OCR A Level Chemistry A: How fast? (H432) – Revision Notes

For full explanations and longer worked examples, use the How fast? study guide.

These notes condense section 5.1.1, How fast?, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. They cover outcomes (a) to (k). H432 is the full A level. OCR’s assessment overview shows Module 5 being assessed in components 01 and 03 (Periodic table, elements and physical chemistry; Unified chemistry).

When you are ready, work through the How fast? practice set. Boltzmann curves and collision theory are summarised in the 3.2.2 notes on rates. For planning, use our H432 course page, the printable outcome list and a free diagnostic.

Definitions (5.1.1(a))

Term	Meaning
Rate of reaction	How fast a reactant’s concentration falls or a product’s rises, per second (mol dm⁻³ s⁻¹)
Order (with respect to a reactant)	Exponent on [A] in the rate equation, for reactant A
Overall order	Sum of the individual orders
Rate constant, k	Proportionality constant in the rate equation; fixed at one temperature
Half-life, t½	Time for [reactant] to fall to half of any starting value
Rate-determining step	Slowest step of a mechanism

Equations you need

Equation	Use
rate = k[A]^m[B]^n	m, n = 0, 1 or 2 in H432
k = rate ÷ ([A]^m[B]^n)	k from one experiment
k = ln 2/t½	First order only (ln 2 = 0.693)
k = Ae^(–Ea/RT)	Arrhenius equation
ln k = –Ea/RT + ln A	Straight-line form: gradient –Ea/R, intercept ln A

R = 8.314 J mol⁻¹ K⁻¹. OCR’s additional guidance says R and the Arrhenius equations are provided on the Data Sheet. Integrated rate equations are not required, and you will not be asked to derive k = ln 2/t½.

Orders from initial rates (b)

Method in steps

1. Pick two experiments where only one concentration changes.
2. Work out the factor for that concentration and the factor for the rate.
3. Rate factor = 1: order 0. Rate factor = concentration factor: order 1. Rate factor = (concentration factor)²: order 2.
4. Repeat for each reactant, then write the rate equation.

Worked reminder. [P] trebles and the rate trebles: first order in P. [Q] doubles and the rate goes up 4 times: second order in Q. So rate = k[P][Q]².

Units of k (c)

Method in steps

1. Rearrange the rate equation so that k is the subject.
2. Substitute units: mol dm⁻³ s⁻¹ on top, (mol dm⁻³) to the power of the overall order underneath.
3. Cancel and write the result with positive powers first.

Overall order	0	1	2	3	4
Units of k	mol dm⁻³ s⁻¹	s⁻¹	dm³ mol⁻¹ s⁻¹	dm⁶ mol⁻² s⁻¹	dm⁹ mol⁻³ s⁻¹

Worked reminder. rate = k[P][Q]², rate = 4.8 × 10⁻⁴ mol dm⁻³ s⁻¹ at [P] = 0.20 and [Q] = 0.10 mol dm⁻³. k = 4.8 × 10⁻⁴ ÷ (0.20 × 0.10²) = 0.24 dm⁶ mol⁻² s⁻¹.

Graph shapes (d) and (g)

Order	Concentration–time graph	Rate–concentration graph
0	Straight line sloping down; constant gradient	Horizontal line
1	Curve with constant half-life	Straight line through origin; gradient = k
2	(not required from this graph)	Upward curve through origin

• Rate at a time t = gradient of the tangent at t on a concentration–time graph. Quote it as a positive rate.
• For zero order, the gradient of the concentration–time line is the rate, and rate = k.
• For first order, divide a tangent rate by the concentration at that point to get k.

Half-life (e) and (f)

• Read two or more successive half-lives (for example 0.80 → 0.40, then 0.40 → 0.20).
• Equal half-lives mean first order. A zero-order reactant has half-lives that get shorter as it is used up.
• Worked reminder. t½ = 46 s, so k = 0.693 ÷ 46 = 1.51 × 10⁻² s⁻¹.

Practical techniques (h)

Continuous monitoring (PAG9): one mixture followed over time by gas volume, mass loss or colorimetry. Gives a concentration–time graph.

Colorimetry: use a filter of the colour the species absorbs. Absorbance ∝ concentration; a calibration curve from standard solutions converts readings to concentrations. Zero the colorimeter with a blank first.

Initial rates (PAG10): separate runs, one concentration changed each time, temperature and the other concentrations kept constant. Initial rate from the tangent at t = 0, or from a clock reaction.

Clock reaction: time t for a fixed small amount of product to form. Initial rate ∝ 1/t. Valid only while the reaction has not proceeded too far, so the rate stays nearly constant over t.

Mechanisms and the rate-determining step (i)

• Species in the rate equation take part in the rate-determining step, or in a step before it.
• The orders tell you how many of each particle react in the slow step and any steps before it.
• A reactant that is zero order enters after the slow step.
• All steps must add up to the overall equation; intermediates cancel.

Worked reminder. 2NO₂(g) + F₂(g) → 2NO₂F(g), rate = k[NO₂][F₂].

• Slow: NO₂ + F₂ → NO₂F + F
• Fast: NO₂ + F → NO₂F

One NO₂ and one F₂ in the slow step match the rate equation. F is the intermediate.

Temperature and k (j)

Higher temperature: Boltzmann curve shifts to higher energy, a greater proportion of molecules have energy ≥ Ea, more successful collisions per second. Concentrations are unchanged, so k increases. Lower temperature does the reverse.

Arrhenius (k)

Method in steps (graph)

1. Convert T to kelvin; calculate 1/T and ln k for each point.
2. Put ln k on the y-axis and 1/T on the x-axis; draw a line of best fit.
3. Take the gradient using two well-separated points on that line.
4. Ea = –gradient × R gives J mol⁻¹; convert to kJ mol⁻¹ by dividing by 10³.
5. ln A = ln k + (Ea/R)(1/T) using a point on the line; A = e^(ln A). A has the units of k.

Worked reminder. Gradient = –8.42 × 10³ K. Ea = 8.42 × 10³ × 8.314 = 7.00 × 10⁴ J mol⁻¹ = 70.0 kJ mol⁻¹.

Worked reminder (direct use). Ea = 45 kJ mol⁻¹, A = 3.0 × 10⁷ dm³ mol⁻¹ s⁻¹, T = 350 K. Ea/RT = 45 000 ÷ (8.314 × 350) = 15.46. k = 3.0 × 10⁷ × e^(–15.46) = 5.8 dm³ mol⁻¹ s⁻¹.

Must-know distinctions

• Rate vs k: the rate drops as reactants are used up; k stays fixed unless T changes.
• Order vs stoichiometry: orders come from experiment, not from the balanced equation.
• Concentration–time vs rate–concentration: the first shows one run over time; the second compares initial rates from several runs.
• Continuous monitoring vs initial rates: one mixture followed over time vs many mixtures each measured at the start.
• Rate-determining step vs overall equation: a species can be in the overall equation and still be zero order.

Quick self-test

1. State the overall order for rate = k[A][B]².
2. Give the units of k for rate = k[C]².
3. Give the units of k for a zero-order reaction.
4. [D] is tripled and the rate rises 9 times. State the order in D.
5. [G] is halved and the rate does not change. State the order in G.
6. A first-order reaction has t½ = 210 s. Calculate k.
7. A first-order reaction has k = 0.0150 s⁻¹. Calculate t½.
8. A rate–concentration graph is a straight line through the origin with gradient 3.6 × 10⁻³ s⁻¹. State the order and k.
9. For 2X + Y → X₂Y, rate = k[X][Y]. Which particles are in the rate-determining step?
10. The Arrhenius graph for a reaction (ln k on y, 1/T on x) has a gradient of –1.05 × 10⁴ K. Calculate Ea in kJ mol⁻¹.
11. Explain why 1/t from a clock reaction can be used as a measure of initial rate.

Answers

1. 3
2. dm³ mol⁻¹ s⁻¹
3. mol dm⁻³ s⁻¹
4. 9 = 3², so second order
5. Zero order
6. k = 0.693 ÷ 210 = 3.30 × 10⁻³ s⁻¹
7. t½ = 0.693 ÷ 0.0150 = 46.2 s
8. First order; k = 3.6 × 10⁻³ s⁻¹
9. One X and one Y; the second X reacts in a later, fast step.
10. Ea = 1.05 × 10⁴ × 8.314 = 8.73 × 10⁴ J mol⁻¹ = 87.3 kJ mol⁻¹
11. Only a small fixed amount of product forms in time t, so the rate is nearly constant and close to the initial rate; rate = amount ÷ t, so rate ∝ 1/t.

Where marks are usually lost

• Writing orders that match the coefficients in the balanced equation.
• Treating a change by a factor of 2.5 or 3 as a doubling when comparing experiments.
• Units of k with the powers the wrong way round, such as mol² dm⁻⁶ s⁻¹.
• Quoting a negative rate from the tangent to a falling reactant curve.
• Measuring each half-life from t = 0 instead of between successive halvings.
• Using k = ln 2/t½ for a zero-order or second-order reaction.
• Putting a zero-order reactant into the proposed slow step.
• Using T, or a temperature in °C, on the x-axis of an Arrhenius plot instead of 1/T in K⁻¹.
• Giving Ea in J mol⁻¹ when kJ mol⁻¹ was asked for, or dropping the minus sign in Ea = –gradient × R.

Next steps

If a section felt shaky, reread it in the How fast? study guide. Then attempt the How fast? practice set under timed conditions.

Official syllabus

Source: section 5.1.1 How fast?, learning outcomes (a) to (k), in the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, published by OCR.