Determining rate equations experimentallyAQA A-Level Chemistry: Revision notes
Section 1
The rate equation is found by experiment
The rate equation links the rate of a reaction to the concentrations of reactants: rate = k[A]ᵐ[B]ⁿ. The powers m and n are the orders of reaction with respect to A and B, and they can only be found by experiment. They are not the stoichiometric coefficients in the balanced equation.
- Zero order: changing the concentration has no effect on the rate, so the reactant does not appear in the rate equation.
- First order: rate is proportional to the concentration, so doubling the concentration doubles the rate.
- Second order: rate is proportional to concentration squared, so doubling the concentration quadruples the rate.
The overall order is the sum of the individual orders. k is the rate constant; it is constant only at a fixed temperature, and it is larger for a faster reaction. Rate is measured in mol dm⁻³ s⁻¹.
Orders cannot be read off the balanced equation. A reactant with a coefficient of 2 may be zero, first or second order.
Section 2
Rate from concentration–time graphs
A concentration–time graph is a curve for most reactions. The rate at any time is the magnitude of the gradient of the tangent drawn to the curve at that time. For a reactant the gradient is negative, so the rate is the gradient with the sign removed.
The initial rate is found from the tangent at t = 0, where the concentrations are known exactly.
Worked example: a tangent at t = 0 passes through (0 s, 0.80 mol dm⁻³) and (200 s, 0.20 mol dm⁻³). Gradient = (0.20 − 0.80) ÷ 200 = −3.0 × 10⁻³, so the initial rate = 3.0 × 10⁻³ mol dm⁻³ s⁻¹.
The shape of the curve also shows the order. A straight line of constant gradient means zero order. A curve whose half-life is constant (the time for the concentration to halve is the same each time) means first order. A half-life that gets longer as the concentration falls suggests second order.
Draw tangents large and carefully, with a ruler, and choose two points on the line far apart to calculate the gradient.
Section 3
Orders from rate–concentration data and graphs
In the initial rate method, the reaction is run several times with different starting concentrations, changing one concentration at a time and measuring the initial rate each time.
How the rate changes when you double one concentration shows the order:
- rate unchanged: zero order
- rate × 2: first order
- rate × 4: second order
A rate–concentration graph shows the same thing. A horizontal line is zero order. A straight line through the origin is first order. A curve with a gradient that increases with concentration is second order (a graph of rate against concentration squared would be straight).
If you change one concentration by a factor other than 2, work out the power: if concentration × 3 gives rate × 9, the order is 2.
Only compare two experiments in which just one concentration has changed. Changing two at once makes the order impossible to deduce.
Section 4
Deriving the rate equation and calculating k
- Find the order with respect to each reactant by comparing experiments.
- Write the rate equation, leaving out any zero order reactants.
- Substitute the data from one experiment and rearrange to get k.
- Work out the units by substituting units into the rearranged equation.
Worked example: 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, [B] = 0.10, rate = 8.0 × 10⁻⁴. Experiment 3: [A] = 0.10, [B] = 0.20, rate = 2.0 × 10⁻⁴.
A doubles and rate × 4, so A is second order. B doubles and the rate is unchanged, so B is zero order. Rate = k[A]². k = 2.0 × 10⁻⁴ ÷ (0.10)² = 2.0 × 10⁻² dm³ mol⁻¹ s⁻¹.
Units of k: overall order 0: mol dm⁻³ s⁻¹; order 1: s⁻¹; order 2: dm³ mol⁻¹ s⁻¹; order 3: dm⁶ mol⁻² s⁻¹.
Always quote units with k. An exam answer for k with no units, or the wrong units, usually loses a mark.
Section 5
Required practical 7: measuring rates
Initial rate method: run the reaction with different starting concentrations and measure the time for a fixed, easily seen change. In the clock reaction (for example sodium thiosulfate with acid, timing the disappearance of a cross, or an iodine clock timing the appearance of the blue-black starch colour) the rate is proportional to 1/time. Keep the total volume, temperature and all other concentrations constant by adding water.
Continuous monitoring method: follow one reaction as it proceeds and plot concentration (or a quantity proportional to it) against time. Methods include a colorimeter (a coloured species, after calibration), a gas syringe or mass loss (a gas is released), conductivity, or titration of samples. In the titration method, samples are removed at intervals and the reaction is stopped (quenched), for example by adding excess sodium hydrogencarbonate to remove acid or by cooling in ice.
Use a thermostatic water bath so that k does not change during the run, and a large excess of any reagent whose concentration should stay effectively constant.
The time for a clock reaction gives an average rate over that period. It only approximates the initial rate if the change is small, and 1/time is proportional to it, not equal to it.
Must know
- The rate equation is found by experiment, not from the balanced equation
- Rate = magnitude of the gradient of the tangent on a concentration–time graph; the initial rate is the tangent at t = 0
- Double a concentration: rate unchanged is zero order, rate × 2 is first order, rate × 4 is second order
- Rate–concentration graphs: horizontal is zero order, straight through the origin is first order, upward curve is second order
- Constant half-life on a concentration–time curve means first order
- k = rate ÷ concentration terms, with units that depend on the overall order
- Practical: change one variable at a time, keep temperature constant, use 1/t for clock reactions, quench samples when titrating
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Determining rate equations experimentally
- The gas-phase reaction 2NO + 2H₂ → N₂ + 2H₂O was studied at constant temperature. Experiment 1: [NO] 0.0020 mol dm⁻³, [H₂] 0.0040 mol dm⁻³, initial rate 8.0 × 10⁻⁷ mol dm⁻³ s⁻¹. Experiment 2: [NO] 0.0040, [H₂] 0.0040 (both mol dm⁻³), initial rate 3.2 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 3: [NO] 0.0020, [H₂] 0.0080 (both mol dm⁻³), initial rate 1.6 × 10⁻⁶ mol dm⁻³ s⁻¹.Calculate the value of the rate constant, k, using Experiment 1, and state its units.2 marks
- A student studies the decomposition of hydrogen peroxide, 2H₂O₂ → 2H₂O + O₂, using a catalyst, at constant temperature. She works out the concentration of H₂O₂ remaining at intervals: 0.800 mol dm⁻³ at 0 s, 0.400 mol dm⁻³ at 150 s, 0.200 mol dm⁻³ at 300 s and 0.100 mol dm⁻³ at 450 s. A tangent drawn to her smooth concentration–time curve at t = 0 passes through the point (0 s, 0.800 mol dm⁻³) and meets the time axis at 215 s.The student states that the reaction is first order with respect to H₂O₂. Use the concentration data to justify her conclusion.2 marks
- In the reaction S₂O₃²⁻(aq) + 2H⁺(aq) → S(s) + SO₂(g) + H₂O(l), a student mixes sodium thiosulfate solution, hydrochloric acid and water in a flask placed over a cross marked on paper. She records the time for the cross to disappear when viewed from above because of the sulfur formed. In three runs the total volume and the hydrochloric acid concentration are identical. The thiosulfate concentrations are 0.100, 0.050 and 0.025 mol dm⁻³ and the times are 25 s, 50 s and 100 s respectively.Describe how the student should vary the concentration of thiosulfate ions so that the experiment is a fair test, and how the results can be used to show the order with respect to thiosulfate.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).