Rate equations and orders of reactionEdexcel International 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 of mol dm⁻³ s⁻¹. The rate equation links the rate to the concentrations of the reactants:
rate = k[A]ᵐ[B]ⁿ
The powers m and n are the orders with respect to A and B, and are 0, 1 or 2 at this level. The overall order is m + n. The rate constant, k, is the constant of proportionality. It is fixed at a given temperature and changes only if the temperature changes or a catalyst is used.
The orders can only be found from experiment. They are not the stoichiometric coefficients in the balanced equation.
Orders come from experimental data, not from the balanced equation. 2NO + 2H₂ gives rate = k[NO]²[H₂], not a rate with power 2 on H₂.
Section 2
What an order means
The order with respect to a substance tells you how the rate changes when that concentration changes, with everything else constant:
- Zero order: changing the concentration has no effect on the rate. The substance does not appear in the rate equation.
- First order: the rate is proportional to the concentration. Doubling the concentration doubles the rate.
- Second order: the rate is proportional to the concentration squared. Doubling the concentration makes the rate four times faster.
The units of k depend on the overall order. For overall order 1 they are s⁻¹, for order 2 they are dm³ mol⁻¹ s⁻¹ and for order 3 they are dm⁶ mol⁻² s⁻¹. Work them out by rearranging the rate equation and substituting units.
To find the units of k, write rate = k × (concentrations) with units, then divide. For rate = k[A][B]: k = mol dm⁻³ s⁻¹ ÷ mol² dm⁻⁶ = dm³ mol⁻¹ s⁻¹.
Section 3
Orders from concentration–time and rate–concentration graphs
A concentration–time graph shows how a concentration falls over time.
- Zero order: a straight line with constant negative gradient, so the rate is constant.
- First order: a curve with a constant half-life. The time for the concentration to halve is the same wherever you start.
- Second order: a curve with half-lives that double each time.
A rate–concentration graph is made by measuring gradients (rates) from tangents on the concentration–time curve.
- Zero order: a horizontal line.
- First order: a straight line through the origin.
- Second order: a curve whose gradient increases with concentration (a straight line only if rate is plotted against concentration squared).
A tangent's gradient at a given concentration gives the rate at that concentration. Draw it carefully with a ruler and use a large triangle.
Section 4
Orders from initial-rate data
The initial rate is the rate at the start of the reaction, before the concentrations have changed much. In a series of experiments, change the concentration of one reactant at a time and keep the others and the temperature constant.
- If doubling the concentration has no effect on the rate, the order is 0.
- If it doubles the rate, the order is 1.
- If it makes the rate four times faster, the order is 2.
Worked example. Rate = 2.4 × 10⁻⁶ mol dm⁻³ s⁻¹ when [A] = 0.020 and [B] = 0.040 mol dm⁻³. The rate doubles when [A] doubles and doubles when [B] doubles, so rate = k[A][B]. Then k = 2.4 × 10⁻⁶ ÷ (0.020 × 0.040) = 3.0 × 10⁻³ dm³ mol⁻¹ s⁻¹.
Section 5
Rate-determining step, activation energy and catalysts
The rate-determining step is the slowest step in a reaction mechanism. It controls the overall rate. Reactants that appear in the rate equation are in the rate-determining step or in a fast step before it.
The activation energy is the minimum energy colliding particles need for a reaction to occur. A catalyst speeds up a reaction by providing an alternative route with a lower activation energy. It is not used up.
A homogeneous catalyst is in the same phase as the reactants (for example aqueous H⁺ or I⁻ in solution). A heterogeneous catalyst is in a different phase (for example solid iron in a gas-phase reaction) and the reaction takes place on its surface.
A catalyst can appear in the rate equation, because it can take part in the rate-determining step, even though it is not in the overall equation.
Must Know
- rate = k[A]ᵐ[B]ⁿ, with orders found from experiment
- Overall order is the sum of the orders; k has units that depend on it
- Constant half-life means first order
- Rate–concentration graph: horizontal means zero order, straight line through the origin means first order
- Initial-rate data: change one concentration at a time and compare
- The rate-determining step is the slowest step; a catalyst gives a lower activation energy route
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Rate equations and orders of reaction
- A technician studies the gas-phase reaction 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) at constant temperature. Experiments show that the rate equation is rate = k[NO]²[H₂].The concentration of NO is doubled and the concentration of H₂ is halved, with the temperature unchanged. Deduce the factor by which the rate changes.2 marks
- Peroxodisulfate ions oxidise iodide ions: S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq). A student measures the initial rate of this reaction in three experiments at constant temperature. Experiment 1: [S₂O₈²⁻] = 0.020 mol dm⁻³, [I⁻] = 0.040 mol dm⁻³, initial rate = 2.4 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 2: [S₂O₈²⁻] = 0.040 mol dm⁻³, [I⁻] = 0.040 mol dm⁻³, initial rate = 4.8 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 3: [S₂O₈²⁻] = 0.020 mol dm⁻³, [I⁻] = 0.080 mol dm⁻³, initial rate = 4.8 × 10⁻⁶ mol dm⁻³ s⁻¹.Calculate the initial rate of the reaction at the same temperature when [S₂O₈²⁻] = 0.050 mol dm⁻³ and [I⁻] = 0.060 mol dm⁻³.2 marks
- A reactant R decomposes in solution at constant temperature. Its concentration is 0.800 mol dm⁻³ at the start, 0.400 mol dm⁻³ after 120 s, 0.200 mol dm⁻³ after 240 s and 0.100 mol dm⁻³ after 360 s.Deduce the order of reaction with respect to R and justify your answer.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).