Rate equations, orders and rate constantsAQA A-Level Chemistry: Revision notes
Section 1
Rate of reaction and the rate equation
The rate of reaction is the change in concentration of a reactant or product per unit time, with units mol dm⁻³ s⁻¹.
The rate is related to the concentrations of the reactants by a rate equation. For a reaction of A and B:
rate = k[A]ᵐ[B]ⁿ
where [A] and [B] are concentrations in mol dm⁻³, m and n are the orders with respect to A and B, and k is the rate constant. The orders are found by experiment and cannot be deduced from the balanced equation.
Reading the orders from the coefficients in the balanced equation. The orders must come from experimental data.
Section 2
Order of reaction
The order of reaction with respect to a reactant is the power to which the concentration of that reactant is raised in the rate equation. At AS and A Level the orders are 0, 1 or 2. The overall order is the sum of the individual orders.
- Zero order: the rate is unaffected by changing the concentration. The reactant does not appear in the rate equation.
- First order: the rate is proportional to concentration. Doubling [A] doubles the rate.
- Second order: the rate is proportional to the concentration squared. Doubling [A] quadruples the rate; tripling [A] multiplies the rate by 9.
For rate = k[A][B]², the order is 1 in A, 2 in B and 3 overall.
Section 3
Rate constant and its units
The rate constant, k, is the constant of proportionality in the rate equation. It has a fixed value for a given reaction at a given temperature. Raising the temperature increases k (and a catalyst also changes it), but changing concentrations does not.
The units of k depend on the overall order and are found by rearranging the rate equation, with rate in mol dm⁻³ s⁻¹ and concentrations in mol dm⁻³:
- 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
Work out the units by substituting units into k = rate / ([A]ᵐ[B]ⁿ) and cancelling. Do not rely on memory.
Section 4
Deducing orders from initial rate data
In an initial rates experiment the concentration of one reactant is changed while the others are kept constant, and the effect on the initial rate is compared.
Worked example. Experiment 1: [X] = 0.10, [Y] = 0.20, rate = 3.0 × 10⁻⁴. Experiment 2: [X] = 0.20, [Y] = 0.20, rate = 6.0 × 10⁻⁴. Experiment 3: [X] = 0.20, [Y] = 0.40, rate = 6.0 × 10⁻⁴ (rate in mol dm⁻³ s⁻¹).
- Experiments 1 and 2: [X] doubles, rate doubles, so X is first order.
- Experiments 2 and 3: [Y] doubles, rate unchanged, so Y is zero order.
The rate equation is rate = k[X].
Compare only two experiments where one concentration changes and the others stay constant. Say which experiments you compared.
Section 5
Calculations with the rate equation
Once the rate equation is known, k and the rate can be calculated.
Finding k. Use the data from one experiment. For the worked example above: k = rate / [X] = 3.0 × 10⁻⁴ / 0.10 = 3.0 × 10⁻³ s⁻¹.
Finding a rate. Substitute k and the concentrations. If [X] = 0.50 mol dm⁻³, rate = 3.0 × 10⁻³ × 0.50 = 1.5 × 10⁻³ mol dm⁻³ s⁻¹.
Finding a concentration. Rearrange the equation, for example [X] = rate / k.
Always raise each concentration to its order (square for second order), and give units for k.
Forgetting to square the concentration of a second-order reactant, or omitting the units of k.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Rate equations, orders and rate constants
- A student investigates the reaction between substances A and B at constant temperature. The reaction has the rate equation rate = k[A][B]², where square brackets show concentrations in mol dm⁻³.State what is meant by the statement that the reaction is first order with respect to A.2 marks
- The initial rate of the reaction between substances P and Q was measured at constant temperature in three experiments. Experiment 1: [P] = 0.10 mol dm⁻³, [Q] = 0.10 mol dm⁻³, initial rate = 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹. Experiment 2: [P] = 0.20 mol dm⁻³, [Q] = 0.10 mol dm⁻³, initial rate = 8.0 × 10⁻⁴ mol dm⁻³ s⁻¹. Experiment 3: [P] = 0.20 mol dm⁻³, [Q] = 0.20 mol dm⁻³, initial rate = 1.6 × 10⁻³ mol dm⁻³ s⁻¹.Define the term rate constant, k, and state one factor that changes its value.2 marks
- In acid solution, propanone reacts with iodine: CH₃COCH₃ + I₂ → CH₃COCH₂I + HI. The initial rate was measured in four experiments, with the concentrations of propanone, H⁺ and I₂ in mol dm⁻³. Experiment 1: 1.00, 0.50, 0.0050, rate = 1.35 × 10⁻⁵ mol dm⁻³ s⁻¹. Experiment 2: 2.00, 0.50, 0.0050, rate = 2.70 × 10⁻⁵. Experiment 3: 1.00, 1.00, 0.0050, rate = 2.70 × 10⁻⁵. Experiment 4: 1.00, 0.50, 0.0100, rate = 1.35 × 10⁻⁵.Deduce the order of reaction with respect to each of propanone, H⁺ and I₂. Give your reasoning.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).