Rate of Reaction and Rate Equations

Spec Mapping — OCR H432 Module 5.1.1 — How fast?, covering the definition of rate of reaction in terms of the change in concentration of a reactant or product per unit time, the rate equation rate = $k[A]^m[B]^n$k[A]m[B]n with experimentally-determined orders, the meaning of order with respect to a reactant and overall order, and the principle that orders cannot be deduced from stoichiometric coefficients (refer to the official OCR H432 specification document for exact wording).

Module 5.1.1 opens with the formal mathematical language that A-Level kinetics will use for the next eight lessons. This lesson lays down three intertwined ideas: (i) rate of reaction as a precise calculus quantity ($-\frac{1}{a}\frac{d[A]}{dt}$−a1​dtd[A]​, not a vague "how fast"), (ii) the rate equation as an experimentally-determined power-law summary of how rate depends on each reactant, and (iii) order of reaction as the exponent on each concentration in that power law. The single most-tested message — and one that earns or loses two AO1 marks on virtually every kinetics question in the OCR H432 corpus — is that orders are determined by experiment, not by reading off stoichiometric coefficients. We meet that idea twice in this lesson and four times more in Module 5.1.1 as a whole; getting it right here is the difference between routine A-band scoring and the systematic mis-application of "rate = $k[\text{NO}]^2[\text{O}_2]$k[NO]2[O2​]" to reactions that have a completely different rate law. By the end of the lesson you will be able to write a rate equation, compute a rate from a given $k$k and concentrations, and predict how rate scales with concentration for orders 0, 1, 2 and 3.

Key Equation: the general rate equation is $\text{rate} = k[A]^m[B]^n$rate=k[A]m[B]n where $k$k is the rate constant (temperature-dependent only, NOT concentration-dependent), $m$m is the order with respect to A, $n$n is the order with respect to B, and the overall order is $m+n$m+n. For a general reaction $aA + bB \rightarrow \text{products}$aA+bB→products the underlying definition of rate is $\text{rate} = -\tfrac{1}{a}\tfrac{d[A]}{dt} = -\tfrac{1}{b}\tfrac{d[B]}{dt} = +\tfrac{1}{p}\tfrac{d[P]}{dt}$rate=−a1​dtd[A]​=−b1​dtd[B]​=+p1​dtd[P]​ with units mol dm$^{-3}$−3 s$^{-1}$−1.

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What do we mean by "rate"?

The rate of reaction is the change in concentration of a reactant or product per unit time. The negative sign convention on reactants ensures that rate is always positive — reactant concentrations fall with time, so $d[A]/dt < 0$d[A]/dt<0, and the minus sign flips this to a positive rate. For a reaction $aA + bB \rightarrow cC + dD$aA+bB→cC+dD the single-valued rate is

$\text{rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$rate=−a1​dtd[A]​=−b1​dtd[B]​=+c1​dtd[C]​=+d1​dtd[D]​

The stoichiometric coefficients in the denominators normalise the four definitions so they give the same numerical rate regardless of which species you monitor. For $2N_2O_5 \rightarrow 4NO_2 + O_2$2N2​O5​→4NO2​+O2​, the rate of disappearance of $N_2O_5$N2​O5​ is twice the rate of formation of $O_2$O2​, but dividing by the coefficients makes them agree.

Units: mol dm$^{-3}$−3 s$^{-1}$−1 is the OCR-standard combination. You may meet mol dm$^{-3}$−3 min$^{-1}$−1 for slow reactions or mol m$^{-3}$−3 s$^{-1}$−1 in physical-chemistry texts — always state the units explicitly with every numerical rate.

Why rate depends on concentration

Collision theory (Module 3.2.2) tells us that particles must collide with sufficient energy ($E \geq E_a$E≥Ea​) and correct orientation for a reaction to occur. Increasing the concentration of a reactant raises the frequency of collisions per unit volume — there are simply more reactive particles per cubic decimetre, so more chances per second for a reactive collision. How sharply rate rises with concentration depends on the mechanism of the reaction: if the slow (rate-determining) step involves two molecules of A, doubling [A] quadruples the rate; if it involves only one molecule of A, doubling [A] simply doubles the rate. This translation from mechanism to mathematical rate-law is the entire job of rate equations.

The rate equation

For a reaction A + B $\rightarrow$→ products, the rate equation (or rate law) takes the form

$\text{rate} = k[A]^m[B]^n$rate=k[A]m[B]n

with:

• $k$k — the rate constant, a proportionality constant that depends on temperature (via the Arrhenius equation, Lesson 9) but not on concentration.
• $m$m — the order with respect to A, found experimentally.
• $n$n — the order with respect to B, found experimentally.
• $m + n$m+n — the overall order of the reaction.

The crucial point (worth two marks every time)

Orders $m$m and $n$n must be determined experimentally. They are not taken from the stoichiometric coefficients of the balanced equation.

Consider three concrete examples that demonstrate the danger of guessing from stoichiometry:

Reaction	Stoichiometric prediction (WRONG to use)	Experimental rate equation
$2NO_2 \rightarrow 2NO + O_2$2NO2​→2NO+O2​	rate = $k[NO_2]^2$k[NO2​]2	rate = $k[NO_2]^2$k[NO2​]2 (second order — matches by coincidence)
$H_2 + I_2 \rightarrow 2HI$H2​+I2​→2HI (moderate T)	rate = $k[H_2][I_2]$k[H2​][I2​]	rate = $k[H_2][I_2]$k[H2​][I2​] (matches at modest T; mechanism changes at high T)
$(CH_3)_3CBr + OH^- \rightarrow (CH_3)_3COH + Br^-$(CH3​)3​CBr+OH−→(CH3​)3​COH+Br−	rate = $k[(CH_3)_3CBr][OH^-]$k[(CH3​)3​CBr][OH−]	rate = $k[(CH_3)_3CBr]$k[(CH3​)3​CBr] — zero order in $OH^-$OH−!

The third example (the $S_N1$SN​1 hydrolysis of tert-butyl bromide) is the classic A-Level mark-scheme trap: candidates who "read off the equation" award themselves an order of 1 for $OH^-$OH− and lose the structural mark for understanding that the slow step does not involve hydroxide at all. The rate law mirrors the mechanism — and the mechanism is hidden until experiment reveals it.

Order of reaction — the exponent table

The order of reaction with respect to a reactant is the power to which the concentration of that reactant is raised in the experimentally determined rate equation. The overall order is the sum of the powers. At A-Level you will meet integer orders 0, 1, 2 (occasionally 3); fractional orders signal radical mechanisms (e.g. rate $= k[H_2][Br_2]^{1/2}$=k[H2​][Br2​]1/2 for $H_2 + Br_2$H2​+Br2​) and are beyond OCR scope.

Order w.r.t. A	Rate dependence	Doubling [A] gives	Tripling [A] gives
0	rate independent of [A]	$\times 1$×1 (no change)	$\times 1$×1
1	rate $\propto [A]$∝[A]	$\times 2$×2	$\times 3$×3
2	rate $\propto [A]^2$∝[A]2	$\times 4$×4	$\times 9$×9
3	rate $\propto [A]^3$∝[A]3	$\times 8$×8	$\times 27$×27

The fold-change ratio $r = (\text{rate ratio})/(\text{[A] ratio})^m$r=(rate ratio)/([A] ratio)m should equal 1 if the correct $m$m is chosen — this is the diagnostic Lesson 6 (initial rates method) will exploit.

Worked example 1 — using a rate equation

Q: For the reaction $2NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(g)$2NO(g)+2H2​(g)→N2​(g)+2H2​O(g), the experimentally-determined rate equation is rate = $k[NO]^2[H_2]$k[NO]2[H2​]. (a) State the order with respect to NO, $H_2$H2​, and overall. (b) If [NO] is doubled, by what factor does the rate change? (c) If [NO] and $[H_2]$[H2​] are both doubled, by what factor does the rate change?

A:

(a) Order w.r.t. NO = 2; order w.r.t. $H_2$H2​ = 1; overall order = 2 + 1 = 3.

(b) Rate $\propto [NO]^2$∝[NO]2, so doubling [NO] multiplies rate by $2^2 = \mathbf{4}$22=4.

(c) Both concentrations contribute: rate change factor = $2^2 \times 2^1 = \mathbf{8}$22×21=8.

Worked example 2 — calculating a rate from $k$k and concentrations

Q: The rate equation is rate = $k[A][B]^2$k[A][B]2 with $k = 2.5 \times 10^{-3}$k=2.5×10−3 mol$^{-2}$−2 dm$^6$6 s$^{-1}$−1. Calculate the rate when $[A] = 0.20$[A]=0.20 mol dm$^{-3}$−3 and $[B] = 0.10$[B]=0.10 mol dm$^{-3}$−3.

A:

$$\begin{aligned} \text{rate} &= k[A][B]^{2} \\ &= (2.5 \times 10^{-3})(0.20)(0.10)^{2} \\ &= (2.5 \times 10^{-3})(0.20)(1.0 \times 10^{-2}) \\ &= 5.0 \times 10^{-6}\;\text{mol dm}^{-3}\,\text{s}^{-1} \end{aligned}$$rate​=k[A][B]2=(2.5×10−3)(0.20)(0.10)2=(2.5×10−3)(0.20)(1.0×10−2)=5.0×10−6mol dm−3s−1​

Units check: $k$k has units mol$^{-2}$−2 dm$^6$6 s$^{-1}$−1; multiplying by [A][B]$^2$2 which has units (mol dm$^{-3}$−3)$^3$3 = mol$^3$3 dm$^{-9}$−9 gives mol$^{-2+3}$−2+3 dm$^{6-9}$6−9 s$^{-1}$−1 = mol dm$^{-3}$−3 s$^{-1}$−1 — correct for rate.

Worked example 3 — finding the missing concentration

Q: For rate = $k[A]^2[B]$k[A]2[B] with $k = 0.040$k=0.040 mol$^{-2}$−2 dm$^6$6 s$^{-1}$−1, a rate of $1.6 \times 10^{-3}$1.6×10−3 mol dm$^{-3}$−3 s$^{-1}$−1 is observed when [A] = 0.20 mol dm$^{-3}$−3. Find [B].

A: Rearrange:

$[B] = \frac{\text{rate}}{k[A]^2} = \frac{1.6 \times 10^{-3}}{0.040 \times (0.20)^2} = \frac{1.6 \times 10^{-3}}{1.6 \times 10^{-3}} = \mathbf{1.0 \text{ mol dm}^{-3}}$[B]=k[A]2rate​=0.040×(0.20)21.6×10−3​=1.6×10−31.6×10−3​=1.0 mol dm−3

Worked example 4 — predicting the new rate

Q: A reaction follows rate = $k[A]^2[B]$k[A]2[B]. The initial rate is $4.0 \times 10^{-4}$4.0×10−4 mol dm$^{-3}$−3 s$^{-1}$−1. What is the new rate if [A] is halved and [B] is tripled?

A: Rate factor = $(1/2)^2 \times 3 = 1/4 \times 3 = 3/4$(1/2)2×3=1/4×3=3/4. New rate = $0.75 \times 4.0 \times 10^{-4} = \mathbf{3.0 \times 10^{-4}}$0.75×4.0×10−4=3.0×10−4 mol dm$^{-3}$−3 s$^{-1}$−1.

The rate constant $k$k — what it tells us

$k$k is a measure of intrinsic reactivity at a given temperature. A large $k$k means a fast reaction at unit concentrations; a small $k$k means a slow reaction. Five facts about $k$k that you must memorise:

1. Depends on temperature — increases with $T$T via $k = Ae^{-E_a/RT}$k=Ae−Ea​/RT (Arrhenius, Lesson 9).
2. Independent of concentration — $k$k is a constant for given $T$T; only concentrations change during a reaction.
3. Has order-dependent units — Lesson 7 derives the units rule (mol$^{1-\text{order}}$1−order dm$^{3(\text{order}-1)}$3(order−1) s$^{-1}$−1 for any overall order).
4. Different for different reactions at the same $T$T — $k$k encodes the activation energy and pre-exponential factor unique to each reaction's transition state.
5. Increased by catalysts — a catalyst lowers $E_a$Ea​, which raises $k$k at any given $T$T without altering [A] or [B].

The lowercase k is reserved for rate constant; uppercase $K$K is the equilibrium constant ($K_c$Kc​, $K_p$Kp​ — Lessons 10–12). Confusing these costs a mark every time it occurs.

Visualising how rate scales with concentration

The mermaid below maps the chain of cause and effect from concentration through to rate, separating the contributions of $Z_\text{coll}$Zcoll​, the Boltzmann fraction $f$f, and the rate constant $k$k:

graph LR
  A["[A] increases"] --> B[More A molecules per dm³]
  B --> C[Higher collision frequency Z_coll]
  C --> D[Larger rate if order in A > 0]
  E[Temperature rises] --> F[Larger Boltzmann fraction f]
  F --> G[Larger rate constant k]
  G --> D
  H[Catalyst added] --> I[Lower Ea]
  I --> F

And here is a simple SVG that compares zero-, first-, and second-order rate-vs-[A] curves on a single axis — Lesson 4 returns to this in much more detail.

The horizontal zero-order line, the straight-line-through-origin for first order (gradient = $k$k), and the upward curve of second order are the three signature shapes you must memorise. Notice that order 2 starts shallower than order 1 at very low [A] (because $[A]^2 < [A]$[A]2<[A] when $[A] < 1$[A]<1) but quickly overtakes it as [A] grows.

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Synoptic Links

Synoptic Links — Connects to:

• ocr-alevel-chemistry-enthalpy-rates-equilibrium / reaction-rates-collision-theory (the qualitative collision-theory account of rate from Module 3.2.2 — this lesson is the quantitative mathematical layer that sits on top of it).
• ocr-alevel-chemistry-enthalpy-rates-equilibrium / boltzmann-distribution (the Boltzmann fraction $e^{-E_a/RT}$e−Ea​/RT explains the exponential temperature-dependence of $k$k that Lesson 9 formalises through Arrhenius).
• ocr-alevel-chemistry-quantitative-rates-equilibrium / rate-constant-k (Lesson 7 — the units of $k$k depend on overall order, derived directly from the rate equation here).
• ocr-alevel-chemistry-quantitative-rates-equilibrium / the-arrhenius-equation (Lesson 9 — quantifies the temperature dependence of $k$k).
• ocr-alevel-chemistry-quantitative-rates-equilibrium / rate-determining-step (Lesson 8 — the link between rate equation and underlying mechanism).

Practical Activity Group anchor: PAG 9 (Continuous-monitoring rate) is the primary practical context — the bromine/methanoic-acid colorimetry and the iodine-clock are the canonical PAG experiments that produce concentration-time data from which rate equations are deduced. PAG 10 (Initial-rate / clock reactions) is the secondary anchor for Lesson 6, where multiple clock-reaction trials at varied [reactant] give the orders directly.

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Specimen question modelled on the OCR H432 paper format

Question (6 marks): The decomposition of dinitrogen pentoxide,

$2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$2N2​O5​(g)→4NO2​(g)+O2​(g)

has the experimentally-determined rate equation rate = $k[N_2O_5]$k[N2​O5​] with $k = 6.2 \times 10^{-4}$k=6.2×10−4 s$^{-1}$−1 at 318 K.

(a) State the order with respect to $N_2O_5$N2​O5​ and the overall order. [1 mark]

(b) Explain why the order with respect to $N_2O_5$N2​O5​ is not 2, even though the stoichiometric coefficient is 2. [2 marks]

(c) Calculate the rate of decomposition when $[N_2O_5] = 0.040$[N2​O5​]=0.040 mol dm$^{-3}$−3, giving units. [3 marks]

AO breakdown

Mark	AO	Awarded for
1	AO1	Order w.r.t. $N_2O_5$N2​O5​ = 1; overall order = 1
2	AO2	Orders are determined experimentally, not from stoichiometry
3	AO2	The slow / rate-determining step involves only one $N_2O_5$N2​O5​ molecule
4	AO2	rate = $6.2 \times 10^{-4} \times 0.040$6.2×10−4×0.040
5	AO2	= $2.48 \times 10^{-5}$2.48×10−5 (to 3 sf)
6	AO3	Correct units: mol dm$^{-3}$−3 s$^{-1}$−1

AO split: AO1 = 1, AO2 = 4, AO3 = 1.

Mid-band response (4/6):

(a) Order with respect to $N_2O_5$N2​O5​ is 1, overall order is 1.

(b) Orders come from experiments, not from the balanced equation. So even though the coefficient is 2, the experimental order is 1.

(c) rate = $k[N_2O_5]$k[N2​O5​] = $6.2 \times 10^{-4} \times 0.040$6.2×10−4×0.040 = $2.5 \times 10^{-5}$2.5×10−5 mol dm$^{-3}$−3 s$^{-1}$−1.

Examiner-style commentary: To reach top-band: explicit AO2 link from order = 1 to the mechanism (only one molecule of $N_2O_5$N2​O5​ in the slow step) is the AO3 move missing here. Marks 1, 2, 4, 5, 6 awarded; M3 (mechanism justification) and the higher-precision sig-fig handling are absent. The candidate also rounded $2.48 \times 10^{-5}$2.48×10−5 to $2.5 \times 10^{-5}$2.5×10−5 without comment — acceptable but not exemplary. To lift this from B-band to A*-band, add one sentence on what "experimentally-determined" tells us about the rate-determining step.

Top-band response (6/6):

(a) Order with respect to $N_2O_5$N2​O5​ = 1; overall order = 1.

(b) Orders in a rate equation are determined experimentally (e.g. via initial-rates or half-life analysis) — they cannot be deduced from stoichiometric coefficients. The stoichiometric coefficient 2 indicates that two $N_2O_5$N2​O5​ molecules are consumed per mole of reaction, but the observed first-order kinetics show that the rate-determining step involves only one $N_2O_5$N2​O5​ molecule (most likely the unimolecular bond-breaking $N_2O_5 \rightarrow N_2O_3 + O_2$N2​O5​→N2​O3​+O2​, followed by a fast $N_2O_3 \rightarrow NO + NO_2$N2​O3​→NO+NO2​ step). The overall stoichiometry adds across the full mechanism, but the rate law tracks only the slow step.

(c) Substituting into rate = $k[N_2O_5]$k[N2​O5​]:

$\text{rate} = (6.2 \times 10^{-4}\,\text{s}^{-1})(0.040\,\text{mol dm}^{-3}) = \mathbf{2.48 \times 10^{-5}\,\text{mol dm}^{-3}\,\text{s}^{-1}}$rate=(6.2×10−4s−1)(0.040mol dm−3)=2.48×10−5mol dm−3s−1

(to 3 sf; units checked by dimensional analysis: s$^{-1}$−1 $\times$× mol dm$^{-3}$−3 = mol dm$^{-3}$−3 s$^{-1}$−1, correct for rate).

Examiner-style commentary: Full 6/6. All AO1, AO2, and AO3 marks awarded. Discriminators between Grade A and Grade A*: (i) explicit naming of the rate-determining step concept (forward-references Lesson 8), (ii) the proposed unimolecular mechanism showing chemical insight beyond the bare mark scheme, (iii) dimensional-analysis check of the units, and (iv) 3-sf precision with explicit comment. This is the type of answer that lifts a paper from a Grade A to an A* on synoptic kinetics questions.

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Common errors and mark-loss patterns

Teacher-voice observations from marking OCR H432 papers:

• Reading orders off the stoichiometry. Almost every cohort produces some candidates who write rate = $k[N_2O_5]^2$k[N2​O5​]2 for the decomposition above. This is the single most-deducted error in 5.1.1 and the easiest one to avoid by stating "order = 1, determined experimentally".
• Confusing $k$k (lowercase) with $K$K (uppercase). $k$k is the kinetic rate constant; $K_c$Kc​ and $K_p$Kp​ are equilibrium constants. Mark schemes treat these as completely separate quantities; a missing or mis-capitalised symbol loses a mark on every question that uses both.
• Forgetting to include units of rate. "rate = $2.5 \times 10^{-5}$2.5×10−5" is incomplete — units mol dm$^{-3}$−3 s$^{-1}$−1 are essential.
• Treating $k$k as concentration-dependent. Candidates sometimes write "$k$k increases when [A] increases" — this is wrong: $k$k depends only on $T$T (and on catalyst presence).
• Wrong fold-change for order 2. "Doubling [A] doubles rate" applied to a second-order reaction is a classic Grade C error; the answer is $\times 4$×4, not $\times 2$×2.
• Missing the negative sign in $d[A]/dt$d[A]/dt. Writing rate = $\frac{d[A]}{dt}$dtd[A]​ instead of $-\frac{d[A]}{dt}$−dtd[A]​ for a reactant gives a negative numerical rate — examiners deduct the AO1 mark for definition.
• Stating rate in min$^{-1}$−1 but $k$k in s$^{-1}$−1 without converting. Mixed time units propagate into wrong $k$k values; always convert to seconds before computing $k$k.

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Practical context — why orders matter in industry

Industrial chemists care about rate equations for three reasons:

1. Reactor sizing. A first-order reaction that takes 10 minutes to reach 90% conversion in a batch reactor needs a much larger continuous-flow reactor than a second-order reaction at the same starting concentrations — because the second-order rate falls quadratically as [A] drops, so the late-stage reaction is much slower.
2. Conversion economics. From Lesson 5, achieving 99% conversion typically takes about $\log_2 100 \approx 6.6$log2​100≈6.6 half-lives for a first-order process; pushing to 99.9% takes $\log_2 1000 \approx 10$log2​1000≈10 half-lives — half as long again. The cost of "ultra-pure" product is dominated by the last few percent of conversion.
3. Side reactions and selectivity. When a desired reaction is first order in A and an undesired side reaction is second order in A, low [A] favours selectivity — pushing the chemist toward a continuous-flow stirred reactor where [A] is held constant at a low value, rather than a batch reactor where [A] starts high.

For the Haber process, which is roughly first order in $N_2$N2​ but second order in $H_2$H2​, the ratio $[H_2]^3 : [N_2] = 3:1$[H2​]3:[N2​]=3:1 at the reactor inlet is chosen to balance reaction rate (favours high $[H_2]$[H2​]) against equilibrium yield (favours low $[H_2]$[H2​] to avoid pushing the equilibrium back). Rate equations directly drive reactor design.

Worked example 5 — combining two rate equations

Q: Two reactions both proceed at 298 K in the same solvent. Reaction A: rate $= k_A[\text{X}]^2$=kA​[X]2 with $k_A = 0.10$kA​=0.10 mol$^{-1}$−1 dm$^3$3 s$^{-1}$−1. Reaction B: rate $= k_B[\text{Y}]$=kB​[Y] with $k_B = 5.0 \times 10^{-3}$kB​=5.0×10−3 s$^{-1}$−1. Initial concentrations are $[\text{X}]_0 = 0.10$[X]0​=0.10 mol dm$^{-3}$−3 and $[\text{Y}]_0 = 0.50$[Y]0​=0.50 mol dm$^{-3}$−3. Which reaction is faster initially?

A: Reaction A: rate$_A = 0.10 \times (0.10)^2 = 1.0 \times 10^{-3}$A​=0.10×(0.10)2=1.0×10−3 mol dm$^{-3}$−3 s$^{-1}$−1. Reaction B: rate$_B = 5.0 \times 10^{-3} \times 0.50 = 2.5 \times 10^{-3}$B​=5.0×10−3×0.50=2.5×10−3 mol dm$^{-3}$−3 s$^{-1}$−1. Reaction B is initially $\sim 2.5\times$∼2.5× faster, despite having a smaller rate constant — because the higher concentration of Y wins the comparison.

Worked example 6 — rate equation determination challenge

Q: A four-reactant rate equation rate = $k[A]^2[B][C]^0[D]$k[A]2[B][C]0[D] is hypothesised. Predict the rate-fold-change if every concentration is doubled.

A: Total fold = $2^2 \times 2^1 \times 2^0 \times 2^1 = 4 \times 2 \times 1 \times 2 = \mathbf{16}\times$22×21×20×21=4×2×1×2=16×.

This worked example illustrates that the overall order (= 2 + 1 + 0 + 1 = 4) determines the rate's sensitivity to a uniform concentration change: doubling everything multiplies rate by $2^{\text{overall order}} = 2^4 = 16$2overall order=24=16. This is a powerful sanity check when interpreting kinetic data.

Going further

The rate equation is a special case of a more general phenomenological law: any reaction has a power-law rate expression at some level of resolution, but the exact form depends on whether you are sampling close to equilibrium (where back-reactions matter), in a complex multi-step mechanism (where pre-equilibrium and steady-state approximations apply), or far from equilibrium (where the simple power law dominates). The Dutch chemist Jacobus Henricus van't Hoff (Nobel Prize 1901) introduced much of the kinetics nomenclature you are using here — "order of reaction", "molecularity", and the distinction between the two — in his 1884 Études de dynamique chimique. Beyond A-Level you will meet the steady-state approximation (used to derive Michaelis–Menten enzyme kinetics, where rate = $V_\text{max}[S]/(K_M + [S])$Vmax​[S]/(KM​+[S]) — a non power-law form that reduces to first order at low [S] and zero order at high [S]) and the pre-equilibrium approximation (used for fast-reversible-then-slow mechanisms). Both are introduced at undergraduate level and have direct A-Level analogues in catalyst surface-saturation (zero order at high [A]) and enzyme kinetics. The deeper conceptual link is that the rate equation, like any thermodynamic potential, is path-dependent in form but state-dependent in value: two completely different mechanisms can give the same observed rate law, but the same mechanism must give the same rate law in any solvent or condition. This is why mechanistic chemists distinguish between microscopic rate constants (specific to elementary steps) and macroscopic rate constants (the experimentally-observed $k$k); the macroscopic $k$k depends on the microscopic ones through the rate-determining step (Lesson 8) and pre-equilibrium constants.

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The historical context — how chemists came to understand rate

Before the late 19th century, rate-of-reaction work was largely descriptive: chemists noted that some reactions were "fast" and others "slow", that warming things up generally helped, and that adding more of a reactant generally accelerated things, but the mathematical form of how rate depends on concentration was not appreciated. The breakthrough came in 1864 when Norwegian chemists Cato Guldberg and Peter Waage proposed the law of mass action: that the rate of a homogeneous reaction is proportional to a power of the concentration of each reactant. By 1884 van't Hoff had refined this into the modern rate equation with experimentally-determined exponents, and the framework you are studying took shape. The Arrhenius equation (Lesson 9) followed in 1889, completing the picture by adding temperature dependence. The deep insight was that rate is governed by the rate-determining step alone — multi-step mechanisms collapse mathematically to a single power law in the [reactants] of the slow step. This is why a balanced equation with five $Br^-$Br− ions can still be first order in $Br^-$Br−: only one $Br^-$Br− enters the slow step. The 20th century then asked the harder question: given an observed rate law, what is the mechanism? — and answered it via the steady-state and pre-equilibrium approximations introduced in advanced kinetics (Atkins ch. 22). The Nobel Prize-winning work on picosecond and femtosecond spectroscopy (Zewail, 1999) finally made it possible to watch elementary steps directly, confirming the mechanistic interpretation of rate laws that had been inferred from bulk kinetics for a century.

A-Level misconceptions

The errors that distinguish Grade A from Grade A*:

1. Orders are part of the mechanism, not the equation. The stoichiometric equation tells you what reacts and in what proportions; the rate equation tells you how the rate-determining step works. They are independent until experiment links them.
2. A coefficient of 2 in the balanced equation does NOT mean second order. $H_2 + I_2 \rightarrow 2HI$H2​+I2​→2HI is first order in each, not second order in I — even though the coefficient is 2.
3. $k$k is temperature-dependent, but not via the rate equation. The Arrhenius equation $k = Ae^{-E_a/RT}$k=Ae−Ea​/RT is the separate route to $T$T-dependence; the rate equation only shows the concentration-dependence at a fixed T.
4. Units of $k$k are NOT universal. For overall order 0, $k$k has units mol dm$^{-3}$−3 s$^{-1}$−1; for overall order 1, s$^{-1}$−1; for overall order 2, mol$^{-1}$−1 dm$^3$3 s$^{-1}$−1; for overall order 3, mol$^{-2}$−2 dm$^6$6 s$^{-1}$−1. Lesson 7 returns to this rule formally.
5. Rate is a single-valued quantity for the whole reaction. It does not matter which species you monitor — provided you divide by the stoichiometric coefficient, the numerical rate is the same.
6. "Doubling concentration always doubles rate" is FALSE. Only true for first order in that species. For zero order, no change; for second order, $\times 4$×4.
7. Order 0 does not mean the reactant is irrelevant. It means the reactant is present in excess, or that its concentration is fixed by some other equilibrium (e.g. a saturated catalyst surface). The reactant is consumed; its concentration does affect the position of equilibrium — but not the rate of the forward step.
8. The rate equation is empirical, not theoretical. No amount of structural reasoning can predict the rate law from first principles for a multi-step mechanism — only experiment can.

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Summary

The rate of reaction is the change in concentration of a reactant or product per unit time, with units mol dm$^{-3}$−3 s$^{-1}$−1. The rate equation takes the experimental form rate = $k[A]^m[B]^n$k[A]m[B]n, with orders $m$m and $n$n determined by experiment — never by stoichiometric coefficients. The overall order is $m + n$m+n. The rate constant $k$k depends on temperature (via the Arrhenius equation, Lesson 9) and on the presence of catalysts, but not on concentration. The next lesson examines what each integer order means physically (0, 1, 2) and how to deduce them from rate-ratio data; Lessons 3–6 then explore the four graphical and experimental signatures of each order, building toward the units rule for $k$k in Lesson 7 and the Arrhenius equation in Lesson 9.

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Reference: OCR A-Level Chemistry A (H432) Module 5.1.1 (refer to the official OCR H432 specification document for exact wording).