AS & A Level Chemistry · Area 1 · AS Level Content

Reaction kinetics

Reaction kinetics explains how fast a reaction goes and why. Particles must collide with at least the activation energy for a collision to be effective, so concentration, pressure, temperature and catalysts change the rate by changing the frequency of effective collisions or the activation energy.

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Syllabus coverage

  • Explain and use the terms rate of reaction, frequency of collisions, effective collisions and non-effective collisions
  • Explain qualitatively, in terms of the frequency of effective collisions, the effect of changes in concentration and in pressure on the rate of a reaction
  • Use experimental data to calculate the rate of a reaction, including the rate at an instant from the gradient of a concentration-time graph
  • Define activation energy, EA, as the minimum energy required for a collision to be effective
  • Sketch and use the Boltzmann distribution to explain the significance of activation energy
  • Explain qualitatively, in terms both of the Boltzmann distribution and of the frequency of effective collisions, the effect of a change in temperature on the rate of a reaction
  • Explain and use the terms catalyst and catalysis
  • Explain that, in the presence of a catalyst, a reaction has a different mechanism, that is one of lower activation energy
  • Explain the catalytic effect in terms of the Boltzmann distribution
  • Construct and interpret a reaction pathway diagram for a reaction in the presence and in the absence of an effective catalyst, and distinguish between homogeneous and heterogeneous catalysts

Part of the AS & A Level Chemistry syllabus 9701.

Rate of reaction: the three graphs you must be able to readThe rate is the gradient of a concentration-time graph, measured with a tangent. The rate falls as the reactants are used up, becausecollisions become less frequent.1. Concentration against timethe gradient at any point is the rate03060901200.001.00time (s)concentration / mol dm-3tangent at t = 0tangent at 60 s: slower2. Rate against concentrationthe gradient of the line is k, at a fixed temperature00.501.0000.020concentration of A (mol dm-3)rate / mol dm-3 s-1rate = k[A]double [A] and the rate doubles,so the reaction is first order in A3. With and without a catalystno catalystwith a catalystBoth curves flatten at the same value:a catalyst changes the rate, not the yield.0601200.001.00time (s)concentration / mol dm-3Measuring a rate from a tangentA tangent is a straight line that touches the curve at onepoint and has the same gradient as the curve there. Draw it att = 0 to get the initial rate.the tangent at t = 0 passes through(0 s, 1.00) and (30 s, 0.46) on the axesinitial rate = (1.00 - 0.46) / 30 = 0.018 mol dm-3 s-1at 60 s a second tangent gives 0.0054 mol dm-3 s-1:a third of the initial raterate = k[A], so the reaction is first order in A
Rate graphs for a reaction: a curve of concentration against time whose gradient falls as the reactants are used up, with a tangent drawn at t = 0 to give the initial rate, and a Boltzmann distribution curve with the activation energy marked and the area beyond EA shaded to show the fraction of molecules that can react.

The rate of reaction and how it is measured

The rate of reaction is the change in concentration of a reactant or a product per unit time. Concentration is measured in mol dm-3 and time in seconds, so a rate is usually quoted in mol dm-3 s-1; when a gas is collected instead, the rate may be quoted in cm3 s-1 or cm3 min-1. A rate is only meaningful if you say when it was measured, because the rate of a reaction is not constant: it is fastest at the very beginning, when the reactants are most concentrated, and it falls steadily as they are used up.

To follow a rate you need a property that changes measurably as the reaction proceeds. You can collect the gas produced in a gas syringe or over water, follow the loss in mass on a balance when a gas escapes, time how long a mixture takes to become cloudy enough to hide a cross on paper, or measure the change in colour with a colorimeter. Each method gives you a set of readings of the same kind: how much of something is there at known times.

From a graph of concentration (or volume, or mass) against time, the rate at any instant is the gradient of the curve at that point. Because the curve is not a straight line, you draw a tangent at the point you want and find the gradient of the tangent: rate = change in concentration / change in time. The gradient at t = 0 is the initial rate, which is the value chemists usually compare when they study how a change in conditions affects a reaction.

What you measureWhat changes as the reaction proceedsHow it is followed
Volume of gasThe number of moles of gas producedGas syringe or inverted burette, read at fixed times
MassThe mass of the flask and its contents falls as a gas escapesBalance, used only when a gas is lost to the air
Colour or cloudinessThe concentration of a coloured species or of a precipitate changesColorimeter, or the time for a cross to disappear
pHThe concentration of hydrogen ions changespH meter, or an indicator for a sharp end-point
A rate is a gradient, not a value read off the axis. Draw the tangent, then divide the change in concentration by the change in time.

Collision theory: frequency and effectiveness of collisions

For a reaction to happen at all, the reacting particles must collide. Collisions in a liquid or a gas are extremely frequent, but almost all of them come to nothing: the particles simply bounce apart. A collision that does not lead to reaction is a non-effective collision, and it fails because the particles were not carrying enough energy when they met (and, in a molecule with a complicated shape, sometimes because they met the wrong way round).

A collision in which the particles do react is an effective collision. For a collision to be effective, the particles must collide with at least a certain minimum energy; anything less and they rebound unchanged. The frequency of collisions is simply how many collisions happen per unit time; what matters for the rate is the frequency of effective collisions, because only those change the reactants into products.

This gives the two levers that control the rate of any reaction. You can increase how often the particles meet, which is what raising the concentration or the pressure does, or you can increase the fraction of those collisions that are effective, which is what raising the temperature or adding a catalyst does. Every rate effect in this topic is one of these two ideas, so say which one you are using before you explain it.

  • Collision: two particles meet. Most collisions in a gas or solution are frequent but useless.
  • Non-effective collision: the particles rebound unchanged, because there was not enough energy (or the orientation was wrong).
  • Effective collision: the collision leads to reaction, which requires energy greater than or equal to the activation energy.
  • Rate depends on the frequency of effective collisions per unit time, not on the total number of collisions.
Two ideas do all the work: how often the particles collide, and what fraction of those collisions are effective.
Collision theory: why a reaction needs energyParticles react only when they collide with at least the activation energy and in the right orientation. Anything that raises howoften particles collide, or how energetic those collisions are, raises the rate.A collision that reactsenough energy, and the right orientationABreactA-Bthe new bond formsenergy is releasedEa, the activation energymore energy than Ea: the particles reactA collision that does not reacttoo little energy, or the wrong orientationABbounce apart: no reactionthe particles keeptheir own bondsEabelow Ea: the particles bounce apartHigher concentrationmore particles in the same volumediluteconcentratedThere are more particles in thesame volume.Collisions happen more often, sothe rate is higher.Each collision has the sameenergy as before.Higher temperatureparticles move fastercoolhot: fasterParticles move faster, so theycollide more often.More of them also have at leastEa, so more collisions succeed.Temperature raises the rate themost.Higher pressurea gas squeezed into less spacelarge volumesmaller volumeGases only. The same number ofparticles is squeezed into asmaller volume.They are closer together, socollisions are more frequent.The rate rises.A catalysta route with a lower EaEano catalystcatalystA catalyst provides analternative route with a lowerEa.A larger fraction of collisionsnow have enough energy.The catalyst is not used up.
Collision theory: fast-moving particles collide many times per second, but only the collisions in which the particles meet with at least the activation energy and in a favourable orientation are effective and lead to products.

Concentration and pressure: getting the particles closer together

Increasing the concentration of a reactant means putting more of its particles into the same volume. The particles are closer together, they meet more often, and the frequency of collisions rises. Because collisions in general are more frequent, the number of collisions per second that carry at least the activation energy also rises, so the frequency of effective collisions rises and the reaction goes faster.

The same argument covers pressure. For a reaction between gases, raising the pressure squeezes the same number of molecules into a smaller volume, which is exactly the same as increasing the concentration. Doubling the pressure of a gaseous reactant doubles its concentration and makes the collisions between its molecules more frequent, so the rate increases.

Two things are worth pinning down. First, concentration and pressure change the rate only; they do not change the activation energy, and they do not change the position of equilibrium, so they cannot be used to raise a yield at equilibrium. Second, in a reaction between two different substances, changing the concentration of one reactant often changes the rate much more than changing the concentration of the other by the same amount, so a rate experiment has to test the two reactants separately.

ChangeEffect on the particlesEffect on collisionsEffect on rate
Higher concentration of a reactant in solutionMore particles in the same volume, so they are closer togetherCollisions between reactant particles are more frequent, so effective collisions per second increaseRate increases
Higher pressure on gaseous reactantsThe same molecules are forced into a smaller volume, so the concentration risesCollisions are more frequent, so effective collisions per second increaseRate increases
Lower concentration of a reactantFewer particles in the same volumeFewer collisions per second, so fewer effective collisionsRate decreases
Concentration and pressure change how often particles collide. They do not change the activation energy, and they do not move an equilibrium.

Activation energy and the Boltzmann distribution

The activation energy, EA, is the minimum energy required for a collision to be effective. It is the energy barrier that the colliding particles must overcome, and it is a property of the reaction: the bonds in the reactants must be stretched and broken far enough for the rearrangement to begin, and that costs energy. A reaction with a high activation energy is slow at room temperature because very few collisions carry that much energy; a reaction with a low activation energy is fast.

The Boltzmann distribution shows how the energies of the molecules of a gas or liquid are spread at a fixed temperature. Plot the number of molecules (or the fraction of molecules) on the vertical axis against kinetic energy on the horizontal axis, and you get a curve with a characteristic shape: it starts at the origin, because no molecule has zero energy; it rises steeply to a single maximum; it then falls away smoothly to the right; and it approaches the horizontal axis without ever touching it, because a tiny number of molecules always have very high energies.

The curve is not symmetrical, and it moves with the temperature. At a higher temperature the maximum becomes lower, the curve becomes flatter, the peak moves to the right, and the tail at high energy gets much fatter. The important quantity is an area: mark the activation energy on the horizontal axis and shade the area under the curve to the right of it. That shaded area is the fraction of collisions that have enough energy to react, and it is the reason activation energy has its significance. For a typical reaction at room temperature only a very small fraction of collisions is found in that shaded region, so only a small fraction is effective, however many collisions there are.

  • The curve starts at the origin: no molecule has zero kinetic energy.
  • It rises to a single peak, then falls away to the right.
  • It never meets the energy axis: a few molecules always have very high energies.
  • The area under the curve to the right of EA is the fraction of molecules that can react.
  • Raising the temperature lowers and flattens the peak and moves it to the right, so the area beyond EA grows.
Activation energy is not the average energy. It is the barrier at the top of the path, and the size of the area beyond it sets how many collisions can get over.

Temperature: why a small rise can double the rate

Raising the temperature speeds a reaction up for two reasons, and Cambridge expects both. First, the molecules move faster, so they meet more often: the frequency of collisions increases. Second, and far more important, the particles have a higher average kinetic energy, so a much larger fraction of them has energy greater than or equal to the activation energy. Both effects increase the frequency of effective collisions, but the second dominates by a long way.

The Boltzmann distribution shows why a modest rise in temperature has such a large effect. At the original temperature the area to the right of EA is a thin sliver. Raise the temperature and the whole curve flattens and shifts to the right, and because the curve is falling steeply near EA, even a small shift to the right adds a large amount of area under the tail. A rise of only 10 K, taking the mixture from about 298 K to about 308 K, therefore produces a much bigger increase in the number of effective collisions than the small increase in collision frequency would explain on its own.

Keep the vocabulary straight in an answer. Temperature does not change the activation energy, and it does not give every molecule enough energy to react: the great majority of molecules still have less than EA. What changes is the proportion of molecules that has at least EA, and therefore the frequency of effective collisions. A common but wrong explanation says that raising the temperature lowers the activation energy; only a catalyst does that.

Temperature raises the fraction of molecules above EA, which the fattened tail of the Boltzmann curve shows. It leaves EA itself untouched.

Catalysts: homogeneous and heterogeneous

A catalyst is a substance that increases the rate of a reaction by providing an alternative reaction pathway of lower activation energy. It is not used up in the reaction and is chemically unchanged at the end, although it may change physically, for example a lump of catalyst may break up into powder. A catalyst works because the new pathway takes the reaction over a lower energy barrier, and it is this new mechanism that is the definition worth remembering: in the presence of a catalyst the reaction has a different mechanism, one of lower activation energy.

The Boltzmann distribution explains the effect at once. The molecules still have the same spread of energies at a given temperature, because the temperature has not changed, but the activation energy on the horizontal axis moves to the left. The area under the curve beyond that smaller activation energy is much larger than before, so a much greater fraction of collisions is effective, and the rate rises. In an uncatalysed reaction the catalyst can be regenerated many times over, so a small amount of catalyst can speed up a very large amount of reaction.

In homogeneous catalysis the catalyst is in the same physical state as the reactants. Aqueous acid catalysing the hydrolysis of an ester is the standard example: the acid is in the same aqueous phase as the ester and it is regenerated at the end, and the new pathway goes through a protonated intermediate that breaks down more readily. Iron(II) and iron(III) ions catalysing the reaction between peroxodisulfate ions and iodide ions work in the same way, the iron ion being reduced and then reoxidised in two steps of lower activation energy than the single uncatalysed step.

In heterogeneous catalysis the catalyst is in a different physical state from the reactants, which is the usual arrangement in industry: a solid catalyst with gaseous or liquid reactants. The mechanism runs on the surface. The reactant molecules are adsorbed onto the active sites of the solid, they react there in a favourable orientation with the activation energy of the surface pathway, and the product molecules are desorbed back into the gas or liquid. Because the reaction happens on the surface, the rate depends on the surface area of the catalyst, which is why industrial catalysts are made as finely divided powders or spread thinly over a honeycomb support.

  • Homogeneous catalysis: catalyst and reactants in the same physical state, often both in solution or both gases.
  • Heterogeneous catalysis: catalyst in a different physical state, usually a solid with gaseous or liquid reactants.
  • Heterogeneous mechanism: adsorption onto active sites, reaction on the surface, desorption of the products.
  • A finely divided catalyst has a larger surface area, so more reactant can be adsorbed per unit time and the rate is greater.
A catalyst changes the route, not the destination: it lowers the activation energy, it is regenerated, and it does not change the enthalpy change or the position of equilibrium.

Reaction pathway diagrams and what they show

A reaction pathway diagram (an energy profile) plots the energy of the reacting species on the vertical axis against the progress of the reaction on the horizontal axis. The reactants are marked at the left and the products at the right, and between them the curve rises to a maximum that represents the highest-energy arrangement along the way. The difference in energy between the reactants and the top of that barrier is the activation energy; the difference between the products and the reactants is the enthalpy change of the reaction.

Draw a second curve on the same axes for the catalysed reaction and the difference stands out. Both curves start at the same energy for the reactants and finish at the same energy for the products, because the catalyst does not change the reactants or the products; the catalysed curve simply takes a different route, usually in more than one step, so it may show two or more humps, and its highest point is lower. The activation energy is smaller with the catalyst, and because the catalyst lowers the barrier of the forward and the reverse reaction by the same amount, it is smaller for the backward reaction too.

That last point explains the behaviour of equilibrium systems. A catalyst changes neither the enthalpy change nor the position of equilibrium, because the products and reactants are unchanged and both directions are accelerated equally. What it does change is the time taken to reach that position: with a lower activation energy, equilibrium is established sooner. So when a question asks why an industrial catalyst such as iron in the Haber process is used, the answer is that it allows a workable rate at a lower temperature and pressure, not that it increases the yield.

Feature of the diagramUncatalysed reactionCatalysed reaction
Starting energyEnergy of the reactantsThe same energy of the reactants
Finishing energyEnergy of the productsThe same energy of the products
Number of stepsUsually shown as a single stepAn alternative route, often with more than one step and several humps
Height of the barrierThe activation energy of the uncatalysed reactionA lower activation energy for both the forward and the reverse reaction
Enthalpy changeDifference between products and reactantsExactly the same difference, so the same enthalpy change
On the diagram the catalysed curve starts and ends at the same energies; only the height of the barrier between them changes.

Key terms you must be able to define

Rate of reaction
The change in concentration of a reactant or a product per unit time, usually measured in mol dm-3 s-1; on a graph of concentration against time it is the gradient of the curve at that instant.
Collision theory
The explanation of reaction rates in terms of the frequency of collisions between particles and the fraction of those collisions that carry at least the activation energy.
Effective collision
A collision between particles that leads to a reaction, because the particles meet with at least the activation energy and in a suitable orientation.
Non-effective collision
A collision between particles that does not lead to a reaction, because the particles do not have enough energy (or are wrongly oriented) and simply move apart unchanged.
Activation energy, EA
The minimum energy required for a collision to be effective; on a reaction pathway diagram it is the energy difference between the reactants and the highest point on the curve.
Boltzmann distribution
The distribution of molecular kinetic energies at a given temperature, drawn as a curve of number of molecules against energy; the area under the curve to the right of the activation energy is the fraction of molecules that can react.
Catalyst
A substance that increases the rate of a reaction by providing an alternative pathway of lower activation energy, and which is not used up and is chemically unchanged at the end of the reaction.
Homogeneous catalysis
Catalysis in which the catalyst is in the same physical state as the reactants, for example an aqueous acid catalysing the hydrolysis of an ester in solution.
Heterogeneous catalysis
Catalysis in which the catalyst is in a different physical state from the reactants, usually a solid catalyst with gaseous or liquid reactants; the reaction takes place on the surface, by adsorption, reaction and desorption, so the surface area matters.

Applied to the real world

Cambridge sets 9701 in named contexts - an industrial plant, a laboratory, a hospital, a car battery, a water treatment works. Use these as models for how to apply the chemistry, then write your own.

In an industrial plant

The Haber process makes ammonia from nitrogen and hydrogen over an iron catalyst at about 450 degrees Celsius and 200 atm. Nitrogen is very unreactive because its triple bond gives the uncatalysed reaction a very high activation energy, so without a catalyst the rate would be hopelessly slow. The iron provides a surface mechanism of lower activation energy: the nitrogen and hydrogen molecules are adsorbed onto the metal, react there and the ammonia formed is desorbed. The catalyst does not raise the yield, because it speeds up the reverse reaction by the same amount; it lets a workable rate be reached at a moderate temperature, which also keeps the equilibrium yield reasonable.

In the atmosphere

Chlorofluorocarbons released from old refrigerants and aerosols are broken down by ultraviolet light in the stratosphere to give chlorine atoms, and each chlorine atom catalyses the destruction of ozone in a cycle: a chlorine atom reacts with ozone to give chlorine monoxide and oxygen, and the chlorine monoxide then reacts with another ozone molecule to regenerate the chlorine atom and give two oxygen molecules. Because the chlorine atom is regenerated, one atom destroys a very large number of ozone molecules. This is homogeneous catalysis: catalyst and reactants are all gases in the same phase, and the catalysed route has a far lower activation energy than the uncatalysed decomposition of ozone.

In a laboratory

The rate of the reaction between sodium thiosulfate solution and dilute hydrochloric acid is followed by timing how long the mixture takes to become cloudy enough to hide a black cross drawn on paper. The sulfur precipitate makes the solution opaque, so the shorter the time, the faster the rate. Repeating the experiment with the same volume of more concentrated thiosulfate solution, and with the flask in water baths at different temperatures, shows the two effects in this topic: more particles in the same volume means more frequent effective collisions, and a higher temperature puts a much larger fraction of the molecules above the activation energy.

In a water treatment works

Drinking water is disinfected by adding chlorine, and the rate at which the chlorine kills bacteria depends on its concentration and on the temperature of the water. A higher chlorine concentration means the chlorine and the bacteria meet more often, so the chlorine demand is met sooner, while cold water slows the disinfection down, which is why a higher dose or a longer contact time is needed in winter. Contact tanks are designed to give the water enough time in contact with the chlorine before it is released to the supply.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What is meant by the rate of a reaction?
The change in concentration of a reactant or a product per unit time, usually in mol dm-3 s-1. On a concentration-time graph it is the gradient of the curve.
Q2What is an effective collision?
A collision that leads to reaction, because the particles collide with at least the activation energy (and in a suitable orientation).
Q3What is a non-effective collision?
A collision that does not lead to reaction because the particles do not have enough energy, so they simply bounce apart unchanged.
Q4What two things decide how fast a reaction goes, in terms of collisions?
How frequently the particles collide, and what fraction of those collisions are effective, that is, have energy greater than or equal to the activation energy.
Q5Why does increasing the concentration of a reactant increase the rate?
There are more particles in the same volume, so they collide more often; therefore the frequency of effective collisions per unit time increases and the rate rises.
Q6Why does increasing the pressure of a gaseous reaction increase the rate?
The same molecules are squeezed into a smaller volume, which raises the concentration, so collisions become more frequent and the frequency of effective collisions increases.
Q7Define activation energy.
The minimum energy required for a collision to be effective; on a reaction pathway diagram it is the energy difference between the reactants and the highest point of the curve.
Q8Describe the shape of the Boltzmann distribution.
A curve of number of molecules against kinetic energy that starts at the origin, rises to a single peak, falls away smoothly to the right and approaches the energy axis without touching it.
Q9What does the area under the Boltzmann curve beyond the activation energy represent?
The fraction of molecules (and so of collisions) that have enough energy to react, that is, the fraction that can make an effective collision.
Q10How does the Boltzmann distribution explain the effect of raising the temperature?
The curve flattens and shifts to the right, so the area beyond EA becomes much larger; a far greater fraction of molecules has enough energy, giving many more effective collisions per unit time. The activation energy itself is unchanged.
Q11What is a catalyst?
A substance that increases the rate of a reaction by providing an alternative pathway of lower activation energy; it is not used up and is chemically unchanged at the end of the reaction.
Q12How does a catalyst affect the position of equilibrium and the enthalpy change?
Neither is changed. The catalyst lowers the activation energy of the forward and the reverse reaction equally, so equilibrium is reached faster but the yield and the enthalpy change are the same.
Q13Distinguish between homogeneous and heterogeneous catalysis.
Homogeneous: the catalyst is in the same physical state as the reactants, for example an aqueous acid. Heterogeneous: the catalyst is in a different physical state, usually a solid with gaseous or liquid reactants, and the reaction happens on its surface by adsorption, reaction and desorption.
Q14On a reaction pathway diagram, what is different about the catalysed curve?
It starts and finishes at the same energies as the uncatalysed curve, but it follows a different route, often with more than one step, whose highest point is lower, so the activation energy is smaller.
0 of 14 marked as known

Exam-style questions

9 questions, 28 marks. Write your answer, then reveal the mark scheme and tick each point you actually made. Answering honestly here is what makes the rest of the site useful.

StateAO11 mark
State what is meant by the activation energy of a reaction.

Mark scheme — tick each point you actually made

Examiner tip: One mark, so one idea. Do not write 'the energy needed to start a reaction off' or 'the energy of the reactants'; the mark is for the minimum energy required for a collision to be effective.
DefineAO12 marks
Define the term rate of reaction and state the units in which it is usually measured for a reaction taking place in solution.

Mark scheme — tick each point you actually made

Examiner tip: The definition must contain both ideas: a change in concentration, and per unit time. Just 'how fast a reaction goes' scores nothing.
CalculateAO23 marks
In a reaction in aqueous solution the concentration of a reactant falls from 0.800 mol dm-3 to 0.200 mol dm-3 in 120 s. (a) Calculate the average rate of the reaction over this period. (b) The graph of concentration against time is much steeper during the first 20 s. State what this shows about the rate during the first 20 s.

Mark scheme — tick each point you actually made

Examiner tip: Divide the change in concentration by the time in seconds and give the unit. A gradient question wants rate = change in concentration / change in time, never concentration divided by time.
ExplainAO24 marks
Explain, in terms of collisions, why the rate of a reaction between two solutions increases when the concentration of one of the reactants is increased.

Mark scheme — tick each point you actually made

Examiner tip: The chain of reasoning is concentration -> more frequent collisions -> more effective collisions per unit time -> faster rate. Writing 'more collisions' and stopping there usually gets one mark, not the full four.
ExplainAO23 marks
A rise in temperature of 10 K often roughly doubles the rate of a reaction, even though it increases the frequency of collisions only slightly. Explain this observation using the Boltzmann distribution.

Mark scheme — tick each point you actually made

Examiner tip: Do not say that the molecules now have more energy than the activation energy, or that the temperature lowers the activation energy. The mark is for the increased fraction of molecules above EA.
OutlineAO24 marks
Outline how the presence of a catalyst increases the rate of a reaction. Refer to the mechanism of the catalysed reaction and to the Boltzmann distribution.

Mark scheme — tick each point you actually made

Examiner tip: 'Provides an alternative route' is not enough on its own: add 'of lower activation energy' and link it to the Boltzmann distribution, where the vertical line at EA moves to the left.
DescribeAO25 marks
Describe the reaction pathway diagram for a reaction with and without a catalyst, and explain what the diagram shows about the activation energy, the enthalpy change and the position of equilibrium.

Mark scheme — tick each point you actually made

Examiner tip: Label the axes and mark both curves at the same starting and finishing energies. A sketch with a catalyst curve that finishes lower than the uncatalysed one loses marks straight away.
SuggestAO34 marks
A student investigates how the concentration of sodium thiosulfate solution affects the rate of its reaction with dilute hydrochloric acid. The mixture is timed until a cross on the paper underneath the flask is no longer visible. Suggest and explain two precautions the student must take to make the results comparable.

Mark scheme — tick each point you actually made

Examiner tip: Every precaution needs a reason. Naming a control without saying what it controls does not score, and a rate comparison must use the same cross, the same viewing position and the same temperature.
DeduceAO22 marks
Nickel is used as a catalyst in the hydrogenation of ethene to ethane. Deduce why the rate of reaction is greater when the nickel is finely divided than when the same mass of nickel is used as a single piece. State what is meant by heterogeneous catalysis.

Mark scheme — tick each point you actually made

Examiner tip: Link the surface area to adsorption and then to the rate. For the definition, the key words are 'different physical state from the reactants', and a solid catalyst with gaseous reactants is the common case.

Exam technique for this topic

  • Never answer a rate question with 'more collisions' alone. The mark is for the frequency of effective collisions per unit time, so write that phrase and explain what makes a collision effective.
  • For a temperature question, describe what happens to the Boltzmann curve: it flattens, its peak moves to the right, and the area under the curve beyond EA becomes much larger. Saying that all the molecules now have energy greater than EA, or that the temperature lowers EA, loses the mark.
  • In a catalyst question, say that the catalyst provides an alternative route of lower activation energy. A catalyst is regenerated and is chemically unchanged at the end, so it is not used up even though it takes part in the mechanism.
  • Do not claim a catalyst increases the yield. It lowers the activation energy of the forward and the reverse reaction equally, so equilibrium is reached sooner but its position, and the value of the equilibrium constant, are unchanged; only the temperature moves an equilibrium.
  • When you calculate a rate, work from the gradient of the graph: draw the tangent, read two points, and divide the change in concentration by the change in time in seconds. Quote the unit, usually mol dm-3 s-1 or cm3 s-1, and if the question asks for the initial rate, take the tangent at t = 0 where the curve is steepest.
  • For a heterogeneous catalyst, link the surface to the rate: adsorption of the reactants onto the active sites, reaction on the surface, and desorption of the products. A finely divided catalyst has a greater surface area, so more molecules can be adsorbed per unit time and the rate is faster.

Why students lose marks here

A student wroteA student wrote: 'A catalyst speeds up a reaction by increasing the energy of the reactant particles, so that they collide with more energy.'
Why it loses marksA catalyst does not change the energy of the particles or the temperature of the mixture. The molecules keep the same spread of energies (the same Boltzmann distribution) and the activation energy is what changes, not the particles.
The answer that scoresA catalyst provides an alternative reaction pathway of lower activation energy. At the same temperature a much larger fraction of the molecules now has energy greater than or equal to that lower activation energy, so the frequency of effective collisions increases and the reaction goes faster. The catalyst takes part in the reaction and is regenerated at the end, so it is chemically unchanged.
A student wroteA student wrote: 'Raising the temperature increases the rate simply because the particles move faster and collide more often.'
Why it loses marksMore frequent collisions are only a small part of the effect and are not enough to explain why a 10 K rise can roughly double the rate. The explanation the examiner wants is about the fraction of collisions that are effective.
The answer that scoresTwo effects operate. The molecules move faster, so collisions are a little more frequent, but far more importantly the higher temperature shifts and flattens the Boltzmann distribution, so a much greater proportion of the molecules has energy greater than or equal to the activation energy. The area under the curve beyond EA increases greatly, so the frequency of effective collisions rises sharply and the rate increases far more than the collision frequency alone would explain.
A student wroteA student wrote: 'Adding a catalyst to a reaction at equilibrium increases the yield of the product, because it speeds up the forward reaction.'
Why it loses marksA catalyst lowers the activation energy of the forward and the reverse reactions by the same amount, so both directions are accelerated equally. It does not change the position of equilibrium and does not change the yield; only a change in temperature (or concentration, or pressure) does that.
The answer that scoresThe catalyst accelerates both directions equally, because it lowers the energy barrier of the forwards and the backwards reaction. The position of equilibrium and the value of the equilibrium constant are unchanged, so the final yield is the same. What changes is the time taken to reach equilibrium: the reaction gets there sooner, which is why industrial catalysts are used to make a workable rate at a moderate temperature.
A student wroteA student wrote: 'The concentration of a reactant was 0.600 mol dm-3 after 30 s, so the rate of the reaction at that time was 0.600 / 30 = 0.0200 mol dm-3 s-1.'
Why it loses marksThe rate at an instant is the gradient of the concentration-time curve at that point, not the concentration divided by the time. Dividing the concentration by the time gives a rough average rate over the whole 30 s, and it is not the rate at 30 s. The curve is steepest at the start, so this value is also too large for the rate at 30 s.
The answer that scoresDraw a tangent to the curve at t = 30 s and find its gradient: rate = change in concentration / change in time, taken between two points on the tangent. For example, if the tangent passes through 0.750 mol dm-3 at 20 s and 0.450 mol dm-3 at 40 s, the rate at 30 s is 0.300 / 20 = 0.0150 mol dm-3 s-1. The unit is mol dm-3 s-1, and the answer should be given to the same number of significant figures as the data.
A laboratory bench set up to follow a rate: a flask in a water bath at a fixed temperature, a gas syringe to collect the gas produced, a stopwatch, and a repeat run at a different concentration to compare gradient with gradient rather than one reading with the time.

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