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Rate equations notes and lesson for A Level chemistry (free PDF download)

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Rate equations notes and lesson for A Level chemistry (free PDF download)

This chemistry grades 11-12 lesson on rate equations notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on rate equations and check yourself with the quiz.

Quick answer: A rate equation, rate = k[A]^m[B]^n, shows how concentration controls reaction speed; orders come from experiments, not from the balanced equation. The rate constant k rises with temperature according to the Arrhenius equation, and the slowest step in a mechanism, the rate-determining step, fixes the form of the rate equation.

Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R2.2

Rate equations notes: the key ideas

What is a rate equation?

Definition

A rate equation links the rate of a reaction to the concentrations of the reactants: rate = k[A]m[B]n. The exponents m and n are the orders of reaction with respect to A and B, the overall order is m + n, and k is the rate constant, which depends only on temperature (and catalyst).

Key law

Orders are found only by experiment. They are not the coefficients in the balanced equation. The rate equation of a reaction is also unrelated to its stoichiometry.

Three graphs of rate against concentration: flat for zero order, straight line for first order, curve for second orderZero order[A]rateFirst order[A]rateSecond order[A]rate
Rate against [A]: flat line (zero), straight line through the origin (first), upward curve (second).
Order in ARate against [A]Doubling [A] changes the rate byUnits of k (overall)
0constant× 1mol dm⁻³ s⁻¹
1straight line through origin× 2s⁻¹
2curve upward× 4dm³ mol⁻¹ s⁻¹

How do you find the order from experimental data?

The initial-rates method changes one concentration at a time. If doubling [A] doubles the rate, the order is 1; if it quadruples the rate, the order is 2; if the rate is unchanged, the order is 0. Continuous monitoring gives the order from the shape of a concentration–time graph, as below.

Concentration against time curves: a first order exponential decay with constant half-life and a zero order straight line0601201802400.000.250.500.751.00Time (s)[A] (mol/dm³)first order: half-life 60 s each timezero order: straight line
Concentration against time for a first order and a zero order reaction.
Formula

First order half-life: t½ = ln 2 / k = 0.693 / k
t½ = half-life (s), k = rate constant (s⁻¹). The half-life of a first order reaction is constant; for zero order it falls as the concentration falls.

Worked example

For A + B → products: experiment 1 has [A] = 0.10 and [B] = 0.10 mol/dm³ with rate 2.4 × 10⁻⁴ mol dm⁻³ s⁻¹; experiment 2 has [A] = 0.20 and [B] = 0.10 with rate 4.8 × 10⁻⁴; experiment 3 has [A] = 0.10 and [B] = 0.30 with rate 2.16 × 10⁻³.

Doubling [A] doubles the rate, so the order in A is 1. Tripling [B] multiplies the rate by 9, so the order in B is 2. Rate = k[A][B]². Then k = 2.4 × 10⁻⁴ / (0.10 × 0.10²) = 0.24 dm⁶ mol⁻² s⁻¹.

How does temperature change the rate constant?

Formula

Arrhenius: k = A e−Ea/RT, so ln k = ln A − Ea / (RT)
k = rate constant, A = pre-exponential factor (same units as k), Ea = activation energy (J/mol), R = 8.314 J/(mol K), T = temperature (K). A graph of ln k against 1/T has gradient −Ea/R.

Arrhenius plot: straight line of ln k against 1 over T sloping downwards with gradient equal to minus Ea over R0.00300.00310.00320.00330.0034-8-6-4-21/T (K⁻¹)ln kgradient = −Ea / R
Arrhenius plot: a straight line with negative gradient −Ea/R.
Worked example

A reaction has k = 1.5 × 10⁻³ s⁻¹ at 300 K and 1.2 × 10⁻² s⁻¹ at 320 K. Calculate Ea.

ln(k₂/k₁) = ln 8 = 2.079. 1/300 − 1/320 = 2.083 × 10⁻⁴ K⁻¹. Ea = 8.314 × 2.079 / 2.083 × 10⁻⁴ = 82 980 J/mol = 83.0 kJ/mol.

What is the rate-determining step?

Many reactions occur in several steps. The slowest step controls the overall rate and is the rate-determining step. Species in the rate equation appear in that step or in the equilibria before it. For example, the hydrolysis of (CH₃)₃CBr by hydroxide has rate = k[(CH₃)₃CBr]. The slow step is (CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻, which involves no OH⁻, and the fast step (CH₃)₃C⁺ + OH⁻ → (CH₃)₃COH follows.

Energy profile with two steps: a high first peak for the slow rate-determining step, an intermediate, then a lower second peakEnergyReaction progressstep 1 (slow): highest peakintermediatestep 2 (fast)reactantsproducts
Two-step mechanism: the step with the highest peak is the slow, rate-determining step.
Common mistake

Do not write the rate equation from the balanced equation, and do not give k a fixed unit. The units of k depend on the overall order and must be worked out from the rate equation.

Lina’s tip

To get the units of k, put the units into the rate equation, mol dm⁻³ s⁻¹ = k × (mol dm⁻³)order, and solve for k.

Key points to remember
  • Rate = k[A]^m[B]^n; overall order = m + n; orders are found experimentally.
  • Units of k: zero order mol dm⁻³ s⁻¹, first order s⁻¹, second order dm³ mol⁻¹ s⁻¹.
  • First order half-life t½ = 0.693 / k and is constant.
  • Arrhenius: ln k = ln A − Ea / RT; gradient of ln k against 1/T is −Ea/R.
  • The rate-determining step is the slowest step and sets the species in the rate equation.

Orders, rate constants and activation energy: frequently asked questions

What is the order of reaction?

The order with respect to a reactant is the power to which its concentration is raised in the rate equation. It tells you how the rate changes when that concentration changes and is found from experiments.

How do you find the units of the rate constant?

Substitute the units of rate and of each concentration into the rate equation and rearrange for k. For a first order reaction the units are s⁻¹, and for a second order reaction dm³ mol⁻¹ s⁻¹.

What does the half-life tell you?

For a first order reaction the half-life is constant and equals 0.693 divided by k. If the half-life changes as the reaction proceeds, the reaction is not first order.

How do you find activation energy from a graph?

Plot ln k against 1/T for several temperatures. The gradient of the straight line equals −Ea/R, so Ea = −gradient × 8.314 J/(mol K).

What is a rate-determining step?

It is the slowest step in a reaction mechanism and controls the overall rate. The reactants that appear in the rate equation take part in this step or in steps before it.

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