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Rate of Reaction and Rate Laws — Notes

6 sections 5 min readBeta

1. Rate of reaction

Chemical kinetics studies reaction rates, the factors that affect them and reaction mechanisms. Ionic reactions (e.g. Ag⁺ + Cl⁻ → AgCl) are very fast because no bonds need breaking; molecular reactions are slower because covalent bonds must break.

Average rate =−Δ[R]/Δt=Δ[P]/Δt= -\Delta[\ce{R}]/\Delta t = \Delta[\ce{P}]/\Delta t; instantaneous rate =−d[R]/dt= -d[\ce{R}]/dt, found as minus the slope of the tangent to the concentration–time curve. For a A+b B→c C+d D\ce{aA + bB -> cC + dD}: rate =−1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dt= -\frac{1}{a}\frac{d[\ce{A}]}{dt} = -\frac{1}{b}\frac{d[\ce{B}]}{dt} = \frac{1}{c}\frac{d[\ce{C}]}{dt} = \frac{1}{d}\frac{d[\ce{D}]}{dt}. Units: mol L⁻¹ s⁻¹.

  • Tangent at t = 15 s
  • Concentration of R
Illustrative data. Instantaneous rate = −(slope of the tangent).

Formulas

  • Rate in terms of each species
    Rate=−1ad[A]dt=1cd[C]dt\text{Rate} = -\frac{1}{a}\frac{d[\ce{A}]}{dt} = \frac{1}{c}\frac{d[\ce{C}]}{dt}

Key terms

Rate of reaction

Change in concentration of a reactant or product per unit time, divided by its stoichiometric coefficient.

Instantaneous rate

The rate at a particular moment, equal to minus the slope of the tangent to the concentration–time curve.

2. Factors affecting rate; law of mass action

Factors: nature of reactants (ionic fast, covalent slower); surface area (powders faster); concentration or pressure (more collisions); temperature (rate roughly doubles per 10 K rise); catalyst (lower activation energy); radiation (photochemical reactions such as H₂ + Cl₂ in sunlight, photosynthesis, fading of dyes).

Law of mass action (Guldberg and Waage): at constant temperature, the rate of a reaction is proportional to the product of the active masses of the reactants, each raised to its stoichiometric coefficient. Active mass = molar concentration; for pure solids and liquids it is taken as 1. Strictly valid for elementary reactions.

Table of factors and their effect on rate
FactorEffect on rate
Larger surface areaIncreases
Higher concentrationIncreases
Rise in temperatureIncreases (about ×2 per 10 K)
CatalystIncreases (lower Ea)
Light (photochemical)Starts or speeds reaction

Key terms

Active mass

Molar concentration of a substance; taken as unity for pure solids and liquids.

Law of mass action

Rate is proportional to the product of the active masses of reactants, each raised to its stoichiometric coefficient.

3. Rate law, order and units of k

Rate law: rate =k[A]x[B]y= k[\ce{A}]^x[\ce{B}]^y, with xx, yy found experimentally (not from coefficients). Rate constant kk: rate at unit concentrations; depends on temperature and catalyst, not on concentration. Order =x+y= x + y: may be 0, a whole number or a fraction.

Method of initial rates: compare runs in which only one concentration changes; rate ratio = (concentration ratio)ⁿ.

Units of k: (mol L−1)1−n s−1(\text{mol L}^{-1})^{1-n}\ \text{s}^{-1} — zero order mol L⁻¹ s⁻¹; first order s⁻¹; second order L mol⁻¹ s⁻¹; third order L² mol⁻² s⁻¹.

Table of units of the rate constant
OrderUnit of k
0mol L⁻¹ s⁻¹
1s⁻¹
2L mol⁻¹ s⁻¹
3L² mol⁻² s⁻¹

Formulas

  • Rate law
    Rate=k[A]x[B]y\text{Rate} = k[\ce{A}]^x[\ce{B}]^y
  • Units of k
    unit of k=(mol L−1)1−n s−1\text{unit of } k = (\text{mol L}^{-1})^{1-n}\,\text{s}^{-1}

Key terms

Rate constant

The proportionality constant in the rate law; equal to the rate when all concentrations are 1 mol L⁻¹.

Order of reaction

The sum of the powers of the concentration terms in the experimental rate law.

4. Zero-order reactions

Rate =k= k. Integrated: [R]=[R]0−kt[\ce{R}] = [\ce{R}]_0 - kt; k=([R]0−[R])/tk = ([\ce{R}]_0 - [\ce{R}])/t. Plot of [R] against t: straight line, slope −k-k. Half-life t1/2=[R]0/2kt_{1/2} = [\ce{R}]_0/2k (proportional to initial concentration). Examples: decomposition of NH₃ on hot Pt at high pressure; enzyme reactions at saturation.

  • Zero order: straight line, slope −k
k = 0.02 mol L⁻¹ min⁻¹ in this example.

Formulas

  • Zero-order integrated equation
    [R]=[R]0−kt[\ce{R}] = [\ce{R}]_0 - kt
  • Zero-order half-life
    t1/2=[R]02kt_{1/2} = \frac{[\ce{R}]_0}{2k}

Key terms

Zero-order reaction

A reaction whose rate is independent of reactant concentration.

5. First-order reactions

Derivation: −d[R]/dt=k[R]-d[\ce{R}]/dt = k[\ce{R}] → d[R]/[R]=−k dtd[\ce{R}]/[\ce{R}] = -k\,dt → ln⁡([R]0/[R])=kt\ln([\ce{R}]_0/[\ce{R}]) = kt, so k=2.303tlog⁡[R]0[R]k = \dfrac{2.303}{t}\log\dfrac{[\ce{R}]_0}{[\ce{R}]} and [R]=[R]0e−kt[\ce{R}] = [\ce{R}]_0e^{-kt}. Plot of log[R] against t: straight line, slope −k/2.303-k/2.303.

Half-life: t1/2=0.693/kt_{1/2} = 0.693/k, independent of [R]0[\ce{R}]_0; kk is also independent of initial concentration. After nn half-lives, [R]=[R]0/2n[\ce{R}] = [\ce{R}]_0/2^n. Time for 99.9% reaction ≈ 10 t1/210\,t_{1/2}.

Examples: radioactive decay, decomposition of N₂O₅, elimination of many drugs. Pseudo-first-order: one reactant in large excess (acid hydrolysis of an ester in water; inversion of cane sugar).

  • First order, half-life 8 min
Each half-life takes the same time, whatever the starting concentration.
  • log[R] = log[R]₀ − kt ÷ 2.303
Illustrative: slope −0.03 min⁻¹, so k = 2.303 × 0.03 = 0.069 min⁻¹.

Formulas

  • First-order integrated equation
    k=2.303tlog⁡[R]0[R]k = \frac{2.303}{t}\log\frac{[\ce{R}]_0}{[\ce{R}]}
  • First-order half-life
    t1/2=0.693kt_{1/2} = \frac{0.693}{k}

Key terms

Half-life

The time taken for the concentration of a reactant to fall to half its initial value.

Pseudo-first-order reaction

A reaction of higher order that behaves as first order because one reactant is in large excess.

6. Molecularity, mechanism and the rate-determining step

Elementary reaction: one step; complex reaction: several steps (mechanism). Molecularity: number of reacting species colliding in an elementary step — 1, 2 or (rarely) 3, never 0 or fractional; defined only for elementary steps. Order: experimental, for the overall reaction; may be 0 or fractional.

The slowest step is the rate-determining step; the rate law follows from it. Intermediates are eliminated using a fast pre-equilibrium. Example: NO₂ + CO → NO + CO₂ at low T: slow step 2NO₂ → NO₃ + NO gives rate = k[NO₂]².

Table comparing order and molecularity
OrderMolecularity
SourceExperimentMechanism
Zero possible?YesNo
Fraction possible?YesNo
Applies toOverall reactionElementary step

Key terms

Molecularity

The number of reacting species taking part in an elementary step.

Rate-determining step

The slowest step in a reaction mechanism, which controls the overall rate.

Exam tips

Common mistakes

  • Writing orders from the coefficients of the overall equation.
  • Forgetting to divide rates by stoichiometric coefficients.
  • Using the percentage reacted instead of the fraction remaining in first-order calculations.
  • Saying the half-life of a first-order reaction depends on the initial concentration.
  • Including an intermediate in the final rate law.

How marks are usually awarded

  • Derivation of the first-order equation (3 marks): differential rate law, integration with limits, final form.
  • Numericals: equation, substitution with units, answer with units of k or time.
  • Initial-rate problems: order with respect to each reactant with reasoning, rate law, k with units.