R2.2 How fast? The rate of chemical change

Chemical kinetics is the study of how fast reactions happen and why. Some reactions are over in a fraction of a second, such as the explosion of a hydrogen–oxygen mixture, while others, such as the rusting of iron, take years. Controlling rates matters in industry, where a slow process makes no profit, in food storage, where the aim is to slow down unwanted reactions, and in the body, where enzymes make reactions fast enough to sustain life.

Rate of reaction

As a reaction proceeds, the concentrations of the reactants fall and the concentrations of the products rise. The rate of reaction measures how quickly these changes happen.

The rate of reaction can be written for any one species in the equation. Over a time interval $\Delta t$:

The concentration of a reactant decreases, so $\Delta[\text{reactant}]$ is negative; the minus sign makes the rate a positive number. A rate is always positive, whichever species is followed.

Because the coefficients in an equation are not all equal, the rate depends on which species is chosen. In the decomposition of hydrogen peroxide,

two moles of ${{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$ are used up for every mole of ${{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$ formed. The rate of formation of oxygen is therefore half the rate at which hydrogen peroxide is used up. For this reason a rate is always stated for a named species: "the rate of formation of oxygen" or "the rate of consumption of hydrogen peroxide".

When a quantity other than concentration is measured, the rate takes the units of that quantity per unit time: $\text{cm}^3\ \text{s}^{-1}$ for a volume of gas, $\text{g s}^{-1}$ for a loss of mass, or $\text{s}^{-1}$ for an absorbance. The time unit may be seconds, minutes or hours, depending on how fast the reaction is.

How the rate changes during a reaction

The graph of concentration against time for a typical reaction shows that the rate is not constant. Both curves are steepest at the start and become flatter as time passes.

The rate falls because the reactant is being used up. As its concentration decreases there are fewer reactant particles in each unit of volume, so they collide less often and fewer react each second. The curve becomes horizontal when the limiting reactant has been used up and the reaction stops. The same pattern is seen whatever property is measured: a graph of gas volume or mass lost against time rises steeply at first and then levels off.

Average rate and instantaneous rate

The average rate over a time interval is the total change in concentration divided by the time taken. On a graph it is the gradient of the straight line (the chord) joining the two points at the ends of the interval.

The rate at one particular moment is the instantaneous rate. It equals the gradient of the tangent to the curve at that time: a straight line that touches the curve at that point, with the same slope as the curve there. The rate at the very start of the reaction, the initial rate, is the gradient of the tangent at $t = 0$.

The concentration of hydrogen peroxide during its catalysed decomposition, ${{2\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{aq})\xrightarrow{}2\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{l}) {}+{} \mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$, was measured at intervals.

Finding a rate from a tangent

A student adds $0.39\ \text{g}$ of powdered calcium carbonate to $50.0\ \text{cm}^3$ of hydrochloric acid, which is in excess, and collects the carbon dioxide in a gas syringe:

The graph of gas volume against time rises steeply and levels off at $96\ \text{cm}^3$, when all the calcium carbonate has reacted. Two tangents are drawn on it, at the start and at 60 s.

To find the gradient of a tangent:

1. Draw a straight line that touches the curve at the chosen time without crossing it, so that the curve lies on the same side of the line just before and just after that point.

2. Extend the line as far as the graph allows. Reading two points far apart reduces the effect of small reading errors.

3. Read the coordinates of two points on the line (not on the curve) and calculate $\text{gradient} = \Delta y / \Delta x$, with units.

A rate measured in $\text{cm}^3\ \text{s}^{-1}$ can be converted into a change in concentration per second by using the ideal gas equation and the stoichiometry of the reaction.

Following the rate of a reaction in the laboratory

Concentrations are rarely measured directly while a reaction is running. Instead, a property that changes as the reaction proceeds is measured at regular intervals, and the reaction is followed through that property. The method must suit the reaction: a gas syringe is no use if no gas is produced, and a colorimeter is no use if nothing coloured is formed or used up. Data loggers with gas-pressure, pH, conductivity or light sensors can record readings many times a second, which allows fast reactions to be followed and gives smooth curves.

Collecting a gas

When a gas is produced, its volume can be collected in a gas syringe and read at regular intervals. Displacing water from an upturned measuring cylinder or burette is an alternative for gases that are not very soluble, such as hydrogen and oxygen; carbon dioxide dissolves slightly in water, so a syringe gives more reliable volumes.

The reactants are usually kept apart until the start, for example by tipping a small tube of solid into the acid or by adding the solid and replacing the bung quickly, so that no gas escapes before the clock starts. Typical reactions followed this way are magnesium or calcium carbonate with hydrochloric acid, and the decomposition of hydrogen peroxide with a catalyst.

Measuring the loss of mass

When a gas escapes from an open flask, the mass of the flask and its contents falls. The loss of mass at each time equals the mass of gas formed. The method works well for a dense gas such as carbon dioxide ($M = 44.01\ \text{g mol}^{-1}$) but poorly for hydrogen ($M = 2.02\ \text{g mol}^{-1}$). The $96\ \text{cm}^3$ of carbon dioxide from $0.39\ \text{g}$ of calcium carbonate has a mass of about $0.17\ \text{g}$; the same volume of hydrogen has a mass of less than $0.01\ \text{g}$, too little to measure on a balance that reads to $0.01\ \text{g}$.

Measuring a colour change: colorimetry

If a reactant or product is coloured, the intensity of colour shows its concentration. A colorimeter shines light of one colour through the solution and measures how much is absorbed. The filter is chosen so that the light is the colour the solution absorbs most strongly, which is the complementary colour on the colour wheel in the data booklet: an orange solution absorbs blue light, so a blue filter is used.

The absorbance is proportional to the concentration of the coloured species in dilute solution, so a graph of absorbance against time has the same shape as the concentration–time graph. Absorbances can be converted into concentrations with a calibration curve made from standard solutions of known concentration. Examples of reactions that can be followed by colorimetry are:

• bromine with methanoic acid, in which the orange colour of bromine fades: ${{\mathrm{Br}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{aq}) {}+{} \mathrm{HCOOH}\mskip2mu (\mathrm{aq})\xrightarrow{}2\,\mathrm{Br}{\vphantom{X}}^{-}\mskip2mu (\mathrm{aq}) {}+{} 2\,\mathrm{H}{\vphantom{X}}^{+}\mskip2mu (\mathrm{aq}) {}+{} \mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$

• iron with copper(II) sulfate solution, in which the blue colour of ${{\mathrm{Cu}{\vphantom{X}}^{2+}\mskip2mu (\mathrm{aq})}}$ fades: ${{\mathrm{Fe}\mskip2mu (\mathrm{s}) {}+{} \mathrm{Cu}{\vphantom{X}}^{2+}\mskip2mu (\mathrm{aq})\xrightarrow{}\mathrm{Fe}{\vphantom{X}}^{2+}\mskip2mu (\mathrm{aq}) {}+{} \mathrm{Cu}\mskip2mu (\mathrm{s})}}$

• propanone with iodine in acid solution, in which the brown colour of iodine fades.

Measuring pH, conductivity or pressure

A reaction that uses up or produces hydrogen ions can be followed with a pH probe. A reaction that changes the number or type of ions in solution can be followed by its electrical conductivity. For example, the hydrolysis of 2-bromo-2-methylpropane produces ions from a neutral molecule, so both the acidity and the conductivity of the solution increase as it proceeds:

In a gas-phase reaction at constant volume and temperature, the pressure is proportional to the total amount of gas, so a change in pressure follows any reaction that changes the number of gas molecules, such as ${{2\,\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{5}}\mskip2mu (\mathrm{g})\xrightarrow{}4\,\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g}) {}+{} \mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$.

Sampling and titration

Samples can be withdrawn from a reaction mixture at intervals and analysed by titration. Each sample is first quenched: the reaction in it is stopped, for example by cooling it rapidly in ice, by diluting it greatly or by removing a catalyst. In the reaction of propanone with iodine, which is catalysed by acid, adding each sample to sodium hydrogencarbonate solution neutralizes the acid catalyst; the iodine left is then titrated with sodium thiosulfate solution. Titration is slower than the other methods but gives concentrations directly.

Timing a fixed amount of reaction

Sometimes it is easier to measure how long it takes for a fixed amount of reaction to happen than to follow the whole reaction. In the reaction of sodium thiosulfate with hydrochloric acid, a pale yellow precipitate of sulfur forms:

The flask stands on a piece of paper marked with a cross, and the time is measured from mixing until the cross can no longer be seen from above. Every run stops at about the same amount of sulfur, so a faster reaction gives a shorter time. The rate is therefore inversely proportional to the time, and $1/t$ (in $\text{s}^{-1}$) is used as a measure of the rate.

Clock reactions, in which a sudden colour change appears after a fixed amount of reaction, work on the same principle. The iodine clock, in which a blue-black colour appears suddenly, is a common way of measuring initial rates.

The two kinds of experiment use time differently. When the volume of gas or the mass is read at fixed times, time is the independent variable and the quantity measured is the dependent variable. When the time for a fixed amount of reaction is measured, time is the dependent variable and the factor being investigated is the independent variable.

Collision theory

The kinetic molecular theory describes matter as particles in constant random motion. Collision theory applies this model to reactions: before two particles can react they must come into contact, and they must do so in a way that allows bonds to break and new bonds to form.

Kinetic energy and temperature

The particles in a gas or a solution move in all directions with a wide range of speeds, colliding constantly with each other and with the walls of the container. Their average kinetic energy depends only on the temperature.

Doubling the absolute temperature doubles the average kinetic energy. Heating a gas from $300\ \text{K}$ to $600\ \text{K}$ ($27\ ^\circ\text{C}$ to $327\ ^\circ\text{C}$) doubles it, but heating from $20\ ^\circ\text{C}$ to $40\ ^\circ\text{C}$ (from $293\ \text{K}$ to $313\ \text{K}$) raises it by only about 7 %. The proportionality holds only on the Kelvin scale, because only there does zero mean zero kinetic energy.

The conditions for a successful collision

In a gas at room temperature and pressure each molecule collides with others billions of times every second. If every collision led to reaction, all gas reactions would be over almost instantly. In fact only a tiny fraction of collisions are successful, meaning that they lead to reaction. Two conditions must be met at the same time:

1. Sufficient energy. The colliding particles must have a combined kinetic energy at least equal to the activation energy, $E_\text{a}$. This energy is needed to stretch and break bonds in the reactants so that new bonds can form. Particles that collide with less energy than $E_\text{a}$ bounce apart unchanged.

2. Proper orientation. The particles must collide with a suitable geometry, so that the atoms that need to bond come into contact.

The importance of orientation is clear in the reaction of nitrogen monoxide with ozone, which takes place in a single collision. An oxygen atom moves from ozone to the nitrogen atom of ${{\mathrm{NO}}}$, so the reaction can happen only if the nitrogen atom meets an end oxygen atom of the ozone molecule.

Orientation, or collision geometry, matters more for large and complex molecules, where the reacting group is only a small part of the molecule. Reactions between single atoms or small ions in solution depend much less on it, because almost any approach brings the reacting particles together correctly. The substitution of a halogenoalkane by a hydroxide ion is a well-known case: the hydroxide ion must approach the carbon atom from the side opposite the halogen atom.

Collision theory leads to a simple rule. The rate of a reaction depends on two things:

• the collision frequency, the number of collisions per unit time in a given volume;

• the fraction of collisions that are successful, which depends mostly on the fraction with $E \geq E_\text{a}$ and also on orientation.

Every factor that changes the rate does so by changing one or both of these. Explaining the effect of a change of conditions always comes back to them.

Activation energy and energy profiles

As two reacting particles approach each other with enough energy, their bonds stretch and begin to break while new bonds begin to form. The arrangement of atoms at the point of highest potential energy along this path is the transition state. The energy needed to reach it from the reactants is the activation energy.

An energy profile shows the potential energy of the reacting system as the reaction proceeds. The horizontal axis, the reaction coordinate, represents the progress from reactants to products; it has no scale. The activation energy is the height of the hump measured from the reactant level, and the enthalpy change $\Delta H$ is the difference between the product and reactant levels.

The activation energy is always positive, whether the reaction is exothermic or endothermic. Even a strongly exothermic reaction needs an initial input of energy to break bonds before the energy released by bond formation can be given out. The size of $E_\text{a}$ is not linked to the size or sign of $\Delta H$: some very exothermic reactions are slow, and some endothermic reactions are fast.

The reverse reaction

The reverse reaction passes through the same transition state, so its activation energy is measured from the product level up to the top of the same hump:

For an exothermic reaction $\Delta H$ is negative, so the reverse activation energy is larger than the forward one by $|\Delta H|$. For an endothermic reaction the reverse activation energy is the smaller of the two.

High activation energy and kinetic stability

A mixture of methane and air can be stored indefinitely at room temperature even though the combustion of methane is strongly exothermic ($\Delta H_\text{c}^{\ominus} = -891\ \text{kJ mol}^{-1}$, data booklet). The activation energy is so high that almost no collisions at room temperature have enough energy. A spark or flame provides that energy to a few molecules; the heat released by their reaction then supplies the activation energy for neighbouring molecules, and the reaction spreads. This kinetic stability is a useful property of a fuel: it can be stored and transported safely and burns only when ignited.

The Maxwell–Boltzmann distribution

The particles in a sample do not all have the same kinetic energy. Collisions constantly transfer energy between particles, so at any moment a few are almost at rest, most have moderate energies and a few have very high energies. The Maxwell–Boltzmann distribution shows how the kinetic energies are spread among the particles at one temperature.

Features of the curve

• The axes are labelled number of particles (or fraction or proportion of particles) on the vertical axis and kinetic energy on the horizontal axis.

• The curve starts at the origin: no particles have zero kinetic energy.

• It rises to a peak at the most probable kinetic energy, the energy that the largest number of particles have.

• It is not symmetrical. It falls more slowly than it rises and has a long tail at high energy, so the mean kinetic energy is a little higher than the most probable value.

• The tail approaches the energy axis but never meets it, because there is no upper limit to the energy a particle can have.

• The area under the curve represents the total number of particles in the sample.

Marking $E_\text{a}$ on the energy axis divides the particles into two groups. The area under the curve to the right of $E_\text{a}$ represents the particles with $E \geq E_\text{a}$, the only ones that can react when they collide. For most reactions at room temperature this area is a very small fraction of the total.

The effect of temperature on the distribution

At a higher temperature the particles have a higher average kinetic energy, so the distribution changes shape. The Maxwell–Boltzmann distributions of the same sample at two temperatures, $T_1$ and a higher $T_2$, differ in four ways.

• The peak of the $T_2$ curve is lower and further to the right: the most probable energy is higher, and the energies are spread over a wider range.

• The $T_2$ curve lies above the $T_1$ curve at high energies, for the whole of the tail.

• The number of particles is unchanged, so the area under the two curves is the same.

• The area beyond $E_\text{a}$ is much larger at $T_2$.

Factors that affect the rate of reaction

The rate of a reaction can be controlled by changing the concentration of a reactant in solution, the pressure of a gas, the surface area of a solid, the temperature, or by adding a catalyst. Light also speeds up some reactions: a mixture of hydrogen and chlorine reacts explosively in bright sunlight, because light energy breaks Cl–Cl bonds. Stirring a mixture of a solid and a liquid keeps fresh solution in contact with the surface of the solid.

Concentration

Increasing the concentration of a reactant in solution increases the number of reactant particles in each unit of volume. The particles collide more frequently, so there are more successful collisions per unit time and the rate increases. The fraction of collisions that are successful is unchanged, because the temperature, and therefore the energy distribution, is the same.

Concentration also explains why rates fall as reactions proceed: the reactants are used up, their concentrations fall, and the collision frequency falls with them.

Pressure

For reactions between gases, increasing the pressure at constant temperature pushes the same number of molecules into a smaller volume. This increases their concentration: at constant temperature, doubling the pressure of a gas doubles its concentration. The molecules collide more frequently, and the rate increases. Pressure has almost no effect on reactions in solution, because liquids and solids are almost incompressible.

Surface area of a solid

When one reactant is a solid, only the particles at its surface can collide with particles of the other reactant. Breaking a solid into smaller pieces exposes more particles at the surface, so collisions with the other reactant become more frequent and the rate increases.

The effect is large. Grinding a 1 cm cube into grains of $0.1\ \text{mm}$ multiplies its surface area by 100. Powdered calcium carbonate fizzes vigorously in acid while marble chips of the same mass react steadily for several minutes. Fine powders of combustible materials such as flour, sugar, coal or wood can burn so fast when dispersed in air that they explode. At the Imperial Sugar refinery in Port Wentworth, Georgia, an explosion fuelled by accumulated sugar dust killed 14 workers in 2008; mills and refineries control dust and sources of sparks for this reason.

Temperature

Raising the temperature increases the rate of almost every reaction, often dramatically. For many reactions near room temperature the rate roughly doubles for a rise of $10\ \text{K}$. There are two reasons:

1. The particles move faster, so they collide more frequently.

2. The particles have a higher average kinetic energy, so a greater proportion of collisions have energy greater than or equal to $E_\text{a}$. On the Maxwell–Boltzmann distribution, the area beyond $E_\text{a}$ is much larger at the higher temperature.

The second effect is much the more important. A $10\ \text{K}$ rise near room temperature increases the collision frequency by less than 2 %, yet it can double the rate, because it roughly doubles the number of particles with enough energy to react.

Low temperatures slow reactions down. Food kept in a refrigerator at about $4\ ^\circ\text{C}$ stays fresh for days rather than hours, because the reactions that cause decay, including those of the enzymes in bacteria, are much slower than at room temperature. Freezing slows them further.

Catalysts

A catalyst increases the rate by providing a different route from reactants to products with a lower activation energy. A greater proportion of collisions then have $E \geq E_\text{a}$ at the same temperature.

Summary of the factors

Reading graphs of rate experiments

Graphs of gas volume or mass loss against time show both how fast a reaction goes and how much product it makes. The gradient at any time gives the rate; the final, level value gives the total amount of product, which is fixed by the amount of the limiting reactant.

• Changing the surface area, the temperature or adding a catalyst changes how quickly the curve rises, but not its final value, because the amount of the limiting reactant is unchanged.

• Changing the amount of the limiting reactant changes the final value. In run C the acid is twice as concentrated, so there is twice the amount of acid (the limiting reactant) and twice the volume of gas.

• Increasing the concentration of a reactant that is in excess increases the initial rate but leaves the final value unchanged.

Designing a fair rate investigation

To find the effect of one factor on the rate, only that factor (the independent variable) is changed, and every other variable that could affect the rate is kept constant: these are the control variables. In the reaction of sodium thiosulfate with hydrochloric acid, for example, the concentration of thiosulfate is changed by diluting a stock solution with different volumes of water.

Each run should be repeated at least three times so that anomalous results can be identified and a mean calculated. Sulfur dioxide is released, so the work is done in a well-ventilated room and the flask is emptied into a fume cupboard sink.

Random and systematic errors on rate graphs

Graphs of experimental data show the two kinds of error in different ways. Random errors make the points scatter on either side of the best-fit line; repeating the measurements and averaging reduces their effect. Systematic errors shift every reading in the same direction by a similar amount, so the points can lie close to a line that is in the wrong place. A best-fit line that misses the origin when the theory says it should pass through it is a sign of a systematic error.

In the early readings of gas volume against time, the points scatter slightly about a straight line, showing small random errors. The line has the expected gradient but meets the volume axis at about $3\ \text{cm}^3$: a syringe that started at $3\ \text{cm}^3$ instead of zero, or air displaced when the bung was inserted, adds the same volume to every reading. Repeating the experiment in the same way would not remove this error; checking the zero of the syringe before each run would. A systematic error of this kind does not change the gradient, so a rate found from the gradient is still reliable.

Catalysts and how they work

Because a catalyst is regenerated, a small amount can convert a very large amount of reactants. A catalyst takes part in the reaction, for example by forming bonds with a reactant, but it is released again in a later step. It changes the route by which the reaction happens (the mechanism), not the reactants or products.

Energy profiles with and without a catalyst

The energy profiles of a catalysed and an uncatalysed reaction share the same reactant level and the same product level. Only the height of the hump changes.

Three consequences follow from these profiles:

• $E_\text{a}$ is lower on the catalysed pathway, so the rate is higher.

• $\Delta H$ is unchanged, because the reactants and products are the same. A catalyst cannot make an endothermic reaction exothermic, or give out more heat.

• The activation energy of the reverse reaction is lowered by exactly the same amount as that of the forward reaction. A catalyst therefore speeds up the forward and reverse reactions equally. In a reversible reaction it shortens the time taken to reach equilibrium but does not change the position of equilibrium or the equilibrium constant.

A catalyst on the Maxwell–Boltzmann distribution

A catalyst does not change the temperature, so the distribution of kinetic energies stays exactly the same. What changes is the energy that a collision needs. With the lower activation energy, $E_\text{a}(\text{cat})$, many more particles lie beyond the threshold.

Homogeneous and heterogeneous catalysts

Catalysts are classified by their physical state, or phase, compared with the reactants.

A heterogeneous catalyst works only at its surface, so it is made with as large a surface area as possible: finely divided, porous, or spread as a thin layer over an inert support. Many important catalysts are transition elements or their compounds, such as iron, platinum, nickel, vanadium(V) oxide and manganese(IV) oxide. Their variable oxidation states and their ability to bond to reactant molecules at their surface make them effective catalysts.

A heterogeneous catalyst can be poisoned: a substance that bonds strongly to its surface blocks the sites where the reactants would bind, so the catalyst stops working. Lead compounds poison the metals in catalytic converters, which is one reason why cars fitted with them must use unleaded petrol.

Enzymes

Enzymes are biological catalysts: protein molecules that catalyse the reactions in living cells. They are highly specific, each usually catalysing one reaction, and they work at body temperature and close to neutral pH. Catalase, found in almost all cells that use oxygen, catalyses the decomposition of hydrogen peroxide, a toxic by-product of metabolism:

It is one of the fastest enzymes known: a single catalase molecule can decompose millions of hydrogen peroxide molecules every second. The reaction is easy to follow in the laboratory by collecting the oxygen in a gas syringe, using a small piece of potato or liver as the source of catalase and comparing it with solid manganese(IV) oxide.

Unlike most catalysts, enzymes are sensitive to temperature. Their rate increases with temperature at first, as for any reaction, but above about 40 °C for most human enzymes the protein begins to lose its shape (it is denatured) and the rate falls sharply.

Case study: the catalytic converter

A petrol engine releases carbon monoxide, unburnt hydrocarbons and nitrogen oxides (${{\mathrm{NO}}}$ and ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}$, together ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{x }}}}$) in its exhaust gases. All three are pollutants, and the reactions that would remove them are slow at the temperature of the exhaust. A catalytic converter, fitted in the exhaust pipe, speeds up these reactions so that most of the pollutants are converted before the gases leave the car:

The catalyst is a thin layer of the platinum-group metals platinum, palladium and rhodium, spread over a ceramic honeycomb with thousands of narrow channels. The honeycomb gives a very large surface area for a small mass of these expensive metals and lets the gases flow through freely.

The converter shows several of the ideas about catalysts at work. The catalyst is heterogeneous: solid metal with gaseous reactants, which react on its surface. It is not used up, so one converter works for many years. It needs to be hot, typically above about 250–300 °C, before it works efficiently, so most of the pollution from a modern car is released in the first minutes after a cold start. And it can be poisoned: lead from leaded petrol coats the metal surface and stops it working.

Reaction mechanisms and elementary steps (HL)

A balanced equation shows the reactants and products of a reaction and their mole ratio. It does not show how the reaction happens at the level of particles. The equation ${{2\,\mathrm{NO}\mskip2mu (\mathrm{g}) {}+{} 2\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})\xrightarrow{}\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g}) {}+{} 2\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{g})}}$, for example, suggests four molecules colliding at once, but a collision of four particles with enough energy and the right orientation is so improbable that the reaction would be immeasurably slow. Most reactions instead take place as a series of simpler events, each involving one, two or occasionally three particles.

Each elementary step is written as an ordinary equation, and the steps of a reaction mechanism must add up to the overall stoichiometric equation. Species that appear on both sides when the steps are added cancel out. The reaction between nitrogen dioxide and carbon monoxide at temperatures below about 500 K is a well-studied example:

Adding the two steps gives ${{2\,\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}} {}+{} \mathrm{NO}{\vphantom{X}}_{\smash[t]{3}} {}+{} \mathrm{CO}\xrightarrow{}\mathrm{NO}{\vphantom{X}}_{\smash[t]{3}} {}+{} \mathrm{NO} {}+{} \mathrm{NO}{\vphantom{X}}_{\smash[t]{2}} {}+{} \mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}}}$. One ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}$ and the ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{3}}}}$ appear on both sides and cancel, leaving the overall equation. The ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{3}}}}$ is made in step 1 and used up in step 2.

Intermediates

In the nitrogen dioxide and carbon monoxide mechanism, ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{3}}}}$ is a reaction intermediate. Intermediates are real chemical species with complete bonds. Many are reactive and short-lived, such as radicals, carbocations or unstable molecules, but because each occupies an energy minimum it can in principle be detected, for example by spectroscopy, and some can be isolated. Detecting an intermediate that a mechanism predicts is strong evidence for that mechanism.

A catalyst behaves in the opposite way to an intermediate: it is used up in an early step and regenerated in a later one. It also does not appear in the overall equation, but it is present at the start and at the end.

Transition states

Every elementary step has exactly one transition state, so a two-step mechanism has two. A transition state exists only for the instant in which the atoms pass over the top of the barrier. It cannot be collected or isolated, because it sits at an energy maximum: any movement of its atoms lowers its energy, either back to the reactants of that step or on to its products. Transition states are written in square brackets with the symbol ‡, and partial bonds are drawn as dashed lines, for example $[{{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mathrm{N}}}\cdots{{\mathrm{O}}}\cdots{{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}]^{\ddagger}$ for the first step of that mechanism.

The rate-determining step (HL)

The steps of a mechanism usually take place at very different rates. The reaction as a whole cannot go faster than its slowest step, in the same way that the number of cars leaving a car park each minute is set by the single barrier at the exit, however quickly the cars reach it. Speeding up a fast step makes no difference to the rate; speeding up the slow step increases the rate of the whole reaction.

In the nitrogen dioxide and carbon monoxide mechanism, step 1 (the collision of two ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}$ molecules) is the rate-determining step. Carbon monoxide only takes part in the fast step 2, after the slow step. Changing the concentration of ${{\mathrm{CO}}}$ therefore has no effect on the rate, as experiments confirm: the reaction is zero order with respect to carbon monoxide.

When the rate-determining step is not the first step

The slow step can come later in the mechanism. The steps before it are then fast and often reversible, so they reach equilibrium and keep up a small, steady concentration of an intermediate (a pre-equilibrium), which the slow step consumes. The oxidation of nitrogen monoxide, which turns colourless ${{\mathrm{NO}}}$ from car exhausts into brown ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}$ in polluted air, is thought to follow this pattern:

Oxygen takes part only in step 2. The rate of step 2 also depends on how much of the intermediate ${{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$ is present, and that depends on the concentration of ${{\mathrm{NO}}}$ through the equilibrium in step 1. The rate of the whole reaction therefore depends on both $[{{\mathrm{NO}}}]$ and $[{{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}]$, even though ${{\mathrm{NO}}}$ does not take part in the slow step.

Energy profiles of multistep reactions (HL)

An energy profile for a reaction with more than one step has one hump for each elementary step. The top of each hump is a transition state; each trough between two humps is an intermediate. The reactant and product levels are fixed by the substances involved, so $\Delta H$ is the same however many steps there are.

The energy profile of the reaction of nitrogen dioxide with carbon monoxide has two humps. The first is much higher than the second, so step 1 has the larger activation energy and is the rate-determining step.

The activation energy of each step is measured from the energy of the species that start that step: for step 1 from the reactants, and for step 2 from the intermediate. A step with a larger activation energy has a smaller fraction of collisions with $E \geq E_\text{a}$, so it is slower.

When the second step is rate-determining, the second hump rises higher above the intermediate than the first hump rises above the reactants. The energy profile of such a reaction has a low first barrier, which is crossed quickly in both directions, and a high second barrier.

The halogenoalkanes give a contrasting pair of profiles. The $\text{S}_\text{N}1$ substitution of a tertiary halogenoalkane has two steps with a carbocation intermediate in the trough, and the first step (forming the carbocation) is rate-determining. The $\text{S}_\text{N}2$ substitution of a primary halogenoalkane is a single step with one transition state and no intermediate. Both mechanisms are drawn with curly arrows in electron-pair sharing reactions (R3.4).

Constructing an energy profile from kinetic data

An energy profile can be sketched from a mechanism together with the rate equation and the sign of $\Delta H$. The rate equation shows which step is slow, and so which hump must have the larger barrier; the sign of $\Delta H$ fixes whether the products lie below or above the reactants.

Molecularity (HL)

A step in which a single particle breaks apart or rearranges is unimolecular; one in which two particles collide is bimolecular; and one in which three particles collide at the same moment is termolecular. The particle diagram compares the three types of step.

Most elementary steps are unimolecular or bimolecular. Termolecular steps are rare, because three particles must arrive at the same point at the same instant, with enough energy and the right orientation. Steps involving four or more particles are not found. This is why a reaction such as ${{2\,\mathrm{NO} {}+{} 2\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\xrightarrow{}\mathrm{N}{\vphantom{X}}_{\smash[t]{2}} {}+{} 2\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}}}$ must have a mechanism of several steps. The molecularity of each step in a mechanism fixes the form of that step's rate:

For an elementary step, and only for an elementary step, the rate is proportional to the concentration of each reacting particle, so the power of each concentration equals the number of that particle in the step. Molecularity is a property of a single step in a proposed mechanism. The order of reaction is a property of the experimental rate equation for the whole reaction. The order of the whole reaction equals the molecularity of the rate-determining step only when the slow step is the first step.

Rate equations (HL)

The rate of a reaction depends on the concentrations of some of the species present. For a reaction between ${{\mathrm{A}}}$ and ${{\mathrm{B}}}$, the relationship found by experiment usually takes the form:

This is the rate equation (also called the rate expression or rate law). $k$ is the rate constant, and the exponents $m$ and $n$ are the orders of reaction with respect to ${{\mathrm{A}}}$ and ${{\mathrm{B}}}$.

A rate equation contains only the species whose concentrations affect the rate. These may include a catalyst, which does not appear in the balanced equation at all, and they may leave out a reactant that does appear in it. The order of reaction with respect to each species and the overall order cannot be predicted from the balanced equation. They depend on the mechanism and can only be found by experiment.

The iodination of propanone in acid solution shows how the rate constant and the orders relate to the species present:

Iodine is a reactant but does not appear in the rate equation; it is zero order. The acid is a catalyst but does appear; it is first order. The mechanism accounts for both facts: the slow step involves propanone and hydrogen ions, forming an intermediate that then reacts rapidly with iodine. The rate of this reaction is often followed by colorimetry, since the yellow-brown colour of iodine fades at a constant rate.

Zero, first and second order

In practice a reactant is most often zero order, first order or second order. The table summarizes what each means for a reactant ${{\mathrm{A}}}$.

In general, if $[{{\mathrm{A}}}]$ is multiplied by a factor $x$, the rate is multiplied by $x^m$, where $m$ is the order. Tripling the concentration of a second-order reactant multiplies the rate by $3^2 = 9$. Some reactions have fractional or negative orders, which arise from more complex mechanisms.

Deducing orders from initial-rate data

In the initial-rates method, several runs of a reaction are carried out at the same temperature, each with a different set of starting concentrations. The initial rate of each run is measured, either from the tangent at $t = 0$ to a concentration–time graph or from the time taken for a small, fixed amount of reaction. The initial rate is used because at the start the concentrations are known exactly and the products have not yet had time to interfere.

The runs are planned so that only one concentration changes between a pair of runs. Comparing that pair isolates the effect of one reactant:

• rate unchanged when the concentration changes: zero order;

• rate changes by the same factor as the concentration: first order;

• rate changes by the square of the factor: second order.

The rate of the reaction ${{\mathrm{S}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{8}}{\vphantom{X}}^{2-}\mskip2mu (\mathrm{aq}) {}+{} 2\,\mathrm{I}{\vphantom{X}}^{-}\mskip2mu (\mathrm{aq})\xrightarrow{}2\,\mathrm{SO}{\vphantom{X}}_{\smash[t]{4}}{\vphantom{X}}^{2-}\mskip2mu (\mathrm{aq}) {}+{} \mathrm{I}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{aq})}}$ was measured at 298 K. A student obtained these initial rates.

When two concentrations change at once between a pair of runs, the effect of the reactant whose order is already known is divided out first. For example, if $[{{\mathrm{A}}}]$ (first order) and $[{{\mathrm{B}}}]$ both double and the rate increases eightfold, the factor due to ${{\mathrm{B}}}$ is $8/2 = 4 = 2^2$, so the reaction is second order in ${{\mathrm{B}}}$.

Measuring initial rates: the iodine clock

A clock reaction gives a sudden, easily seen change after a certain amount of reaction has taken place. The peroxodisulfate–iodide reaction becomes a clock when a small, fixed amount of sodium thiosulfate and some starch are added to the mixture. The thiosulfate immediately turns the iodine produced by the slow reaction back into iodide. Only when all the thiosulfate has been used up does free iodine remain, and the solution suddenly turns blue-black with the starch. The flow diagram sets out the sequence of reactions in the iodine clock.

Every run uses the same amount of thiosulfate, so the colour appears when the same small amount of iodine has formed. The average rate up to that point is that fixed amount divided by the time $t$, and because so little of the reactants has been used, it is close to the initial rate. The initial rate is therefore proportional to $1/t$, and $1/t$ can be used directly to find orders.

In a typical method, solutions of potassium iodide, sodium thiosulfate and starch are mixed in a conical flask, potassium peroxodisulfate solution is added from a pipette, and a stopwatch is started as it is added. The concentration of one reactant is varied by using different volumes of its solution, and the total volume is kept the same by adding water, so that the other concentrations are unchanged. The flasks stand in a water bath to keep the temperature constant. A graph of $1/t$ against $[{{\mathrm{I}{\vphantom{X}}^{-}}}]$ that is a straight line through the origin shows first order with respect to iodide.

The main random error is judging the moment the colour appears, which matters most for the shortest times; placing the flask on a white tile and repeating each run reduces it. Systematic errors include a change in the temperature of the solutions and a delay between adding the last solution and starting the stopwatch. The same apparatus, used at several temperatures with the concentrations fixed, gives the rate constants needed for an Arrhenius plot.

Graphs of zero, first and second order reactions (HL)

Concentration–time graphs

When the concentration of one reactant is measured as a reaction proceeds (with any other reactants in large excess, so that their concentrations hardly change), the shape of the concentration–time graph depends on the order with respect to that reactant. The concentration–time graph compares three reactions that start with the same concentration and the same initial rate.

• Zero order: a straight line with a constant negative gradient. The rate (the gradient) does not change as $[{{\mathrm{A}}}]$ falls, until the reactant runs out.

• First order: a curve whose gradient falls in proportion to $[{{\mathrm{A}}}]$. Halving the concentration halves the gradient. The concentration falls by half in equal intervals of time: the half-life is constant.

• Second order: a curve that starts steeply and then flattens more than the first-order curve, because halving $[{{\mathrm{A}}}]$ divides the rate by four. Each half-life is longer than the one before (it doubles each time).

A constant half-life is the quickest test for a first-order reactant. The first-order decay graph for dinitrogen pentoxide, ${{2\,\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{5}}\mskip2mu (\mathrm{g})\xrightarrow{}4\,\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g}) {}+{} \mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$, shows the concentration halving every 20 minutes.

Order can also be found from a single concentration–time curve by drawing tangents at several points. Each gradient gives the rate at that concentration. If the rate at $0.050\ \text{mol dm}^{-3}$ is half the rate at $0.100\ \text{mol dm}^{-3}$, the reaction is first order; if it is a quarter, second order.

Rate–concentration graphs

When the initial rate is plotted against the concentration of one reactant, each order gives a characteristic line, as the rate–concentration graph shows.

A second-order reactant gives a curve on this graph, which is hard to tell apart from other curves by eye. Plotting the rate against $[{{\mathrm{A}}}]^2$ gives a straight line through the origin, which confirms second order. In the same way, a straight line through the origin on a graph of rate (or $1/t$) against $[{{\mathrm{A}}}]$ confirms first order.

The rate constant and its units (HL)

The rate constant does not depend on concentration. As the reactants are used up the rate falls, but $k$ stays the same, because the rate equation already accounts for the change in concentration. The rate constant changes only when the fraction or frequency of successful collisions changes: it increases with temperature, and it is larger when a catalyst provides a pathway with a lower activation energy. A value of $k$ is therefore always quoted with the temperature at which it applies.

A larger rate constant means a faster reaction at the same concentrations. Because $k$ has different units for different orders, values can only be compared between reactions of the same overall order.

Deriving the units of k

The units of $k$ are whatever makes both sides of the rate equation have units of $\text{mol dm}^{-3}\ \text{s}^{-1}$. Rearranging the rate equation for $k$ and substituting units gives them directly. For a reaction that is second order overall:

In general, for an overall order $n$ the units are $\text{mol}^{1-n}\ \text{dm}^{3(n-1)}\ \text{s}^{-1}$. Each extra order adds one $\text{mol}^{-1}\ \text{dm}^{3}$ to the units.

Solving problems with the rate equation

Once the rate equation and $k$ are known, the rate equation can be rearranged to find a rate at new concentrations, or the concentration needed to give a certain rate.

Linking rate equations to mechanisms (HL)

The rate equation is set by the rate-determining step. The species that appear in it are those that take part in the slow step and those in any fast steps before it, because these control how much of an intermediate the slow step can use. Species that react only in fast steps after the slow step do not appear. The order with respect to each reactant then reflects the number of its particles in the slow step and the steps before it.

The slow step is the first step

When the first step is slow, the rate equation is simply the rate of that elementary step. For ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}} {}+{} \mathrm{CO}\xrightarrow{}\mathrm{NO} {}+{} \mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}}}$, the slow step is ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}} {}+{} \mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}\xrightarrow{}\mathrm{NO}{\vphantom{X}}_{\smash[t]{3}} {}+{} \mathrm{NO}}}$, a bimolecular step involving two ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}$ molecules, so:

The reaction is second order in ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}$ and zero order in ${{\mathrm{CO}}}$, which agrees with experiment. The halogenoalkanes give a second pair of examples. The hydrolysis of a tertiary halogenoalkane such as 2-bromo-2-methylpropane by aqueous alkali has rate $= k[{{(\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}){\vphantom{X}}_{\smash[t]{3}}\mathrm{CBr}}}]$, because its slow step is the unimolecular breaking of the C–Br bond. The hydrolysis of a primary halogenoalkane such as 1-bromobutane has rate $= k[{{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{Br}}}][{{\mathrm{OH}{\vphantom{X}}^{-}}}]$, because its single step is bimolecular. The units of $k$ are $\text{s}^{-1}$ and $\text{mol}^{-1}\ \text{dm}^{3}\ \text{s}^{-1}$ respectively.

The slow step comes after a fast equilibrium

When a fast, reversible step comes first, the slow step uses an intermediate. An intermediate cannot appear in the final rate equation, because its concentration cannot be measured or controlled directly. It is replaced by the reactants that form it in the fast step. For the oxidation of nitrogen monoxide:

The slow step gives rate $= k_2[{{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}][{{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}]$. The fast equilibrium in step 1 keeps $[{{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}]$ proportional to $[{{\mathrm{NO}}}]^2$, since two ${{\mathrm{NO}}}$ molecules form each ${{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$. Substituting gives the rate equation in terms of reactants only:

This matches the experimental rate equation. The two ${{\mathrm{NO}}}$ molecules count towards the order even though they react in the fast step, because that step comes before the slow step. In short, the orders equal the total number of each reactant particle in the slow step and all the steps before it, once intermediates are traced back to the reactants that form them.

Evaluating a proposed mechanism

A mechanism is a model, put forward to explain the evidence. A proposed mechanism must pass two tests, set out in the flow chart: its steps must add up to the stoichiometric equation, and the rate equation it predicts must agree with the experimental one. A mechanism that also predicts an intermediate that is then detected gains further support.

Why a mechanism is only ever a possible mechanism

Kinetic data can show that a mechanism is wrong, but they cannot prove that one is right, because different mechanisms can predict the same rate equation. The reaction ${{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g}) {}+{} \mathrm{I}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})\xrightarrow{}2\,\mathrm{HI}\mskip2mu (\mathrm{g})}}$ has rate $= k[{{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}}}][{{\mathrm{I}{\vphantom{X}}_{\smash[t]{2}}}}]$, and for decades it was taken as the standard example of a single bimolecular step. In 1967 J. H. Sullivan used light from a mercury lamp to split more of the iodine molecules into atoms, and found that the rate depended on the concentration of iodine atoms. The reaction proceeds, at least largely, through iodine atoms formed in a fast equilibrium, ${{\mathrm{I}{\vphantom{X}}_{\smash[t]{2}}\xrightleftharpoons{}2\,\mathrm{I}}}$, which gives the same rate equation under ordinary conditions. Mechanisms are therefore accepted provisionally, as the simplest explanation consistent with all the evidence, and are revised when new evidence appears.

Temperature and the rate constant: the Arrhenius equation (HL)

When only the temperature is raised, the concentrations do not change, so the faster rate is an increase in $k$. The relationship is not linear: the graph of $k$ against temperature rises ever more steeply.

The Arrhenius equation

The temperature dependence of the rate constant is described by the Arrhenius equation (data booklet):

Here $E_\text{a}$ is the activation energy in $\text{J mol}^{-1}$, $R = 8.31\ \text{J K}^{-1}\ \text{mol}^{-1}$ is the gas constant, $T$ is the absolute temperature in kelvin, and $A$ is the Arrhenius factor, which has the same units as $k$. The two parts of the equation correspond to the two conditions for a successful collision.

• $\text{e}^{-E_\text{a}/RT}$ is the fraction of collisions with energy at least equal to $E_\text{a}$. It is always less than 1. It increases steeply as $T$ rises, matching the larger area beyond $E_\text{a}$ under the Maxwell–Boltzmann curve at a higher temperature, and it is smaller for a reaction with a larger $E_\text{a}$.

• $A$ takes into account the frequency of collisions with the proper orientation.

The Arrhenius equation accounts for the size of the temperature effect. For a reaction with $E_\text{a} = 50\ \text{kJ mol}^{-1}$, $\text{e}^{-E_\text{a}/RT} = \text{e}^{-20.2} \approx 1.7\times10^{-9}$ at 298 K, and the exponential term nearly doubles for a 10 K rise, while $A$, which depends on the collision frequency, changes by less than 2 %.

The Arrhenius factor hardly changes with temperature over a normal range. It is largest when almost every collision has a suitable orientation, as when two simple atoms or radicals combine, and much smaller for reactions between large molecules, where only a collision at one particular site can lead to reaction.

The linear form and the Arrhenius plot

Taking natural logarithms of both sides of the Arrhenius equation gives its linear form (data booklet):

This has the form $y = mx + c$, with $y = \ln k$ and $x = 1/T$. A graph of $\ln k$ against $1/T$ (an Arrhenius plot) is therefore a straight line with:

• gradient $= -\dfrac{E_\text{a}}{R}$, so $E_\text{a} = -\text{gradient} \times R$;

• intercept on the $\ln k$ axis (where $1/T = 0$) $= \ln A$, so $A = \text{e}^{\text{intercept}}$

The line slopes downwards because a larger $1/T$ means a lower temperature and a smaller $k$. The gradient has units of K, since $E_\text{a}/R$ has units of $\text{J mol}^{-1} \div \text{J K}^{-1}\ \text{mol}^{-1} = \text{K}$. A steeper line means a larger activation energy: the rate constant of a reaction with a high $E_\text{a}$ is more sensitive to temperature. A catalysed reaction, with a lower $E_\text{a}$, gives a less steep line.

Laboratory temperatures span only a narrow range of $1/T$ (from about $3.0$ to $3.4\times10^{-3}\ \text{K}^{-1}$ for 20–60 °C), far from $1/T = 0$. The intercept is found either by extrapolating the line or, more reliably, by substituting the gradient and one point on the line into the linear form and solving for $\ln A$.

A student measured the rate constant of a second-order reaction at five temperatures.

Using two temperatures

When rate constants are known at only two temperatures, writing the linear form for each and subtracting eliminates $\ln A$:

This form is not printed in the data booklet, but it follows in one line from the linear form. It gives $E_\text{a}$ from two measurements, or $k$ at a new temperature when $E_\text{a}$ is known. It is less reliable than a graph through several points, because an error in either value of $k$ goes straight into the answer.

Measuring rate constants at several temperatures uses the same methods as any rate experiment, with each reactant solution brought to the temperature of a thermostatted water bath before mixing. With a clock reaction at fixed concentrations, $k$ is proportional to $1/t$, so a graph of $\ln(1/t)$ against $1/T$ has the same gradient, $-E_\text{a}/R$, and gives $E_\text{a}$ without calculating $k$ itself.