Eduqas A-level Chemistry (A410QS) · PI3 — Chemical kinetics
Mini-Lesson
PI3 · Chemical kinetics
Eduqas PI3 makes rates quantitative. You will build rate equations from data, find orders and the rate constant k, read half-lives off concentration-time graphs, use the rate equation to deduce the rate-determining step, and calculate activation energies with the Arrhenius equation.
Work through each screen, answer every question as you go — the multiple-choice checks, the calculations and the sorting games — and collect ⭐ stars. Press Start when you are ready.
PI3 · rate equations
The rate equation
Rate is the change in concentration of a reactant or product per unit time (mol dm⁻³ s⁻¹).
rate = k[A]m[B]nm = order with respect to A · n = order with respect to B · overall order = m + n
The single most important idea in this topic: the orders m and n cannot be read off the balanced equation. They must be found experimentally — because the rate equation reflects the mechanism, not the stoichiometry.
Units of k depend on the overall order. Rearrange k = rate ÷ [A]ⁿ and cancel:
Zero order overall: mol dm⁻³ s⁻¹
First order overall: s⁻¹
Second order overall: dm³ mol⁻¹ s⁻¹
Third order overall: dm⁶ mol⁻² s⁻¹
Quick check
Units of k
?A reaction is second order overall. What are the units of the rate constant k?
PI3 · initial rates
Finding orders by the initial-rates method
Change one concentration at a time and see what the rate does. Compare two experiments where everything else is held constant:
Double [X], rate unchanged → zero order (2⁰ = 1)
Double [X], rate doubles → first order (2¹ = 2)
Double [X], rate × 4 → second order (2² = 4)
Triple [X], rate × 9 → second order (3² = 9)
Worked example
Exp 1: [A] = 0.10, [B] = 0.10, rate = 2.0 × 10⁻³
Exp 2: [A] = 0.20, [B] = 0.10, rate = 8.0 × 10⁻³ → [A] doubled, rate × 4 → second order in A
Exp 3: [A] = 0.10, [B] = 0.20, rate = 2.0 × 10⁻³ → [B] doubled, rate unchanged → zero order in B
So rate = k[A]², overall second order.
Calculate
Your turn — order in A
1When [A] is doubled (with [B] held constant) the rate increases four-fold. What is the order with respect to A?
(order)
Hint: 2ⁿ = 4.
Calculate
Your turn — order in B
2When [B] is doubled (with [A] held constant) the rate is unchanged. What is the order with respect to B?
(order)
Hint: 2ⁿ = 1, so n must be zero — B does not appear in the rate equation.
Calculate
Your turn — the rate constant
3The rate equation is rate = k[A]². When [A] = 0.20 mol dm⁻³ the rate is 8.0 × 10⁻³ mol dm⁻³ s⁻¹. Calculate k, in dm³ mol⁻¹ s⁻¹.
The half-life, t½, is the time for the concentration to fall to half its value.
Zero order: the concentration-time graph is a straight line with constant gradient; each successive half-life gets shorter.
First order: an exponential decay curve — the half-life is constant. This is the diagnostic test.
Second order: a steeper curve that tails off; each half-life gets longer.
For a first-order reaction: k = ln 2 ÷ t½ln 2 = 0.693 · so k has units of s⁻¹, independent of concentration
A rate-concentration graph is just as useful: zero order gives a horizontal line, first order a straight line through the origin, second order an upward curve.
Calculate
Your turn — read the half-life
4In a first-order reaction, [A] falls from 0.80 mol dm⁻³ to 0.10 mol dm⁻³ in 60 s. Calculate the half-life, in seconds.
s
Hint: 0.80 → 0.40 → 0.20 → 0.10 is THREE half-lives. So t½ = 60 ÷ 3.
Calculate
Your turn — k from the half-life
5Using t½ = 20 s, calculate the rate constant k for this first-order reaction, in s⁻¹. Give your answer to 3 significant figures. (ln 2 = 0.693)
s⁻¹
Hint: k = ln 2 ÷ t½ = 0.693 ÷ 20.
PI3 · mechanism
The rate-determining step
Most reactions happen in a sequence of steps. The slowest step is the rate-determining step (RDS) — it is the bottleneck.
The golden rule: only the species involved in the RDS (or in steps before it) appear in the rate equation, and the order with respect to each equals the number of its molecules in the RDS.
Worked example — S(N)1 vs S(N)2
A tertiary halogenoalkane hydrolyses with rate = k[RBr] — OH⁻ does not appear. So the RDS involves only RBr: the C–Br bond breaks first, forming a carbocation. That is S(N)1, two steps.
A primary halogenoalkane hydrolyses with rate = k[RBr][OH⁻] — both appear, so both are in the RDS: a single-step S(N)2 with backside attack.
Quick check
Which step is rate-determining?
?A reaction has the rate equation rate = k[A][B]. A proposed mechanism is: Step 1: A + B → C. Step 2: C + B → D. Which step is the rate-determining step?
PI3 · Maxwell-Boltzmann
Temperature, catalysts and the energy distribution
The Maxwell-Boltzmann distribution plots the number of molecules against their kinetic energy. It starts at the origin, rises to a peak (the most probable energy), and has a long tail that never touches the axis. The area under the curve is the total number of molecules.
Raise T: the peak moves right and lower; the area is unchanged; the shaded area beyond Ea grows dramatically. A 10 K rise typically doubles the rate — far more than the ~3% rise in mean energy could explain by collision frequency alone. It is the fraction of successful collisions that soars.
Add a catalyst: the curve does not move. Ea shifts left to Ea(cat), so more molecules lie beyond it.
Quick check
What happens on heating?
?How does the Maxwell-Boltzmann distribution change when the temperature is increased?
PI3 · Arrhenius
The Arrhenius equation
k = Ae−Ea/RTA = pre-exponential (frequency) factor · Eₐ in J mol⁻¹ · R = 8.314 J K⁻¹ mol⁻¹
Take natural logs to get a straight line:
ln k = ln A − (Eₐ ÷ R)(1 ÷ T)plot ln k (y) against 1/T (x): gradient = −Eₐ ÷ R · intercept = ln A
Sign check: the gradient of an Arrhenius plot is always negative (a higher T gives a smaller 1/T but a larger k). So Ea = −gradient × R, which comes out positive. Divide by 1000 to get kJ mol⁻¹.
Calculate
Your turn — Eₐ from a gradient
6An Arrhenius plot of ln k against 1/T is a straight line of gradient −6.00 × 10³ K. Calculate the activation energy in kJ mol⁻¹. (R = 8.314 J K⁻¹ mol⁻¹)
kJ mol⁻¹
Hint: gradient = −Eₐ ÷ R, so Eₐ = 6.00 × 10³ × 8.314 = 49 884 J mol⁻¹. Now convert to kJ mol⁻¹.
Calculate
Your turn — the rate at a higher temperature
7A reaction has Ea = 50.0 kJ mol⁻¹. By what factor does k increase when the temperature rises from 300 K to 310 K? Give your answer to 3 significant figures. (R = 8.314 J K⁻¹ mol⁻¹)