Lecture 6: Diffusion and Reaction kinetics 12-10-2010

• Lecture plan: – diffusion • thermodynamic forces • evolution of concentration distribution – reaction rates and methods to determine them – reaction mechanism in terms of the elementary reaction steps – basic reaction types – problems Diffusion

• Thermodynamic force If the chemical potential depends on position, the maximum nonexpansion work

μ ⎛⎞∂μ dw== d dx ⎜⎟ w ⎝⎠∂x pT, = Δμ μ μ + Δμ Comparing with dw= − Fdx

⎛⎞∂μ F =−⎜⎟ Thermodynamic force ⎝⎠∂x pT, Diffusion

• Thermodynamic force of concentration gradient

μ = μ0 + RTaln

For ideal solution ⎛⎞∂ ln a RT⎛⎞∂ c FRT=− F =− ⎜⎟ ⎜⎟ cx∂ ⎝⎠∂x pT, ⎝⎠pT,

Fick’s law of diffusion: ∂c Particles flux: J~~~ drift velocity F ∂x Diffusion • The Einstein relation ∂c sAtcΔ JD=− Jsc= = ∂x AtΔ

∂c sc=− D ∂x D ∂cDF s =− = cx∂ RT In the case of charged particle (ion) in electric field we know drift speed vs force relation for ion mobility, so we can deduce diffusion constant DzeN E uRT suE== A D = Einstein relation RT zF

−82 −92 For example: for μ =⋅510msV / we find D =⋅110ms / Diffusion

• The Nernst-Einstein equation zD22F Molar conductivity of ions in the solution λμ= z F= RT F2 Λ=()ννzD22 + zD m +++ −−−RT

• The Stokes-Einstein equation ezE μE= zeRTzekN T kT f D ==A = Frictional force fzfzeNfF μRT A D = zF kT Using Stokes’s law D = No charge involved -> applicable to 6πηa all molecules Diffusion equation

• How concentration distribution evolves with time due to diffusion

∂−c JAdt J′′ Adt J J =− = ∂t Aldt Aldt l ∂cc∂∂∂∂′ c⎛⎞ c ∂2 c ′ ⎛⎞ JJ−=−+ D D =−+ D D⎜⎟ c +⎜⎟ l = Dl 2 ∂xx∂∂∂∂ xx⎝⎠⎝⎠ x ∂ x ∂∂cc2 = D ∂∂tx2

Diffusion with convection ∂−c JAdt J′′ Adt J J =− = ∂t Aldt Aldt l ⎛⎞⎛⎞⎛⎞∂cc∂ JJ−=−′′ cvcvc =−+⎜⎟⎜⎟ c⎜⎟ lvvl = ⎝⎠⎝⎠⎝⎠∂xx∂ ∂∂cc = v due to convection only ∂∂tx ∂ccc∂∂2 =−D v ∂txx∂∂2 Solution of diffusion equation

∂∂∂ccc2 =−D v ∂∂∂txx2

•2nd order differential equation: two boundary condition are required for spatial dependence and single for time dependence

Example 1: one surface of a container (x=0) is coated

with N0 molecules at time t=0

n 2 cxt(,)= 0 e−xDt/4 ADt()π 1/2

Example 2: dissolution infinitely small solid containing

N0 molecules at time t=0 in 3D solvent n 2 crt(,)= 0 e−rDt/4 8(π Dt )3/2 Diffusion probabilities

• Probability to find a particle at a given slab of thickness dx is proportional to the concentration there: p()xcxANdxN= ()A / 0 • The mean distance traveled by the particles:

∞∞1/2 cxANdx() 1 2 ⎛⎞Dt xxedx==A −xDt/4 =2 ∫∫1/2 ⎜⎟ 00NDt0 ()ππ⎝⎠

1/2 x21/2= (2Dt )

Diffusion in unstirred solution, D=5x10-10 m2/S Random Walk

• Apparently diffusion can be modeled as a random walk, where particle is jumping distance λ in a time τ. Direction of the jump is chosen randomly • One dimensional walk:

1/2 ⎛⎞2τ −x22τ /2tλ pe= ⎜⎟ ⎝⎠πt Comparing with the solution of diffusion equation

λ 2 D = Einstein-Smoluchowski equation 2τ

Connection between microscopic and macroscopic parameters. RATES OF CHEMICAL REACTIONS Rates of chemical reactions

AB+ 23⎯⎯→+ CD

Instantaneous rate of consumption of a reactant: −dR[]/ dt Instantaneous rate of formation of a product: dP[]/ dt

From stoichiometry dD[] 1[] dC dA [] 1[] dB ==−=− dt32 dt dt dt Rate of the reaction: 1 dn dξ v ==i ν i dt dt In case of heterogeneous reaction the rate will be defined as mol/m2s Reaction order Reaction rate is generally dependent on temperature, pressure, concentration of species, the phases where reaction occurs etc. However, an empirical relation called a rate law exists stating that: vkAB= [][]...ab algebraic dependence on the reagent concentration raised to some power reaction order with respect to species A: α overall reaction order: α + β.... rate constant, depends only on T and not on the concentration

The power is generally not equal to the stoichiometric coefficients, has to be determined from the experiment

Order of a reaction:

−1 vk= Zero order. ⎡⎣M ⋅ s ⎤⎦

−−11 vkAB= [][] First order in A, first order in B, overall second order. ⎣⎡M ⋅ s ⎦⎤

vkAB= [][]1/2 Halforder in A, first order in B, overall tree-halves order. Measuring the rates of chemical reactions

• Experimental measuring progress of the reaction

– Monitoring pressure in the reaction involving gases

2NO25 ()g ⎯⎯→+ 4NO 2 ()g O 2 ()g 1 nnn(1−ααα ) 2 2 3 pp=+(1 ) 0 2 α

– Absorption at particular wavelength (e.g. Br2 below)

Hg22()+⎯⎯Br ()g → 2HBr ()g

– Conductance of the ionic solution

+− (CH33 ) CCl () aq+⎯⎯ H 2 O () aq→++ ( CH 33 ) COH () aq H () aq Cl () aq Measuring the Rates of chemical reactions

concentration as concentration as function of position function of time

•Flow method •Stopped-flow method (down to 1 ms range)

•Flash photolysis (down to 10 fs = 10-14 s range) •Chemical quench flow •Freeze quench method Measuring the rates of chemical reactions

• Determination of the rate law Usual technique is the isolation method, where all the components except one are present in large amounts (therefore their concentration is constant)

vkAB= [][]0 = kA′ [] It usually accompanied by the method of initial rates, when several initial concentration of A measured (again assuming that the concentrations are constant) a vkA00==+′′[ ] log v 0 log ka log[ A 0 ] Measuring the rates of chemical reactions

• Example 2()()()()Ig+⎯⎯ Arg→+ I2 g Arg

A B

α = 2 β =1 Integrated rate laws

• First order reaction. Slope -> k – Let’s find concentration of reagent A after time t At dA[] dA[] =−kA[] =−kdt ∫ []A ∫ dt A0 0

⎛⎞[]A −kt ln ⎜⎟=−kt []AAe= [0 ] ⎝⎠[]A0 – Half-life – time required for concentration to drop by ½. ⎛⎞1 []A ⎜⎟2 0 ln 2 kt1/2=−ln⎜⎟ = ln 2 t 1/2 = ⎜⎟[]A0 k ⎝⎠ – Time constant – time required for concentration to drop by 1/e: 1 τ = k Integrated rate laws • Second-order reactions

AtdA[] dA[] 2 =−kdt =−kA[] ∫ []A 2 ∫ dt A0 0

11 []A Second order −=kt []A = 0 []AA [0 ] 1[]+ kt A0 – Half-life – depends on initial concentration First order

11[]A0 []At01/2== 21[][]+ kt1/2 A 0 k A 0

The concentration of the reagent drops faster in the 1st order reaction than in the 2nd order reaction Reaction approaching equilibrium • Generally, most kinetics measurements are made far from equilibrium where reverse reactions are not important. Close to equilibrium the amount of products is significant and reverse reaction should be considered. A ⎯⎯→=BvkA[] BAvkB⎯⎯→='[ ]

dA =−kA[] + k′ [] B [][] A + B = [] A dt 0 dA =−kA[] + k′ ([] A − []) A dt 0 kke′ + −+()kkt′ []A = []A kk+ ′ 0

kk′ []B eq k []AABAKeq= [],[]00 eq === [], kk++′ kk′′[] Aeq k

kkab Generally: K = ××... kkab′ ′ The temperature dependence of reaction rate

• The rate constants of most reaction increase with the temperature. • Experimentally for many reactions k follows Arrhenius equation E lnkA=− ln a RT

A – pre-exponential (frequency) factor,

Ea – activation energy

High activation energy means that rate constants depend strongly on the temperature, zero would mean reaction independent on temperature The temperature dependence of reaction rate

• Stages of reaction: – Reagents

– Activation complex Transition state – Product

kAe= −ERTa /

Fraction of collision with required energy

•Activation energy is the minimum kinetic energy reactants must have to form the products. •Pre-exponential is rate of collisions reaction coordinate: e.g. •Arrhenius equation gives the rate of changes of interactomic successful collisions. distances or angles Elementary reactions

• Most reactions occur in a sequence of steps called elementary reactions. • Molecularity of an elementary reaction is the number of molecules coming together to react (e.g. uni-molecular, bimolecular)

Uni-molecular: first order in the reactant

dA[] A ⎯⎯→=−PkA[] dt

Bimolecular: first order in the reactant

dA[] A +⎯⎯BP→=− kAB[][] dt

Proportional to collision rate

H +⎯⎯Br2 →+ HBr Br Consecutive elementary reactions

A ⎯kk⎯ab→⎯⎯IP→

dA[] =−kA[] dt a dI[] =−kA[] kI [] dt ab dP[] = kI[] dt b

−kta Solution for A should be in a form: []A = []Ae0

dI[] −−−ktaabka kt kt +=kIba[] kAe []00 [] I = ( e − e )[] A dt kba− k

−−ktab kt ⎛⎞keab− ke [][][]A ++IPA = []00 [] P =+⎜⎟ 1 [] A ⎝⎠kkba− Consecutive elementary reactions

dI[] = 0 • The (quasi) steady-state approximation dt dI[] =−=kA[] kI []0 Then dt ab dP[] and =≈kI[] kA [ ] dt ba

−kta []PeA≈−( 1) []0 Consecutive elementary reactions

• Pre-equilibria

′ AB+ ←⎯kkaa, ⎯→⎯⎯ I k b→ P

This condition arises when k’a >>kb. []I k Then K ==a [][]A Bka′

dP[] Second order form with composite rate and ==kIbb[] kKAB [][] dt constant k Unimolecular reactions • As the molecule acquires the energy as a result of collision why the reaction is still a first order?

• Lindemann-Hinshelwood mechanism

dA[]* AA+⎯⎯ka →+ A*2 A = kA[] dt a * ′ dA[] A +⎯⎯AAAkAA**ka →+ =−′[][] dt a dA[]* AP**⎯⎯kb →=− kA[] dt b

If the last step is rate-limiting the overall reaction will have first order kinetics Lindemann-Hinshelwood mechanism

* * dA[] dA[] 2* * ka *2 ′ AA+⎯⎯→+ A A = kA[] =−kAaa[] kA [ ][] A − kA b [ ] dt a dt 2 * * kA[] **k′ dA[] []A = a A +⎯⎯AAA→+ =− kAAa′[][] dt kkAba+ ′[] * dP kk[] A2 **kb dA[] * ba AP⎯⎯→=− kAb[] ==kAb[] dt dt kba+ k′[] A

If the rate of deactivation is much higher that unimolecular decay than:

dP kk[] A2 kk [] A =≈ba ba dt kba+ k′′[] A k a

The Lindemann-Hinshelwood mechanism can be tested by reducing the pressure (slowing down the activation step) so the reaction will switch to the second order. The activation energy of the composite reaction

Let’s consider Lindemann-Hinshelwood mechanism and apply Arrhenius-like temperature dependence to each rate constant

Ae−−Eaa()/ a RT Ae E ()/ b RT kk()ab( ) AA ′ ke==ab = a b −+−{()EaaaaEaEa () ()}/ RT ()EaRT′ ()/ ′′− a kAaaAea′

Overall activation energy can be positive or negative

Eaaaa()+> Eb () Ea′ ()

Eaaaa()+< Eb () Ea′ () Problems • P20.35. The diffusion coefficient of particular RNA molecule is 1.0x10-11 m2/s. Estimate time required for a molecule to diffuse 1 um from nucleus to the cell wall

• E21.11a The second-order rate constant for the reaction

- - CH3COOC2H5(aq) + OH (aq) → CH3CO2 (aq) + CH3CH2OH(aq) is 0.11 dm3 mol-1 s-1. What is the concentration of ester after (a) 10 s, (b) 10 min when ethyl acetate is added to sodium hydroxide so that the initial -3 concentrations are [NaOH] = 0.050 mol dm and [CH3COOC2H5] = 0.100 mol dm-3? • E21.17a The effective rate constant for a gaseous reaction that has a Lindemann–Hinshelwood mechanism is 2.50 × 10-4 s-1 at 1.30 kPa and 2.10 × 10-5 s-1 at 12 Pa. Calculate the rate constant for the activation step in the mechanism.