Uni-Sine
ChemistryRates and equilibrium

The rate equation

A rate equation shows how the rate of a reaction depends on the concentrations of the reactants: rate = k[A]ᵐ[B]ⁿ. The square brackets mean concentration 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, not from the balanced equation. If rate = k[A][B]², doubling [A] doubles the rate, and doubling [B] makes it four times faster.

The rate equation

\text{rate} = k[\text{A}]^m[\text{B}]^nThe general form of a rate equation
TermMeaning
rateHow fast the reaction goes, in mol dm⁻³ s⁻¹
[A], [B]Concentrations of the reactants, in mol dm⁻³
m, nOrders of reaction with respect to A and B
m + nOverall order of the reaction
kRate constant, which changes with temperature

AQA's A-level specification gives the rate equation in this form, restricts the orders to 0, 1 and 2, and says that the rate equation is an experimentally determined relationship. Only reactants that affect the rate appear in it, and its orders do not have to match the numbers in the balanced equation.

Orders of reaction

Order with respect to ADouble [A] and the rate...On a rate against [A] graph
0stays the samea horizontal line
1doublesa straight line through the origin
2goes up 4 timesa curve rising more and more steeply

A reactant that is zero order does not appear in the rate equation at all, because [A]⁰ = 1. More generally, multiplying a concentration by x multiplies the rate by xⁿ, where n is the order.

y = 2{ x}^{2}
0.20.40.60.811.21.41.61.822468[A] = 0.5: rate 0.5double: rate × 4
Drag to pan, scroll to zoom. x from 0 to 2, y from -1.6 to 9.6.

Here x is the concentration of the reactant in mol dm⁻³ and y is the rate in arbitrary units, for rate = 2[A]². From 0.5 to 1.0 mol dm⁻³ the rate rises from 0.5 to 2: four times as much.

Finding orders by the initial rates method

AQA expects students to deduce orders from rate-concentration data. Run the reaction several times, changing one concentration at a time, and measure the initial rate each time.

Example 1: deducing a rate equation

Experiment[A] / mol dm⁻³[B] / mol dm⁻³Initial rate / mol dm⁻³ s⁻¹
10.0500.203.0 × 10⁻⁵
20.100.206.0 × 10⁻⁵
30.0500.401.2 × 10⁻⁴
  • From 1 to 2, [A] doubles with [B] fixed, and the rate doubles: first order in A.
  • From 1 to 3, [B] doubles with [A] fixed, and the rate goes up 4 times: second order in B.

So the rate equation is rate = k[A][B]², and the overall order is 1 + 2 = 3.

Example 2: calculating k and its units

Rearrange the rate equation and use Experiment 1:

k = \frac{\text{rate}}{[\text{A}][\text{B}]^2} = \frac{3.0 \times 10^{-5}}{0.050 \times 0.20^2} = 0.015

For the units, cancel: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ = mol⁻² dm⁶ s⁻¹. So k = 0.015 mol⁻² dm⁶ s⁻¹. AQA's specification asks students to use given rate data to deduce a rate equation, then calculate the rate constant including its units.

Example 3: using the rate equation

For the same reaction at the same temperature, find the rate when [A] = 0.10 mol dm⁻³ and [B] = 0.30 mol dm⁻³.

\text{rate} = 0.015 \times 0.10 \times 0.30^2 = 1.35 \times 10^{-4}\ \text{mol dm}^{-3}\ \text{s}^{-1}

Now try one in the course. Use the rate equation, then press Mark to check it there.

Use rate = k[A] or k[A] squared to work out a rate or the rate constant.

Press Mark to have your answer marked in the Reaction Kinetics course.

Units of the rate constant

The units of k depend on the overall order. Work them out by rearranging rate = k[ ]ⁿ and cancelling:

Overall orderUnits of k
0mol dm⁻³ s⁻¹
1s⁻¹
2mol⁻¹ dm³ s⁻¹
3mol⁻² dm⁶ s⁻¹

Concentration-time graphs

You can also find orders from how a concentration falls with time:

  • Zero order: the concentration falls in a straight line. AQA mentions finding the rate constant of a zero-order reaction from the gradient of a concentration-time graph.
  • First order: a curve with a constant half-life: the concentration halves in the same time, whatever it starts at.
  • Any order: drawing a tangent to the curve gives the rate at that moment, as in our rate of reaction formula guide.

The rate equation and the rate-determining step

AQA's specification says the orders can provide information about the mechanism, including the rate-determining (slowest) step. A reaction often happens in several steps, and the slowest one controls the overall rate. The species in the rate equation, with their orders, are the ones involved in or before that slow step.

For rate = k[A][B]², a likely slow step involves one A and two B. A reactant that is zero order is only involved after the slow step, so changing its concentration does not change the rate.

Temperature and the rate constant

The rate equation holds at one temperature. Raising the temperature does not change the orders, but it makes k bigger, so the reaction goes faster at the same concentrations. The link between k and temperature is the Arrhenius equation, k = Ae−Ea/RT. The Rates and Equilibrium Calculator finds orders from initial-rate data, works with the rate equation, and handles the Arrhenius equation too.

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Common mistakes

  • Taking orders from the balanced equation. Orders come only from experiments.
  • Changing two concentrations at once when comparing experiments. Compare two runs where only one concentration changes.
  • Wrong units for k. Cancel the units properly; they depend on the overall order.
  • Forgetting zero-order reactants. They are left out of the rate equation, but they still take part in the reaction.
  • Thinking k changes with concentration. k only changes with temperature (or a catalyst).

Rate equation questions with answers

1. rate = k[X]²[Y]. What is the overall order?
2 + 1 = 3
2. For rate = k[X]²[Y], what happens to the rate if [X] is tripled?
3² = 9 times faster
3. Give the units of k for a second-order reaction.
mol⁻¹ dm³ s⁻¹
4. For rate = k[P] with k = 5.0 × 10⁻³ s⁻¹, find the rate when [P] = 0.40 mol dm⁻³.
rate = 5.0 × 10⁻³ × 0.40 = 2.0 × 10⁻³ mol dm⁻³ s⁻¹
5. Doubling [Z] has no effect on the rate. What is the order with respect to Z, and does Z appear in the rate equation?
Zero order, so Z does not appear in the rate equation.

Practise kinetics in a short timed paper

A-level Chemistry: Year 2 gives you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so rate equations turn up regularly but not in every paper. The Reaction Kinetics course has sections on rate equations and finding orders from data.

Paper resets in

Common questions

What is a rate equation?
An equation linking the rate of a reaction to the concentrations of the reactants: rate = k[A]ᵐ[B]ⁿ.
How do you find the order of a reaction?
By experiment: change one reactant's concentration and see how the initial rate changes. Rate doubles when concentration doubles means first order; × 4 means second order; no change means zero order.
Can you get the rate equation from the balanced equation?
No. The orders must be found by experiment, and they often differ from the numbers in the balanced equation.
What affects the rate constant?
Temperature, and a catalyst. Concentration does not change k.
What are the units of k?
It depends on the overall order: s⁻¹ for first order, mol⁻¹ dm³ s⁻¹ for second order, and so on.
What is the rate-determining step?
The slowest step in a reaction mechanism. The species in the rate equation are involved in or before it.

Revision cards

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Where to go next

  • Solve it

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  • Learn it

    Reaction Kinetics

    Rates, catalysts and collision theory: what makes reactions fast or slow

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    A-level Chemistry: Year 2

    A new exam-style paper every day, with instant marking and a mark scheme

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