Edexcel A-level Chemistry (9CH0) · Topic 16: Kinetics II
Mini-Lesson
Kinetics II
This mini-lesson covers the whole of Edexcel Topic 16: the rate equation and orders of reaction; finding orders from initial-rate data and from concentration–time graphs; the constant half-life of a first-order reaction; the rate constant k and its units; the rate-determining step and what it tells you about the mechanism; and the Arrhenius equation.
Work through each screen, answer the questions as you go (most are calculations) and collect ⭐ stars. Every calculation is worked through for you first. Press Start when you're ready.
The rate equation
rate = k[A]m[B]n
The rate of a reaction is the change in concentration of a reactant or product per unit time — units mol dm⁻³ s⁻¹. How it depends on concentration is given by the rate equation:
rate = k[A]m[B]nk = the rate constant · m = order with respect to A · n = order with respect to B · m + n = the overall order
The order with respect to a reactant is the power to which its concentration is raised in the rate equation.
The overall order is the sum of the individual orders.
The orders can only be found by experiment. They are not the balancing numbers in the equation — a reactant with a coefficient of 2 can perfectly well be first order, or even zero order.
A reactant that is zero order does not appear in the rate equation at all, because [A]⁰ = 1. Changing its concentration has no effect on the rate.
k depends only on temperature (and on whether a catalyst is present). Changing a concentration changes the rate, never k.
Quick check
What does "second order in A" mean?
1A reaction is second order with respect to A. What happens to the rate if [A] is tripled and everything else is kept the same?
Initial rates
Finding orders from initial-rate data
The initial rate is the rate at t = 0, found from the gradient of the tangent to a concentration–time graph at the start. Run the reaction several times, changing one concentration at a time, and compare:
Experiment
[A] / mol dm⁻³
[B] / mol dm⁻³
Initial rate / mol dm⁻³ s⁻¹
1
0.100
0.100
2.0 × 10⁻³
2
0.200
0.100
8.0 × 10⁻³
3
0.100
0.200
4.0 × 10⁻³
Experiments 1 → 2: [B] is constant. [A] is doubled and the rate goes ×4. Since 2m = 4, m = 2.
Experiments 1 → 3: [A] is constant. [B] is doubled and the rate goes ×2. Since 2n = 2, n = 1.
Golden rule: only compare two experiments in which exactly one concentration has changed. If ×2 gives ×1 the order is 0; ×2 → the order is 1; ×4 → 2; ×8 → 3.
Calculate
Your turn — order with respect to A
2From the table (experiments 1 and 2 above), state the order with respect to A.
order
Hint: [B] is held at 0.100. [A] goes 0.100 → 0.200 (×2) and the rate goes 2.0 × 10⁻³ → 8.0 × 10⁻³ (×4). Solve 2m = 4.
Calculate
Your turn — order with respect to B
3From the table (experiments 1 and 3), state the order with respect to B.
order
Hint: [A] is held at 0.100. [B] goes 0.100 → 0.200 (×2) and the rate goes 2.0 × 10⁻³ → 4.0 × 10⁻³ (×2). Solve 2n = 2.
Calculate
Your turn — overall order
4The rate equation is therefore rate = k[A]²[B]. State the overall order of the reaction.
order
Hint: the overall order is the sum of the individual orders: 2 + 1.
The rate constant
Finding k — and working out its units
Once you have the rate equation, substitute any one experiment's data and rearrange for k. Then derive the units from scratch every time:
k = rate ÷ ([A]m[B]n)units of k = (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)overall order
Worked example — units for a second-order reaction
units of k = (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)² = (mol dm⁻³ s⁻¹) ÷ (mol² dm⁻⁶)
= mol1−2 dm−3+6 s⁻¹ = mol⁻¹ dm³ s⁻¹
Overall order
Units of k
0
mol dm⁻³ s⁻¹
1
s⁻¹
2
mol⁻¹ dm³ s⁻¹
3
mol⁻² dm⁶ s⁻¹
Never memorise the table blindly — one slip and you lose the mark. Write the rate units over the concentration units and cancel.
Calculate
Your turn — the value of k
5Using rate = k[A]²[B] and experiment 1 ([A] = 0.100, [B] = 0.100, rate = 2.0 × 10⁻³ mol dm⁻³ s⁻¹), calculate k in mol⁻² dm⁶ s⁻¹.
6A reaction has the rate equation rate = k[X][Y]. What are the units of k?
Match it
Match the overall order to the units of k
Tap an overall order on the left, then the units of k that go with it.
Overall order
Units of k
Calculate
Your turn — predicting a rate
7For the same reaction (rate = k[A]²[B], k = 2.0 mol⁻² dm⁶ s⁻¹), calculate the rate when [A] = 0.300 and [B] = 0.150 mol dm⁻³. Give your answer in mol dm⁻³ s⁻¹.
mol dm⁻³ s⁻¹
Hint: rate = 2.0 × (0.300)² × 0.150 = 2.0 × 0.0900 × 0.150. Square [A] before you multiply.
Graphs
Orders from concentration–time and rate–concentration graphs
The other route to an order is to follow one concentration continuously and look at the shape of the graph.
Zero order: the concentration–time graph is a straight line with a constant negative gradient — the rate never changes until the reactant runs out. The rate–concentration graph is a horizontal line.
First order: the concentration–time graph is an exponential decay with a constant half-life. The rate–concentration graph is a straight line through the origin (rate ∝ [A]), and its gradient is k.
Second order: the concentration–time graph falls steeply then has a long tail (the half-life gets longer each time). The rate–concentration graph is an upward-curving line; plotting rate against [A]² gives a straight line.
First order: each successive half-life takes the same time. That constancy is the diagnostic test.Half-life
The half-life of a first-order reaction
The half-life (t½) is the time taken for the concentration of a reactant to fall to half of its value. For a first-order reaction — and only for a first-order reaction — t½ is independent of concentration, so it is constant throughout.
t½ = ln 2 ÷ k ⟹ k = ln 2 ÷ t½ln 2 = 0.693 · k has units s⁻¹ when t½ is in seconds
Worked example
A first-order decomposition has t½ = 45 s.
k = 0.693 ÷ 45 = 1.54 × 10⁻² s⁻¹
Use the constancy as a test. Read three successive half-lives off a concentration–time graph. If they are equal (within experimental error), the reaction is first order in that reactant.
Calculate
Your turn — k from a half-life
8A first-order reaction has a half-life of 120 s. Calculate the rate constant k, giving your answer in units of 10⁻³ s⁻¹ to 3 significant figures. (ln 2 = 0.693.)
× 10⁻³ s⁻¹
Hint: k = ln 2 ÷ t½ = 0.693 ÷ 120 = 5.78 × 10⁻³ s⁻¹. Type just the number in front.
Calculate
Your turn — counting half-lives
9A different first-order reaction has t½ = 25 s. How long does it take for [A] to fall from 0.80 mol dm⁻³ to 0.10 mol dm⁻³?
s
Hint: count the halvings — 0.80 → 0.40 → 0.20 → 0.10. That is 3 half-lives, each of 25 s.
Sort it
Zero, first or second order?
Tap an experimental observation about reactant X, then tap the order with respect to X that it shows.
0️⃣ Zero order
1️⃣ First order
2️⃣ Second order
Mechanisms
The rate-determining step
Most reactions happen in a sequence of simple steps — the mechanism. The slowest step is the rate-determining step (RDS): it is the bottleneck, so the overall rate can be no faster than it.
Only species involved in the RDS (or in a step before it) appear in the rate equation.
The order with respect to a species tells you how many of its molecules are involved up to and including the RDS.
A species in the balanced equation that is absent from the rate equation (zero order) takes part only in a fast step after the RDS.
A catalyst can appear in the rate equation even though it is not in the overall equation — that proves it is involved at or before the RDS.
Worked example — nucleophilic substitution
(CH₃)₃CBr + OH⁻ → (CH₃)₃COH + Br⁻ has the experimental rate equation rate = k[(CH₃)₃CBr] — it is zero order in OH⁻.
So the slow step must be the C–Br bond breaking alone: (CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻ (slow). The OH⁻ then attacks the carbocation in a fast step. This is the SN1 mechanism — the kinetics are the evidence for it.
Quick check
Which mechanism fits the data?
10The overall equation is A + 2B → C and the experimental rate equation is rate = k[A][B]. Which mechanism is consistent with this?
Quick check
The missing reactant
11Reactant Q appears in the balanced equation but is zero order, so it does not appear in the rate equation. What does that tell you?
Arrhenius
The Arrhenius equation
The rate constant depends on temperature and on the activation energy — and the relationship is exponential, which is why a modest temperature rise has such a dramatic effect on rate.
k = Ae−Ea/RTA = the pre-exponential (frequency) factor · Ea = activation energy in J mol⁻¹ · R = 8.31 J K⁻¹ mol⁻¹ · T = temperature in K
Taking natural logarithms of both sides gives the logarithmic form — the one you plot:
ln k = ln A − Ea ÷ (RT)compare with y = mx + c: plot ln k (y) against 1/T (x)
The gradient of that straight line is −Ea/R, so Ea = −gradient × R. The gradient is negative, so Ea comes out positive.
The y-intercept is ln A.
Watch the units: R is in J K⁻¹ mol⁻¹, so Ea comes out in J mol⁻¹. Divide by 1000 for kJ mol⁻¹.
A catalyst lowers Ea, which makes the exponential term bigger and so increases k — it changes k without changing the temperature.
Calculate
Your turn — Ea from a gradient
12A plot of ln k against 1/T is a straight line of gradient −6.05 × 10³ K. Calculate the activation energy in kJ mol⁻¹ to 3 significant figures. (R = 8.31 J K⁻¹ mol⁻¹.)
kJ mol⁻¹
Hint: gradient = −Ea/R, so Ea = −gradient × R = 6.05 × 10³ × 8.31 = 50 275 J mol⁻¹. Now convert to kJ mol⁻¹.
Calculate
Your turn — a 10 K temperature rise
13For a reaction with Ea = 50.0 kJ mol⁻¹, calculate the factor by which k increases when the temperature rises from 300 K to 310 K. Give your answer to 3 significant figures. (R = 8.31 J K⁻¹ mol⁻¹.)
×
Hint: subtract the two log forms — ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) = (50 000 ÷ 8.31) × (1/300 − 1/310) = 6017 × 1.075 × 10⁻⁴ = 0.647. Now take e0.647.
Quick check
Why does a small temperature rise do so much?
14Your answer above shows that k roughly doubles for a 10 K rise near room temperature, even though the mean molecular kinetic energy rises by only about 3%. What is the best explanation?
Quick check
What a catalyst does to k
15A catalyst is added to a reaction at constant temperature. According to the Arrhenius equation, what happens to k?
Recap
The big ideas to know
Rate equation: rate = k[A]m[B]n; orders m and n come from experiment, never from the balanced equation
Overall order: m + n. Zero order means a reactant does not appear in the rate equation at all
Initial rates: change one concentration at a time — ×2 gives ×1 (order 0), ×2 (order 1), ×4 (order 2), ×8 (order 3)
Units of k: derive them — (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)overall order. Order 1 → s⁻¹; order 2 → mol⁻¹ dm³ s⁻¹; order 3 → mol⁻² dm⁶ s⁻¹
Half-life: constant only for a first-order reaction; t½ = ln 2 ÷ k
Graphs: zero order → straight conc–time line; first order → exponential decay, rate ∝ [A] (gradient = k); second order → rate ∝ [A]²
Rate-determining step: the slowest step; only species in it (or before it) appear in the rate equation, and the order = how many of them are involved
Arrhenius: k = Ae−Ea/RT; ln k = ln A − Ea/(RT); plot ln k vs 1/T → gradient = −Ea/R, intercept = ln A
You've now covered Topic 16: Kinetics II of the Edexcel A-level Chemistry (9CH0) specification. Press Finish to see your score.
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