OxfordAQA A-Level Chemistry: Rate equations (9620) – Revision Notes

These notes condense section 3.1.11 Rate equations of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. They cover 3.1.11.1 Rate equations, 3.1.11.2 Determination of rate equation and Required practical 8. Section 3.1.11 belongs to International A2, so it is International A-level only. You meet it in the Unit 4 paper (Organic 2 and Physical 2), and the Unit 5 paper (Practical and synoptic) may test any topic.

Detailed teaching and longer examples are in the rate equations study guide. Then test yourself with the rate equations practice questions. The energy-distribution ideas are in the kinetics revision notes (3.1.6). The course hub links every topic, the printable checklist lets you tick off each point, and the free 10-minute diagnostics point you at weak areas.

Definitions

• Rate equation: Rate = k[A]^m [B]^n. It is found by experiment.
• Order with respect to a reactant: the power of that reactant’s concentration in the rate equation. In 9620 each order is 0, 1 or 2.
• Overall order: the sum of all the individual orders.
• Rate constant, k: the number linking rate to the concentration terms. It has one value for a reaction at each temperature.
• Rate-determining (rate-limiting) step: the slowest step of a mechanism.
• Initial rate: the rate at the moment the reactants are mixed (t = 0).

3.1.11.1 Equations at a glance

Equation	Tells you	Trap
Rate = k[A]^m [B]^n	how rate depends on concentration	orders are not the balancing numbers
k = Ae^(–Ea/RT)	how k depends on temperature	J mol⁻¹ for Ea, kelvin for T
ln k = –Ea/RT + ln A	the straight-line form for a graph	y = ln k, x = 1/T, slope = –Ea/R

According to the specification, you are given these equations and R when a question needs them. Use R = 8.31 J K⁻¹ mol⁻¹. A, the Arrhenius constant, carries the units of k.

Finding the units of k

1. Rearrange: k = rate ÷ (the concentration terms).
2. Top: mol dm⁻³ s⁻¹. Bottom: (mol dm⁻³) to the power of the overall order.
3. Cancel and write any leftover units with negative powers.

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

Small reminder: Rate = k[V][L]² is overall order 3, so k is in mol⁻² dm⁶ s⁻¹.

Concentration factors

Raise each concentration factor to its order, then multiply the results together.

Order	conc × 1.5	conc × 0.5	conc × 4
0	rate × 1	rate × 1	rate × 1
1	rate × 1.5	rate × 0.5	rate × 4
2	rate × 2.25	rate × 0.25	rate × 16

Small reminder: Rate = k[V]²[L]. Halve [V] and triple [L]: factor = 0.5² × 3 = 0.75.

Temperature and k

• Higher temperature → larger k → faster rate at the same concentrations.
• Changing a concentration changes the rate, not k.
• Why: a larger fraction of collisions have energy equal to or greater than Ea, so more collisions succeed per second. The effect is large for a small temperature change.
• In calculations, rearrange k = Ae^(–Ea/RT) as needed: ln(A/k) = Ea/RT.

Method: Ea (and A) from data

1. Turn every T into 1/T (keep at least five significant figures) and every k into ln k.
2. Plot ln k on the y-axis, 1/T on the x-axis; draw the best straight line.
3. Read two points on the line that are far apart and find the gradient.
4. Ea = –gradient × R (J mol⁻¹). Divide by 1000 for kJ mol⁻¹.
5. For A: ln A = ln k + Ea/RT at any point on the line; A = e^(ln A).

Small reminder: line through (3.50 × 10⁻³, –6.20) and (3.10 × 10⁻³, –3.10). Gradient = 3.10 ÷ (–4.0 × 10⁻⁴) = –7.75 × 10³ K, so Ea = 7.75 × 10³ × 8.31 = 6.44 × 10⁴ J mol⁻¹ = 64.4 kJ mol⁻¹.

3.1.11.2 Graph shapes

Graph	Order 0	Order 1	Order 2
Concentration against time	straight, sloping down	curve, levelling off	curve, levelling off (long tail)
Rate against concentration	flat line	straight line through origin	curve through origin, getting steeper
Rate against concentration²	not used	not used	straight line through origin

• To get the rate at time t, draw a tangent to the concentration–time curve and take its gradient, ignoring the minus sign.
• Initial rate = tangent gradient at t = 0, or the steady fall in early readings ÷ time taken.
• Order 0: Rate = k, so k = size of the concentration–time gradient.

Method: orders from initial rates

1. Choose two runs where only one concentration changes.
2. Compare the concentration factor with the rate factor.
3. Same factor → order 1. Factor squared → order 2. Rate unchanged → order 0.
4. If two concentrations change, divide out the reactant you have already solved.
5. Write the rate equation, leaving out order-0 reactants.
6. Calculate k from one run, work out its units, and check it against another run.

Orders and the mechanism

• Every reactant in the rate equation reacts in the slow step or before it.
• The order of a reactant = how many of its particles take part up to and including that step.
• Order 0 → that reactant joins in after the slow step.
• Any mechanism you suggest must give the right rate equation and add up to the overall equation.

Required practical 8

Initial rate method

• Several runs, each changing one starting concentration; temperature and everything else fixed.
• Keep the total volume the same by adding water.
• Clock reaction: time to a sudden colour change, when a fixed small amount of reaction has happened, so rate ∝ 1/t.

Continuous monitoring method

• One run, measured repeatedly: gas volume, mass, colorimeter absorbance, or quenched samples titrated.
• Plot concentration (or a proportional reading) against time; tangents give rates.

The written papers test the procedures and require an evaluation of the techniques. They may ask you to interpret specimen results.

Must-know distinctions

• Rate vs k: rate falls as reactants are used up; k stays the same unless the temperature changes.
• Order vs stoichiometry: an order is measured; a balancing number is not an order.
• Initial rate vs average rate: a tangent at t = 0 is not a change over a long interval.
• ln vs log₁₀: the straight-line Arrhenius form uses ln.
• Order 0 vs no reaction: a zero-order reactant is still consumed.

Quick self-test

1. Give the units of k when Rate = k[L][T].
2. Rate = k[V]². [V] falls to 0.4 of its starting value. By what factor does the rate change?
3. Rate = k[L][Z]. [L] is multiplied by 5 and [Z] by 0.2. What happens to the rate?
4. A first-order reaction has k = 2.4 × 10⁻³ s⁻¹. Calculate the rate when the reactant concentration is 0.35 mol dm⁻³.
5. A = 3.0 × 10¹¹ s⁻¹ and Ea = 88 kJ mol⁻¹. Calculate k at 340 K.
6. The graph of ln k (y) against 1/T (x) has a gradient of –1.12 × 10⁴ K. Calculate Ea in kJ mol⁻¹.
7. What is the y-intercept of a plot of ln k against 1/T?
8. Runs: [V] = 0.10, [L] = 0.10, rate 5.0 × 10⁻⁵; [V] = 0.30, [L] = 0.10, rate 1.5 × 10⁻⁴; [V] = 0.30, [L] = 0.20, rate 6.0 × 10⁻⁴ (concentrations in mol dm⁻³, rates in mol dm⁻³ s⁻¹). Write the rate equation and find k.
9. A plot of rate against [Z] is a flat line. What does that tell you about Z and the mechanism?
10. Does adding more of a reactant change k? Explain.

Answers

1. Overall order 2, so mol⁻¹ dm³ s⁻¹.
2. 0.4² = 0.16 (the rate falls to 16 % of its value).
3. 5 × 0.2 = 1, so the rate is unchanged.
4. Rate = 2.4 × 10⁻³ × 0.35 = 8.4 × 10⁻⁴ mol dm⁻³ s⁻¹.
5. Ea/RT = 88 000 ÷ (8.31 × 340) = 31.15; k = 3.0 × 10¹¹ × e^(–31.15) = 8.9 × 10⁻³ s⁻¹.
6. Ea = 1.12 × 10⁴ × 8.31 = 9.31 × 10⁴ J mol⁻¹ = 93.1 kJ mol⁻¹.
7. ln A.
8. V: ×3 gives rate ×3, first order. L: ×2 gives rate ×4, second order. Rate = k[V][L]²; k = 5.0 × 10⁻⁵ ÷ (0.10 × 0.10²) = 0.050 mol⁻² dm⁶ s⁻¹.
9. Zero order in Z: Z is not in the rate equation and reacts after the rate-determining step.
10. No. k depends on temperature (and the reaction), not on concentration; the rate changes instead.

Where marks are usually lost

• Writing an order equal to the balancing number in the equation.
• Getting the units of k wrong by not cancelling for the actual overall order.
• Putting Ea into the Arrhenius equation in kJ mol⁻¹, or T in °C.
• Plotting ln k against T rather than 1/T.
• Losing the sign: the gradient is negative, but Ea is positive.
• Choosing gradient points from scattered data instead of from the best-fit line.
• Treating 1/t as a time rather than as a measure of rate in a clock experiment.
• Claiming a zero-order reactant “does not react”.
• Picking a mechanism with a suitable slow step that fails to balance to the overall equation.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.1.11 Rate equations (3.1.11.1 Rate equations and 3.1.11.2 Determination of rate equation, International A2), with Required practical 8.