HomeLearning HubIB DP ChemistryR2.2 Rate of chemical change
R2.2

How fast? The rate of chemical change

Reactivity 2 · How much, how fast and how far? · SL and HL

🎯What you need to be able to do

  • Determine rates of reaction, including from the tangent to a graph of concentration, volume or mass against time.
  • Explain, using collision theory, the relationship between kinetic energy, temperature and collision geometry.
  • Predict and explain the effect on rate of changing pressure, concentration, surface area, temperature and the presence of a catalyst.
  • Construct and interpret Maxwell–Boltzmann energy distribution curves to explain the effect of temperature and of activation energy on the proportion of successful collisions.
  • Sketch and explain energy profiles with and without a catalyst, for both endothermic and exothermic reactions.
  • AHL Evaluate proposed reaction mechanisms, recognise intermediates and transition states, and interpret molecularity.
  • AHL Deduce a rate equation from experimental data, identify the order with respect to each reactant, and sketch and analyse zero, first and second order graphs.
  • AHL Solve problems involving the rate constant, including deducing its units.
  • AHL Analyse the Arrhenius equation and its linear form, and determine \( E_{\mathrm{a}} \) and the Arrhenius factor from experimental data.

📚The chemistry

Measuring rate

Rate of reaction is the change in concentration of a reactant or product per unit time, with units mol dm−3 s−1.

Concentration is rarely measured directly. In practice you follow something proportional to it and plot it against time:

  • Volume of gas collected in a syringe — for any reaction producing a gas;
  • Loss of mass on a balance — also for gas-producing reactions, provided the gas is dense enough for the change to be measurable;
  • Change in colour, measured with a colorimeter or spectrophotometer — if a reactant or product is coloured;
  • Change in pH, using a probe;
  • Time to reach a fixed point — the classic “disappearing cross” sulfur precipitation, where rate is taken as \( 1/t \);
  • Change in electrical conductivity, if the number or charge of ions changes.

From the graph, the rate at any instant is the gradient of the tangent at that point. The rate is greatest at \( t = 0 \), when the reactants are at their highest concentration, and falls as they are consumed — which is why the initial rate, the gradient of the tangent at time zero, is the only fair value to quote when comparing different starting concentrations.

Collision theory

Particles react only when they collide, and a collision leads to reaction only if both of these hold:

Sufficient energy
the colliding particles must have combined kinetic energy at least equal to the activation energy \( E_{\mathrm{a}} \), the minimum needed to break the existing bonds
Correct orientation
the particles must be aligned so that the reacting parts of each molecule meet — collision geometry. Most collisions fail on this alone

The temperature link comes from S1.1: absolute temperature is proportional to the average kinetic energy of the particles.

The factors that change rate

  • Concentration — more particles per unit volume, so a greater collision frequency, so more successful collisions per second.
  • Pressure (for gases) — the same argument. Increasing pressure squeezes the same number of molecules into a smaller volume, increasing concentration and hence collision frequency.
  • Surface area (for a solid) — more particles are exposed at the surface where reaction can occur, so collision frequency rises. Powdering a solid can increase the rate dramatically, which is why flour and custard powder can cause dust explosions.
  • Temperature — two effects, and both should be mentioned. Particles move faster, so collisions are more frequent; and, much more importantly, a far greater proportion of collisions have energy at least equal to \( E_{\mathrm{a}} \). The second effect dominates by a wide margin, which is why a rise of only 10 K can roughly double a reaction rate.
  • Catalyst — below.
“Raising the temperature increases the rate because the particles move faster and collide more often.” True but nowhere near sufficient, and this half-answer is extremely common. The frequency of collisions rises only a few per cent for a 10 K rise, while the rate roughly doubles. The dominant reason is that the proportion of particles with energy greater than \( E_{\mathrm{a}} \) increases sharply. Any temperature explanation that does not mention the activation energy is incomplete.

The Maxwell–Boltzmann distribution

At any temperature, particles do not all have the same energy: they are spread over a distribution. The Maxwell–Boltzmann curve plots the number (or fraction) of particles against kinetic energy. Its features, all of which earn marks:

  • it starts at the origin — no particle has zero energy;
  • it rises to a peak (the most probable energy) then falls;
  • it is asymmetric, with a long tail that approaches but never touches the energy axis — there is no upper limit on energy;
  • the total area under the curve equals the total number of particles, so it must not change when you redraw it at a different temperature;
  • \( E_{\mathrm{a}} \) is marked as a vertical line, and the area to the right of it represents the particles able to react.

At higher temperature the curve flattens and shifts to the right: the peak is lower and further along, and the tail is raised. The area to the right of \( E_{\mathrm{a}} \) increases markedly — that is the picture of why heating speeds up a reaction. A catalyst does not change the curve at all; it moves the \( E_{\mathrm{a}} \) line to the left, which again increases the area to its right.

Catalysts

A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy. It is not consumed and does not appear in the overall equation.

Three consequences worth stating explicitly:

  • On an energy profile, the catalysed route is drawn as a lower peak — sometimes with two smaller humps if the catalysed mechanism has more steps. The reactant and product levels are unchanged.
  • A catalyst therefore does not change \( \Delta H \), and it does not change the position of equilibrium or the value of \( K \) — it speeds up the forward and reverse reactions equally, so equilibrium is simply reached sooner (R2.3).
  • Biological catalysts are called enzymes. The different mechanisms of homogeneous and heterogeneous catalysts will not be assessed.

AHL Mechanisms, molecularity and the rate-determining step

Most reactions occur as a series of elementary steps, and the overall equation is just their sum. The slowest step determines the overall rate and is called the rate-determining step — and it need not be the first step.

Two species to distinguish carefully:

Intermediate
produced in one step and consumed in a later one — a real species with a finite lifetime.
On an energy profile it sits in a trough between two peaks.
Transition state
the arrangement at the top of an energy barrier, with bonds partly broken and partly formed.
It sits at a peak and cannot be isolated.

Molecularity is the number of particles taking part in an elementary step: unimolecular (one), bimolecular (two), termolecular (three). Termolecular steps are rare, because a simultaneous three-particle collision with correct geometry is very improbable — which is itself a good reason to be suspicious of any proposed mechanism containing one.

A proposed mechanism is acceptable only if it is consistent with both the stoichiometry and the kinetic data: the steps must add to the overall equation, and the rate equation predicted by the rate-determining step must match the one measured.

AHL Rate equations and order

For a reaction of A and B:

\[ \text{rate} = k\,[\mathrm{A}]^{m}\,[\mathrm{B}]^{n} \]

\( m \) and \( n \) are the orders with respect to A and B, and \( m + n \) is the overall order. Only integer values are assessed.

Orders can only be determined experimentally. They are not the stoichiometric coefficients, and assuming they are is the central error of this topic. They reflect the composition of the rate-determining step, not the overall equation, which is exactly why kinetics is evidence about mechanism.

Finding orders from a table of initial rates: compare two experiments in which only one concentration changes. If doubling \( [\mathrm{A}] \) leaves the rate unchanged the order is zero; if it doubles the rate the order is one; if it quadruples the rate the order is two.

Zero order
concentration–time: straight line, negative gradient
rate–concentration: horizontal line
First order
concentration–time: curve with constant half-life
rate–concentration: straight line through the origin
Second order
concentration–time: steeper curve, half-life increasing
rate–concentration: upward curve (a parabola)

The constant half-life of a first-order reaction is the quickest diagnostic from a concentration–time graph: read off the time to fall from 1.0 to 0.5, then from 0.5 to 0.25, and if they are equal the reaction is first order.

AHL The rate constant and its units

\( k \) is constant for a given reaction at a given temperature and increases sharply as temperature rises. Its units depend on the overall order, and are deduced by rearranging the rate equation:

Zero order
mol dm−3 s−1
First order
s−1
Second order
dm3 mol−1 s−1
Third order
dm6 mol−2 s−1

The pattern: each order above the first adds dm3 mol−1. Derive them rather than memorising, by writing \( k = \dfrac{\text{rate}}{[\mathrm{A}]^m[\mathrm{B}]^n} \) and cancelling.

AHL The Arrhenius equation

\[ k = A\,e^{-E_{\mathrm{a}}/RT} \qquad\qquad \ln k = -\frac{E_{\mathrm{a}}}{R}\cdot\frac{1}{T} + \ln A \]

Both forms are in the data booklet. \( A \) is the Arrhenius factor, which accounts for the frequency of collisions with the correct orientation — the geometry half of collision theory made quantitative.

Qualitatively: raising \( T \), or lowering \( E_{\mathrm{a}} \), makes the exponent less negative and so increases \( k \). Because the dependence is exponential, a small change in either produces a large change in rate.

The linear form is what you use with data. Plot \( \ln k \) on the y-axis against \( 1/T \) on the x-axis:

  • the gradient is \( -E_{\mathrm{a}}/R \), so \( E_{\mathrm{a}} = -\text{gradient} \times R \) — and because the gradient is negative, \( E_{\mathrm{a}} \) comes out positive, as it must;
  • the intercept on the \( \ln k \) axis is \( \ln A \), so \( A = e^{\text{intercept}} \).

\( R \) is in J K−1 mol−1, so \( E_{\mathrm{a}} \) comes out in J mol−1 — divide by 1000 for kJ mol−1.

✏️Worked example

AHL The reaction \( 2\mathrm{NO}(g) + \mathrm{O_2}(g) \rightarrow 2\mathrm{NO_2}(g) \) was studied at constant temperature. Concentrations are in mol dm−3 and rates in mol dm−3 s−1.
Experiment 1
[NO] = 0.010
[O2] = 0.010
rate = \( 2.5 \times 10^{-5} \)
Experiment 2
[NO] = 0.020
[O2] = 0.010
rate = \( 1.0 \times 10^{-4} \)
Experiment 3
[NO] = 0.020
[O2] = 0.030
rate = \( 3.0 \times 10^{-4} \)
(a) Deduce the order with respect to each reactant and the overall order.
(b) Write the rate equation and calculate \( k \), including units.
(c) Suggest a two-step mechanism consistent with the rate equation, and identify the rate-determining step and any intermediate.
(d) Explain, using a Maxwell–Boltzmann distribution, why raising the temperature increases the rate far more than the increase in collision frequency alone would suggest.

(a) Compare experiments in which only one concentration changes.

1 to 2: \( [\mathrm{O_2}] \) is constant; \( [\mathrm{NO}] \) doubles; the rate goes from \( 2.5 \times 10^{-5} \) to \( 1.0 \times 10^{-4} \), a factor of four. Since \( 2^m = 4 \), \( m = 2 \): second order in NO.

2 to 3: \( [\mathrm{NO}] \) is constant; \( [\mathrm{O_2}] \) triples; the rate goes from \( 1.0 \times 10^{-4} \) to \( 3.0 \times 10^{-4} \), a factor of three. Since \( 3^n = 3 \), \( n = 1 \): first order in O2.

Overall order \( = 2 + 1 = \) 3.

(b)

\[ \text{rate} = k\,[\mathrm{NO}]^2[\mathrm{O_2}] \]

Substituting experiment 1:

\[ k = \frac{2.5 \times 10^{-5}}{(0.010)^2(0.010)} = \frac{2.5 \times 10^{-5}}{1.0 \times 10^{-6}} = 25 \]

Units: rate is mol dm−3 s−1 and the concentration term is (mol dm−3)3, so

\[ k = 25\ \mathrm{dm^6\,mol^{-2}\,s^{-1}} \]

(c) The rate equation contains two NO and one O2, so the rate-determining step must involve those three particles — but a termolecular step is improbable. A two-step mechanism with a fast pre-equilibrium fits:

\[ \text{Step 1 (fast):} \quad 2\mathrm{NO} \rightleftharpoons \mathrm{N_2O_2} \]
\[ \text{Step 2 (slow, rate-determining):} \quad \mathrm{N_2O_2} + \mathrm{O_2} \rightarrow 2\mathrm{NO_2} \]

The steps add to the overall equation ✓. \( \mathrm{N_2O_2} \) is the intermediate — produced in step 1 and consumed in step 2 — and since its concentration depends on \( [\mathrm{NO}]^2 \) through the fast equilibrium, the slow step gives rate \( \propto [\mathrm{NO}]^2[\mathrm{O_2}] \), matching the data ✓. Step 2 is the rate-determining step, and it is bimolecular.

(d) Sketch the distribution at \( T_1 \) and at a higher \( T_2 \) on the same axes, with \( E_{\mathrm{a}} \) marked as a vertical line. At \( T_2 \) the curve is flatter and displaced to the right, with the same total area beneath it.

The number of collisions per second rises only modestly with temperature, because average speed goes as \( \sqrt{T} \). What changes dramatically is the area under the curve to the right of \( E_{\mathrm{a}} \): because the high-energy tail is raised, the proportion of particles with at least the activation energy increases sharply. Since only those collisions can react, the rate rises far more than collision frequency alone would predict — typically doubling for a 10 K rise. The Arrhenius equation says the same thing algebraically: \( k \) depends on \( T \) exponentially.

Check it. Verify the rate constant on a different experiment from the one used to find it. Experiment 3: \( k[\mathrm{NO}]^2[\mathrm{O_2}] = 25 \times (0.020)^2 \times 0.030 = 25 \times 4.0 \times 10^{-4} \times 0.030 = 3.0 \times 10^{-4} \) ✓, matching the tabulated rate exactly. If it does not match, an order is wrong. Check the units separately by cancelling rather than recalling: overall order 3 always gives dm6 mol−2 s−1. And check the mechanism by adding the steps — if the intermediate does not cancel, the mechanism is not consistent with the overall equation.
Reading the orders off the stoichiometric coefficients. Here they happen to agree — 2 for NO, 1 for O2 — which makes the habit feel safe, and it is not. Orders are properties of the rate-determining step and can only be found experimentally. Plenty of reactions are zero order in a reactant that appears in the equation, and some rate equations contain a species (a catalyst, say) that does not appear in the overall equation at all. If a question gives you a table of rates, it is asking you to derive the orders; if it does not, you cannot know them.

📝Practise

Work through these on paper, then reveal the answer.

1. Explain, in terms of collision theory, why powdered calcium carbonate reacts faster with hydrochloric acid than the same mass of marble chips, and why increasing the acid concentration also increases the rate.
Surface area: the reaction can only occur where acid particles meet solid carbonate, which is at the surface of the solid. Powdering the same mass exposes a far greater total surface area, so a much larger number of carbonate particles are available to be collided with at any instant. The frequency of collisions between \( \mathrm{H^{+}} \) ions and the solid therefore increases, so the number of successful collisions per second — and hence the rate — increases. Note the mass of solid is the same, so the total volume of gas eventually produced is unchanged; only the time taken differs. Concentration: a more concentrated acid contains more \( \mathrm{H^{+}} \) ions per unit volume, so again the collision frequency with the surface rises and more successful collisions occur per second. Neither change alters the proportion of collisions that succeed — that depends on the activation energy and temperature, which are unchanged.
2. Sketch Maxwell–Boltzmann distributions for a gas at two temperatures on the same axes, label the axes and \( E_{\mathrm{a}} \), and use the sketch to explain the effect of temperature on rate. State two features your sketch must have to be correct.
Axes: number (or fraction) of particles on the y-axis, kinetic energy on the x-axis. Draw the lower-temperature curve with a higher, narrower peak, and the higher-temperature curve flatter and shifted to the right, with a raised tail. Mark \( E_{\mathrm{a}} \) as a vertical line well to the right of both peaks. Explanation: only particles with energy at least equal to \( E_{\mathrm{a}} \) can react on collision, and these are represented by the area under the curve to the right of the \( E_{\mathrm{a}} \) line. At the higher temperature that area is much larger, so a far greater proportion of collisions are successful and the rate increases sharply. Two essential features: (i) both curves must start at the origin and approach the energy axis asymptotically without touching it — no particle has zero energy and there is no maximum energy; (ii) the total area under both curves must be the same, because the total number of particles has not changed — heating redistributes energy, it does not create particles. A sketch in which the hotter curve is simply taller is wrong.
3. Explain how a catalyst increases the rate of reaction, and state two things it does not change. Sketch the energy profiles with and without a catalyst for an exothermic reaction.
A catalyst provides an alternative reaction pathway with a lower activation energy. Because \( E_{\mathrm{a}} \) is lower, a greater proportion of the colliding particles have sufficient energy to react — on a Maxwell–Boltzmann diagram the \( E_{\mathrm{a}} \) line moves to the left, so the area to its right increases — and the rate rises. The catalyst is not consumed and does not appear in the overall equation. Two things it does not change: (i) the enthalpy change \( \Delta H \), because the reactants and products are unaltered and only the route between them differs; (ii) the position of equilibrium and the value of \( K \), because a catalyst lowers the activation energy of the forward and reverse reactions equally, so both are accelerated by the same factor and equilibrium is simply reached sooner. It also does not change the shape of the Maxwell–Boltzmann curve itself. Sketch: potential energy against reaction coordinate; reactants high on the left, products lower on the right, one curve with a tall peak (uncatalysed) and a second with a lower peak (catalysed) starting and finishing at exactly the same two levels.
4. AHL For a reaction of X and Y, doubling [X] at constant [Y] doubles the rate; tripling [Y] at constant [X] leaves the rate unchanged. Deduce the rate equation, the overall order, and the units of \( k \). What does the result say about Y?
Doubling \( [\mathrm{X}] \) doubles the rate, so \( 2^m = 2 \) and \( m = 1 \): first order in X. Tripling \( [\mathrm{Y}] \) leaves the rate unchanged, so \( 3^n = 1 \) and \( n = 0 \): zero order in Y. Rate equation: rate \( = k[\mathrm{X}] \) — Y is omitted, since any concentration raised to the power zero is 1. Overall order 1. Units: \( k = \dfrac{\text{rate}}{[\mathrm{X}]} = \dfrac{\mathrm{mol\,dm^{-3}\,s^{-1}}}{\mathrm{mol\,dm^{-3}}} = \mathbf{s^{-1}} \). What it says about Y: Y does not appear in the rate-determining step. It must take part in a step that occurs after the slow step (or in a fast pre-equilibrium whose position does not depend on it), so changing its concentration cannot affect the overall rate. This is precisely why kinetics is evidence about mechanism: the rate equation reveals which species are involved up to and including the slowest step, and a species from the balanced equation being absent from the rate equation is informative, not an error.
5. AHL A first-order reaction has \( k = 3.5 \times 10^{-3}\ \mathrm{s^{-1}} \) at 300 K and \( 2.8 \times 10^{-2}\ \mathrm{s^{-1}} \) at 330 K. Determine the activation energy. (\( R = 8.31\ \mathrm{J\,K^{-1}\,mol^{-1}} \).)
Use the linear Arrhenius form at two points and subtract, which eliminates \( \ln A \): \( \ln k_2 - \ln k_1 = -\dfrac{E_{\mathrm{a}}}{R}\left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right) \). Left side: \( \ln\!\left(\dfrac{2.8 \times 10^{-2}}{3.5 \times 10^{-3}}\right) = \ln 8.00 = 2.079 \). Right side bracket: \( \dfrac{1}{330} - \dfrac{1}{300} = 3.0303 \times 10^{-3} - 3.3333 \times 10^{-3} = -3.030 \times 10^{-4}\ \mathrm{K^{-1}} \). So \( 2.079 = -\dfrac{E_{\mathrm{a}}}{8.31} \times (-3.030 \times 10^{-4}) \), giving \( E_{\mathrm{a}} = \dfrac{2.079 \times 8.31}{3.030 \times 10^{-4}} = 5.70 \times 10^{4}\ \mathrm{J\,mol^{-1}} = \mathbf{57.0\ kJ\,mol^{-1}} \). Two checks: the activation energy must be positive (if you get a negative value, a reciprocal has been subtracted the wrong way round), and a value of a few tens of kJ mol−1 is typical — an answer of 57 J mol−1 or 57 000 kJ mol−1 means the final unit conversion went wrong.
6. AHL Distinguish between an intermediate and a transition state, and explain how each appears on an energy profile for a two-step reaction in which the first step is rate-determining.
An intermediate is a real chemical species that is produced in one elementary step and consumed in a later one. It has a finite lifetime, has ordinary bonds, and can in favourable cases be detected or even isolated. A transition state (activated complex) is the arrangement of atoms at the maximum of an energy barrier, in which bonds are partly broken and partly formed. It exists for the duration of a molecular vibration, cannot be isolated, and is not a species in the ordinary sense. On the energy profile of a two-step reaction there are two peaks separated by a trough. Each peak is a transition state; the trough between them is the intermediate, which lies at a local energy minimum — higher in energy than the reactants and products, but stable enough to have a real existence. If the first step is rate-determining, the first peak is the higher of the two, since the largest barrier is the one that controls the overall rate. The activation energy for the whole reaction is measured from the reactant level up to that highest peak.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • PhET — Reactions and Rates, which lets you set the activation energy and temperature and watch the Maxwell–Boltzmann distribution and the reaction respond together. The single best resource on this page.
  • RSC Learn Chemistry — the iodine clock and the sodium thiosulfate “disappearing cross” practicals, both excellent IA starting points for an order-of-reaction investigation.
  • Your data booklet — the Arrhenius equation, its linear form and the value of \( R \). None of these needs to be memorised; knowing which page they are on does.