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H2 Chemistry · JC 1 and 2

Reaction Kinetics

Reaction kinetics is the quantitative study of reaction rates. In H2 Chemistry you go beyond collision theory to rate equations, orders of reaction and rate constants, deduced from experimental data rather than from the balanced equation. You use half-lives, link the rate equation to the mechanism through the rate-determining step, and explain the effects of temperature and catalysts using the Boltzmann distribution. Kinetics also underpins the organic mechanisms you meet later, such as SN1 and SN2.

What the syllabus expects

  • Explain and use the terms rate of reaction, rate equation, order of reaction, rate constant and half-life
  • Deduce orders of reaction from initial rates data and from concentration-time and rate-concentration graphs
  • Calculate the rate constant and state its units
  • Use the constant half-life of a first-order reaction, including t½ = ln 2 ÷ k
  • Relate the rate equation to the rate-determining step of a proposed mechanism
  • Explain the effects of temperature and catalysts using the Boltzmann distribution, including homogeneous and heterogeneous catalysis

Rate equations and orders

For a reaction A + B gives products, the rate equation has the form rate = k[A]^m[B]^n. The powers m and n are the orders with respect to A and B, and m + n is the overall order. The rate constant, k, is constant at a given temperature. Orders are usually 0, 1 or 2 and must be found by experiment. They cannot be read from the coefficients in the balanced equation. If a reactant is zero order, changing its concentration has no effect on rate. If it is first order, doubling its concentration doubles the rate. If it is second order, doubling it multiplies the rate by four. The units of k depend on the overall order: mol dm⁻³ s⁻¹ for zero order, s⁻¹ for first order, mol⁻¹ dm³ s⁻¹ for second order and mol⁻² dm⁶ s⁻¹ for third order.

Finding orders from data and graphs

In the initial rates method, several experiments are run with different starting concentrations, and the initial rate is measured each time. Compare two experiments in which only one concentration changes, and see how the rate responds. Graphs also reveal order. On a rate against concentration graph, zero order gives a horizontal line, first order a straight line through the origin, and second order an upward curve. On a concentration against time graph, zero order gives a straight line with constant negative gradient, while first order gives a curve with a constant half-life. When one reactant is in large excess, its concentration hardly changes, so the reaction behaves as if it does not depend on that reactant. This pseudo-order approach lets you find the order with respect to one reactant at a time.

Half-life and first-order reactions

The half-life, t½, is the time taken for the concentration of a reactant to fall to half its value. For a first-order reaction the half-life is constant, whatever the starting concentration, and t½ = ln 2 ÷ k, which is about 0.693 ÷ k. So after one half-life, half remains, after two, a quarter remains, and after three, an eighth remains. A constant half-life on a concentration against time graph is a quick test for first order. For zero-order reactions the half-life gets shorter as the concentration falls, and for second-order reactions it gets longer, so successive half-lives that are not equal tell you the reaction is not first order.

Mechanisms and the rate-determining step

Most reactions happen in a series of elementary steps called the mechanism. The slowest step is the rate-determining step, and it controls the overall rate. The rate equation includes the species that take part in the rate-determining step and any steps before it, so species that only appear in fast steps after it are zero order. A proposed mechanism must add up to the overall equation and must be consistent with the experimental rate equation. Intermediates are formed in one step and used up in a later one, while a catalyst is used in an early step and regenerated later. Organic examples show this clearly. The hydrolysis of a tertiary halogenoalkane by hydroxide ions is first order, rate = k[RX], because the slow step is the breaking of the C-X bond, while the hydrolysis of a primary halogenoalkane is second order, rate = k[RX][OH⁻], because both species collide in the single step.

Temperature, catalysts and the Boltzmann distribution

The Boltzmann distribution shows how the kinetic energies of gas particles are spread at a given temperature. The area under the curve beyond the activation energy, Ea, represents the particles with enough energy to react. At a higher temperature the curve flattens and shifts to the right, with the same total area. The number of particles with energy at least Ea increases significantly, so the frequency of effective collisions rises and k increases. A catalyst provides an alternative pathway with a lower Ea. It does not change the curve, but a larger area now lies beyond the new, lower Ea. A homogeneous catalyst is in the same phase as the reactants, for example Fe²⁺ ions catalysing the reaction between peroxodisulfate and iodide ions. A heterogeneous catalyst is in a different phase, such as iron in the Haber process, which works by adsorbing reactants on its surface, weakening their bonds and then releasing the products by desorption. Enzymes are biological catalysts with specific active sites.

Worked example

For the reaction A + B gives products, the initial rates were: Experiment 1, [A] = 0.10 and [B] = 0.10 mol dm⁻³, rate = 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹. Experiment 2, [A] = 0.20 and [B] = 0.10, rate = 4.0 × 10⁻⁴. Experiment 3, [A] = 0.20 and [B] = 0.30, rate = 3.6 × 10⁻³. Deduce the rate equation and calculate k with units.

  1. Experiments 1 and 2: [B] is constant and [A] doubles, and the rate doubles from 2.0 × 10⁻⁴ to 4.0 × 10⁻⁴, so the reaction is first order with respect to A.
  2. Experiments 2 and 3: [A] is constant and [B] triples, and the rate rises from 4.0 × 10⁻⁴ to 3.6 × 10⁻³, which is 9 times, or 3², so the reaction is second order with respect to B.
  3. Rate equation: rate = k[A][B]², which is third order overall.
  4. Using experiment 1: k = rate ÷ ([A][B]²) = 2.0 × 10⁻⁴ ÷ (0.10 × 0.10²) = 2.0 × 10⁻⁴ ÷ 1.0 × 10⁻³ = 0.20.
  5. Units: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ = mol⁻² dm⁶ s⁻¹.
  6. Check with experiment 3: 0.20 × 0.20 × 0.30² = 0.20 × 0.20 × 0.090 = 3.6 × 10⁻³, which matches.

Answer: rate = k[A][B]², with k = 0.20 mol⁻² dm⁶ s⁻¹.

Common mistakes

  • Taking orders from the coefficients in the balanced equation. Orders come only from experimental data or from the rate-determining step of a known mechanism.
  • Giving the wrong units for k. Work them out from the overall order each time rather than guessing.
  • Saying a higher temperature increases rate mainly because particles collide more often. The main reason is that many more particles have energy at least equal to Ea.
  • Saying a catalyst shifts the Boltzmann distribution. The curve stays the same, and the catalyst lowers Ea so a larger area lies beyond it.
  • Confusing an intermediate with a catalyst. An intermediate is made and then used up, while a catalyst is used and then regenerated.
  • Assuming every reaction has a constant half-life. Only first-order reactions do.

Exam tips

  • In initial rates questions, quote the pair of experiments you are comparing and state the factor by which each concentration and the rate change. That reasoning earns the marks, not just the order.
  • Check your final rate equation and k against an experiment you did not use to calculate them.
  • When drawing a Boltzmann distribution, start the curve at the origin, never let it touch the energy axis again, and keep the area the same at a higher temperature.

Practice questions

1. Easy: A first-order reaction has a half-life of 30 s. Calculate k and the fraction of reactant left after 90 s.

Show answer

k = ln 2 ÷ 30 = 0.693 ÷ 30 = 0.0231 s⁻¹. 90 s is three half-lives, so (1/2)³ = 1/8 of the reactant remains.

2. Medium: For 2NO(g) + O₂(g) gives 2NO₂(g), rate = k[NO]²[O₂]. Give the units of k and state the effect on the rate of doubling [NO] while halving [O₂].

Show answer

Units: mol⁻² dm⁶ s⁻¹, since the reaction is third order overall. Doubling [NO] multiplies the rate by 2² = 4, and halving [O₂] multiplies it by 1/2, so the rate doubles overall.

3. Harder: For 2NO₂(g) + F₂(g) gives 2NO₂F(g), the rate equation is rate = k[NO₂][F₂]. A proposed mechanism is step 1: NO₂ + F₂ gives NO₂F + F (slow), step 2: NO₂ + F gives NO₂F (fast). Show that the mechanism is consistent with the rate equation and identify the intermediate.

Show answer

Adding the steps gives 2NO₂ + F₂ gives 2NO₂F, after cancelling F, so the steps add up to the overall equation. The slow step involves one NO₂ and one F₂, which matches rate = k[NO₂][F₂]. The second NO₂ appears only in the fast step after the rate-determining step, so it does not appear in the rate equation. F is the intermediate, since it is formed in step 1 and used up in step 2.

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