A-Level Chemistry kinetics: rate equations, orders, half-life and mechanisms

Where kinetics sits in each specification

The key ideas

Particles react only when they collide with energy at least equal to the activation energy and with the right orientation. Raising the temperature shifts the Boltzmann distribution so a much larger fraction of molecules has energy above Ea, which is why rate rises sharply with temperature. A catalyst provides an alternative route with a lower activation energy.

The rate equation, rate = k[A]^m[B]^n, is found by experiment, not from the balanced equation. Order 0 means the concentration has no effect on rate; order 1 means rate is proportional to concentration; order 2 means rate is proportional to concentration squared. On a concentration-time graph, a zero-order reaction is a straight line, and a first-order reaction has a constant half-life. On a rate-concentration graph, zero order is horizontal, first order is a straight line through the origin, and second order is a curve.

The species in the rate equation are the ones involved in or before the rate-determining step, the slowest step of the mechanism. For example, the hydrolysis of 2-bromo-2-methylpropane has rate = k[(CH3)3CBr], showing the slow step involves only the halogenoalkane: evidence for an SN1 mechanism. For Edexcel, the Arrhenius equation k = Ae^(-Ea/RT) gives ln k = ln A - Ea/(RT), so a graph of ln k against 1/T has gradient -Ea/R.

Worked example 1: find the rate equation and k from initial rates

1. Experiment 1: [A] = 0.10, [B] = 0.10 mol dm^-3, rate = 2.0 × 10^-4 mol dm^-3 s^-1. Experiment 2: [A] = 0.20, [B] = 0.10, rate = 8.0 × 10^-4. Experiment 3: [A] = 0.10, [B] = 0.20, rate = 4.0 × 10^-4.
2. Experiments 1 and 2: [A] doubles with [B] constant and the rate goes up four times, so the reaction is second order in A.
3. Experiments 1 and 3: [B] doubles with [A] constant and the rate doubles, so it is first order in B.
4. Rate equation: rate = k[A]^2[B], overall third order.
5. k = rate ÷ ([A]^2[B]) = 2.0 × 10^-4 ÷ (0.10^2 × 0.10) = 0.20 dm^6 mol^-2 s^-1.

Worked example 2: half-life of a first-order reaction

1. A concentration-time graph shows the concentration falls from 0.80 to 0.40 mol dm^-3 in 35 s, and from 0.40 to 0.20 in another 35 s.
2. The half-life is constant at 35 s, so the reaction is first order.
3. k = 0.693 ÷ t½ = 0.693 ÷ 35 = 0.0198 s^-1.
4. After three half-lives (105 s) the concentration is 0.80 ÷ 2^3 = 0.10 mol dm^-3.

Worked example 3 (Edexcel IAL): activation energy from an Arrhenius plot

1. A graph of ln k against 1/T is a straight line with gradient -6000 K.
2. From ln k = ln A - Ea/(RT), the gradient equals -Ea/R.
3. Ea = 6000 × 8.31 = 49 900 J mol^-1, so Ea ≈ 49.9 kJ mol^-1.

Common mistakes that cost marks

• Taking orders from the coefficients in the balanced equation.
• Getting the units of k wrong. Work them out from the rate equation each time.
• Saying higher temperature increases rate 'because there are more collisions' without the key point: more collisions have energy above Ea.
• Drawing the Boltzmann curve touching the energy axis at high energy, or starting it away from the origin.
• Saying a catalyst lowers the activation energy of the same reaction; it provides an alternative route with lower Ea.
• Saying the rate-determining step is the fastest step.

Exam technique and how a tutor helps

For initial-rate tables, write a sentence for each order: 'When [A] doubles at constant [B], the rate quadruples, so the reaction is second order with respect to A.' If both concentrations change between two experiments, deal with the known order first. Show the units of k worked out, not just stated.

In mechanism questions, check that the steps add up to the overall equation, that the slow step contains the species in the rate equation, and that any intermediate is used up later.

In one-to-one lessons a tutor works through data sets from past papers with the student, including the awkward ones where two concentrations change together, and checks each unit and each order statement. For Edexcel students, the tutor also goes through Core Practicals 9 and 10 so the student can answer questions about the method and its errors.

Self-check: can you do these?

• What are the units of k for a first-order reaction? (Answer: s^-1)
• What does a horizontal line on a rate-concentration graph show? (Answer: zero order)
• A first-order reaction has k = 0.010 s^-1. Find t½. (Answer: 69.3 s)
• Why does a small rise in temperature cause a large rise in rate? (Answer: many more molecules have energy at least Ea)
• rate = k[(CH3)3CBr]. Is the mechanism SN1 or SN2? (Answer: SN1)