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Chemical Kinetics — Notes

5 sections 6 min readBeta

1. Rate of a reaction and factors affecting it

The rate of a reaction is the change in concentration of a reactant or product per unit time (unit mol L−1 s−1\text{mol L}^{-1}\ \text{s}^{-1}; for gases, atm s−1\text{atm s}^{-1}). For R→P\mathrm{R} \rightarrow \mathrm{P}: average rate =−Δ[R]Δt=Δ[P]Δt= -\dfrac{\Delta[\ce{R}]}{\Delta t} = \dfrac{\Delta[\ce{P}]}{\Delta t}; the instantaneous rate is the slope of the tangent to the concentration–time curve, −d[R]dt-\dfrac{d[\ce{R}]}{dt}.

For a A+b B→c C+d D\ce{aA + bB -> cC + dD}, rate =−1aΔ[A]Δt=−1bΔ[B]Δt=1cΔ[C]Δt=1dΔ[D]Δt= -\dfrac{1}{a}\dfrac{\Delta[\ce{A}]}{\Delta t} = -\dfrac{1}{b}\dfrac{\Delta[\ce{B}]}{\Delta t} = \dfrac{1}{c}\dfrac{\Delta[\ce{C}]}{\Delta t} = \dfrac{1}{d}\dfrac{\Delta[\ce{D}]}{\Delta t}.

Factors: concentration (or pressure) of reactants, temperature, catalyst, surface area of solids, nature of reactants (and, for photochemical reactions, light).

  • Tangent at t = 10 s
  • Concentration of R
Illustrative data. The rate is minus the slope of the tangent.

Formulas

  • Rate in terms of species
    Rate=−1aΔ[A]Δt=1cΔ[C]Δt\text{Rate} = -\dfrac{1}{a}\dfrac{\Delta[\ce{A}]}{\Delta t} = \dfrac{1}{c}\dfrac{\Delta[\ce{C}]}{\Delta t}

Key terms

Rate of reaction

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

Instantaneous rate

Rate at a particular moment: the slope of the tangent to the concentration–time curve.

2. Rate law, order and molecularity

Rate law: rate =k[A]x[B]y= k[\ce{A}]^x[\ce{B}]^y, with xx and yy found experimentally (they need not equal the coefficients). kk is the rate constant: the rate at unit concentrations; it depends on temperature, not on concentration.

Order =x+y= x + y; it may be 0, a whole number or a fraction. Unit of kk for order nn: (mol L−1)1−n s−1(\text{mol L}^{-1})^{1-n}\ \text{s}^{-1} — zero order mol L−1 s−1\text{mol L}^{-1}\ \text{s}^{-1}, first s−1\text{s}^{-1}, second L mol−1 s−1\text{L mol}^{-1}\ \text{s}^{-1}.

Method of initial rates: compare runs in which only one concentration changes: r2r1=([A]2[A]1)x\dfrac{r_2}{r_1} = \left(\dfrac{[\ce{A}]_2}{[\ce{A}]_1}\right)^x.

Molecularity is the number of reacting species colliding in an elementary step: 1, 2 or (rarely) 3; never zero or fractional. For a complex reaction, the slowest step is rate-determining; e.g. iodide-catalysed decomposition of HX2OX2\ce{H2O2}: slow HX2OX2+IX−→HX2O+IOX−\ce{H2O2 + I- -> H2O + IO-}, fast HX2OX2+IOX−→HX2O+IX−+OX2\ce{H2O2 + IO- -> H2O + I- + O2}, so rate =k[HX2OX2][IX−]= k[\ce{H2O2}][\ce{I-}].

Table comparing order and molecularity
OrderMolecularity
BasisExperimentMechanism (elementary step)
Values0, whole, fractional1, 2, 3 only
Complex reactionsDefined overallFor each step

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 law

Expression of the rate in terms of reactant concentrations raised to experimentally determined powers.

Order of reaction

Sum of the powers of the concentrations in the rate law.

Molecularity

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.

3. Integrated rate equations and half-life

Zero order: [R]=[R]0−kt[\ce{R}] = [\ce{R}]_0 - kt; [R][\ce{R}] vs tt is a straight line of slope −k-k; t1/2=[R]02kt_{1/2} = \dfrac{[\ce{R}]_0}{2k}. Examples: 2 NHX3→NX2+3 HX2\ce{2NH3 -> N2 + 3H2} on hot platinum at high pressure; decomposition of HI on gold.

First order: ln⁡[R]=ln⁡[R]0−kt\ln[\ce{R}] = \ln[\ce{R}]_0 - kt, or k=2.303tlog⁡[R]0[R]k = \dfrac{2.303}{t}\log\dfrac{[\ce{R}]_0}{[\ce{R}]}; log⁡[R]\log[\ce{R}] vs tt is a straight line of slope −k/2.303-k/2.303; t1/2=0.693kt_{1/2} = \dfrac{0.693}{k}, independent of [R]0[\ce{R}]_0. After nn half-lives, [R]=[R]0/2n[\ce{R}] = [\ce{R}]_0/2^n. Examples: radioactive decay, decomposition of NX2OX5\ce{N2O5}, hydrogenation of ethene. For a gas-phase reaction A(g)→B(g)+C(g)\mathrm{A(g)} \rightarrow \mathrm{B(g)} + \mathrm{C(g)} followed by total pressure, k=2.303tlog⁡p02p0−ptk = \dfrac{2.303}{t}\log\dfrac{p_0}{2p_0 - p_t}.

Pseudo-first-order: one reactant in large excess (e.g. water in the acid hydrolysis of ethyl ethanoate, inversion of cane sugar) so that rate ≈k′[other reactant]\approx k'[\text{other reactant}].

  • First order, half-life 10 min
Each half-life takes the same time, whatever the starting concentration.
  • Zero order: straight line, slope −k
k = 0.02 mol L⁻¹ min⁻¹ in this example.

Formulas

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

Key terms

Half-life

Time for the concentration of a reactant to fall to half its initial value.

Pseudo-first-order reaction

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

4. Temperature dependence: the Arrhenius equation

For many reactions a 10 K rise roughly doubles the rate. Arrhenius equation: k=Ae−Ea/RTk = Ae^{-E_\text{a}/RT}, where AA is the frequency factor and EaE_\text{a} the activation energy; e−Ea/RTe^{-E_\text{a}/RT} is the fraction of molecules with energy at least EaE_\text{a}. ln⁡k=ln⁡A−EaRT\ln k = \ln A - \dfrac{E_\text{a}}{RT}: a plot of ln⁡k\ln k against 1/T1/T is a straight line of slope −Ea/R-E_\text{a}/R.

Two temperatures: log⁡k2k1=Ea2.303R(T2−T1T1T2)\log\dfrac{k_2}{k_1} = \dfrac{E_\text{a}}{2.303R}\left(\dfrac{T_2 - T_1}{T_1T_2}\right), with R=8.314 J K−1 mol−1R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}.

Activation energy is the extra energy reactants need to form the activated complex at the top of the energy profile. At higher temperature, the Maxwell–Boltzmann distribution spreads out and the fraction of molecules beyond EaE_\text{a} rises sharply.

Catalyst: provides an alternative path of lower EaE_\text{a}; is unchanged at the end; does not change ΔH\Delta H, ΔG\Delta G or KK; speeds up forward and backward reactions equally.

Schematic, not to scale.
  • Temperature T
  • Temperature T + 10 K
Illustrative curves; the temperature gap is exaggerated so that the difference is visible.
  • slope = −Ea ÷ (1000 R)
Illustrative line with slope −8, so Ea = 8 × 1000 × 8.314 J mol⁻¹ ≈ 66.5 kJ mol⁻¹.

Formulas

  • Arrhenius equation
    k=A e−Ea/RTk = A\,e^{-E_\text{a}/RT}
  • Two temperatures
    log⁡k2k1=Ea2.303R(T2−T1T1T2)\log\dfrac{k_2}{k_1} = \dfrac{E_\text{a}}{2.303R}\left(\dfrac{T_2 - T_1}{T_1T_2}\right)

Key terms

Activation energy

Minimum extra energy that reacting molecules must have to form the activated complex.

Activated complex

Unstable arrangement of atoms at the top of the energy barrier.

Catalyst

A substance that increases the rate by providing a path of lower activation energy and is unchanged at the end.

5. Collision theory

Reactant molecules behave as hard spheres; a reaction happens only through effective collisions: those with energy at least the threshold energy and with proper orientation. Rate =P ZAB e−Ea/RT= P\,Z_\text{AB}\,e^{-E_\text{a}/RT}, where ZABZ_\text{AB} is the collision frequency and PP the probability (steric) factor.

Raising the concentration increases ZABZ_\text{AB}; raising the temperature increases ZABZ_\text{AB} slightly and the energetic fraction greatly; a catalyst lowers EaE_\text{a}. Limitation: molecules are treated as structureless spheres, so PP is an empirical correction.

Schematic: real molecules are three-dimensional.

Formulas

  • Collision theory
    Rate=P ZAB e−Ea/RT\text{Rate} = P\,Z_\text{AB}\,e^{-E_\text{a}/RT}

Key terms

Effective collision

A collision with enough energy and the correct orientation to form products.

Threshold energy

Minimum total energy colliding molecules must have for an effective collision.

Steric factor

Probability factor P that allows for the fraction of collisions with suitable orientation.

Exam tips

Common mistakes

  • Taking the order from the coefficients of the overall equation.
  • Forgetting to divide by the coefficient (and the minus sign) when relating rates of different species.
  • Using t1/2=0.693/kt_{1/2} = 0.693/k for a zero-order reaction.
  • Inverting the ratio inside the logarithm in k=2.303tlog⁡[R]0[R]k = \dfrac{2.303}{t}\log\dfrac{[\text{R}]_0}{[\text{R}]}.
  • Using kJ with R=8.314 J K−1 mol−1R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}, or °C instead of K in the Arrhenius equation.
  • Saying a catalyst shifts the equilibrium or changes ΔH\Delta H.
  • Calling molecularity zero or fractional.

How marks are usually awarded

  • 1-mark items: correct option, order, unit of k or definition.
  • Numericals (2–3 marks): correct formula (½–1) + substitution (1) + answer with unit (½–1).
  • Initial-rate problems: order w.r.t. each reactant (1 each) + rate law and k with unit (1).
  • 5-mark answers: derivation of the first-order equation or half-life, a labelled graph, and a numerical.