Reaction kinetics

Cambridge International AS & A Level Chemistry 9701 · A Level

Every objective in this topic is printed under "A Level subject content" in the 9701 syllabus and carries the tier "A Level". None of it is AS, and it is a separate topic from the AS topic 8 of the same name. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper and the whole 9701 bank on this site comes from it, so no practice here is tagged to this topic.

The vocabulary

• Rate equation: an expression relating the rate of reaction to the concentrations of the reactants, of the form rate = k[A]^m[B]^n.
• Order of reaction with respect to a reactant: the power to which that reactant's concentration is raised in the rate equation.
• Overall order: the sum of the individual orders, m + n.
• Rate constant k: the constant of proportionality in the rate equation, whose value depends on the reaction and on the temperature.
• Half-life t½: the time taken for the concentration of a reactant to fall to half its value.
• Rate-determining step: the slowest step in a multi-step mechanism, which therefore controls the overall rate.
• Intermediate: a species formed in one step and used up in a later step, which appears nowhere in the overall equation.

The single most important point about a rate equation is that it cannot be deduced from the stoichiometric equation. It must be determined experimentally, and the reason is that the balanced equation describes the overall change while the rate equation reflects only the steps up to and including the slowest one.

Determining orders from initial rates

The initial rates method varies one concentration at a time and observes what happens to the rate at the very start, before the concentrations have had time to change.

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Worked example. For a reaction between A and B:

Comparing 1 and 2, [B] is constant and [A] doubles:

4.00 / 2.00 = 2

so the rate doubles and the order with respect to A is 1.

Comparing 2 and 3, [A] is constant and [B] doubles:

16.0 / 4.00 = 4

so the rate quadruples and the order with respect to B is 2.

The rate equation is therefore

rate = k[A][B]²

and the overall order is 3.

To find k, substitute any one experiment.

The denominator, using experiment 1:

0.100 × 0.100² = 0.001

k = 2.00 / 0.001 = 2000

so k = 2.00 × 10³. The units of k must be worked out from the equation rather than remembered, by rearranging so that the rate is in mol dm⁻³ s⁻¹ and each concentration in mol dm⁻³. For an overall third-order reaction that gives mol⁻² dm⁶ s⁻¹.

The general result is worth noting: for overall order n, the units of k are mol^(1-n) dm^(3n-3) s⁻¹, which gives s⁻¹ for first order, mol⁻¹ dm³ s⁻¹ for second order, and mol dm⁻³ s⁻¹ for zero order.

Reading the graphs

Two kinds of graph appear and they must not be confused.

Concentration-time graphs. The gradient at any point is the rate at that moment.

• Zero order: a straight line with constant negative gradient. The rate does not change as the reactant is used up.
• First order: a curve with a constant half-life, so successive halvings take equal times.
• Second order: a curve that falls more steeply at first and then tails off much more slowly, with a half-life that doubles each time.

Rate-concentration graphs, obtained by taking gradients from the graph above.

• Zero order: a horizontal line. Rate is independent of concentration.
• First order: a straight line through the origin. Rate is proportional to concentration, and the gradient is k.
• Second order: an upward curve, since rate is proportional to the square of concentration. Plotting rate against concentration squared straightens it, with gradient k.

Being able to name the order from either graph, and to say which feature identifies it, is the most reliably examined skill in the topic.

The half-life method

For a first-order reaction the half-life is constant, independent of the starting concentration. That is the fastest test on a concentration-time graph: measure the time from the start to half, then from half to a quarter, and if they are equal the reaction is first order.

The reason is worth seeing. Rate is proportional to concentration, so at half the concentration the reaction proceeds at half the rate, and the two effects cancel exactly.

For a first-order reaction:

k = 0.693 / t½

Worked example. A first-order reaction has a half-life of 120 s.

k = 0.693 / 120 = 0.005775

so k is 5.78 × 10⁻³ s⁻¹.

Worked example, using half-lives. A first-order reactant starts at 0.800 mol dm⁻³ with a half-life of 25 s. What is its concentration after 100 s?

100 / 25 = 4

so four half-lives have passed:

0.800 / 2⁴ = 0.05

giving 0.0500 mol dm⁻³.

Mechanisms and the rate-determining step

Only species involved in or before the rate-determining step appear in the rate equation. That single rule answers most mechanism questions in both directions.

From mechanism to rate equation. Take the mechanism

Step 1 (slow): A + B → X

Step 2 (fast): X + C → D

Step 1 is the rate-determining step and involves one A and one B, so

rate = k[A][B]

and C does not appear, even though it is in the overall equation. X is the intermediate, formed in step 1 and used up in step 2.

From rate equation to mechanism. Suppose the overall reaction is A + 2B → products and the experimental rate equation is rate = k[A][B]. The rate-determining step must involve one A and one B, so a mechanism consistent with both is a slow step A + B → X followed by a fast step X + B → products.

Identifying the rate-determining step. Match the molecularity of each proposed step against the rate equation. If rate = k[A]²[B] then the slow step must involve two A and one B, either directly or as the sum of a fast pre-equilibrium and a slow step.

Telling an intermediate from a catalyst. Both appear in the mechanism and not in the overall equation, and the distinction is the order in which they appear:

• An intermediate is formed first and used up later.
• A catalyst is used up first and reformed later.

The classic application is nucleophilic substitution of halogenoalkanes. The S_N1 mechanism of a tertiary halogenoalkane has a slow first step in which only the halogenoalkane ionises, so it is first order overall, rate = k[RX]. The S_N2 mechanism of a primary halogenoalkane has a single step involving both species, so it is second order, rate = k[RX][Nu⁻]. Measuring the order therefore distinguishes the two mechanisms, which is why kinetics is evidence about mechanism rather than merely a description of speed.

Temperature and the rate constant

Raising the temperature increases the rate of a reaction, and the way it does so is worth stating precisely: the concentrations are unchanged, so the increase must come from an increase in k. The rate constant is the only temperature-dependent quantity in the rate equation.

The reason is the Boltzmann distribution from AS topic 8. At a higher temperature a greater proportion of molecules have energy at least equal to the activation energy, so a greater proportion of collisions is successful. The molecules also collide more frequently, but that effect is much the smaller of the two, which is why a rise of ten degrees can roughly double a rate when the collision frequency rises by only a couple of per cent.

Catalysts

A catalyst increases the rate of a reaction by providing an alternative route of lower activation energy, and is chemically unchanged at the end.

• A homogeneous catalyst is in the same phase as the reactants.
• A heterogeneous catalyst is in a different phase, usually a solid catalysing a gas or solution reaction.

Heterogeneous catalysis

The mechanism has three named stages and all three are required:

1. Adsorption. Reactant molecules form bonds to active sites on the catalyst surface, which brings them close together and in the right orientation, so the concentration at the surface is high.
2. Bond weakening. Forming those bonds to the surface weakens the bonds within the reactant molecules, lowering the activation energy for the reaction between them.
3. Desorption. The product molecules are released from the surface, freeing the active sites for more reactant.

The strength of adsorption has to be intermediate. Too weak and the molecules do not stay long enough or the bonds are not weakened; too strong and the products never leave, so the surface is blocked.

Two examples are named:

• Iron in the Haber process. Nitrogen and hydrogen adsorb onto the iron surface, the very strong nitrogen triple bond is weakened, the atoms combine, and ammonia desorbs.
• Platinum, palladium and rhodium in a catalytic converter. Oxides of nitrogen and carbon monoxide from the exhaust adsorb onto the metal surface, and react to give nitrogen and carbon dioxide, which desorb:

2CO + 2NO → 2CO₂ + N₂

The catalyst is spread thinly over a ceramic honeycomb to give a very large surface area from a small mass of expensive metal.

Homogeneous catalysis

A homogeneous catalyst works by being used up in one step and reformed in a later one, so the net change to it is zero.

Oxides of nitrogen in the oxidation of atmospheric sulfur dioxide. NO₂ oxidises SO₂ and is itself reduced, then is regenerated by atmospheric oxygen:

NO₂ + SO₂ → NO + SO₃

2NO + O₂ → 2NO₂

Adding the steps in the right proportion cancels the nitrogen species entirely, leaving 2SO₂ + O₂ → 2SO₃. This is part of the chemistry of acid rain.

Iron ions in the peroxodisulfate and iodide reaction. The uncatalysed reaction

S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂

is very slow, because it requires two negative ions to collide, and they repel one another. Adding Fe²⁺ or Fe³⁺ provides a route through two steps, each between ions of opposite charge:

2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻

2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂

The iron is reformed, so a trace of it catalyses the whole reaction. Note that it works either way round, starting from Fe²⁺ or from Fe³⁺, which is only possible because the transition element has two accessible oxidation states. That is the link to topic 28.

Common mistakes

• Reading the orders off the balanced equation instead of determining them from data.
• Saying a zero-order reactant is not involved in the reaction. It is involved, just not before or in the rate-determining step.
• Quoting units for k from memory rather than deriving them from the rate equation.
• Confusing a concentration-time graph with a rate-concentration graph. A straight line means zero order on the first and first order on the second.
• Saying a constant half-life proves second order. It is the signature of first order.
• Including a species from a fast step after the rate-determining step in the rate equation.
• Confusing an intermediate with a catalyst. Formed then used up is an intermediate; used up then reformed is a catalyst.
• Saying a higher temperature increases the rate because the concentrations rise. Concentrations are unchanged; k rises.
• Explaining a temperature rise only as more frequent collisions, leaving out the greater proportion of molecules exceeding the activation energy.
• Saying a catalyst lowers the activation energy of the original reaction. It provides an alternative route with a lower activation energy.
• Leaving out desorption from an account of heterogeneous catalysis, so the catalyst never becomes available again.

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