 gives a negative Ea — activation energy is positive.
The thermal decomposition of but-3-en-1-ol was investigated at different temperatures and the rate constant calculated at each. Part of the results table is shown.
| T / K | 1/T / K−1 | k / s−1 | ln k |
|---|---|---|---|
| 553 | 1.81 × 10−3 | 4.6 × 10−4 | −7.68 |
| 563 | 1.78 × 10−3 | 8.4 × 10−4 | −7.08 |
| 573 | ? | 15.6 × 10−4 | ? |
| 583 | 1.72 × 10−3 | 28.0 × 10−4 | −5.88 |
| 593 | 1.69 × 10−3 | 49.9 × 10−4 | −5.30 |
Calculate the two missing values at 573 K.
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1/T = 1 ÷ 573 = 1.7452 × 10−3 = 1.75 × 10−3 K−1; ln k = ln(15.6 × 10−4) = −6.46 1 mark
Both values are needed. 0.00175 is allowed, but 1.74 × 10−3 is not — that comes from rounding 573 or truncating too early. Other numbers of significant figures (1.7 × 10−3, −6.5) are not accepted: match the precision of the rest of the column.
A ligand-substitution reaction was studied at several temperatures and ln k plotted against 1/T. The points (1/T = 3.41 × 10−3 K−1, ln k = −17.7) and (1/T = 3.10 × 10−3 K−1, ln k = −13.6) both lie on the line of best fit. Use them to calculate the activation energy, Ea, in kJ mol−1 (R = 8.31 J K−1 mol−1).
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Gradient = Δ(ln k) ÷ Δ(1/T) = (−13.6 − (−17.7)) ÷ ((3.10 − 3.41) × 10−3) = 4.1 ÷ (−3.1 × 10−4) = −13 226 K 1 mark
Gradient = −Ea/R, so Ea = −gradient × R = 13 226 × 8.31 = 109 900 J mol−1 1 mark
Ea = 110 kJ mol−1 1 mark
The mark scheme quotes a gradient of −13 125 (giving 109 kJ mol−1) and accepts gradients from −12 876 to −13 598 and Ea values from 107 to 114 kJ mol−1. A negative activation energy is never accepted — the gradient is negative, Ea is not.
For the thermal decomposition of but-3-en-1-ol in Q37, a graph of ln k against 1/T is a straight line through the tabulated points, including (1.81 × 10−3 K−1, −7.68) and (1.69 × 10−3 K−1, −5.30). Calculate a value for Ea, in kJ mol−1 (R = 8.31 J K−1 mol−1).
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Gradient = (−5.30 − (−7.68)) ÷ ((1.69 − 1.81) × 10−3) = 2.38 ÷ (−1.2 × 10−4) = −19 833 K 1 mark
Ea = −gradient × R = 19 833 × 8.31 = 164 800 J mol−1 1 mark
Ea = 165 kJ mol−1 1 mark
The mark scheme expects a gradient near −19 900 (allowing −19 400 to −20 400) and awards full marks for any Ea from 161 to 170 kJ mol−1. Use two points as far apart on the line as possible — a short gradient triangle magnifies read-off error.
For the reaction 2HI(g) → H2(g) + I2(g), a graph of ln k against 1/T is a straight line. Two points on the line are (1/T = 0.00128 K−1, ln k = −2.8) and (1/T = 0.00180 K−1, ln k = −14.1). Calculate a value for the activation energy, Ea, in kJ mol−1 (R = 8.31 J K−1 mol−1).
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Gradient = (−14.1 − (−2.8)) ÷ (0.00180 − 0.00128) = −11.3 ÷ 5.2 × 10−4 = −21 731 K 1 mark
Ea = −gradient × R = 21 731 × 8.31 = 180 583 J mol−1 1 mark
Ea = 181 kJ mol−1 1 mark
The mark scheme allows a gradient of −21 330 to −22 130. Do not forget the final division by 1000 — the answer is asked for in kJ mol−1, but R is in J K−1 mol−1.
An experiment measured the time, t, taken for a solution of iodine to react completely when added to an excess of an acidified solution of butanone. The experiment was repeated at different temperatures and 1/t was plotted against temperature. The graph rises from left to right and becomes steeper and steeper. Describe and explain the shape of this graph.
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As the temperature increases, the rate (1/t) increases — the time for the reaction to finish decreases. 1 mark
The increase is exponential — the rate rises by a greater and greater factor for each equal rise in temperature. 1 mark
Because many more particles have energy ≥ Ea. 1 mark
“Higher collision frequency” on its own is not accepted, and neither is “more successful collisions” on its own — the third mark needs the idea that many more particles now have energy greater than or equal to the activation energy. This is the Maxwell–Boltzmann picture behind the exponential in k = Ae−Ea/RT.
From the graph in Q41, the value of 1/t at 35 °C is 0.030 s−1. Deduce the time taken for the reaction at 35 °C.
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t = 1 ÷ 0.030 = 33 s 1 mark
The plotted quantity is 1/t, so the time is its reciprocal — not the read-off itself.
Source: AQA A-Level Chemistry past papers.
From data to rate equation
Most long rate-equation questions follow one sequence. Work it in order and lay out every line.
- Deduce each order from the data (initial rates or graph shape).
- Write the rate equation from the orders.
- Calculate k by rearranging and substituting one experiment — then work out its units from the overall order.
- Identify the slow step from the species in the rate equation.
- Examiner reports warn against guessing the slow step from a mechanism diagram — decide it from the species in the rate equation.
- Rearrange for k the right way up (k = rate ÷ concentration terms); an upside-down calculation is a common error.
- Always give the units of k, derived from the overall order.
Interactive — build the mark-scheme answer
“Propanone reacts with bromine in alkaline conditions: CH3COCH3 + Br2 + OH− → CH3COCH2Br + Br− + H2O. The rate equation is rate = k[CH3COCH3][OH−]. When [CH3COCH3] = 1.50 × 10−2 mol dm−3 and [OH−] = 2.50 × 10−2 mol dm−3 the initial rate is 2.75 × 10−11 mol dm−3 s−1. Calculate k, give its units, and state which step is rate-determining and why.” [5 marks] — select every statement that earns a mark, and nothing that doesn’t. Order doesn’t matter: AQA credits each point on its own.
the answer, assembled
Capstone question
Propanone reacts with bromine in alkaline conditions with rate = k[CH3COCH3][OH−]. In one experiment [CH3COCH3] = 1.50 × 10−2 and [OH−] = 2.50 × 10−2 mol dm−3, and the initial rate is 2.75 × 10−11 mol dm−3 s−1. Calculate k and give its units.
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k = rate / ([CH3COCH3][OH−]) = 2.75 × 10−11 / (1.50 × 10−2 × 2.50 × 10−2) 1 mark
k = 7.33 × 10−8 1 mark
Overall order 2, so units = mol−1 dm3 s−1 1 mark
Source: AQA A-Level Chemistry past papers.
- rate = k[A]m[B]n; orders m, n are 0/1/2, found by experiment; overall order = m + n. Units of k depend on the overall order (order 1 → s−1; order 2 → mol−1 dm3 s−1).
- Orders from initial rates: compare two experiments where only one concentration changes; how the rate scales gives the order (×2 rate → order 1; ×4 → order 2; no change → order 0).
- Orders from graphs. Concentration–time: zero order is a straight line; first order has a constant half-life. Rate–concentration: zero order is horizontal; first order is a straight line through the origin; second order is an upward curve.
- Rate-determining step: the slow step contains only the species in the rate equation (to their order). Species absent from the rate equation come after the slow step.
- Arrhenius: k = Ae−Ea/RT. Raising T increases k. Rearranged: ln k = −Ea/R × (1/T) + ln A; a plot of ln k against 1/T is a straight line of gradient −Ea/R.