Orders of Reaction (0, 1, 2)

Spec Mapping — OCR H432 Module 5.1.1 — Orders, rate equations and rate constants, covering the qualitative and quantitative meaning of zero, first and second orders of reaction, the determination of orders from comparison of initial rates while holding all other concentrations constant, the limitations of pseudo-order experimental designs (large excess of one reactant), and the principle that orders are mechanistic in origin and therefore must be measured rather than calculated (refer to the official OCR H432 specification document for exact wording).

Module 5.1.1 places enormous emphasis on the physical interpretation of orders 0, 1, and 2 — these are not just exponents in a power law, they are direct read-outs of how many molecules of each reactant appear in the rate-determining step. This lesson moves beyond the formal "rate = $k[A]^m[B]^n$k[A]m[B]n" introduced in Lesson 1 and asks: what does it mean for a reaction to be zero order in $A$A? When does this happen? Why does an $S_N1$SN​1 hydrolysis end up first order even though the balanced equation has two reactants? And how do we extract orders from a table of experimental initial-rate data without algebraic guesswork? You will learn the ratio method (the workhorse of A-Level rate problems), the logarithmic method for non-integer concentration ratios, and the four characteristic patterns that link "doubling [A]" to a change in rate. These tools, combined with the graphical signatures in Lessons 3–4 and the half-life test in Lesson 5, form the diagnostic toolkit for any rate-equation question in the OCR H432 paper.

Key Equation: for any reactant $A$A in a rate equation, the order $m$m satisfies $\frac{\text{rate}_2}{\text{rate}_1} = \left(\frac{[A]_2}{[A]_1}\right)^m \quad \text{(when all other [reactants] are held constant)}$rate1​rate2​​=([A]1​[A]2​​)m(when all other [reactants] are held constant) Rearranging: $m = \dfrac{\log(\text{rate}_2/\text{rate}_1)}{\log([A]_2/[A]_1)}$m=log([A]2​/[A]1​)log(rate2​/rate1​)​. For integer ratios (×2, ×3, ×4), the ratio method gives $m$m by inspection; for non-integer ratios, take logarithms.

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The three integer orders — what they mean physically

Zero order — rate independent of [A]

If a reaction is zero order with respect to A,

$\text{rate} = k[A]^0 = k$rate=k[A]0=k

then changing [A] has no effect on the rate. This counter-intuitive result occurs when some other step governs the rate — typically because A is not a participant in the rate-determining step (RDS), or because A is present in such excess that its concentration is effectively constant. Two physical scenarios:

1. Saturated catalytic surface. $2NH_3(g) \rightarrow N_2(g) + 3H_2(g)$2NH3​(g)→N2​(g)+3H2​(g) on a hot platinum wire at high pressure: every Pt site is occupied by an $NH_3$NH3​ molecule, so adding more $NH_3$NH3​ does not increase the rate — it just creates a queue.
2. Saturated enzyme active site. $E + S \rightarrow E + P$E+S→E+P at $[S] \gg K_M$[S]≫KM​: every enzyme molecule is binding substrate, so adding more $S$S does not speed up turnover — the rate is set by the enzyme's $k_\text{cat}$kcat​.

In a zero-order rate equation, the reactant is consumed (the stoichiometry still operates), but the rate of consumption is independent of how much A is present.

First order — rate proportional to [A]

If a reaction is first order with respect to A,

$\text{rate} = k[A]^1 = k[A]$rate=k[A]1=k[A]

then doubling [A] doubles the rate, tripling [A] triples the rate, halving [A] halves the rate. Rate is directly proportional to concentration. This is the signature of a rate-determining step that involves one molecule of A (unimolecular elementary step). Two physical scenarios:

1. Radioactive decay (every radionuclide is first order — rate = $\lambda N$λN, where $\lambda = k$λ=k).
2. $S_N1$SN​1 nucleophilic substitution: $(CH_3)_3CBr + OH^- \rightarrow (CH_3)_3COH + Br^-$(CH3​)3​CBr+OH−→(CH3​)3​COH+Br− — the slow step is the unimolecular ionisation $(CH_3)_3CBr \rightarrow (CH_3)_3C^+ + Br^-$(CH3​)3​CBr→(CH3​)3​C++Br−, so rate = $k[(CH_3)_3CBr]$k[(CH3​)3​CBr] — first order overall, zero order in $OH^-$OH−. The hydroxide is a reactant, but it is not in the RDS.

Second order — rate proportional to $[A]^2$[A]2 or $[A][B]$[A][B]

If a reaction is second order, either with respect to a single reactant A,

$\text{rate} = k[A]^2$rate=k[A]2

(doubling [A] quadruples the rate, tripling [A] gives a 9× rate increase), or first-order in each of two reactants,

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

(doubling [A] or [B] alone doubles the rate; doubling both quadruples the rate). Second-order kinetics indicate a rate-determining step with two molecules (either two of A, or one each of A and B — both are bimolecular). The textbook example is the $S_N2$SN​2 hydrolysis $OH^- + CH_3Br \rightarrow CH_3OH + Br^-$OH−+CH3​Br→CH3​OH+Br− with rate = $k[CH_3Br][OH^-]$k[CH3​Br][OH−] — the rate-determining step is a single concerted bimolecular collision.

Comparison table — the diagnostic backbone

Order w.r.t. A	Rate equation (simple)	Doubling [A] gives	Tripling [A] gives	Physical meaning
0	rate = $k$k	$\times 1$×1	$\times 1$×1	[A] does NOT appear in slow step
1	rate = $k[A]$k[A]	$\times 2$×2	$\times 3$×3	One molecule of A in slow step
2	rate = $k[A]^2$k[A]2	$\times 4$×4	$\times 9$×9	Two molecules of A in slow step

Memorising the fold-change column (×1, ×2, ×4 for doubling; ×1, ×3, ×9 for tripling) is the single most useful trick for A-Level kinetics — almost every initial-rates table can be diagnosed by inspecting these ratios.

Determining orders from data — the ratio method

The workhorse algorithm for OCR initial-rates questions:

1. Pick two experiments where exactly one [reactant] changes.
2. Compute the ratio of [that reactant] (e.g. $[A]_2/[A]_1 = 2$[A]2​/[A]1​=2).
3. Compute the ratio of rates ($\text{rate}_2/\text{rate}_1$rate2​/rate1​).
4. Match against the doubling/tripling table:
   • rate ratio = 1 → order = 0
   • rate ratio = (concentration ratio) → order = 1
   • rate ratio = (concentration ratio)$^2$2 → order = 2
5. Repeat for each reactant.
6. Write the rate equation and substitute one experiment's data to find $k$k.

Worked example 1 — two reactants, four experiments

Consider $A + B \rightarrow \text{products}$A+B→products with the following initial-rate data:

Expt	[A] / mol dm$^{-3}$−3	[B] / mol dm$^{-3}$−3	initial rate / mol dm$^{-3}$−3 s$^{-1}$−1
1	0.10	0.10	$2.0 \times 10^{-4}$2.0×10−4
2	0.20	0.10	$4.0 \times 10^{-4}$4.0×10−4
3	0.30	0.10	$6.0 \times 10^{-4}$6.0×10−4
4	0.10	0.20	$8.0 \times 10^{-4}$8.0×10−4

Step 1 — order w.r.t. A (compare Expts 1 and 2): [B] is constant. [A] doubles (0.10 → 0.20), rate doubles ($2.0 \times 10^{-4} \to 4.0 \times 10^{-4}$2.0×10−4→4.0×10−4). $\text{ratio} = 2$ratio=2, so order = 1.

Step 1 check (compare Expts 1 and 3): [A] triples (0.10 → 0.30), rate triples ($\times 3$×3). Consistent with first order.

Step 2 — order w.r.t. B (compare Expts 1 and 4): [A] is constant. [B] doubles, rate quadruples ($2.0 \times 10^{-4} \to 8.0 \times 10^{-4}$2.0×10−4→8.0×10−4). Order = 2.

Step 3 — rate equation: rate = $k[A][B]^2$k[A][B]2, overall order = 3.

Step 4 — find $k$k using Expt 1:

k &= \frac{\text{rate}}{[A][B]^2} = \frac{2.0 \times 10^{-4}}{(0.10)(0.10)^2} = \frac{2.0 \times 10^{-4}}{1.0 \times 10^{-3}} \\ &= 0.20\;\text{mol}^{-2}\,\text{dm}^{6}\,\text{s}^{-1} \end{aligned}$$ Units check: (mol dm$^{-3}$ s$^{-1}$)/(mol dm$^{-3}$)$^3$ = mol$^{-2}$ dm$^6$ s$^{-1}$ — correct for overall order 3. ### Worked example 2 — zero-order detection | Expt | [X] | [Y] | rate / mol dm$^{-3}$ s$^{-1}$ | |---|---|---|---| | 1 | 0.10 | 0.10 | $3.0 \times 10^{-3}$ | | 2 | 0.20 | 0.10 | $3.0 \times 10^{-3}$ | | 3 | 0.10 | 0.20 | $1.2 \times 10^{-2}$ | Compare 1 and 2: [X] doubles, rate **unchanged** → order w.r.t. X = **0**. Compare 1 and 3: [Y] doubles, rate quadruples → order w.r.t. Y = **2**. Rate equation: **rate = $k[Y]^2$** (X does not appear). $k = (3.0 \times 10^{-3})/(0.10)^2 = 0.30$ mol$^{-1}$ dm$^3$ s$^{-1}$. ### Worked example 3 — non-integer ratios via the log method When [reactant] changes by a non-integer factor (1.5×, 2.5×, etc.), the visual ratio method fails. Take logarithms: $$m = \frac{\log(\text{rate}_2/\text{rate}_1)}{\log([A]_2/[A]_1)}$$ **Example:** [A] increases by 1.5×, rate increases by 2.25×. $$m = \frac{\log(2.25)}{\log(1.5)} = \frac{0.352}{0.176} = \mathbf{2.00}$$ Order = 2. Notice that $1.5^2 = 2.25$, confirming the result by direct substitution. **Second example:** [A] increases by 2.5×, rate increases by 6.25×. $$m = \frac{\log(6.25)}{\log(2.5)} = \frac{0.796}{0.398} = \mathbf{2.00}$$ Again order 2: $2.5^2 = 6.25$. ### Worked example 4 — three reactants | Expt | [P] | [Q] | [R] | rate | |---|---|---|---|---| | 1 | 0.10 | 0.10 | 0.10 | $1.0 \times 10^{-3}$ | | 2 | 0.20 | 0.10 | 0.10 | $4.0 \times 10^{-3}$ | | 3 | 0.10 | 0.20 | 0.10 | $2.0 \times 10^{-3}$ | | 4 | 0.10 | 0.10 | 0.30 | $1.0 \times 10^{-3}$ | Expts 1 & 2: [P] doubles, rate × 4 → order(P) = 2. Expts 1 & 3: [Q] doubles, rate × 2 → order(Q) = 1. Expts 1 & 4: [R] triples, rate unchanged → order(R) = 0. Rate equation: rate = $k[P]^2[Q]$, overall order = 3. ## Pseudo-order experiments — when [B] is held effectively constant A practical trick in physical-chemistry lab work: if [B] is present in **large excess** (typically $[B]_0 > 10 \times [A]_0$), then [B] hardly changes during the reaction, and the rate equation effectively collapses to $$\text{rate} = k_\text{obs}[A]^m \quad \text{where} \quad k_\text{obs} = k[B]^n$$ This is called a **pseudo-$m$-order** experiment: the reaction *behaves* as $m$th order in A even though the true order is higher. Repeating the experiment at several [B] values, plotting $k_\text{obs}$ vs $[B]^n$, gives the true rate constant $k$. This trick simplifies analysis but is not formally on the OCR specification — you may encounter it in University-style chemistry past papers and Olympiad questions. ## Visualising the diagnostic flow ```mermaid graph TD A["Initial-rate data table"] --> B{Change one concentration at a time} B --> C[Compute ratio of reactant] C --> D[Compute ratio of rates] D --> E{rate ratio = 1?} E -->|Yes| F[Order = 0] E -->|No| G{rate ratio = conc ratio?} G -->|Yes| H[Order = 1] G -->|No| I{rate ratio = conc ratio squared?} I -->|Yes| J[Order = 2] I -->|No| K[Use log method or higher order] F --> L[Write rate equation] H --> L J --> L K --> L L --> M[Substitute one experiment to find k with units] ``` <svg viewBox="0 0 640 320" xmlns="http://www.w3.org/2000/svg" role="img" aria-label="Three orders: zero, first, second concentration-time decay shapes"> <rect x="0" y="0" width="640" height="320" fill="#fafafa" stroke="#bbb" stroke-width="1"/> <text x="320" y="22" font-family="Helvetica, Arial, sans-serif" font-size="14" font-weight="bold" fill="#222" text-anchor="middle">[A] vs t — signature shapes of three orders</text> <line x1="60" y1="280" x2="600" y2="280" stroke="#333" stroke-width="1.4"/> <line x1="60" y1="50" x2="60" y2="280" stroke="#333" stroke-width="1.4"/> <text x="330" y="305" font-family="Helvetica, Arial, sans-serif" font-size="12" fill="#333" text-anchor="middle">time t</text> <text x="30" y="170" font-family="Helvetica, Arial, sans-serif" font-size="12" fill="#333" transform="rotate(-90 30 170)">[A]</text> <line x1="60" y1="70" x2="380" y2="270" stroke="#0b3d91" stroke-width="2.4"/> <text x="220" y="148" font-family="Helvetica, Arial, sans-serif" font-size="11" font-weight="bold" fill="#0b3d91">order 0 (linear)</text> <path d="M 60 70 Q 130 130 200 180 Q 300 240 600 275" fill="none" stroke="#28793a" stroke-width="2.4"/> <text x="350" y="220" font-family="Helvetica, Arial, sans-serif" font-size="11" font-weight="bold" fill="#28793a">order 1 (exponential, constant t1/2)</text> <path d="M 60 70 Q 100 200 200 240 Q 350 270 600 278" fill="none" stroke="#a8323a" stroke-width="2.4"/> <text x="350" y="265" font-family="Helvetica, Arial, sans-serif" font-size="11" font-weight="bold" fill="#a8323a">order 2 (lengthening t1/2)</text> </svg> --- ## Synoptic Links > **Synoptic Links** — *Connects to:* > - `ocr-alevel-chemistry-quantitative-rates-equilibrium / rate-of-reaction-and-rate-equations` (Lesson 1 — the rate equation framework that defines what order means). > - `ocr-alevel-chemistry-quantitative-rates-equilibrium / concentration-time-graphs` (Lesson 3 — the [A]-t shape characteristic of each order). > - `ocr-alevel-chemistry-quantitative-rates-equilibrium / rate-concentration-graphs` (Lesson 4 — the rate-[A] plot that distinguishes orders most cleanly). > - `ocr-alevel-chemistry-quantitative-rates-equilibrium / half-life` (Lesson 5 — constant half-life is the signature of first-order kinetics). > - `ocr-alevel-chemistry-quantitative-rates-equilibrium / rate-determining-step` (Lesson 8 — the *meaning* of orders in terms of the elementary mechanism). > - `ocr-alevel-chemistry-basic-organic / haloalkanes-and-substitution` (the $S_N1$ vs $S_N2$ distinction is a kinetic discriminator: $S_N1$ first-order in halide, $S_N2$ second-order overall). *Practical Activity Group anchor:* **PAG 10 (Initial-rate / clock reactions)** is the canonical source of the multi-experiment initial-rate tables analysed here. The iodine-clock and peroxydisulphate–iodide systems both produce 4–6 experiments at varied [reactant] from which orders are deduced. **PAG 9 (Continuous-monitoring rate)** provides the [A]-t curves whose half-life and tangent analysis (Lessons 3, 5) give complementary order information. --- ## Specimen question modelled on the OCR H432 paper format **Question (9 marks):** The reaction $BrO_3^-(aq) + 5Br^-(aq) + 6H^+(aq) \rightarrow 3Br_2(aq) + 3H_2O(l)$ was studied using the initial-rates method. The following data were obtained at 298 K: | Expt | $[BrO_3^-]$ / mol dm$^{-3}$ | $[Br^-]$ / mol dm$^{-3}$ | $[H^+]$ / mol dm$^{-3}$ | initial rate / mol dm$^{-3}$ s$^{-1}$ | |---|---|---|---|---| | 1 | 0.10 | 0.10 | 0.10 | $1.2 \times 10^{-3}$ | | 2 | 0.20 | 0.10 | 0.10 | $2.4 \times 10^{-3}$ | | 3 | 0.10 | 0.30 | 0.10 | $3.6 \times 10^{-3}$ | | 4 | 0.10 | 0.10 | 0.20 | $4.8 \times 10^{-3}$ | (a) Determine the order with respect to each reactant, giving your reasoning. **[3 marks]** (b) Write the rate equation and state the overall order. **[2 marks]** (c) Calculate $k$ with units. **[3 marks]** (d) Explain why the stoichiometric coefficient of 5 for $Br^-$ does not appear as an exponent in the rate equation. **[1 mark]** ### AO breakdown