Rate equations and orders of reactionEdexcel 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 equation links the rate to the concentrations of the reactants: rate = k[A]ᵐ[B]ⁿ. It can only be found by experiment and cannot be read from the stoichiometric equation.
The order with respect to a substance is the power to which its concentration is raised in the rate equation. The overall order is the sum of the individual orders (m + n). The rate constant, k, is the constant of proportionality. It is constant at a given temperature and increases when the temperature rises.
Do not take the orders from the coefficients in the balanced equation. The orders are only found from experimental data.
Section 2
Zero, first and second order
At Edexcel A Level the orders met are 0, 1 and 2.
- Zero order: changing the concentration has no effect on the rate. The substance does not appear in the rate equation.
- First order: doubling the concentration doubles the rate (rate ∝ [A]).
- Second order: doubling the concentration quadruples the rate (rate ∝ [A]²), and trebling it increases the rate ninefold.
For example, rate = k[NO]²[H₂] is second order in NO, first order in H₂ and third order overall.
Describe an order with numbers: say 'doubling [A] quadruples the rate, so second order', rather than only naming the order.
Section 3
Units of the rate constant
Rearrange the rate equation to make k the subject, then substitute units. All concentrations are in mol dm⁻³ and the rate is in mol dm⁻³ s⁻¹.
- Overall zero order: k has units mol dm⁻³ s⁻¹
- Overall first order: k has units s⁻¹
- Overall second order: k has units dm³ mol⁻¹ s⁻¹
- Overall third order: k has units dm⁶ mol⁻² s⁻¹
Worked example. For rate = k[NO]²[H₂], k = 2.0 × 10⁻⁷ ÷ (0.010² × 0.010) = 0.20. The units are mol dm⁻³ s⁻¹ ÷ (mol³ dm⁻⁹) = dm⁶ mol⁻² s⁻¹.
Quoting k without units, or with the units for a different overall order, loses the mark.
Section 4
Deducing orders from initial-rate data
In an initial-rate experiment the concentration of one reactant is changed while the others are kept constant, and the rate is measured at the start of the reaction.
- Choose two experiments in which only one concentration changes.
- Compare the factor by which the concentration changes with the factor by which the rate changes.
- Rate × 1 for concentration × 2 means zero order; rate × 2 means first order; rate × 4 means second order.
- Write the rate equation, then substitute any experiment to calculate k.
If two concentrations change together, multiply the effects: doubling both reactants while the rate rises eightfold fits first order in one and second order in the other.
Section 5
Concentration–time graphs and half-life
On a concentration–time graph the gradient at any point is the rate of reaction, found by drawing a tangent.
- Zero order: a straight line with a negative gradient, so the rate is constant.
- First order: an exponential decay curve. The half-life, t½, the time for the concentration of a reactant to halve, is constant.
- Second order: a curve that falls steeply and then levels off, with successive half-lives that double each time.
For a first-order reaction t½ = ln 2 ÷ k, so k = 0.693 ÷ t½. For example, a half-life of 120 s gives k = 5.78 × 10⁻³ s⁻¹. After n half-lives the concentration is the original divided by 2ⁿ.
A constant half-life, measured at different starting concentrations, is the quickest evidence of first order.
Section 6
Rate–concentration graphs
A rate–concentration graph plots the rate (found from tangents) against the concentration of one reactant.
- Zero order: a horizontal line, as the rate does not depend on concentration.
- First order: a straight line through the origin; the gradient is k.
- Second order: a curve with increasing gradient; a plot of rate against [A]² is a straight line through the origin.
Section 7
Key terms linked to rate equations
The rate-determining step is the slowest step in a reaction mechanism, and it controls the overall rate. Activation energy is the minimum energy colliding particles need for a reaction to occur. A homogeneous catalyst is in the same physical state as the reactants; a heterogeneous catalyst is in a different state, such as a solid metal surface with gaseous reactants. Catalysts provide an alternative route with a lower activation energy and are not used up.
A zero-order dependence on a gas on a metal surface occurs because the surface is saturated with the reactant, so the rate depends on the number of active sites.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Rate equations and orders of reaction
- The reaction between peroxodisulfate ions and iodide ions is S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq). Initial rates were measured 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⁻] = 0.040 mol dm⁻³, [I⁻] = 0.080 mol dm⁻³, initial rate = 9.6 × 10⁻⁶ mol dm⁻³ s⁻¹.Calculate the value of the rate constant, k, for this reaction, including its units.2 marks
- Dinitrogen pentoxide, N₂O₅, decomposes in solution at constant temperature. The initial concentration of N₂O₅ is 0.80 mol dm⁻³. It falls to 0.40 mol dm⁻³ after 120 s and to 0.20 mol dm⁻³ after a further 120 s. The rate constant for the reaction at this temperature is 5.78 × 10⁻³ s⁻¹.Calculate the initial rate of reaction, including units.2 marks
- Nitrogen monoxide reacts with hydrogen: 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g). Initial rates were measured at constant temperature. Experiment 1: [NO] = 0.010 mol dm⁻³, [H₂] = 0.010 mol dm⁻³, initial rate = 2.0 × 10⁻⁷ mol dm⁻³ s⁻¹. Experiment 2: [NO] = 0.020 mol dm⁻³, [H₂] = 0.010 mol dm⁻³, initial rate = 8.0 × 10⁻⁷ mol dm⁻³ s⁻¹. Experiment 3: [NO] = 0.010 mol dm⁻³, [H₂] = 0.030 mol dm⁻³, initial rate = 6.0 × 10⁻⁷ mol dm⁻³ s⁻¹.Use the data to deduce the order of reaction with respect to NO and with respect to H₂, and write the rate equation.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).