Nine teaching hours at both levels, outcomes 2.2.1 to 2.2.5, and six additional higher level hours, outcomes 2.2.6 to 2.2.13.
Guiding question. How can the rate of a reaction be controlled?
Reactivity 2.2 · How fast? The rate of chemical change
1Rate of reaction and how it is measured 2.2.1 SL + HL
The syllabus statement: The rate of reaction is expressed as the change in concentration of a particular reactant/product per unit time. The skill: determine rates of reaction, including rates from tangents of graphs of concentration, volume or mass against time.
Some reactions are over before they can be watched. The nitroglycerine in dynamite decomposes in a fraction of a millisecond, and the sodium azide in an airbag in about thirty milliseconds. Others are so slow that they seem not to happen at all: iron rusts over months, and the body of the “Iceman” found in an Alpine glacier had barely decomposed in over five thousand years because the cold slowed every reaction that would have destroyed it. Between these extremes are the reactions that chemists, engineers and cooks control deliberately — setting cheese, curing concrete, making ammonia. Controlling a rate starts with measuring it.
Rate of reaction: the change in concentration of a reactant or product per unit time.
Square brackets denote concentration in mol dm−3, so rate has units of mol dm−3 s−1. The minus sign for a reactant makes the rate positive, because the concentration of a reactant decreases. When the property followed is a gas volume or a mass, a rate can also be quoted in cm3 s−1 or g s−1.
The definition refers to a particular reactant or product, because the coefficients in the equation make the substances change at different rates. For N2(g) + 3H2(g) → 2NH3(g), hydrogen is used up three times as fast as nitrogen and ammonia forms twice as fast as nitrogen is consumed, so a rate must always say which substance it refers to.
Average rate, instantaneous rate and the tangent
If the concentration of a product is plotted against time, the curve is steepest at the start and flattens as the reactants are used up; when the limiting reactant has gone, the line becomes horizontal and the rate is zero. The average rate over an interval is the change in concentration divided by the time taken — the gradient of the straight line joining two points on the curve. The instantaneous rate at a particular moment is the gradient of the tangent to the curve at that moment. The initial rate is the gradient of the tangent at t = 0, when the concentrations are exactly those that were mixed.
Question. Using Figure R2.5, determine the rate of reaction at t = 100 s.
| Method | Draw the tangent at t = 100 s with a ruler, touching the curve at that point only. Choose two points far apart on the tangent, not on the curve. |
| Points | (0 s, 0.588 mol dm−3) and (200 s, 0.000 mol dm−3) |
| Gradient | (0.000 − 0.588) ÷ (200 − 0) = −2.94 × 10−3 mol dm−3 s−1 |
| Rate | The reactant concentration falls, so rate = −gradient = 2.94 × 10−3 mol dm−3 s−1 |
| Check | The initial tangent is steeper (8.0 × 10−3 mol dm−3 s−1): the rate falls as the reactant is used up. The answer has two or three significant figures, no more, because a hand-drawn tangent is not more precise than that. |
Following a reaction in the laboratory
Concentrations are seldom measured directly while a reaction is running. Instead, a property that changes in proportion to the extent of reaction is followed with time, and the rate is deduced from it. The choice depends on what the reaction does.
| Method | When it is suitable, and an example | What is measured |
|---|---|---|
| Volume of gas | A gas is produced: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) | Volume in a gas syringe or an inverted burette over water, at intervals of time |
| Loss of mass | A gas escapes from an open flask on a balance; best for a dense gas such as CO2 | Mass of flask and contents, at intervals of time (a cotton-wool plug stops spray escaping) |
| Colour (colorimetry) | A reactant or product is coloured: purple MnO4− decolorized by ethanedioic acid; brown I2 formed or consumed | Absorbance, which is proportional to the concentration of the coloured species |
| pH | H+ or OH− is used or produced; for example, lactic acid formed as milk is set for cheese | pH with a probe and data logger |
| Electrical conductivity | The number or type of ions changes | Conductivity with a probe |
| Formation of a precipitate | Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + SO2(g) + S(s) + H2O(l) | Time for a cross under the flask to disappear; rate is proportional to 1/time |
| Sampling and titration | No convenient physical property; samples withdrawn and the reaction stopped (quenched), e.g. by cooling or dilution | Concentration of a reactant in each sample |
The guide asks for experiments in which time is the dependent variable and in which it is the independent variable. In the “disappearing cross” experiment, the extent of reaction is fixed (enough sulfur to hide the cross) and the time taken is measured: time is dependent, and 1/t is used as a measure of the average rate. In a gas-syringe experiment, times are chosen in advance and the volume is read at each: time is independent, and the rate at any moment comes from a tangent.
Exam focus · what the published papers show
Definitions of rate lose marks for vagueness. One report found definitions poor with many referring to a measure of time rather than a change in concentration; another notes that ‘the time for the reaction to go to completion’ was not an uncommon response; a third that some stated to calculate rate as [FeSO4]/time rather than change in concentration. A mark scheme for a question asking how a rate could be measured specifies a property measured with time and adds Do not accept measure rate of change — it asks what would be measured, not a restatement of the word “rate”.
For initial rate the tangent is essential. Reports observe that only a few suggested drawing a tangent at t = 0, that correct use of the terms ‘tangent’ and ‘gradient’ was rare, and that in coursework students opted for establishing average rate instead of using tangent at initial times. On one multiple-choice item the tangent at t = 0 gave the correct answer, however, 52% of the candidates chose A as the answer.
2Collision theory: energy and orientation 2.2.2 SL + HL
The syllabus statement: Species react as a result of collisions of sufficient energy and proper orientation. The skill: explain the relationship between the kinetic energy of the particles and the temperature in kelvin, and the role of collision geometry.
Collision theory is the kinetic molecular theory of Structure 1.1 applied to reactions. For two particles to react, three things must happen: they must collide; they must collide with at least a minimum amount of energy; and they must collide in an orientation that allows the bonds that are to break and form to do so. Only collisions that satisfy all three are successful collisions, and they are a small fraction of the total. In a gas at room temperature each molecule collides billions of times a second; if every collision led to reaction, every gas-phase reaction would be over almost instantly.
Kinetic energy and temperature
The particles in a sample have a range of speeds and therefore a range of kinetic energies, Ek = ½mv2. The absolute temperature, in kelvin, is proportional to the average kinetic energy of the particles. Doubling the kelvin temperature doubles the average kinetic energy; raising the temperature from 20 °C to 40 °C does not, because 293 K to 313 K is an increase of only 7 %. At a given temperature, all gases have the same average kinetic energy, so lighter particles move faster on average than heavier ones.
A higher temperature affects the rate in two ways. Particles move faster, so they collide more frequently; and a larger proportion of the collisions have enough energy to react. The second effect is by far the larger. Average speed is proportional to √T, so raising the temperature from 298 K to 308 K increases collision frequency by a factor of √(308/298) ≈ 1.02 — about 2 % — yet for many reactions the rate roughly doubles. The explanation, the distribution of energies, is the subject of 2.2.4.
Collision geometry
A collision with ample energy still fails if the particles meet the wrong way round. When two hydrogen iodide molecules collide to form hydrogen and iodine, 2HI → H2 + I2, the two hydrogen atoms must come close enough for an H–H bond to begin to form while the H–I bonds lengthen; a collision in which the iodine atoms strike each other end-on does not lead to reaction however fast the molecules move. The more complex the molecules, the smaller the fraction of collisions with a suitable geometry. Orientation does not depend on temperature, which is why it appears in the Arrhenius factor, A, at higher level (2.2.13).
Collision frequency: the number of collisions per unit time. Rates depend on how often collisions happen, not on how many happen in total.
Successful collision: a collision with energy greater than or equal to the activation energy and with the correct orientation, which results in reaction.
Activation energy, Ea: the minimum energy that colliding particles need for a successful collision leading to a reaction (2.2.4).
The single most repeated comment on this topic is about one word. Reports record many of the oft repeated mistakes (number of collisions rather than collision frequency); that candidates lost marks as they did not refer to the reduced ‘frequency’ of collisions; and that explanations referring to ‘more collisions’ should be developed further to include ‘more frequent collisions’ or ‘number of collisions per unit time’.
A mark scheme for a collision-theory item awards the energy point as E ≥ Ea and the second point for correct orientation. “More collisions” without a time reference earns nothing, because a slower reaction eventually produces just as many collisions.
3Factors that change the rate 2.2.3 SL + HL
The syllabus statement: Factors that influence the rate of a reaction include pressure, concentration, surface area, temperature and the presence of a catalyst. The skill: predict and explain the effects of changing conditions on the rate of a reaction.
Each factor changes either how often particles collide, or what fraction of collisions succeed, or both. An explanation earns its marks by naming which, and by linking it to the rate.
| Change | Particle-level explanation | Everyday or industrial example |
|---|---|---|
| Increase concentration of a solution | More particles per unit volume, so collisions are more frequent. The fraction of collisions with E ≥ Ea is unchanged. | Concentrated acid dissolves marble faster than dilute acid. |
| Increase pressure of a gas | The same particles are in a smaller volume, so the concentration is higher and collisions are more frequent. Pressure matters only for reactions with gaseous reactants. | Industrial gas-phase processes, such as ammonia synthesis, run at high pressure. |
| Increase surface area of a solid | Only particles at the surface can collide. Smaller pieces of the same mass expose more particles, so collisions are more frequent. | Kindling lights before logs; fine coal or flour dust suspended in air can explode. |
| Increase temperature | Particles have a higher average kinetic energy: a much larger fraction of collisions have E ≥ Ea, and collisions are also slightly more frequent. | Refrigeration slows the reactions that spoil food; cooking speeds reactions up. |
| Add a catalyst | Provides an alternative pathway with a lower Ea, so a larger fraction of collisions are successful (2.2.5). Collision frequency and particle energies are unchanged. | Catalytic converters; iron in the Haber process; enzymes. |
Surface area is often misunderstood because it varies inversely with particle size. For a fixed mass of solid, smaller particles mean a larger surface area. A cube of side 1 cm has a surface area of 6 cm2; cut into cubes of side 1 mm, the same volume has a surface area of 60 cm2. Powdered calcium carbonate reacts faster with acid than lumps of the same mass for this reason.
What the graphs show
The effect of a change is seen most clearly on a graph of the amount of product against time. Two features must be read separately: the initial gradient, which shows the initial rate, and the final value, which is fixed by the amount of the limiting reactant (2.1.3). Changing temperature, surface area or adding a catalyst changes only the gradient; the curve reaches the same final value sooner. Changing the amount of limiting reactant changes the final value as well.
Question. Excess magnesium ribbon reacts with 50.0 cm3 of 1.00 mol dm−3 hydrochloric acid and the volume of hydrogen is recorded. Sketch, relative to the original curve, the curve obtained with 50.0 cm3 of 0.50 mol dm−3 acid at the same temperature.
| Limiting reactant | Magnesium is in excess, so the acid is limiting in both runs. |
| Final volume | Half the amount of HCl gives half the amount of H2: the curve levels off at half the original volume. |
| Initial gradient | A lower concentration of H+ means less frequent collisions with the metal surface: a smaller initial gradient. |
| Shape | The new curve starts at the origin, lies below the original throughout, and is horizontal at the end. |
If instead the question had used 100.0 cm3 of 0.50 mol dm−3 acid, the amount of HCl would be unchanged: the same final volume, reached more slowly. The published distractors test exactly this distinction between concentration and amount.
Designing a fair rate experiment
To investigate one factor, every other factor must be held constant: the same temperature (a water bath), the same concentrations and volumes of solutions, the same mass and particle size of solid, the same total volume, and the same method of measuring. When a solid is used, “surface area” is best controlled by using the same mass of the same batch of chips or powder. Graphs also provide evidence about errors. Points scattered on both sides of a smooth curve show random error; a whole curve displaced from the origin — for example, gas collected from a syringe that was not zeroed, or time lost before the bung is inserted — suggests a systematic error.
Counting collisions
Two reactants, blue and red, move at random in a closed box. The model counts blue–red collisions per second and flashes those with relative kinetic energy at least Ea and a suitable orientation. Change one condition at a time and predict each number first.
Counts are averaged over the last few seconds of simulated time, so let the numbers settle after each change. A simplified two-dimensional model: the trends, not the values, are the point.
Exam focus · what the published papers show
On one item 65% of the candidates chose the correct combination to give the greatest rate of reaction; on a parallel item the most common distractor offered ‘smaller surface area of same mass of CaCO3(s)’, and the report concludes: It seems these candidates confused ‘surface area’ with ‘particle size’. Smaller particles, larger surface area, faster rate.
Two multiple-choice items combine rate with stoichiometry. On one, candidates had to consider two factors to answer this question: the slower initial rate of reaction, and the larger amount of product when the reaction is complete. On the other, answered correctly by 58 %, it was surprising that A, which stated that the initial rate was the same, was the most commonly chosen distractor even though the concentrations of the acids were different.
In sketches, almost all students managed to draw a curve with a greater initial gradient that produced the same volume of gas, but marks were lost for curves that passed through a maximum or went above or below the maximum volume. Another report reminds candidates that a question about collision theory needs collisions in the answer, and that some candidates referred to ‘more’ collisions without reference to time.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
3.26 g of iron powder are added to 80.0 cm3 of 0.200 mol dm−3 copper(II) sulfate solution: Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s).
(i) Sketch a graph of the concentration of iron(II) sulfate, FeSO4, against time as the reaction proceeds. [2]
(ii) Outline how the initial rate of reaction can be determined from the graph in part (c)(i). [2]
(iii) Explain, using the collision theory, why replacing the iron powder with a piece of iron of the same mass slows down the rate of the reaction. [2]
Bromine, Br2(l), and methanoic acid, HCOOH(aq), react in the presence of sulfuric acid: Br2(l) + HCOOH(aq) → 2HBr(aq) + CO2(g). Suggest an experimental method that could be used to determine the rate of reaction. [2]
Iodomethane can be converted into methanol: CH3I + HO− → CH3OH + I−. Outline the requirements for a collision between reactants to yield products. [2]
Solutions and mark-scheme guidance · Set R2C
R2C.1 A
The tangent at t = 0 follows the first points: concentration falls from 1.6 to 1.4 mol dm−3 in about 2 s, a gradient of about 0.1 mol dm−3 s−1.
R2C.2 B
Average rate = change in concentration ÷ time = (0.10 − 0.02) ÷ 800 = 1 × 10−4 mol dm−3 s−1. The report notes that the arithmetic with powers of ten without a calculator was the difficulty.
R2C.3 C
The reaction is exothermic (temperature change) and the blue Cu2+ fades while brown copper forms (colour change). No gas is involved, so there is no volume change.
R2C.4 B
Iodine, I2(aq), is coloured; the other species are colourless. A colorimeter follows its formation. No gas is produced and there is no mass change from an open flask.
R2C.5 A
Higher concentration and larger surface area both increase collision frequency. The distractor D confuses “smaller surface area” with smaller particles.
R2C.6 A
At higher pressure the gas is more concentrated, so collisions are more frequent. Pressure does not change the energy of collisions (that needs temperature) or Ea.
R2C.7 B
Curve Y has a smaller initial gradient but a larger final volume: more H2O2 in total, at a lower concentration. Adding dilute (0.1 mol dm−3) peroxide does both. Removing the catalyst (D) would give the same final volume.
R2C.8 B
Na2CO3 is limiting in both (the acid is in excess), so the same 1.0 g gives the same mass of CO2. The rates differ because the acid concentrations differ; A was the most popular wrong answer.
R2C.9 [6]
(i) Initial concentration zero AND concentration increases with time ✔; decreasing gradient as the reaction proceeds ✔ (levelling off).
(ii) Draw a tangent to the curve at time = 0 ✔; the rate equals the gradient of the tangent ✔. The report notes that weaker candidates omitted the tangent.
(iii) The piece has a smaller surface area ✔; lower frequency of collisions OR fewer collisions per unit time ✔. The scheme says: Do not accept just “fewer collisions”.
R2C.10 [2]
Measure the change in mass OR pressure OR volume of CO2 OR intensity of colour (bromine is brown) OR conductivity OR pH ✔ with time ✔. Equipment is accepted for the first mark: balance, pressure probe, gas syringe, colorimeter, conductivity or pH meter. The scheme does not accept “measure rate of change” for the second mark.
R2C.11 [2]
Energy ≥ activation energy (E ≥ Ea) ✔; correct orientation OR correct geometry of the reacting particles ✔.
4Activation energy and the Maxwell–Boltzmann distribution 2.2.4 SL + HL
The syllabus statement: Activation energy, Ea, is the minimum energy that colliding particles need for a successful collision leading to a reaction. The skill: construct Maxwell–Boltzmann energy distribution curves to explain the effect of temperature on the probability of successful collisions.
Activation energy on an energy profile
As two particles approach and bonds begin to break and form, the potential energy of the system rises to a maximum before falling to the energy of the products. The arrangement of atoms at the maximum is the transition state (or activated complex), in which old bonds are partly broken and new bonds partly formed. The height of the maximum above the reactants is the activation energy of the forward reaction. On an energy profile the Ea arrow always starts at the reactant level and ends at the top of the curve, whether the reaction is exothermic or endothermic. The activation energy of the reverse reaction is measured from the products to the same maximum, so Ea(reverse) = Ea(forward) − ΔH.
Activation energy explains why many thermodynamically favourable reactions do not happen at room temperature. Methane and oxygen can be mixed safely; a spark is needed to give some molecules enough energy to react, after which the energy released keeps the reaction going. Nitroglycerine, by contrast, has such a low activation energy that a knock can supply it — which is why it was stabilized in dynamite.
The Maxwell–Boltzmann distribution
In a gas at a fixed temperature, the particles do not all have the same kinetic energy. Collisions constantly transfer energy, so at any moment a few particles are almost stationary, most have moderate energies, and a few have very high energies. The Maxwell–Boltzmann distribution is the graph of the number (or fraction) of particles against their kinetic energy.
Axes: x — kinetic energy (of particles); y — number of particles (or fraction, or probability) with that energy. These are not the axes of an energy profile.
Shape: starts at the origin (no particles have zero energy); rises to a peak at the most probable energy; is asymmetric, with a long tail to high energy that approaches but never meets the x-axis.
Area: the total area under the curve represents the total number of particles, so it does not change when the temperature changes.
Ea: a vertical line on the energy axis; the area to its right is the number of particles with E ≥ Ea.
Why a small rise in temperature has a large effect
At a higher temperature the average kinetic energy is greater, so the curve shifts to the right. Because the total number of particles is fixed, the area must stay the same, so the peak becomes lower and the curve becomes broader. The activation energy does not change — it is a property of the reaction pathway, not of the particles. The crucial change is in the tail: the fraction of particles with energy at least equal to Ea increases greatly, because Ea usually lies far out in the tail where the curve is most sensitive to temperature. This increase in the proportion of successful collisions, together with the small increase in collision frequency, explains the large increase in rate.
Three linked statements, in this order: (1) at higher temperature the particles have greater average kinetic energy; (2) a greater proportion of particles (or collisions) have energy ≥ Ea, shown by the larger area beyond Ea; (3) there are more frequent collisions, and so more successful collisions per unit time. A sketch with both curves labelled and Ea marked supports points (1) and (2).
How fast? The Maxwell–Boltzmann distribution
Only molecules past the barrier can react. Raising the temperature and adding a catalyst both increase that shaded fraction, but they do it from opposite ends.
Fraction of molecules against kinetic energy. The area under each curve is the same — the total number of molecules does not change when you heat it, only how the energy is shared out. The axis is held at 0–60 kJ mol−1 so the two curves can be compared: at 300 K almost every molecule sits below 15 kJ mol−1, which is exactly why a barrier of even 25 kJ mol−1 is crossed by so few of them.
The curve never touches the axis at high energy and never starts above zero at zero energy — two things examiners look for in a sketched distribution, along with the peak moving right and down as temperature rises.
Exam focus · what the published papers show
Maxwell–Boltzmann sketches lose marks for avoidable reasons. Reports list curves that did not start at the origin, were not labelled or the activation energy was missing, and note that too many curves did not start at the origin and lacked correct labels. A mark scheme for a two-temperature sketch requires the peak at T1 to right of AND lower than T2 — here T1 is the higher temperature — and lines begin at origin AND T1 must finish above T2.
The other recurring error is drawing the wrong kind of graph: in some answers, enthalpy diagrams were drawn in place of Maxwell Boltzmann distribution curve, and a report comments on some confusion between Maxwell-Boltzmann diagrams and energy profile diagrams. On temperature explanations, one found most students failing to mention ‘activation energy’ in their answer or failing to annotate the graph, and another that ‘More kinetic’ energy was often all that was offered for the first mark.
One report is explicit: Understanding that increasing temperature does not alter Ea is a key concept of Maxwell-Boltzmann curve explanations. On a later item, incorrect responses show that a significant proportion believe that activation energies vary with temperature. Temperature changes the particles; a catalyst changes the pathway. Only a catalyst moves the Ea line.
5Catalysts 2.2.5 SL + HL
The syllabus statement: Catalysts increase the rate of reaction by providing an alternative reaction pathway with lower Ea. The skills: sketch and explain energy profiles with and without catalysts, and construct Maxwell–Boltzmann curves to explain the effect of different values of Ea. Biological catalysts are called enzymes.
Hydrogen peroxide decomposes slowly at room temperature: 2H2O2(aq) → 2H2O(l) + O2(g). Add a pinch of manganese(IV) oxide and the solution froths with oxygen within seconds; at the end the black powder can be filtered off, dried and weighed, and its mass is unchanged. A catalyst is a substance that increases the rate of a reaction without itself undergoing a permanent chemical change.
How a catalyst works
A catalyst does not give the particles more energy and does not make collisions more frequent. It provides a different reaction pathway — a different sequence of steps, usually involving the catalyst temporarily — whose activation energy is lower. The catalyst takes part in the reaction and is regenerated at the end, which is why it is not used up. With a lower Ea, a larger fraction of collisions at the same temperature have enough energy to react, so the rate increases.
Three consequences follow from the energy profile. First, ΔH is unchanged: the reactants and products are the same substances at the same energies. Secondly, the reverse activation energy is lowered by exactly the same amount as the forward one, so a catalyst speeds up the forward and reverse reactions by the same factor; this is why a catalyst does not change the position of an equilibrium or the value of K (Reactivity 2.3). Thirdly, the profile of a real catalysed reaction usually has more than one maximum, because the alternative pathway has more than one step; a single lower hump is the simplified sketch expected at this level.
Both increase the fraction of collisions with E ≥ Ea, but on the Maxwell–Boltzmann diagram they look completely different. Temperature changes the curve (a second, flatter curve shifted right) and leaves Ea where it is. A catalyst leaves the curve alone and moves the Ea line to the left. When a question gives a single curve and asks for the effect of a catalyst, the answer is a second vertical line, not a second curve.
Types of catalyst and enzymes
A homogeneous catalyst is in the same phase as the reactants — for example, aqueous H+ ions catalysing esterification. A heterogeneous catalyst is in a different phase, usually a solid with gaseous or liquid reactants: iron in the Haber process, platinum, palladium and rhodium in a car's catalytic converter, zeolites in the cracking of hydrocarbons. Heterogeneous catalysts are preferred in industry because they are easily separated from the products and reused; a large surface area, achieved with a powder or a fine coating on a support, makes them more effective. The mechanisms by which the two types act are not assessed.
Without a catalyst, the reaction of nitrogen with hydrogen would need temperatures of thousands of degrees to proceed at a useful rate, at which almost no ammonia would survive at equilibrium. With an iron catalyst, around 450 °C is enough. A catalytic converter oxidizes carbon monoxide and unburned hydrocarbons and reduces nitrogen monoxide within the fraction of a second that exhaust gas spends in it.
Enzymes are biological catalysts, usually proteins. The reactant (the substrate) binds to a specific region of the enzyme, the active site, whose shape and chemical groups fit it closely; this makes enzymes highly specific. Lipase, for example, catalyses the hydrolysis of fats and is used in biological washing powders. Enzymes work under mild conditions, but because their activity depends on their three-dimensional structure, they are denatured at high temperature and outside a narrow pH range, and the rate of an enzyme-catalysed reaction falls sharply above an optimum temperature.
Exam focus · what the published papers show
A report notes that candidates lost a mark in their explanation of catalyst action as they did not refer to an alternative pathway; on a later item most wrote that activation energy was lowered by the catalyst, they forgot to mention the pathway and only 30% of candidates managed to get the mark; on another, the average was 40 % because some referred to lowering activation energy, but without reference to alternate pathway. The complete statement is: a catalyst provides an alternative pathway (or mechanism) with a lower activation energy, so a greater proportion of collisions have E ≥ Ea.
On one multiple-choice item only 40% of students gained the mark and rather disturbingly over 50% seem to think that a catalyst increases the energy of particles in the system. Even worse, this misconception appears to occur over the whole ability range with the item having a negative discrimination index! Another report records stating that molecules have higher kinetic energy when a catalyst is added as a common mistake. The distribution of energies depends only on temperature.
Asked to show a catalyst on a Maxwell–Boltzmann curve, about half of the candidates annotated the Maxwell-Boltzmann distribution to show the effect of the catalyst. Some left it blank and some sketched a new distribution that would be obtained at a higher temperature instead. An earlier report describes the same error: some weaker students confused it with the effect of temperature and constructed a second curve. On energy profiles the common error was incorrectly labelling the activation energies (missing labels or arrows starting from the product to the top of the energy profile).
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
1-chloropentane reacts with aqueous sodium hydroxide. The reaction was repeated at a lower temperature.
(i) Sketch labelled Maxwell–Boltzmann energy distribution curves at the original temperature (T1) and the new lower temperature (T2). Axes: fraction of particles against kinetic energy. [2]
(ii) Explain the effect of lowering the temperature on the rate of the reaction. [2]
Nickel catalyses the conversion of propanone to propan-2-ol (with H2, Ni, heat).
(a) Outline how a catalyst increases the rate of reaction. [1]
(b) Explain why an increase in temperature increases the rate of reaction. [2]
Sketch a Maxwell–Boltzmann distribution curve for a chemical reaction showing the activation energies with and without a catalyst. [3]
Solutions and mark-scheme guidance · Set R2D
R2D.1 B
A catalyst provides an alternative pathway of lower Ea. Temperature, concentration and collision frequency change the rate without changing Ea.
R2D.2 C
At a higher temperature the peak moves to higher energy and becomes lower, because the area (number of particles) is unchanged.
R2D.3 B
Ea is a value of kinetic energy on the x-axis: B. C is the area representing particles with E ≥ Ea; A and D are values on the probability axis.
R2D.4 B
The forward activation energy runs from the reactant level to the top of the curve: B. C is the reverse activation energy; A and D are not measured from the reactants.
R2D.5 C
The reverse reaction starts at Z. With a catalyst the maximum is the lower (inner) curve, so the arrow runs from Z to the lower maximum: C. D is the uncatalysed reverse Ea.
R2D.6 B
A catalyst increases the initial gradient but cannot change the final volume, which is set by the amount of H2O2.
R2D.7 C
A catalyst changes the mechanism (alternative pathway) and is regenerated. It does not increase the energy of collisions (statement I).
R2D.8 [4]
(i) Peak at T1 to the right of AND lower than T2 ✔; lines begin at the origin AND T1 finishes above T2 at high energy ✔.
(ii) Rate is lower AND (average) kinetic energy lower, OR less frequent collisions, OR fewer collisions per unit time ✔; rate lower AND a smaller fraction of molecules/collisions have E ≥ Ea ✔. “Fewer collisions” without reference to time, frequency or probability is not accepted.
R2D.9 [3]
(a) Provides an alternative pathway/mechanism AND lower Ea ✔.
(b) Greater proportion of molecules with E ≥ Ea ✔; greater frequency of collisions OR more collisions per unit time ✔.
R2D.10 [3]
Both axes correctly labelled ✔ (x: kinetic energy or speed, not potential energy; y: number, fraction or probability of particles); correct shape starting at the origin ✔ (not touching the x-axis at high energy; two curves are not accepted); Ea(catalyst) < Ea(without catalyst) marked on the x-axis ✔.
Additional higher level · outcomes 2.2.6 to 2.2.13
6Reaction mechanisms and the rate-determining step 2.2.6 HL only
The syllabus statement: Many reactions occur in a series of elementary steps. The slowest step determines the rate of the reaction. The skills: evaluate proposed reaction mechanisms and recognize reaction intermediates, and distinguish between intermediates and transition states, including examples where the rate-determining step is not the first step.
An overall equation describes the start and the finish of a reaction, not the route. The reaction of nitrogen dioxide with carbon monoxide, NO2(g) + CO(g) → NO(g) + CO2(g), looks like a single exchange of an oxygen atom. Yet experiment shows that at temperatures below about 500 K its rate does not depend on the concentration of carbon monoxide at all. A single collision between NO2 and CO cannot explain that. The reaction must happen in more than one step.
Elementary step: a single step in a mechanism, describing one collision (or one decomposition) at the molecular level.
Reaction mechanism: the sequence of elementary steps by which a reaction occurs. The steps add up to the overall equation.
Intermediate: a species formed in one step and used up in a later step. It does not appear in the overall equation. It exists for a finite time and can sometimes be detected.
Transition state (activated complex): the arrangement of highest potential energy in an elementary step, with bonds partly broken and partly formed. It exists only at the instant of the collision and cannot be isolated.
Rate-determining step (RDS): the slowest step in the mechanism; it limits the rate of the overall reaction.
For NO2 + CO the accepted mechanism has two steps:
step 2 (fast): NO3 + CO → NO2 + CO2
Adding the steps and cancelling the species that appear on both sides — one NO2 and the NO3 — gives the overall equation. NO3 is an intermediate: made in step 1 and consumed in step 2. Because step 1 is slow, NO3 is used up by step 2 as fast as it forms, and the overall rate is the rate of step 1. Carbon monoxide takes part only after the slow step, so its concentration has no effect on the rate, and the experimental rate equation is rate = k[NO2]2.
The rate-determining step is sometimes compared with the slowest worker on a production line: speeding up any other worker does not increase the output. Only the species that take part in the RDS, or in steps before it, appear in the rate equation.
When the rate-determining step is not the first step
The guide requires examples in which a fast step comes first. Nitrogen monoxide is oxidized by oxygen, 2NO(g) + O2(g) → 2NO2(g), and the experimental rate equation is rate = k[NO]2[O2]. A mechanism consistent with this is:
step 2 (slow): N2O2 + O2 → 2NO2
The slow step involves the intermediate N2O2 and one O2. An intermediate cannot appear in a rate equation, because its concentration cannot be controlled or measured independently. It is replaced by the species that formed it: each N2O2 comes from two NO, so [N2O2] depends on [NO]2, and the rate equation becomes rate = k[NO]2[O2]. The general rule: the rate equation contains every reactant that appears in the RDS or in any step before it, and no intermediates.
Question. Propanone reacts with iodine in acidic solution: CH3COCH3(aq) + I2(aq) → CH3COCH2I(aq) + H+(aq) + I−(aq). The experimental rate equation is rate = k[CH3COCH3][H+]. Explain what the rate equation shows about the mechanism.
| Zero order in I2 | Iodine is not involved in the rate-determining step or any step before it; it reacts in a fast step after the RDS. |
| First order in propanone and H+ | One propanone and one H+ are involved in or before the RDS. |
| H+ is a catalyst | It appears in the rate equation but not as a reactant in the overall equation: it is used in an early step and released again later. Because the reaction also produces H+, it speeds up as it proceeds (it is autocatalytic). |
| Consistent mechanism | A slow step between propanone and H+ forms an intermediate, which reacts rapidly with I2. The steps must add to the overall equation. |
A mechanism is consistent with the data if its steps add up to the overall equation (stoichiometry) and the RDS, with any earlier steps, predicts the experimental rate equation (kinetics). Consistency does not prove a mechanism: another mechanism might fit the same data, which is why mechanisms are only ever “possible” or “proposed”.
7Energy profiles of multistep reactions 2.2.7 HL only
The syllabus statement: Energy profiles can be used to show the activation energy and transition state of the rate-determining step in a multistep reaction. The skill: construct and interpret energy profiles from kinetic data.
Each elementary step has its own transition state, so a two-step reaction has two maxima on its energy profile. Between them is a minimum, a trough, which represents the intermediate: it is a real species at a local energy minimum, not a transition state. The step with the larger activation energy — the higher barrier measured from the trough or level before it — is the slowest step, the RDS.
| Intermediate | Transition state | |
|---|---|---|
| Position on profile | A trough (local minimum) between two maxima | A maximum |
| Bonds | Fully formed bonds; a definite species | Bonds partly broken and partly formed |
| Lifetime | Finite, sometimes long enough to detect | Only at the instant of collision; cannot be isolated |
| Examples | NO3; N2O2; a carbocation in SN1 (Reactivity 3.4) | The five-coordinate carbon in SN2, drawn in square brackets with ‡ |
Constructing a profile from kinetic data means placing the higher hump where the data say the RDS is. If the rate equation shows that the first step is slow, the first maximum is higher; if the rate equation includes species from a fast first step, the second barrier is the larger one. For a catalysed reaction, the catalyst reacts in an early step and is regenerated in a later one, so it appears as a reactant in one step and a product in another.
8Molecularity 2.2.8 HL only
The syllabus statement: The molecularity of an elementary step is the number of reacting particles taking part in that step. The skill: interpret the terms ‘unimolecular’, ‘bimolecular’ and ‘termolecular’.
| Molecularity | Particles in the step | Example of an elementary step and its rate equation |
|---|---|---|
| Unimolecular | One particle decomposes or rearranges | N2O4 → 2NO2; rate = k[N2O4] |
| Bimolecular | Two particles collide (same or different) | NO2 + NO2 → NO3 + NO; rate = k[NO2]2 |
| Termolecular | Three particles collide simultaneously | 2NO + O2 → 2NO2 written as one step; rate = k[NO]2[O2] |
Termolecular steps are rare because the chance of three particles colliding at the same instant, with enough energy and the right orientations, is very small; steps involving four or more particles are not considered. Molecularity applies only to elementary steps. The overall equation for a reaction may contain many molecules, but that says nothing about any single collision: the equation 2NO + O2 → 2NO2 is consistent both with one termolecular step and with the two bimolecular steps shown in 2.2.6. Molecularity is a whole number describing a step; order (2.2.10) is an experimental quantity describing a rate equation.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
The following mechanism is proposed for a reaction: A + B → C + D (slow); D + B → A + E (fast).
(a) Classify substances B and D as reactant, product, catalyst, or intermediate, based on the proposed mechanism. [2]
(b) Deduce the rate expression. [1]
(c) In experiment 1, [A] = 0.200 and [B] = 0.200 mol dm−3 and the initial rate is 1.20 mol dm−3 s−1. Calculate the initial rate for experiment 2, in which [A] = 0.300 and [B] = 0.200 mol dm−3, under the same conditions. [1]
Nitrogen dioxide and carbon monoxide react: NO2(g) + CO(g) → NO(g) + CO2(g), ΔH = −226 kJ. Experimental data show the reaction is second order with respect to NO2 and zero order with respect to CO.
(i) State the rate expression for the reaction. [1]
(ii) The proposed mechanism is Step I: NO2 + NO2 → NO + NO3; Step II: NO3 + CO → NO2 + CO2. Identify the rate determining step giving your reason. [1]
(iii) State one method that can be used to measure the rate for this reaction. [1]
(iv) An energy profile shows Ea = 132 kJ for the forward reaction. Calculate the activation energy for the reverse reaction. [1]
(v) Sketch the relationship between the rate of reaction and the concentration of NO2. [1]
Solutions and mark-scheme guidance · Set R2E
R2E.1 B
A transition state is at an energy maximum: higher in potential energy than reactants and products, and unstable (it cannot be isolated).
R2E.2 D
Add the steps and cancel B, C and D: 2A → 2E + F. The rate is set by the slow first step: rate = k[A]2.
R2E.3 C
In C the slow first step involves one NO and one O2, predicting rate = k[NO][O2], not second order in NO. A, B and D each put two NO and one O2 in or before the slow step.
R2E.4 A
The slow first step, NO + F2, gives rate = k[NO][F2]; the steps sum to 2NO + F2 → 2NOF.
R2E.5 B
Three elementary steps give three maxima (two intermediates in troughs between them), and because the first step is slow, its activation energy — the first hump — must be the largest. Nearly 73 % matched the profile to the mechanism.
R2E.6 C
The slow step is AB + A; AB comes from A + B in the fast step before it. Replacing AB: rate = k[A]2[B], second order in A.
R2E.7 C
Species in fast steps before the slow step affect the rate and appear in the rate equation. A zero-order species can still react (after the RDS); a catalyst does take part.
R2E.8 [4]
(a) B: reactant ✔; D: intermediate ✔ (formed in step 1, used in step 2). A is the catalyst, used and regenerated; the report notes some listed B or D as the catalyst.
(b) rate = k[A][B] ✔ (the slow step).
(c) [A] × 1.5, first order: rate = 1.20 × 1.5 = 1.80 mol dm−3 s−1 ✔.
R2E.9 [5]
(i) rate = k[NO2]2 ✔. (ii) Step I AND CO does not appear in the rate expression (only two NO2 do) ✔. (iii) (IR or UV–visible) spectroscopy OR colorimetry OR colour change over time ✔ — NO2 is brown; the number of gas moles does not change, so pressure would not work. (iv) Ea(reverse) = 226 + 132 = 358 kJ ✔; the report notes that many gave the forward value or subtracted. (v) An upward curve (parabola) through the origin ✔.
9Rate equations from experimental data 2.2.9 HL only
The syllabus statement: Rate equations depend on the mechanism of the reaction and can only be determined experimentally. The skill: deduce the rate equation for a reaction from experimental data.
For a reaction A + B → products, the rate equation (or rate expression) has the form
where k is the rate constant and m and n are the orders with respect to A and B. The orders are not the coefficients in the equation. They reflect the mechanism, and since the mechanism cannot be read from the overall equation, the rate equation can only be found by experiment. A catalyst may appear in the rate equation even though it is not in the overall equation, and a reactant may be absent from it (zero order).
The initial-rates method
A series of experiments is carried out in which the initial concentration of one reactant is changed while the others are kept constant, and the initial rate is found each time (from a tangent at t = 0, or from 1/t in a clock reaction). Comparing pairs of experiments in which only one concentration changes gives the order with respect to that reactant:
- concentration doubles, rate unchanged: zero order;
- concentration doubles, rate doubles: first order;
- concentration doubles, rate quadruples: second order.
In general, if the concentration is multiplied by a factor x and the rate by xn, the order is n. Initial rates are used because at t = 0 the concentrations are exactly those mixed, and no products are present to interfere.
Question. The following illustrative data were obtained for 2A + B → C at constant temperature. Deduce the rate equation and calculate the rate constant, with units.
| Experiment | [A] / mol dm−3 | [B] / mol dm−3 | Initial rate / mol dm−3 s−1 |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10−4 |
| 2 | 0.20 | 0.10 | 8.0 × 10−4 |
| 3 | 0.10 | 0.30 | 6.0 × 10−4 |
| Order in A | Experiments 1 → 2: [A] × 2, [B] constant, rate × 4 = 22. Second order in A. |
| Order in B | Experiments 1 → 3: [B] × 3, [A] constant, rate × 3 = 31. First order in B. |
| Rate equation | rate = k[A]2[B]; third order overall. |
| Rate constant | From experiment 1: k = 2.0 × 10−4 ÷ (0.102 × 0.10) = 0.20 mol−2 dm6 s−1 |
| Check | Experiment 2: 0.20 × 0.202 × 0.10 = 8.0 × 10−4 ✔. Here the orders happen to match the coefficients; that is a coincidence, not a rule. |
Find the rate equation from initial-rate data
Compare two experiments in which only one concentration changes. What the rate does then is the order with respect to that reactant — and nothing else in the table matters for that step.
The units of k are not memorised — they fall out of the equation. Overall order n gives k in mol1−n dm3n−3 s−1, and the model derives them that way each time.
Rate against the concentration of the reactant you are testing, with the other held constant. A horizontal line is zero order, a straight line through the origin is first order, and a curve that quadruples when the concentration doubles is second order.
10Order of reaction and its graphs 2.2.10 HL only
The syllabus statement: The order of a reaction with respect to a reactant is the exponent to which the concentration of the reactant is raised in the rate equation. The skill: sketch, identify and analyse graphical representations of zero, first and second order reactions; only integer orders are assessed.
Order with respect to a reactant: the exponent to which the concentration of that reactant is raised in the rate equation. It can describe the number of particles of that reactant taking part in the rate-determining step (and steps before it).
Overall order: the sum of the orders with respect to each reactant. For rate = k[A]2[B], the overall order is 3.
Two kinds of graph identify an order. A rate–concentration graph is the more direct: for zero order, a horizontal line (rate independent of concentration); for first order, a straight line through the origin (rate ∝ concentration); for second order, an upward curve through the origin (rate ∝ concentration2). A concentration–time graph shows how the reactant is used up: for zero order, a straight line with constant negative gradient, because the rate does not change as the reactant is used; for first order, a curve whose gradient falls in proportion to the concentration, with a constant half-life; for second order, a curve that falls steeply at first and then flattens more markedly than the first-order curve, because the rate falls with the square of the concentration.
For a first-order reaction, the time taken for the concentration to fall to half its value — the half-life — is the same whatever the starting concentration. This gives a simple test of a concentration–time graph: read the time for the concentration to halve from several starting points; if the times are equal, the reaction is first order. Radioactive decay is the best-known first-order process. For zero order the half-life gets shorter as the concentration falls, and for second order it gets longer.
Exam focus · what the published papers show
Graph identification is generally secure: more than 78% of candidates correctly identified a first order reaction graph, and on another item candidates were able to distinguish between pairs of graphs indicating the same order of reaction. Explaining a straight-line graph is less secure: ‘linear’ was a common answer – which could also be zero order. It was important to talk about ‘proportionality’ (preferable) or ‘linear increase’. A linear graph through the origin is first order; a linear horizontal graph is zero order.
Once the orders are found, marks are lost in writing them down: the orders were usually successfully deduced but many omitted to give the overall rate expression; some left out the rate constant, k, in the rate expression, others wrote the equilibrium constant expression instead; and a reactant found to be first order was often omitted from the rate expression. On zero order, a report asks for the mechanism, not a restatement: many stated that a reactant concentration having no effect indicated that the reaction that was zero order in that species, rather than describing the underlying mechanistic reason — that the reactant is not involved in the rate-determining step.
11The rate constant and its units 2.2.11 HL only
The syllabus statement: The rate constant, k, is temperature dependent and its units are determined from the overall order of the reaction. The skill: solve problems involving the rate equation, including the units of k.
The rate constant is constant only at a fixed temperature. It does not depend on concentration — that dependence is carried by the concentration terms — but it increases, usually steeply, with temperature, and a catalyst increases it by providing a pathway of lower Ea. This is the difference between k and the equilibrium constant K of Reactivity 2.3, which a catalyst does not change.
The units of k are whatever makes the rate equation dimensionally correct, with rate in mol dm−3 s−1. Rearranging, k = rate ÷ (concentration terms), so for overall order n the units are (mol dm−3 s−1) ÷ (mol dm−3)n = mol1−n dm3(n−1) s−1.
| Overall order | Example rate equation | 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] | mol−1 dm3 s−1 |
| 3 | rate = k[A]2[B] | mol−2 dm6 s−1 |
| 4 | rate = k[A][B][C]2 | mol−3 dm9 s−1 |
Question. For rate = k[X][Y]2, k = 3.6 × 10−2 mol−2 dm6 s−1 at 298 K. Calculate the rate when [X] = 0.050 mol dm−3 and [Y] = 0.20 mol dm−3, and state the factor by which the rate changes if both concentrations are halved.
| Substitution | rate = 3.6 × 10−2 × 0.050 × 0.202 |
| Rate | = 7.2 × 10−5 mol dm−3 s−1 |
| Halving both | (½)1 × (½)2 = ⅛: the rate falls to one-eighth, because the overall order is 3. |
| Units check | mol−2 dm6 s−1 × mol dm−3 × mol2 dm−6 = mol dm−3 s−1 ✔ |
On one item over 34% confused the units with those of a first order reaction; on another, half the candidates were able to deduce the units of the rate constant correctly; and one report notes that the units were frequently wrong or omitted. Work the units out from the overall order every time. A related confusion: on an item asking whether a catalyst changes the rate constant, 35% gave C suggesting that a catalyst has no effect on the rate constant. Perhaps candidates had muddled rate constant, k, with the equilibrium constant, K.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
Bromate and bromide ions react in acidic solution: BrO3−(aq) + 5Br−(aq) + 6H+(aq) → 3Br2(l) + 3H2O(l).
| Experiment | [BrO3−] / mol dm−3 | [Br−] / mol dm−3 | [H+] / mol dm−3 | Initial rate / mol dm−3 s−1 |
|---|---|---|---|---|
| 1 | 0.10 | 0.10 | 0.10 | 8.0 × 10−4 |
| 2 | 0.20 | 0.10 | 0.10 | 1.6 × 10−3 |
| 3 | 0.20 | 0.20 | 0.10 | 3.2 × 10−3 |
| 4 | 0.10 | 0.10 | 0.20 | 3.2 × 10−3 |
(a) Determine the rate expression for the reaction. [2]
(b) Determine the value and unit of the rate constant using the rate expression in (a). [2]
Calcium carbonate reacts with hydrochloric acid: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g). The results of a series of experiments in which the concentration of HCl was varied are shown.
(i) Suggest why point D is so far out of line assuming human error is not the cause. [1]
(ii) Draw the best fit line for the reaction excluding point D. [1]
(iii) Suggest the relationship that points A, B and C show between the concentration of the acid and the rate of reaction. [1]
(iv) Deduce the rate expression for the reaction. [1]
(v) Calculate the rate constant of the reaction, stating its units. [2]
The thermal decomposition of dinitrogen monoxide, 2N2O(g) → 2N2(g) + O2(g), is followed by measuring the total pressure with time at constant temperature (arbitrary units).
This decomposition obeys the rate expression −d[N2O]/dt = k[N2O].
(i) Deduce how the rate of reaction at t = 2 would compare to the initial rate. [1]
(ii) It has been suggested that the reaction occurs as a two-step process: Step 1: N2O(g) → N2(g) + O(g); Step 2: N2O(g) + O(g) → N2(g) + O2(g). Explain how this could support the observed rate expression. [2]
Solutions and mark-scheme guidance · Set R2F
R2F.1 C
Doubling [Cl2] doubles the rate (first order); doubling [NO] multiplies the rate by 4 (second order).
R2F.2 A
Rate is proportional to 1/time. Halving [X] doubles the time (rate halves): first order. Halving [Y] multiplies the time by 4 (rate ÷ 4): second order.
R2F.3 D
Doubling [C4H9Br] doubles the rate; doubling [OH−] has no effect. Rate = k[C4H9Br]: the unimolecular (SN1) mechanism of a tertiary halogenoalkane. The report notes that many recognised zero order in OH− but did not link it to SN1.
R2F.4 B
Zero order in X: no effect. Second order in Y: 32 = 9.
R2F.5 D
Third order overall: k = rate ÷ [X]2[Y] = mol dm−3 s−1 ÷ mol3 dm−9 = mol−2 dm6 s−1. Half the candidates answered correctly.
R2F.6 D
Identify the order shown by each graph separately and look for the pair that agree. Zero order: straight, falling concentration–time line; horizontal rate–concentration line. First order: concentration–time curve with constant half-life; straight rate line through the origin. Second order: steeper curve that flattens more; upward-curving rate line. Only pair D shows the same order in both graphs.
R2F.7 [4]
(a) 1 → 2: [BrO3−] × 2, rate × 2: first order; 2 → 3: [Br−] × 2, rate × 2: first order; 1 → 4: [H+] × 2, rate × 4: second order ✔. Rate = k[BrO3−][Br−][H+]2 ✔ (square brackets required).
(b) k = 8.0 × 10−4 ÷ (0.10 × 0.10 × 0.102) = 8.0 ✔ mol−3 dm9 s−1 ✔ (fourth order overall). The orders are not the coefficients 1, 5 and 6.
R2F.8 [6]
(i) Any one: the reaction is so fast at high concentration that it is difficult to measure accurately; the calcium carbonate has been used up / is limiting (HCl is in excess); so many CO2 bubbles inhibit contact of acid with the solid; insufficient change in conductivity/pH ✔. The report notes that many suggested that HCl is no longer limiting.
(ii) A straight line through the origin, as close to A, B and C as possible ✔. (iii) (Directly) proportional ✔ — “rate increases as concentration increases” is not accepted. (iv) rate = k[HCl] (or k[H+]) ✔; the report notes that some included [CaCO3]. (v) k = gradient ≈ 0.02 ✔ s−1 ✔, remembering the × 10−3 on the rate axis.
R2F.9 [3]
(i) Half the initial rate ✔ — at t = 2 half the N2O has gone (the pressure has risen by half its total increase), and the rate is proportional to [N2O].
(ii) Step 1 is slower than step 2 / is the RDS ✔; step 1 is unimolecular, involving only one N2O, so the rate is first order in N2O ✔ (if step 2 were the RDS, the rate would be second order in N2O). The report notes that few mentioned unimolecularity and most gained only one mark.
12The Arrhenius equation 2.2.12 HL only
The syllabus statement: The Arrhenius equation uses the temperature dependence of the rate constant to determine the activation energy. The skills: describe the qualitative relationship between temperature and the rate constant and analyse graphical representations of the Arrhenius equation, including its linear form.
Section 2.2.4 explained qualitatively why a rise in temperature increases the rate: a larger fraction of particles have energy ≥ Ea. The Arrhenius equation makes this quantitative. The fraction of collisions with enough energy is proportional to e−Ea/RT, and the rate constant is this fraction multiplied by a factor that accounts for how often suitably oriented collisions occur:
k: rate constant (units depend on the order). A: the Arrhenius factor, or pre-exponential factor, with the same units as k. Ea: activation energy, in J mol−1 in the equation. R: gas constant, 8.31 J K−1 mol−1. T: absolute temperature in K.
Both forms of the equation are given in the data booklet. Because R is in joules, Ea comes out in J mol−1 and is divided by 1000 for kJ mol−1.
The qualitative relationship
The rate constant increases exponentially, not linearly, with temperature: a graph of k against T curves upwards ever more steeply. A reaction with a larger activation energy is more sensitive to temperature, because its Ea lies further out in the tail of the Maxwell–Boltzmann distribution, where the fractional change with temperature is greatest. A catalyst lowers Ea and so increases k at every temperature. As T becomes very large, e−Ea/RT approaches 1 and k approaches A, the value it would have if every suitably oriented collision succeeded.
The linear form
Taking natural logarithms of both sides gives the linear form:
This has the form y = mx + c. A plot of ln k (on the y-axis) against 1/T (on the x-axis, in K−1) is a straight line with gradient −Ea/R and intercept ln A. The gradient is negative: as 1/T increases (lower temperature), ln k decreases. A steeper line means a larger activation energy.
Question. A plot of ln k against 1/T for a reaction is a straight line passing through the points (3.00 × 10−3 K−1, −6.10) and (3.40 × 10−3 K−1, −8.50). Determine the activation energy in kJ mol−1.
| Gradient | (−8.50 − (−6.10)) ÷ (3.40 × 10−3 − 3.00 × 10−3) = −2.40 ÷ 4.0 × 10−4 = −6.00 × 103 K |
| Relationship | gradient = −Ea/R, so Ea = −gradient × R |
| Calculation | Ea = 6.00 × 103 × 8.31 = 4.99 × 104 J mol−1 = 49.9 kJ mol−1 |
| Check | Positive, as activation energies always are; a typical magnitude for a reaction that is measurable at room temperature. The gradient has units of K because ln k has none. |
Two temperatures without a graph
If rate constants are known at only two temperatures, writing the linear form for each and subtracting eliminates ln A:
This is not a separate equation to memorize; it follows from the data-booklet form in one line. The ratio of rates at two temperatures, with everything else equal, can be used in place of k2/k1.
Question. The rate of a reaction increases by a factor of 3.0 when the temperature is raised from 300 K to 320 K, with all concentrations unchanged. Calculate the activation energy.
| Relationship | ln(k2/k1) = (Ea/R)(1/T1 − 1/T2), with k2/k1 = rate ratio = 3.0 |
| Temperatures | 1/300 − 1/320 = 3.3333 × 10−3 − 3.1250 × 10−3 = 2.083 × 10−4 K−1 |
| Calculation | Ea = ln 3.0 × 8.31 ÷ 2.083 × 10−4 = 1.0986 × 8.31 ÷ 2.083 × 10−4 = 4.38 × 104 J mol−1 = 43.8 kJ mol−1 |
| Check | The familiar rule that a 10 K rise doubles the rate holds only for activation energies of roughly 50 kJ mol−1 near room temperature; here a 20 K rise triples the rate, consistent with a slightly smaller Ea. |
13The Arrhenius factor, A 2.2.13 HL only
The syllabus statement: The Arrhenius factor, A, takes into account the frequency of collisions with proper orientations. The skill: determine the activation energy and the Arrhenius factor from experimental data.
The Arrhenius equation separates the two requirements of collision theory. The exponential term is the fraction of collisions with enough energy. The factor A contains everything else: how frequently the particles collide and what fraction of collisions have a suitable orientation. It is sometimes written as A = pZ, with Z the collision frequency and p the steric (orientation) factor, which is small for large or complex molecules. A is treated as independent of temperature over the ranges used in practice (the small rise in collision frequency with temperature is negligible beside the exponential term), and it has the same units as k.
From an Arrhenius plot, the intercept on the ln k axis is ln A, so A = eintercept. Because the measured values of 1/T are all near 0.003 K−1, the line must be extrapolated a long way to reach 1/T = 0, so it is usually more accurate to calculate the intercept from the gradient and one point: ln A = ln k + Ea/RT.
Question. Using the line in Worked example R2.14, determine ln A and A. The rate constant has units of s−1.
| Relationship | ln A = ln k − gradient × (1/T) |
| Substitution | Using (3.00 × 10−3, −6.10): ln A = −6.10 − (−6.00 × 103 × 3.00 × 10−3) = −6.10 + 18.0 = 11.9 |
| Answer | A = e11.9 = 1.5 × 105 s−1 |
| Check | The second point gives −8.50 + 20.4 = 11.9 ✔. Same units as k for a first-order reaction. |
Exam focus · what the published papers show
On a graph item worth three marks, average performance was 43%. Students tended to make the question more difficult than was necessary. Instead of calculating the slope of the line to calculate the activation energy, candidates used the activation energy equation with ln A found in the data booklet. The mark scheme awards the gradient, then Ea = −gradient × R, and accepts a range of answers because points are read from a hand-drawn line. Another report lists the difficulty directly: calculation involving the use of natural log ln in the determination of the activation energy of a reaction from graphed empirical data, and a third that candidates made rather obvious mistakes when calculating the activation energy such as failing to convert the temperature to Kelvin.
Describing the method is weaker still: the majority of candidates struggled in describing how the activation energy for a reaction could be determined, and those who recognized the equation could not identify the measurements needed, what to graph, or how to use the formula for only 2 different temperatures without graphing. The method: measure the rate (or k) at several temperatures with all concentrations fixed; plot ln k against 1/T; Ea = −gradient × R.
A report records that while most candidates knew that temperature influences a rate constant, more candidates thought that the rate constant increased proportionally with temperature rather than exponentially. Doubling the temperature in °C does not double k, and doubling the temperature in K multiplies k by far more than two for any real activation energy.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
The rate constant for a reaction doubles when the temperature is increased from 25.0 °C to 35.0 °C. Calculate the activation energy, Ea, in kJ mol−1, for the reaction. [2]
A series of experiments were carried out at different temperatures and the rate constant for the reaction of propanone with iodine was calculated at each. The processed data are shown.
Determine the activation energy for this reaction, stating the units. [3]
Calcium carbonate reacts with hydrochloric acid. Describe how the activation energy of this reaction could be determined. [3]
Solutions and mark-scheme guidance · Set R2G
R2G.1 A
The pre-exponential factor A accounts for collision frequency and the fraction with the correct orientation.
R2G.2 B
ln k = −(Ea/R)(1/T) + ln A: the gradient is −Ea/R. The y-intercept is ln A, not A; the graph does not pass through the origin.
R2G.3 C
k depends on temperature through e−Ea/RT. Ea, A (over normal ranges) and R are constants. The report notes that a significant proportion believe Ea varies with temperature.
R2G.4 A
In y = mx + c form, the intercept c is ln A.
R2G.5 A
Gradient = −2 ÷ 0.06 (read from the plotted line); Ea = −gradient × R = (2 × 8.31) ÷ 0.06.
R2G.6 [2]
ln(k2/k1) = (Ea/R)(1/T1 − 1/T2) with T2 = 308 K AND T1 = 298 K ✔. ln 2 = (Ea/8.31)(1/298 − 1/308) gives Ea = 52.9 kJ mol−1 ✔.
R2G.7 [3]
Two construction points on the line and the gradient: e.g. (−1.0 − (−3.0)) ÷ (0.0032 − 0.0035) = −6700 K ✔ (range 6400–7000). Ea = −gradient × R = 6700 × 8.31 = 56 000 J mol−1 = 56 ✔ (range 53–59) kJ mol−1 ✔. The average was 43 %: candidates used the equation with ln A instead of the slope.
R2G.8 [3]
Alternative 1: carry out the reaction at several temperatures ✔; plot ln k (or ln rate) against 1/T ✔; Ea = −gradient × R ✔. Alternative 2: carry out the reaction at two temperatures ✔; determine two rate constants (or the ratio of rates) ✔; use ln(k2/k1) = (Ea/R)(1/T1 − 1/T2) ✔. All other conditions (concentrations, mass and size of chips) are kept the same.
Review · Reactivity 2.2
14Misconceptions, the examiner’s view, and the question types
- “Rate is the time taken for the reaction.” Rate is a change in concentration (or another property) per unit time; time taken is inversely related to average rate.
- “More collisions” explains a faster rate. Rate depends on collision frequency — collisions per unit time — and on the proportion with E ≥ Ea.
- “Smaller surface area because the particles are smaller.” For a fixed mass, smaller particles mean a larger surface area.
- “Raising the temperature lowers the activation energy.” Ea is a property of the pathway; only a catalyst changes it.
- “A catalyst gives the particles more energy.” It provides an alternative pathway with lower Ea; the energy distribution depends only on temperature.
- “At a higher temperature the Maxwell–Boltzmann peak is higher.” The peak is lower and further right; the area is unchanged.
- “A catalyst changes ΔH.” Reactant and product levels are unchanged.
- HL “Orders come from the coefficients.” Orders come only from experiment; they reflect the mechanism.
- HL “Intermediates can appear in the rate equation.” Replace an intermediate by the species that formed it.
- HL “k is proportional to T.” It increases exponentially; ln k is linear in 1/T.
- HL “Molecularity and order are the same.” Molecularity describes one elementary step; order is measured for the overall rate equation.
Evidence base: the IB Diploma chemistry subject reports quoted in this chapter.
Answered well: choosing a suitable method to follow a reaction (almost all candidates suggested a suitable experimental method); identifying the forward activation energy on a profile (88 % on one item); recognizing that a catalyst provides an alternative pathway in a direct multiple-choice item (76 %) and adding a catalysed profile to a given one; applying the correct combination of conditions for the fastest rate; HL deducing orders from initial-rate data (88 % and 72 % on two items), identifying first-order graphs (over 78 %), linking a mechanism to a rate equation or a profile (73 % on two items).
Found difficult: (1) defining rate as a change in concentration per unit time; (2) using a tangent at t = 0 for initial rate rather than an average; (3) collision frequency rather than number of collisions; (4) confusing surface area with particle size; (5) Maxwell–Boltzmann sketches — not starting at the origin, unlabelled, Ea missing, confused with energy profiles; (6) the belief that temperature changes Ea, and that a catalyst gives particles more energy (over half on one item, across the whole ability range); (7) leaving out “alternative pathway” (30 % scored on one item); HL (8) writing the full rate equation with k and every reactant; (9) units of k (about half correct on one item); (10) intermediates left in rate expressions; (11) the exponential dependence of k on T; (12) finding Ea from the gradient of an Arrhenius plot (43 % average) and describing how to measure it.
What successful answers did: named the property measured with time; drew and used a tangent; wrote “frequency of collisions” and “proportion of particles with E ≥ Ea” as separate points; drew Maxwell–Boltzmann curves from the origin with labelled axes and Ea marked; distinguished a catalyst (move the Ea line) from temperature (move the curve); HL compared pairs of experiments one variable at a time; derived units of k from the overall order; and used Ea = −gradient × R with T in kelvin.
Nine question types cover the sub-topic.
| If the question asks… | …then |
|---|---|
| Determine a rate from a graph | Tangent at the stated time (t = 0 for initial rate); gradient from two distant points on the tangent; units. |
| Suggest how to measure the rate | Name the property (gas volume, mass, absorbance, pH) and say it is measured at intervals of time. |
| Explain the effect of a condition | Frequency of collisions and/or proportion with E ≥ Ea; link to rate. Temperature needs both. |
| Sketch a new curve on a product–time graph | Initial gradient from the rate; final value from the limiting reactant. |
| Sketch Maxwell–Boltzmann curves | From origin; labelled axes; higher T peak lower and to the right; long tail; Ea marked; catalyst = new Ea line. |
| Energy profile with a catalyst | Lower maximum, same reactant and product levels; Ea arrows from reactants. |
| HL Orders, rate equation, k | Pairs of experiments; rate = k[ ]m[ ]n; substitute one experiment; units from overall order. |
| HL Mechanism consistent with data? | Steps add up to the equation; RDS and earlier steps give the rate equation; no intermediates. |
| HL Arrhenius | ln k against 1/T; Ea = −gradient × R; T in K; J → kJ; ln A from the intercept or a point. |
15Quick check
- Define rate of reaction, and give its usual units.
- Why is the tangent at t = 0 used to find the initial rate?
- Explain, using collision theory, why powdered zinc reacts faster with acid than a single piece of the same mass.
- State two ways in which the Maxwell–Boltzmann curve at a higher temperature differs from the curve at a lower temperature, and one way in which it is the same.
- How does a catalyst appear on a Maxwell–Boltzmann diagram?
- Why does a catalyst not change ΔH?
- HL Distinguish between an intermediate and a transition state.
- HL Doubling [A] multiplies the rate by 4 and tripling [B] has no effect. Write the rate equation and give the units of k.
- HL What is the gradient of a plot of ln k against 1/T?
- HL What does the Arrhenius factor, A, represent?
- The change in concentration of a reactant or product per unit time; mol dm−3 s−1.
- At t = 0 the concentrations are those mixed and known, and the instantaneous rate is the gradient of the curve at that point; a chord would give only an average.
- The powder has a larger surface area, so more zinc atoms are exposed and collisions with H+ are more frequent.
- Differences: peak lower, peak further right (and a larger area beyond Ea). Same: area under the curve (number of particles), and both start at the origin.
- As an additional Ea line to the left of the uncatalysed one, on the same curve.
- The reactants and products, and so their enthalpies, are unchanged; only the pathway between them changes.
- An intermediate is a species formed in one step and used in a later one, at a trough on the profile; a transition state is the highest-energy arrangement in one step, at a maximum, and cannot be isolated.
- rate = k[A]2; second order overall; mol−1 dm3 s−1.
- −Ea/R (negative).
- The frequency of collisions with the correct orientation; it has the same units as k.
16Summary and knowledge organiser
Essential knowledge
- Rate = change in concentration per unit time (mol dm−3 s−1); instantaneous rate from a tangent; initial rate from the tangent at t = 0.
- Reactions are followed by gas volume, mass loss, colour, pH, conductivity, precipitate or sampling.
- Successful collisions need E ≥ Ea and correct orientation; T (K) ∝ average kinetic energy.
- Concentration, pressure and surface area raise collision frequency; temperature mainly raises the proportion with E ≥ Ea; a catalyst lowers Ea via an alternative pathway.
- Maxwell–Boltzmann: from the origin, peak, long tail; higher T → lower, broader, shifted right, same area.
- Catalysts: unchanged at the end; ΔH unchanged; forward and reverse Ea lowered equally; enzymes are biological catalysts.
- HL Mechanisms: elementary steps sum to the equation; RDS is slowest; intermediates at troughs, transition states at maxima; molecularity 1, 2 or 3.
- HL rate = k[A]m[B]n from experiment only; zero, first, second order graphs; units of k from overall order.
- HL k = Ae−Ea/RT; ln k = −(Ea/R)(1/T) + ln A; gradient −Ea/R; A = frequency of correctly oriented collisions.
Examination checklist
- Write “frequency of collisions” or “collisions per unit time”, never just “more collisions”.
- Maxwell–Boltzmann: start at the origin, label both axes, mark Ea, keep the areas equal.
- Catalyst answers: alternative pathway + lower Ea + greater proportion of successful collisions.
- Draw a tangent, and read its gradient from points on the tangent.
- HL Include k and every reactant (with its order) in the rate equation; derive units of k.
- HL Ea = −gradient × R; T in kelvin; J ÷ 1000 for kJ.
Knowledge organiser
| Outcome | Key facts and relationships | Must-remember distinctions and common errors |
|---|---|---|
| Rate 2.2.1 | Δ[ ]/Δt; mol dm−3 s−1; tangent for instantaneous rate. | Not “time taken”; tangent, not chord; measure a property with time. |
| Collisions 2.2.2 | E ≥ Ea and correct orientation; T ∝ average Ek. | Frequency, not number; T in kelvin. |
| Factors 2.2.3 | c, p, surface area → frequency; T → proportion ≥ Ea (and frequency); catalyst → lower Ea. | Smaller particles = larger surface area; final amount set by limiting reactant. |
| Ea and M–B 2.2.4 | Curve from origin, long tail, area constant; higher T: lower, right-shifted peak. | Ea unchanged by T; not an energy profile. |
| Catalysts 2.2.5 | Alternative pathway, lower Ea; regenerated; enzymes. | Do not add energy; ΔH and K unchanged; new Ea line, not new curve. |
| Mechanisms 2.2.6–2.2.8 HL | Steps sum to equation; RDS slowest; molecularity of a step. | Intermediate (trough) ≠ transition state (peak); no intermediates in rate equations. |
| Rate equations 2.2.9–2.2.11 HL | rate = k[A]m[B]n; order = exponent; units mol1−n dm3(n−1) s−1. | Orders ≠ coefficients; zero order horizontal rate line; first order constant half-life. |
| Arrhenius 2.2.12–2.2.13 HL | ln k = −Ea/R · 1/T + ln A. | Exponential, not linear; Ea in J; A = orientation × frequency. |