2015, Fall Semester Physical Chemistry II (Class 001 / 458 ...

2015, Fall SemesterPhysical Chemistry II (Class 001 / 458.202)

Professor

김대형 : 302-816, 880-1634, [email protected]

Classroom : 302-508

Class time : Tuesday, Thursday 9:30 ~ 10:45

Teaching Assistant

도경식: 311-315, [email protected]

김동찬: 311-516, [email protected]

Textbook

Atkins, Physical Chemistry, 9th ed. 2010 Oxford Univ. Press [Chapter 20 ~ 23]

Atkins, Physical Chemistry, 10th ed. 2014 Oxford Univ. Press [Chapter 19 ~ 22]

Week Lecture Chapter Dates

1 Orientation, Molecular motion in gases 20 9/1, 9/3

2 Molecular motion in liquids 20 9/8, 9/10

3 Diffusion 20 9/15, 9/17

4 Diffusion, Empirical chemical kinetics 20, 21 9/22, 9/24

5 No Class (추석), Accounting for the rate laws, Ch 20 problems 보강 (10/2 6:00) 21 9/29, 10/1

6 Accounting for the rate laws 21 10/6, 10/8

7 Examples of reaction mechanisms, 개교기념일 (수업) 21 10/13, 10/15

8 Reactive encounters, Ch 21 problems 보강 (10/23 6:00) 22 10/20, 10/22

9 Activated complex theory 22 10/27, 10/29

10 Dynamics of molecular collisions, Midterm (11/5, 8:00~11:00) 22 11/3, 11/5

11 Homogeneous catalysis 23 11/10, 11/12

12 Ch 22 Summary (practice problems), No Class 23 11/17, 11/19

13 No Class, Homogeneous catalysis 23 11/24, 11/26

14 Ch 23 Summary (practice problems), No Class 23 12/1, 12/3

15 Supplementary class, Final (12/10, 8:00~11:00) 23 12/8, 12/10

Overview of Chapter 22“Reaction dynamics”

1. Reactive encounters

(1) Collision theory

(2) Diffusion controlled reactions

(3) The material balance equation

2. Transition state theory

(1) The Eyring equation

(2) Thermodynamic aspects

3. The dynamics of molecular collisions

(1) Reactive collisions

(2) Potential energy systems

(3) Some results from experiments and calculations

3. The dynamics of electron transfer

(1) Electron transfer in homogeneous systems

(2) Electron transfer processes at electrodes

Ch. 22. Molecular reaction dynamics* How to build reaction rate laws in Ch 21 A differential equation and its solution

* Account of temperature dependent reaction rate constant change Arrhenius law

* In Ch 22, quantitative account (mathematical calculation) of reaction rate (constant)

* Reaction molecules i) must meet each other in appropriate reaction coordinates and ii)

have certain minimum energy, which is larger than the energy barrier, to react each

other, and these requirements are reflected in the reaction rate (constant).

* Reaction rates in gas phase:

explained by collision theory (kinetics), in which the condition of collision determines

the reaction probability (the simplest quantitative account of reaction rates, which can

be applied to only simple gas phase species)

* Reaction rates in liquid phase:

(i) diffusion controlled case → explained by diffusion equation, in which the diffusion of

reaction molecules determines the reaction probability

(ii) activation-controlled case → explained by transition state theory, in which the

activation of reactants determines the reaction probability (it is assumed that reactant

molecules form an activated complex, which can be discussed by the population in the

energy level. Also this inspires a thermodynamic approach to reaction rates, in which

the rate constant is expressed in terms of thermodynamic parameters.)

Reactive encounters22.1 Collision theory (for gas phase)

* For bimolecular elementary reaction: A+B → P, v = kr[A][B] (from stoichiometry)

* Following three rules are supposed in the collision theory.

(1) Qualitative explanation of reaction rate “v” ∝ Rate of collisions (high collision frequency) high chance of reaction fast reaction

∝ Mean speed of molecules (ĉ = ∫vf(v)dv = [8RT/πM]½ ∝(T/M)1/2) (M: molar mass)

∝ collision cross-section (σ) (large cross-section high collision probability)

∝ number densities (NA, NB) (more particles high collision probability)

Reaction rate, v = kr[A][B] ∝ σ(T/M)1/2NANB ∝ σ(T/M)1/2[A][B]

(2) Other considerations

2-1) For successful collision, kinetic E should exceed a minimum value Ea (activation E)

~ Not every collision leads to reaction. Only certain proportion (E>Ea, Boltzmann factor).

Rate constant ∝Boltzmann factor (exp(-Ea/RT)), a probability term due to activation E

kr ∝ σ(T/M)1/2exp(-Ea/RT)

2-2) Reactants may need to collide in certain relative orientation, “steric requirement (P)”

kr ∝ P σ(T/M)1/2 exp(-Ea/RT)

kr ∝ (steric requirement) (encounter rate/frequency) (minimum E requirement)

v ∝ P σ(T/M)1/2 exp(-Ea/RT) [A][B] (We will justify this in next slides)

(a) Collision rates (collision frequency) in gases

*Reaction rate ( kr) depends on the frequency of molecular collision (collision density).

*Definition of collision density, ZAB = number of (A, B) collisions / (volume interval)

*Gas collision frequency (eqn 20.11a), z = σ črelŇ (where črel = [8kT/πμ]½ , Ň = N/V)

for A, B molecules, ZAB = σ (8kT/πμ)1/2 NA2 [A][B]

for A, A molecules, ZAA = σ (4kT/πmA)1/2 NA2 [A]2

(NA: Avogadro #, μ = mAmB/(mA+mB): reduced mass,

σ = πd2: collision cross-section, d=(dA+dB)/2, See Fig. 22.1)

Justification 22.1 collision density

Collision frequency, z, for a single A gas molecule of mass mA, in other A molecules

z = σčrelŇA

The total collision density is “collision frequency” “number density of A molecules”

For A-A, ZAA = 1/2zŇA (the factor ½ is due to avoid double counting of collision)

For A-B, ZAB = zŇB (the factor ½ is discarded since we are considering A and B)

For A-A, μ = mAmA/(mA + mA) = mA/2, ŇA = NA[A] (NA:Avogadro #), črel = (8kT/πμ)1/2

ZAA = 1/2σčrelŇA2 = 1/2σ(8kT/πμ)1/2 ŇA

2 = σ(4kT/πmA)1/2 NA2[A]2

ZAB = σčrelŇAŇB = σ(8kT/πμ)1/2 ŇAŇB = σ(8kT/πμ)1/2 NA2[A][B]

End of Justification

Justification

Fig. 22.1. Collision cross-section, σ. Between

projectile molecule(A) & target molecule(B)

v ∝ P σ(T/M)1/2 exp(-Ea/RT) [A][B]

Exercise (#1) Calculate the collision frequency, z, and the collision density, Z, in ammonia,

R=190pm, at 25°C and 100kPa. What is the percentage increase when the temperature is raised by

10K at constant volume? (MNH3 = 17.03 g/mol)

The collision frequency is 𝒛 = 𝝈𝒄𝒓𝒆𝒍ℵ [20.11a]

where 𝒄𝒓𝒆𝒍 =𝟖𝒌𝑻

𝝅𝝁

𝟏/𝟐= 𝟒

𝒌𝑻

𝝅𝒎

𝟏/𝟐[20.10] 𝝈 = 𝝅𝒅𝟐 section 20.1b = 𝟒𝝅𝑹𝟐, and ℵ =

𝒑

𝒌𝑻

Therefore, 𝒛 = 𝝈𝒄𝒓𝒆𝒍ℵ = 𝟒𝝅𝑹𝟐 ×𝒑

𝒌𝑻× 𝟒

𝒌𝑻

𝝅𝒎

𝟏

𝟐= 𝟏𝟔𝒑𝑹𝟐 𝝅

𝒎𝒌𝑻

𝟏

𝟐

= 𝟏𝟔 × 𝟏𝟎𝟎 × 𝟏𝟎𝟑𝑷𝒂 × (𝟏𝟗𝟎 × 𝟏𝟎−𝟏𝟐𝒎)𝟐

×𝝅

(𝟏𝟕. 𝟎𝟑 × 𝟏. 𝟔𝟔𝟏 × 𝟏𝟎−𝟐𝟕𝒌𝒈)𝟐× 𝟏. 𝟑𝟖𝟏 × 𝟏𝟎−𝟐𝟑𝑱𝑲−𝟏 × (𝟐𝟗𝟖𝑲)

𝟏/𝟐

= 𝟗. 𝟒𝟗 × 𝟏𝟎𝟗𝒔−𝟏

The collision density for like molecules is

𝒁 =𝒛ℵ

𝟐𝟐𝟐. 𝟕𝒂 =

𝒛

𝟐

𝒑

𝒌𝑻=𝟗. 𝟒𝟗 × 𝟏𝟎𝟗𝒔−𝟏

𝟐

𝟏𝟎𝟎 × 𝟏𝟎𝟑𝑷𝒂

𝟏. 𝟑𝟖𝟏 × 𝟏𝟎−𝟐𝟑𝑱𝑲−𝟏 × (𝟐𝟗𝟖𝑲)

= 𝟏. 𝟏𝟓 × 𝟏𝟎𝟏𝟓𝒔−𝟏𝒎−𝟑

For the percentage increase at constant volume, note that ℵ is constant at constant volume, so

the only constant-volume temperature dependence on z(and on Z) is in the speed factor

𝒛 ∝ 𝑻𝟏/𝟐 so𝟏

𝒛

𝝏𝒛

𝝏𝑻 𝒗=

𝟏

𝟐𝑻and

𝟏

𝒁

𝝏𝒁

𝝏𝑻 𝒗=

𝟏

𝟐𝑻

Therefore, 𝜹𝒛

𝒛=

𝜹𝒁

𝒁=

𝜹𝑻

𝟐𝑻=

𝟏

𝟐

𝟏𝟎𝑲

𝟐𝟗𝟖𝑲= 𝟎. 𝟎𝟏𝟕

so both z and Z increase by about 1.7%

(ℵ =𝒑

𝒌𝑻=

𝒑𝑵𝑨

𝑹𝑻=

𝒏𝑵𝑨

𝑽=

𝑵

𝑽)

(b) Energy requirement* Rate of change in the molar concentration of A molecules

collision density probability that a collision occurs with sufficient energy (not all

collisions proceed to the reaction; energy should exceed activation energy, εa)

* Probability that a collision occurs with sufficient energy is considered through

the collision cross-section, which is a function of kinetic energy, σ(ε).

σ(ε)=0, if kinetic energy<εa (εa: threshold value; Ea = Naεa: molar activation energy)

* For a collision with a specific relative speed “vrel” (not the relative mean speed, črel),

In A-B collision, the rate of change of number density of A is dŇA/dt =-σ(ε)vrelŇAŇB

or d[A]/dt = -σ(ε)vrelNA[A][B] (since [A]=ŇA/NA), by the successful reaction.

* If the relative kinetic energy is ε = ½ μvrel2, then vrel = (2ε/μ)1/2, as a function of ε.

* There is an wide range (distribution) of kinetic energy, present in a sample.

We need to use an average value using a Boltzmann distribution of energies.

d[A]/dt = -{∫σ(ε)vrelf(ε)dε}NA[A][B] = -kr[A][B] Boltzmann Weighting

The rate constant, kr = NA∫σ(ε)vrelf(ε)dε (where vrel = (2ε/μ)1/2)

* And the unknown σ(ε) is defined as, i) “collision cross-section below εa”, σ(ε) = 0

& ii) “collision cross-section above εa”, σ(ε) = [1–(εa/ε)]σ (if ε=εa, σ(ε)=0)

( Cross-section rises from 0 (when ε < εa) and becomes constant σ when ε >> εa)

Justification in the next slide…

v ∝ P σ(T/M)1/2 exp(-Ea/RT) [A][B]

2

1

2

22

cos*

d

advvv relrelBrelA

2

22

d

adBA

22

max 1 da a

22

max 1 da a

a1

Justification 22.2 collision cross-section

Collision cross-section σ(ε) = [1 – (εa/ε)]σ

Consider two colliding molecules A and B (right )

Relative kinetic energy, ε = ½ μvrel2

μ = mAmB/(mA+mB)

If we consider the vector component,

From ε = ½ μvrel2,

There is an energy threshold for reaction, which means that there is a maximum

value of a, above which reactions do not occur.

If we set a = amax and εA-B = εa, amax becomes: (by rearrangement)

Multiply π at both sides of equation:

Since σ = πd2 and σ(ε) = πamax2:

End of Justification

Justification

Justification 22.3 rate constant

Now kr = NA∫σ(ε)vrelf(ε)dε can be integrated using σ(ε) = [1–(εa/ε)]σ & vrel = (2ε/μ)1/2

→ Result: kr = Naσĉrelexp(-Ea/RT)

The Maxwell-Boltzmann distribution of speeds (f(v) = 4π(M/2πRT)3/2 v2 e-Mv2/2RT) can

be expressed with kinetic energy ε as follows:

4π(M/2πRT)3/2 = 4π(μ/2πkT)3/2

ε= ½ μv2 v = (2ε/μ)1/2 v2 = (2ε/μ) 2vdv = 2dε/μ dv = dε/vμ = dε/(2με)1/2

ε= ½ μv2 e-Mv2/2RT = e-μv2/2kT = e-ε/kT

Boltzmann distribution (f(v) = 4π(M/2πRT)3/2 v2 e-Mv2/2RT) becomes,

f(v)dv = 4π(μ/2πkT)3/2 (2ε/μ) e-ε/kT dε/(2με)1/2 = 2π(1/πkT)3/2ε½ e-ε/kT dε = f(ε) dε

Integrate rate constant, kr = NA∫σ(ε) vrel f(ε)dε (integration from 0 to )

∫ σ(ε)vrelf(ε)dε = 2π(1/πkT)3/2∫σ(ε) (2ε/μ)1/2 ε1/2exp(-ε/kT)dε

= (8/πμkT)1/2 (1/kT) ∫ ε σ(ε)exp(-ε/kT)dε

∫ ε σ(ε)exp(-ε/kT)dε = σ ∫ ε (1- εa/ ε) exp(-ε/kT)dε = (kT)2 σ exp(-εa/kT)

∴ ∫σ(ε)vrelf(ε)dε = σ[8kT/πμ]1/2e-εa/kT ∴ kr = Naσĉrelexp(-Ea/RT) (since εa/kT = Ea/RT)

*Arrhenius form, Aexp(-Ea/RT): minimum kinetic energy (Ea) required for reaction

*Pre-exponential factor, A=Naσĉrel: a measure of the rate at which the reaction occurs

End of Justification

Justification

v ∝ P σ(T/M)1/2 exp(-Ea/RT) [A][B]

(c) Steric requirement

Table 22.1 Let’s compare the pre-exponential factor from theory and experiments

* Pre-exponential factor, A: a measure of the rate at which the reaction occurs

(1) The first is in fair agreement. Ok

(2) The second and third show much smaller A in experiments than theory (slower rate)

Collision energy is not only factor for reaction.. Some other things are important..

(3) The fourth has larger A in experiment than theory (faster rate)

Reaction occurs more quickly than the particle collision..

* To solve these discrepancies, we introduce a steric factor, “P”.

• Steric factor, P: expressing reactive cross-section (σ*)

• Reactive cross section, σ* = Pσ (steric factor collision cross-section)

• P can be smaller or larger than 1. Fig. 22.3. Collision cross-section, σ vs. Reactive

collision cross-section, σ* (in above case, P<1)

v ∝ P σ(T/M)1/2 exp(-Ea/RT) [A][B]

Then the rate constant becomes (from kr = NAσĉrelexp(-Ea/RT))

kr = PNAσĉrelexp(-Ea/RT) = PNAσ[8kT/πμ]1/2exp(-Ea/RT) = σ*NA[8kT/πμ]1/2exp(-Ea/RT)

(1) When P is smaller than 1 (σ > σ*=Pσ, 2nd and 3rd case in the table)

ex) 22.1 Estimating a steric factor (#1)

Estimate steric factor for H2 + C2H4 → C2H6 (third case in the table 22.1) at 628K,

when pre-exponential factor A = 1.24106 dm-1mol-1s-1

Solution)

From kr = PNAσĉrel exp(-Ea/RT)

A = PA = PNAσĉrel = PNAσ[8kT/πμ]1/2

μ = mAmB/(mA+mB) = 3.1210-27 kg (for H2 and C2H4)

T = 628K [8kT/πμ]1/2 = 2.66103 ms-1 (at 628K)

From table 20.1, σ(H2) = 0.27nm2, σ(C2H4) = 0.64nm2

σ(mean of H2 and C2H4) = 0.46nm2

A = NAσ[8kT/πμ]1/2 = 7.371011 dm3mol-1s-1

From experimental data and above values → steric factor P = A/A =1.7 10-6

Fig. 22.3. Collision cross-section, σ vs.

Reactive collision cross-section , σ*)

σ*

σ

(2) When P is larger than 1 (σ < σ*=Pσ, 4th case in the table)

P = 4.8 (> 1) for K+Br2→ KBr+Br (fourth case in table 22.1)

→ This is explained by the “harpoon mechanism”.

The “distance that a reaction occurs” is much larger than the “distance for

deflection of molecule’s path by non-reactive collision”. i.e. σ* > σ

harpoon mechanism

When K and Br2 are close enough:

→ An electron (the harpoon) flips across from K to Br2

→ Now those two are no longer neutral

→ They are two ions (due to electron transfer)

→ Coulombic attraction (line on the harpoon)

→ Ions move together due to this harpoon (i.e. coulombic attraction)

→ Reaction takes place

Reaction rate ↑ due to harpoon effect..

* The harpoon extends the cross-section for the reactive encounter. i.e. σ* > σ

* The reaction rate can be greatly underestimated, if the collision is considered a

simple mechanical contact between K and Br2.

ex) 22.2 Estimating a steric factor (#2)

Estimate P for the harpoon mechanism by calculating distance for electron transfer.

Solution)

Consider all contributions to the energy interactions in K + Br2 → K+ + Br2-

(I, positive) Ionization energy (retard reaction) of K to K+: I (constant)

(ii, negative) Electron affinity (accelerate reaction) of Br2 to Br2-: Eea (constant)

(iii, negative) Coulombic interaction (accelerate reaction) energy between ions, K+ and

Br2-: -e2/4πε0R (when separation is R)

→ by R↓, electron flips across, when the sum of these 3 contributions (one + and two –

effects to total E) changes from + to – (i.e. when sum = 0) (i.e., -e2/4πε0R ↓ by R ↓)

The net E change when the transfer occurs at a separation R, E = I–Eea–e2/4πε0R

Since I > Eea, E becomes negative only when R decreases below critical value R*

To calculate R*, let’s set E = 0, then e2/4πε0R* = I – Eea R* = e2/4πε0(I – Eea)

When R = R*= e2/4πε0(I – Eea), the harpoon shoots across from K to Br2.

Using R*, reactive cross-section σ* = πR*2

Steric factor P = σ*/σ = R*2/d2 = {e2/4πε0d(I – Eea)}2 = 4.2 (where ε0 is dielectric

constant of vacuum, d = R(K) + R(Br2) = 400pm, I = 420 kJmol-1, Eea = 250 kJ mol-1)

(d) The RRK model

Lindemann-Hinshelwood mechanism’s limit (Ch 21.8, next slide)

In unimolecular gas phase reaction, plot of 1/kr vs 1/[A] is linear.

But experiments deviate at large kr (or high concentration [A])

(theoretically calculated kr is overestimated than the reality)

Rice-Ramsperger-Kassel (RRK) model (Modified P)

Improved model of LH mechanism is proposed by RRK

: Although a molecule might have enough energy to react, that energy is distributed

over the all the modes of motion of the molecule, and reaction will occur only when

enough of that energy has migrated into a particular location/mode (such as a specific

bond) in the molecule.

* New steric factor: (for E E*, as s increases, P decreases; P<<1, kr)

s: number of modes of motion over which the energy may be dissipated

(as s increases, less chance of reaction, since difficult to focus on a particular mode)

E*: energy required for the bond of interest to break

E: energy available in collision (E should be larger than E*)

* Kassel form of the unimolecular rate constant: (for E E*)

kb: the rate constant used in the original LH theory

1

1

s

E

EP

b

s

b kE

EEk

1

1)(

(a) Lindemann-Hinshelwood mechanism

L-H mechanism explains 1st order unimolecular reaction, in which the

Reactant is excited by collision with another reactant in a bimolecular step:

Activation: A + A → A* + A d[A*]/dt = ka[A]2

Deactivation: A + A* → A + A d[A*]/dt = -ka′[A][A*]

Unimolecular decay: A* → P d[A*]/dt = -kb[A*]

Let’s use “Steady-state approximation” (Intermediate [A*] is constant)

d[A*]/dt = ka[A]2 - ka′[A][A*] – kb[A*] 0

[A*] = ka[A]2/(kb + ka′[A])

d[P]/dt = - d[A*]/dt = kb[A*] = kakb[A]2/(kb + ka′[A])

d[P]/dt = k[A], k = kakb[A]/(kb + ka′[A])

* effective rate constant k

1/k = ka′/kakb + 1/ka[A]

According to theory Plot 1/k vs. 1/[A] is straight line

But deviation at high concentration [A]

Remind: From Ch. 21.8

Energy dependence of rate constant kb(E) for various

s (number of modes of motion), See Fig. 22.5.

*If “s” (number of modes of motion) is large, the rate

constant is small, since it takes long time for the

distributed energy through all oscillations of a large

molecule to migrate and accumulate in the specific

critical mode that is involved in the reaction.

*As E (energy available for collision) becomes much

larger than E* (energy required) (E >> E*), kb(E) kb,

i.e. independent of energy and number of oscillations

in the molecule. This is because there is enough energy to

accumulate immediately in the critical mode.

b

s

b kE

EEk

1

1)(

Fig. 22.5. Energy dependence

of the rate constant for three

different s

22.2 Diffusion-controlled reactions (liquid phase)

• “Encounters between reactants in the solution” is different from

“Encounters in gases”

• Differences between liquid and gas phase reactions are as follows:

(1) Lower encounter frequency in solution due to solvent that prevents movement

(Diffusion is more difficult in solution than in gas phase.)

(2) Since molecules in solution migrate slowly away from a location, two reactant

molecules that encounter each other stay near each other much longer than in a gas.

This lingering of one molecule near another on account of the hindering presence

of solvent molecules is called “the cage effect”,

(3) Encounter pairs surrounded by the solvent has much more complicated activation

energy than encounters in the gas phase, since we need to consider entire local

assembly of “reactants” and “solvent”.

(a) Classes of reaction*2nd order (1st order for each) formation of encounter pair AB, A+B→AB, v = kd[A][B]

(kd: determined by the diffusional characteristics of A and B)

*Encounter pair [AB]: either “breaking up w/o reaction” or “proceeding reaction to P”

AB→A+B, v = kd´[AB] (separation) or AB→ P, v = ka[AB] (production)

*Concentration of encounter pair AB

d[AB]/dt = kd[A][B] – kd´[AB] – ka[AB] 0 (steady state approximation for AB)

[AB] = kd[A][B]/(ka+kd´)

d[P]/dt = ka[AB] = kakd[A][B]/(ka+kd´) = k2[A][B] k2 = kakd/(ka+kd´)

Two kinds of limits:

(i) kd´<<ka (rate of separation << product reaction rate, products are formed ASAP)

→ k2 = kd: diffusion-controlled limit (formation of encounter pair AB)

i.e. reaction rate is governed by reactants diffusion in solvent (little reaction activation E)

ex) radical-atom reaction: combination of radicals involve very little energy (E)

(ii) ka<<kd´ (rate of separation >> product reaction rate, products are formed slowly)

→ k2 = kaK (K=kd/kd´: equilibrium constant of A+B AB): activation controlled limit

i.e. reaction rate is governed not by diffusion but by activation for AB → P reaction

~ the reaction proceeds at the rate at which the energy accumulates in the encounter pair

AB, from the surrounding solvent.

(b) Diffusion and reactionkd of diffusion controlled rxn: kd=4π R* D NA

D: the sum of the diffusion coefficients of two reactant species in the solution

R*: two reactant molecules react if they come within a distance of R*

Justification 22.4. Solution of the radial diffusion equation (diffusion controlled reaction)

*First, we will see diffusion of B molecules to all A molecules for reaction.

Diffusion equation for 1D (x direction) (in Ch 20.9, Fick’s 2nd law), ∂c/∂t = D ∂2c/∂x2

Corresponding diffusion equation to three dimension: DB∇2[B] = ∂[B]/∂t

At steady-state ∂[B]/∂t = 0 → DB∇2[B]r = 0 (a quantity that varies with distance r)

For spherical symmetrical system, ∇2 can be replaced by radial derivatives (next slide).

diffusion equation at steady state: d2[B]r/dr2+(2/r)(d[B]r/dr) = 0

(ignore 2 term: by the symmetry of diffusion)

The general solution, [B]r = a + b/r (can be verified by submission to differential eqn)

Two B.C.: [B]r=[B] bulk value at r→∞, [B]r=0 at r=R* (where reaction occurs)

a=[B], b=-R*[B] (using BC)

[B]r = (1 – R*/r)[B] (for r R*) Fig 22.6

Now we know concentration distribution of B around

one A molecule at steady state.

We need to estimate flux of B to A using B concentration.

Fig. 22.6. The

concentration

profile for reaction

in solution when a

molecule B diffuses

to another reactant

molecule and reacts

if it reaches R*

Rate of reaction = diffusion rate = 4πR*2 J (J: flux) (since diffusion controlled)

~ Area of spherical surface of radius R* Flux of reactant B towards A at R*

From Fick’s first law (eq 20.19, J=-Ddc/dx), the flux towards A is proportional to the

concentration gradient. (J (d[B]r/dr)r=R*, negative sign is changed since we are

interested in in-flux)

J = DB(d[B]r/dr)r=R* = DB([B]R*/r2)r=R* = DB[B]/R* (since [B]r=(1–R*/r)[B])

Rate of reaction = 4πR*2J = 4πR*DB[B] (flux of B toward “one A”)

Rate of diffusion-controlled reaction = Average flow of B molecules to “all A” molecules

# of A molecules in the entire volume = NA[A]V (V = volume)

Global flow of all B particles toward all A particles = 4πR*DBNA [A][B]V

Since A and B both are moving DB can be replaced by sum of DA and DB.

d[AB]/dt = 4πR*DNA[A][B] = kd[A][B], D = DA+DB

Diffusion controlled rate constant kd = 4πR*D NA

Using Stokes-Einstein equation (Ch 20.8) we can relate diffusion coefficients DA and DB

with hydrodynamic radius RA and RB in a medium of viscosity η

DA = kT/6πηRA, DB = kT/6πηRB

If we assume RA=RB= ½ R* (RA+RB=R*), then D = DA+DB = 2kT/3πηR*

kd = 4πR*D NA = 8RT/3η (R=kNA is a gas constant)

End of Justification

Exercise (#2) A typical diffusion coefficient (D) for small molecules in aqueous

solution at 25°C is 5 10-9 m2 s-1. If the critical reaction distance (R*) is 0.4 nm,

what value is expected for the second-order rate constant for a diffusion-

controlled reaction?

The rate constant for a diffusion-controlled bimolecular reaction is

𝒌𝒅 = 𝟒𝝅𝑹∗𝑫𝑵𝑨[22.18]

where

𝑫 = 𝑫𝑨 + 𝑫𝑩 = 𝟐 × 𝟓 × 𝟏𝟎−𝟗𝒎𝟐𝒔−𝟏 = 𝟏. 𝟎 × 𝟏𝟎−𝟖𝒎𝟐𝒔−𝟏

𝒌𝒅 = 𝟒𝝅 × 𝟎. 𝟒 × 𝟏𝟎−𝟗𝒎 × 𝟏. 𝟎 × 𝟏𝟎−𝟖𝒎𝟐𝒔−𝟏 × 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑𝒎𝒐𝒍−𝟏

= 𝟑 × 𝟏𝟎𝟕𝒅𝒎𝟑𝒎𝒐𝒍−𝟏𝒔−𝟏

Exercise (#3) Two neutral species, A and B, with diameters 588 pm and 1650 pm,

respectively, undergo the diffusion-controlled reaction A+B →P in solvent of viscosity

2.37 10-3 kg m-1 s-1 at 40 °C. Calculate the initial rate d[P]/dt if the initial

concentrations of A and B are 0.150 mol dm-3 and 0.330 mol dm-3, respectively.

Since the reaction is diffusion controlled, the rate-limiting step is bimolecular and therefore

second order; hence𝒅[𝑷]

𝒅𝒕= 𝒌𝒅 𝑨 [𝑩]

where

𝒌𝒅 = 𝟒𝝅𝑹∗𝑫𝑵𝑨 = 𝟒𝝅𝑹∗ 𝑫𝑨 + 𝑫𝑩 𝑵𝑨 = 𝟒𝝅𝑵𝑨 𝑹𝑨 + 𝑹𝑩 ×𝒌𝑻

𝟔𝝅𝜼

𝟏

𝑹𝑨+

𝟏

𝑹𝑩

=𝟐𝑹𝑻

𝟑𝜼𝑹𝑨 + 𝑹𝑩 ×

𝟏

𝑹𝑨+

𝟏

𝑹𝑩= 𝟑. 𝟕𝟖 × 𝟏𝟎𝟗𝒅𝒎𝟑𝒎𝒐𝒍−𝟏𝒔−𝟏

(RA+RB=R*, Stokes-Einstein equation, DA = kT/6πηRA, DB = kT/6πηRB)

Therefore, the initial rate is

𝒅[𝑷]

𝒅𝒕= 𝟑. 𝟕𝟖 × 𝟏𝟎𝟗𝒅𝒎𝟑𝒎𝒐𝒍−𝟏𝒔−𝟏 × 𝟎. 𝟏𝟓𝟎𝒎𝒐𝒍 𝒅𝒎−𝟑 × 𝟎. 𝟑𝟑𝟎𝒎𝒐𝒍 𝒅𝒎−𝟑

= 𝟏. 𝟖𝟕 × 𝟏𝟎𝟖𝒎𝒐𝒍 𝒅𝒎−𝟑𝒔−𝟏

Assume R*=RA+RB

22.3 Materials balance equation

(a) The formulation of the equation (of diffusion, convection and reaction)- Net rate at which J molecules enter by diffusion and convection

∂[J]/∂t = D∂2[J]/∂x2 - v∂[J]/∂x (from equation 20.56)

diffusion convection

- Material balance equation considering chemical reactions

∂[J]/∂t = D∂2[J]/∂x2 - v∂[J]/∂x – kr[J]

diffusion convection chemical rxn

(b) Solution of the equations

- The material balance equation is difficult to solve.

- Differential equations are solvable only for some special cases

* In unstirred (no convection) case: ∂[J]/∂t = D∂2[J]/∂x2– kr[J]

* If the solution of the diffusion equation without reaction is [J],

[J]=n0 exp(-x2/4Dt)/A(πDt)1/2 (no reaction, i.e. kr=0)

Then the solution with reaction, [J]*, is [J]*=[J]exp(-krt)

(can be verified by substitution to the differential equation)

- Most of real cases rely on the numerical solution..

Fig. 22.7. Concentration profile

for a diffusing and reacting

system (no convention term)

Transition state theory (Activated complex theory)*Activated complex between reactants is formed when they collide.

*Transition state theory is applied for activation controlled case (not diffusion control).

*Transition state theory identifies the reaction rate constants using mathematical

models of events that take place during the chemical reaction. The overall reaction rate

is determined by how rapidly the complex passes through the transition state.

22.4 Eyring equation (for A+B C‡→ P)*Transition state theory pictures the reaction as proceeding through

the formation of an activated complex, C‡, in a rapid pre-equilibrium

A + B C‡, then equil. const. K‡ = (pC‡/pө) / (pA/pө)(pB/pө) = pC‡p

ө/pApB

Partial pressure, pJ = RT[J], where [J]: molar concentration

(pө: for dimensionless; Above model rxn is a gas phase rxn.)

From equilibrium constant of transition: K‡ = [C‡] pө /(RT[A][B])

Concentration of activation complex: [C‡] = (RT/pө) K‡ [A][B]

The activated complex (C‡) decays into products, P, with a rate constant k‡,

C‡→P, v=k‡[C‡], where k‡ is a rate constant for unimolecular decay of activated complex

Reaction rate, v=k‡[C‡]= k‡ (RT/pө) K‡ [A][B] = kr[A][B], kr = (RT/pө)k‡K‡

The reaction rate constant of activation controlled case is kr and we need to find

unimolecular decay rate constant “k‡” and pre-equilibrium constant “K‡” to estimate the

overall reaction rate constant “kr”.

Let’s see how to calculate kr (i.e. k‡ and K‡) using transition state theory.

Fig. 22.8. A reaction profile

for an exothermic reaction.

Activated complex is near

the potential maximum.

A + B

C‡

P

(a) The rate of unimolecular decay of the activated complex (C‡→P)

Activated complex can form products if it passes through the transition state.

Form activated complex, A + B C‡, and then unimolecular decay, C‡→P

If vibration-like motion along the reaction coordinate occurs with a frequency ν‡ and this

vibration converts activated complexes to products, the rate of passage of the complex

through the transition state is proportional to the vibrational frequency along the

reaction coordinate (k‡ ν‡, higher frequency means higher vibrational energy and then

leads to more unimolecular decay reaction).

(First, we are interested in the vibration, since it is unimolecular decay)

Rate constant for passage through the transition state in terms of transmission

coefficient and vibration frequency can be defined as:

k‡ = κν‡,

where k‡ : rate constant for unimolecular decay of activated complex

κ : transmission coefficient (since not all complex passes through the

transition state)

Normally κ is assumed to be 1 without additional information, which means that almost

every visit to the transition state leads on to products. Now we know k‡. The next step is

to find out K‡.

(b) The concentration of the activated complex (A + B C‡)

Let’s use the results from statistical thermodynamics.

From statistical thermodynamics for K‡ (Ch 16.8, Eqn 16.52a)

Then, Equilibrium constant K‡ = (NAqC‡ө/qA

өqBө)exp(-ΔrE0/RT)

(where pө = 1 bar and ΔrE0 = E0(C‡) - E0(A) – E0(B)

qJө: standard molar partition function (Ch 16.2)

ΔrE0: energy difference between the zero point level C‡ and A+B)

First, let’s see the vibration partition function of C‡.

Partition function for the vibration, q = 1/(1 – e-hν‡/kT)

(ν‡ is vibration frequency, same in rate constant k‡ = κν‡)

This frequency ν‡ is much lower than ordinary molecular vibration, since the

oscillation corresponds to the vibration that makes the complex falling apart and

thereby the force constant is very small. (See harmonic oscillation in Fig 22.9, F=-kx)

hν‡/kT << 1 (in q = 1/(1 – e-hν‡/kT))

the exponential can be expanded and the partition function for vibration reduces:

q = 1/{1 – (1 – hν‡/kT + …)} kT/hν‡ (since we can ignore higher order terms)

qC‡ (kT/hν‡) qC‡

where q, partition function for all other modes of the complex except for the vibration

mode, and q, total particle function, i.e. total partition function = kT/hν‡other modes

Fig. 22.9. In activated complex

close to the transition state, there

is a broad, shallow dip in the

potential E along the reaction

coordinate. The complex vibrate

harmonically in the well.

The equilibrium constant K‡ is

K‡ = (kT/hν‡)K‡, where K‡= (NAqC‡ө/qA

өqBө)exp(-ΔE0/RT)

(Since C‡ is diatomic cluster, vibration is involved, while A and B are not)

(K‡ : an equilibrium constant except for the vibrational mode of activated complex C‡)

(c) The rate constant

We can now combine all parts of calculation (k‡ and K‡) into overall reaction rate

constant kr (in the previous slide):

kr=k‡(RT/pө)K‡=κν‡(RT/pө)(kT/hν‡)K‡=κ(kT/h)(RT/pө)K‡=κ(kT/h)KC‡ Eyring Eqn

KC‡ = (RT/pө)K‡, also K‡ = (NAqC‡ө/qA

өqBө)exp(-ΔE0/RT)

← Eyring equation given in terms of partition function of A, B, C

We have an explicit expression for calculating the second order rate constant for a

bimolecular reaction (A+B C‡→ P) in terms of the molecular parameters for the

reactants and the activated complex and the quantity κ (transmission coefficient).

We have considered the vibration motion. We need to consider other motions such as

translational motion, rotation, electronic.., for KC‡. However, the difficulty is in the

calculation of the partition function of activated complex, C‡. (we need assumption for

its size, shape, structure. Let’s see simplified case in the next slide.)

(d) The collision of structureless particles

*Let’s assume that we have two structureless particles. Consider collision of two

structureless particles A, B → formation of activated complex (like diatomic cluster)

*Since reactants J (A & B) are structureless atoms, the only contribution to their partition

functions is translational term only:

qJө = Vm

ө/ΛJ3, where ΛJ = h/(2πmJkT)1/2, Vm

ө = RT/pө

The activated complex is a diatomic cluster. It has one vibrational mode but does not

appear in qC‡. (qC‡ is the partition function without vibrational mode)

*The standard molar partition function of activated complex: qC‡ө = (2IkT/ħ2)(Vm

ө/ΛC‡3)

The moment of inertia, I = μr2, where μ is effective mass, μ= mAmB/(mA+mB).

Rate constant, kr = κ(kT/h)(RT/pө)(NAqC‡ө/qA

өqBө)exp(-ΔE0/RT), (from Eyring eqn)

kr = κ (kT/h)(RT/pө)(NAΛA3ΛB

3/ΛC‡3Vm

ө)(2IkT/ħ2)exp(-ΔE0/RT)

= κ (kT/h)NA(ΛAΛB/ΛC‡)3(2IkT/ħ2)exp(-ΔE0/RT)

= κ NA(8kT/πμ)1/2πr2exp(-ΔE0/RT)

By identifying σ*=κ πr2 and ΔE0=Ea we arrive the same expression of kr =

NAσcrelexp(-Ea/RT) for the gas phase bimolecular reaction from collision theory (eq 22.13).

diatomic cluster

(A, B: consider translation energy but ignore vibration, rotation and electronic terms; C‡: vibration, translation, and rotation, not electronic)

(Even though rotation of single particle is ignored, that of diatomic complex should be considered)

(translation partition function: movement of the center of mass)

ħ = h/2π

22.5 Thermodynamic aspects*There is difficulty of knowing structure of activated complex and therefore the statistical

thermodynamic version of transition theory has difficulties. (we needed to consider a

simple case, structureless particles.)

More general, empirical approach using concepts of the equilibrium between the

reactants and the activated complex using thermodynamics is made, in which the

activation process is expressed in terms of thermodynamic functions.

(a) Activation parameters

Gibbs energy of activation (Definition in terms of K‡)

kr=k‡(RT/pө)K‡=κν‡(RT/pө)(kT/hν‡)K‡=κ(kT/h)(RT/pө)K‡=κ(kT/h)KC‡ Eyring Eqn

∴Rate constant, kr = κ (kT/h)(RT/pө)exp(-ΔG‡/RT) (Eyring eqn in terms of Gibbs E)

G = H–TS → G can be divided into “entropy of activation” & “enthalpy of activation”

ΔG‡ = ΔH‡ - TΔS‡

∴kr = κ(kT/h)(RT/pө)exp(ΔS‡/R)exp(-ΔH‡/RT)

From definition of activation energy, Ea = RT2(∂lnkr/∂T) = RT2(2/T+ΔH‡/RT2)

Ea = ΔH‡+2RT, ΔH‡ = Ea – 2RT, insert ΔH‡ to kr,

Arrhenius form: kr = κ(kT/h)(RT/pө)exp(ΔS‡/R)exp(2)exp(-Ea/RT)

pre-exponential factor: A=κ(kT/h)(RT/pө)e2eΔS‡/R

‡Kln RT- = ΔG‡

∂(ln[T2exp(-ΔH‡/RT)])/∂T = ∂ [2lnT-ΔH‡/RT] /∂T = 2/T+ΔH‡/RT2

*Entropy of activation (ΔS‡) is negative, since two reactants combines into one complex.

If there is additional entropy decrease, then A will be smaller than the value expected by

the simple collision theory. This additional decrease can be identified as a steric factor.

*Additional reduction of entropy, ΔSsteric‡: origin of the steric factor of collision theory

(Reactive cross section, σ* = Pσ (steric factor collision cross-section))

P = exp(ΔSsteric‡/R) : Steric factor in terms of entropy of activation

The more complex steric requirements of encounters

Difficult to activate two reacting molecules into the complex

The entropy of the complex is smaller

The more negative ΔSsteric‡,

The smaller steric factor, P

The smaller reaction rate constant than theory

*In experiment, correlation analysis between equilibrium constant and reaction rate is

often carried out.

- Find correlation of K and kr by plotting lnK and lnkr (in many cases, it is linear)

- Linear free energy relation: As the reaction becomes thermodynamically more

favorable, i.e. becomes more stable product, Gibbs energy decreased, ΔG‡ < 0,

(kr = κ (kT/h)(RT/pө)exp(-ΔG‡/RT), smaller ΔG‡)

- The rate constant increases (more products).

(b) Reactions between ions

*What if reactants have charges? i.e. ions..

Using the thermodynamic theory, which is simpler than statistical thermodynamics,

the rate law can be obtained:

The rate law, d[P]/dt = k‡[C‡] (rate law of unimolecular decay of activated complex

C‡), is combined with the thermodynamic equilibrium constant, K = aC‡/aAaB =

Kγ([C‡]/[A][B]), where Kγ = γC‡/γAγB (ratio of activity coefficient)

*Then, insert [C‡]= K/Kγ[A][B] using equilibrium constant to reaction rate law:

d[P]/dt = kr[A][B] kr = k‡K/Kγ

If kr0 is defined as rate constant when activity coefficient (γC‡,γA,γB)=1 (i.e. kr

0=k‡K)

d[P]/dt = kr[A][B] kr = kr0/Kγ (kr

0=k‡K; ideal rate constant)

*At low concentrations, the activity coefficient can be expressed with I (I: ionic

strength) by Debye-Hückel limiting law (Ch 5.13, Eq 5.75): logγJ = -AzJ2I1/2

logγA = -AzA2I1/2, logγB = -AzB

2I1/2, logγC‡ = -A(zA+zB)2I1/2

,where A: 0.509 at 298 K in water; zA, zB : Charge # of A and B; zA+zB : Charge # of C‡

logkr = logkr0–logKγ = logkr

0–logγC‡+logγA+logγB = logkr0–A{zA

2+zB2–(zA+zB)2}I1/2

logkr = logkr0 + 2AzAzBI1/2 (kr is a function of I, Kinetic Salt Effect: See next slide)

ideal term nonideality

Kinetic salt effect

kr is a function of ionic strength (I) of the solution.

* Fig 22.13 shows the variation of the rate constant of a reaction

between ions depending on the ionic strength (I) of the solution.

*From “logkr = logkr0 + 2AzAzBI1/2”

If reactant ions have same sign (reaction between cations or

between anions), then zAzB > 0.

→ by increasing I (by adding ions), rate constant kr increases

→ vice versa, then zAzB < 0, and thereby negative sign (slope)

*The formation of a single, highly charged ionic complex from two less highly charged

ions (zAzB > 0) is favorable (increased kr) in a high ionic strength environment, since

the new ionic complex has a denser ionic atmosphere and interacts with that

atmosphere more strongly.

*Ions of opposite charge react more slowly (decreased kr) in solutions of high ionic

strength (zAzB < 0). This is because charges cancel out each other and the complex has

less charges and thus less favorable interactions with its atmosphere than the

separated ions.

Fig. 22.13. Experimental

results of kinetic salt effect.

ex) 22.3 Analyzing the kinetic salt effect

The rate constant for the base (OH-) hydrolysis of [CoBr(NH3)5]2+ varies with

ionic strength. What can be deduced about charge of the activated complex?

I 0.005 0.01 0.015 0.02 0.025 0.03

kr/kr0 0.718 0.631 0.562 0.515 0.475 0.447

Sol) Let’s assume that it follows the kinetic salt effect, logkr = logkr0 + 2AzAzBI1/2

From logkr = logkr0 + 2AzAzBI1/2

Plot logkr/kr0 vs I1/2

It shows the straight line.

follows kinetic salt effect…

From the slope: 2AzAzB=-2.04

Since A=0.509 at 298K

zAzB = -2

Since zA = -1 for OH- and zB = +2 [CoBr(NH3)5]2+, perfectly follows the kinetic

salt effect and OH- and [CoBr(NH3)5]2+ forms the activated complex.

The Dynamics of Molecular Collisions

22.6 Reactive collisions

(a) Experimental probes of reactive collisions

* There are several techniques for the study of reactive collisions.

* Molecular beams: crossed-beam experiment (See Fig 22.15)

Two state-selected molecules are generated in separate sources

and directed perpendicular to one another. The detector detects

scattered molecules (chemical products) after collision.

* Molecular energy (translational E, vibrational E, different orientations) of incoming

beams is adjustable to study how they affects the outcoming product molecules.

* Several methods to examine (detect) the energy distribution of products

(1) Infrared chemiluminescence

(2) Laser-induced fluorescence

(3) Multiphoton ionization (MPI)

(4) Resonance MPI (REMPI)

(b) State-to-state dynamics

* In the collision theory (Ch 22.1), 2nd order rate constant, kr, could be expressed by

Boltzmann-weighted average of the reactive collision cross-section (a function of

energy, σ(ε)) and the relative speed of approach (also energy function) (eqn 22.10):

kr = NA∫σ(ε)vrelf(ε)dε = <σ vrel> NA (< > means Boltzmann average)

Now, we have “molecular beam studies” that provide more sophisticated version of

collision cross-section, the state-to-state cross-section (σnn´).

state-to-state (n to n) rate constant (knn´): knn´ = <σnn´vrel> NA

The reaction rate constant kr is the sum of the state-to-state rate constants over all

final states (for all successful reactions) and over a Boltzmann weighted sum of initial

states (initial characteristic distribution of populations at T).

kr = Σ knn´ (T) fn(T)

Where fn(T) is the Boltzmann distribution factor for “initial state n” at temperature T

nn´

(We are using “Boltzmann Average” twice…)

(n to n : initial n, final n)

22.7 Potential energy surfaces

*Potential energy surface: Potential energy as a function of the

relative positions of all the atoms taking part in the reaction

(Potential energy surface can be constructed by experimental data &

quantum chemical calculations)

Ex) Collision between a “H atom (A) and H2 molecule (B, C)”

Collinear approach (approach of an atom along H-H axis) needs less

energy for reaction than others. We will see the collinear case:

H (A) → ← H-H (B-C) H-H (A-B) ← → H (C)

*Two parameters for nuclear separation: RAB(HA-HB), RBC(HB-HC)

(1) At start: RAB is infinite, RBC is H2 equilibrium bond length

(2) At end: RAB is H2 equilibrium bond length, RBC is infinite

*Total energy of three-atom system depends on relative separations

(See Fig 22.17 & Fig 22.18, potential E change with RAB and RBC)

- When RAB is very large, the variation in potential energy at the

potential E surface as RBC changes are those of isolated H2 molecule

(H-HB-C) as its bond length is altered.

- When RBC is very large, potential E is for isolated HAHB molecule.

Fig. 22. 17. Potential E surface

at different RAB and RBC

Fig. 22. 18. Contour diagram of

potential E surface in Fig. 22.17.

low E

low E

high E

↑↓

↓

↑

HAHB molecule

HBHC molecule

0

*The actual path of atoms in the course of the encounter depends on their total

Energy (total E = kinetic E + potential E). However, let’s consider least potential

energy path to get initial idea for available reaction paths

*See Fig. 22. 19. for each route: HA approaches to HBHC

- Route A: If HBHC bond length is constant (constant RBC) during the initial approach

of HA(decrease in RAB), the potential E of H3 cluster rises along route A.

- Route B: HBHC bond length (RBC) increases while HA is still far away (large RAB).

*Route A and B take atoms to regions of high potential energy (refer to Fig. 22.20).

*Still feasible if initial kinetic energy is large enough (but not optimum route).

- Route C: The path of least potential energy → floor of valley (C‡: saddle point →

activated complex)

- Transition state: in terms of trajectories on potential surfaces, the transition state

can be identified with a critical geometry that every trajectory that goes through this

geometry goes on to react (See Fig. 22.20)

Fig. 22. 19. Various

trajectories of Potential

E surface: Route A, B

and C

Fig. 22. 20. Transition stateis a set of configurations (here,

marked by the line across the

saddle point) through which

successful reactive trajectories

must pass.C‡

22.8 Some results from experiments & calculations*To travel successfully from reactants to products → the incoming molecules must

possess enough kinetic energy to climb to the saddle point of the potential surface

~ The shape of the surface can be explored by changing the relative speed of approach

(kinetic/translational energy) and the degree of vibrational excitation (vibration

energy) of beams that collide (refer to slide 36).

~ Then observe (1) whether reaction occurs or not and (2) whether the products emerge

in a vibrationally excited state or not → See Fig. 22.21 and the next slide for details.

Question: Is it better to smash reactants i) with a lot of translational kinetic energy or

to ensure instead that they approach ii) in highly excited vibrational states for

successful reaction? → See Fig. 22.21. and the next slide for details

Fig. 22. 21. Some successful (w/ *) and unsuccessful (w/o *) encounters (See details of each case in the next slide)

(a) Successful encounter: C1* corresponds to the path along the foot of the valley,

with translational kinetic energy but no vibrational energy of BC molecule

(b) Successful encounter: C2* corresponds to an approach of A atom to vibrating BC

molecule → Formation of vibrating AB as C departs

(c) Unsuccessful encounter: C3 corresponds to an approach of A to a non-vibrating

BC molecule but with insufficient translational kinetic E

(d) Unsuccessful encounter: C4 corresponds to an approach of A to a vibrating BC

molecule, but with insufficient energy and improper phase of vibration for reaction

Fig. 22. 21. Some successful (w/ *) and unsuccessful (w/o *) encounters

* Four different cases observed in experiments

(a) Direction of attack and separation

* example 1) H atom approaches H2 from different angles (See Fig. 22.22.)

Potential barrier is least in the case of collinear attack/approach of the reactant.

(Fig. 22.22.) However, other approach lines (including perpendicular direction to the

bond) could be also feasible (but requires high E) and contribute to overall rate.

* example 2) Cl atom approaches HI (See Fig. 22.23.)

Lowest E barrier occurs for approaches within a cone of half angle 30°surrounding

the H atom; This result is related to the calculation of the steric factor in collision

theory. (Not every collision is successful since collisions only lying within reactive

cone are successful.)

Fig. 22. 22. Anisotropy of

potential E as H approaches to

H2 with different angles.

Collinear attack has the lowest

potential barrier to reaction.

Fig. 22. 23. Potential E barrier

for the approach of Cl to HI

molecule. Successful encounter

happens only when Cl

approaches within a cone

surrounding the H atom.

*If collisions are sticky, reactants collide and stick together when they orbit around

each other. ~ Then products emerge in random directions since all memory of

approach directions are lost during rotation. (rotation speed > reaction speed)

*Since a rotation takes about ~1 ps, if collision (reaction) is << ~1 ps (rotation speed <

reaction speed), the product ends with a specific direction (not random), mainly in the

direction that the reactant approach through least energy barrier regions.

* Examples of anisotropy in product direction (specific direction)

(1) Collision of K and I2 The product will be thrown off in the forward direction

(consistent with harpoon mechanism – coulombic attraction – that fixes the

approaching direction over the long range)

(2) Collision of K and CH3I Reaction happens only when reactants approach very

closely. K bumps into a brick wall, and the KI product bounces out in the backward

direction.

* These examples, i.e. detection of anisotropy in the angular distribution of products,

indicate that 1) specific distance and orientation of approach needed for the successful

reaction and 2) event is completed within 1 ps (reaction time < time for rotation).

(b) Attractive and repulsive surfaces

Some reactions are very sensitive to whether the energy has been pre-consumed as a

vibrational mode or left as the relative translational kinetic energy of the colliding

molecules.

Examples

If two HI molecules are hurled together with more than twice the activation energy of

the reaction, then no reaction occurs if all the energy is translational. (need vibrational

energy)

For F + HCl Cl + HF, the reaction is about five times as efficient when HCl is in

first vibrational excited state than, although HCl has the same total energy, in its

vibrational ground state. (vibrational energy is important in this case also)

The origin of these requirements can be found on the potential surface (See the next

slide)

Attractive surface: saddle point occurs early in the reaction coordinate (See Fig. 22.24)

If the original molecule is vibrationally excited, a collision follows along the path C, which

does not reach saddle point. However, if the same amount of E is solely as translational

kinetic E, then the system moves along the path C* through the saddle point. After the

saddle point, products emerge in a vibrationally excited state.

Reactions of attractive potential surface is more efficient if the energy is relative

translational motion.

Repulsive surface: saddle point occurs late in the reaction coordinate (See Fig. 22.25)

In path C, the collision energy is largely in translation, which takes to the valley and

reflected back to reactant region (unsuccessful collision). In path C*, some of initial

energy is vibrational, which weaves side to side of valley and leads to the saddle point and

then products. Products have high translational kinetic energy (unexcited vibration).

Reactions of repulsive potential surface is more efficient in excessive vibrational motion.

* A reaction that is attractive in one direction is repulsive in the reverse direction.

Fig. 22. 24. Attractive potential

energy surface. A successful

encounter (C*) involves high

translational kinetic energy

and results in a vibrationally

excited product.

Fig. 22. 25. Repulsive potential

energy surface. A successful

encounter (C*) involves a

initial vibrational excitation

and the product has high

translational kinetic energy.

(c) Classical trajectories

Clear understanding of reactions can be obtained using classical mechanics to calculate

the trajectories of the atoms taking place in a reaction from a set of initial conditions

(velocity, relative orientation, internal energy).

*Calculation of positions of three atoms in H + H2 → H2 + H

(1) X axis: time. Y axis: separation between molecules.

(2) Right illustration shows the vibration of an original

molecule (B-C) and an attacking atom (A)

(3) Direct mode process: reaction happens very quickly.

*Calculation in the reaction: KCl + NaBr → KBr + NaCl

(1) X axis: time. Y axis: separation between molecules.

(2) Complex mode process: activated complex (a collision

cluster) survives for an extended period (5ps), during

which atoms oscillate (15 oscillations). Reaction is slow.

Fig. 22. 26. The calculated trajectories for a

reactive encounter between A and vibrating

BC, leading to vibrating AB molecule.

Fig. 22. 27. The calculated trajectories for a

complex-mode reaction, in which the collision

cluster has a long lifetime.

vibration

BC

vibration

AB

approach

depart

extended vibration

lifetime of the complex

short lifetime

(d) Quantum mechanical scattering theory

*The motion of atoms, electrons, and nuclei is governed by quantum mechanics.

~ Complete quantum mechanical calculations of trajectories and rate constants are

difficult because it is necessary to take into account all the allowed electronic,

vibrational, and rotational states populated by each atom and molecule in the system

~ It is common to define a “channel” as a group of molecules in well defined quantum

mechanically allowed states. Then at a given temperature, there are many channels

that represent the reactants and many channels that represent possible products.

Some transition between channels being allowed, but others not allowed.

~ Not every transition leads to a chemical transition, since there are so many distinct

channels.

*Cumulative reaction probability: P(E) = Pij(E), where Pij(E) is the probability for a

transition between a reacting channel i and a product channel j. The summation is

over all possible transitions that lead to product.

*The rate constant in terms of cumulative

reaction probability can be defined:

(QR(T): partition function density of reactants,

= partition function / volume)

kr(T) = hQR(T)

We will end the chapter by applying the concepts of transition state theory

and quantum theory to the study of electron transfer.

First we will see theory for factors governing the rates of electron transfer.

22.9 Electron transfer in homogeneous systems

(homogeneous system - solution mixture)

Then we will discuss electron transfer processes on the surfaces of electrodes.

22.10. Electron transfer processes at electrodes

(heterogeneous system - between solution and solid electrode)

Dynamics of Electron Transfer

* Consider electron transfer from a doner species D to an acceptor species A

~ In the first step of mechanism, D and A must diffuse through the solution

and collide to form a complex DA.

(1) First is the formation of encounter complex (reversible)

(2) Next is transfer of electron (reversible) from D to A

(ket and ket: rate constant for forward and reverse electron transfer)

(3) Complex breaks apart and diffuse through the solution (irreversible)

* The rate constant is

See justification in next slide..

22.9 Electron transfer in homogeneous systems

𝐃 + 𝐀 ↔ 𝐃+ + 𝐀− 𝒗 = 𝒌𝒓 𝐃 [𝐀]

𝐃 + 𝐀 ↔ 𝐃𝐀 (𝒌𝒂, 𝒌′𝒂)

𝐃𝐀 ↔ 𝐃+𝐀− (𝒌𝒆𝒕, 𝒌′𝒆𝒕)

𝐃+𝐀− → 𝐃+ + 𝐀− (𝒌𝒅)

𝟏

𝒌𝒓=

𝟏

𝒌𝒂+

𝒌′𝒂𝒌𝒂𝒌𝒆𝒕

𝟏 +𝒌′𝒆𝒕𝒌𝒅

There are two reaction intermediates, which we can apply steady state approximation

: DA and D+A-

(1) steady-state approximation for D+A- complex:

(2) steady-state approximation for DA complex and use the result from (1)

(3) Insert [D+A-] into overall reaction rate law

Justification 22.5 of rate constant of electron transfer in solution

*The rate of overall reaction = the rate of formation of separated ions (products)

End of Justification

𝒗 = 𝒌𝒓 𝐃 𝐀 = 𝒌𝒅[𝐃+𝐀−]

𝒅[𝐃+𝐀−]

𝒅𝒕= 𝒌𝒆𝒕 𝐃𝐀 − 𝒌′𝒆𝒕 𝐃

+𝐀− − 𝒌𝒅 𝐃+𝐀− = 𝟎 ∴ 𝐃𝐀 =𝒌′𝒆𝒕 + 𝒌𝒅

𝒌𝒆𝒕[𝐃+𝐀−]

𝒅[𝐃𝐀]

𝒅𝒕= 𝒌𝒂 𝐃 𝐀 − 𝒌′𝒂 𝐃𝐀 − 𝒌𝒆𝒕 𝐃𝐀 + 𝒌′𝒆𝒕 𝐃

+𝐀− = 𝒌𝒂 𝐃 𝐀 −𝒌′𝒂 + 𝒌𝒆𝒕 𝒌′𝒆𝒕 + 𝒌𝒅

𝒌𝒆𝒕− 𝒌′𝒆𝒕 𝐃+𝐀− = 𝟎

∴ 𝐃+𝐀− =𝒌𝒂𝒌𝒆𝒕

𝒌′𝒂𝒌′𝒆𝒕 + 𝒌′𝒂𝒌𝒅 + 𝒌𝒅𝒌𝒆𝒕𝐃 [𝐀]

𝒗 = 𝒌𝒓 𝐃 𝐀 = 𝒌𝒅[𝐃+𝐀−]

∴ 𝒌𝒓 =𝒌𝒅𝒌𝒂𝒌𝒆𝒕

𝒌′𝒂𝒌′𝒆𝒕 + 𝒌′𝒂𝒌𝒅 + 𝒌𝒅𝒌𝒆𝒕→

𝟏

𝒌𝒓=

𝟏

𝒌𝒂+

𝒌′𝒂𝒌𝒂𝒌𝒆𝒕

𝟏 +𝒌′𝒆𝒕𝒌𝒅

If we assume that the main decay route for D+A- complex is dissociation of

the complex in separated ions, i.e. “kd >> ket” (competing reactions of D+A-)

(1) If ket>>ka kr ka(competing reaction of encounter complex DA), which

means the rate of product formation is controlled by diffusion of D and A in

solution to form a DA complex

(2) If ket<<ka kr (ka/ka

)ket or kr KDAket (where, KDA=ka/ka), which means

that the process is controlled by the activation energy of electron transfer in the

DA complex.

Using transition state theory, we can write: ket κv‡ exp(-Δ‡G/RT) (eq. 22.59)

where κ is the transmission coefficient, v‡ is the vibrational frequency and Δ‡G is

the Gibbs energy of activation.

From now, we will write theoretical expressions (mathematical model) of “κv‡”

and “Δ‡G”. (theory developed by By Marcus, Hush, Levich and Dogonadze)

i) Electron transfer via tunneling influences the magnitude of κv‡.

ii) The complex DA and solvent molecules around it should undergo structural

rearrangement, whose reaction energy determines Δ‡G.

𝟏

𝒌𝒓=

𝟏

𝒌𝒂+

𝒌′𝒂𝒌𝒂𝒌𝒆𝒕

𝟏 +𝒌′𝒆𝒕𝒌𝒅

→𝟏

𝒌𝒓≈

𝟏

𝒌𝒂𝟏 +

𝒌′𝒂𝒌𝒆𝒕

(i.e. most of complex D+A- complex forms products)

(i.e. rate of electron transfer >> rate of dissociation of DA : A,B diffusion controlled)

(i.e. rate of dissociation of DA >> rate of electron transfer : electron-transfer activation controlled)

For the diffusion, we have a good kinetic model. For the electron transfer activation, we need to use the transition state theory..

(now we need to compare the electron transfer with DA dissociation)

(theory in thermodynamics rather than quantum mechanics)

*Two Key aspects in theory of electron transfer

(By Marcus, Hush, Levich and Dogonadze)

(a) Electrons are transferred by tunneling through a

potential energy barrier (partly determined by

ionization energy, I, for DA to D+A-).

(b) The complex DA and the solvent molecules

surrounding it undergo structural rearrangements prior

to electron transfer. The energy associated with these

rearrangements and the standard reaction Gibbs energy

determine Δ‡G.

(a) The role of electron tunneling

(1) The potential energy (Gibbs energy) of two

complexes (DA and D+A-) can be represented by

harmonic oscillators (See Fig 22.28).

(2) Electronic transition is so fast that they can be

regarded as taking place in a stationary nuclear

framework (Frank-Condon principle Ch. 13.2c).

Therefore the geometry should be at q* by thermal

fluctuation to transfer an electron.

Fig. 22. 28. Gibbs energy surfaces of

complex DA and D+A- for the electron

transfer. Displacement coordinate q

corresponds to changing geometries of

the system. qR and qP shows minima of

Gibbs energy of reactant and product.

Two parabola (reactant and product)

meet at q*. Δ‡G, ΔrG, shows

activation, standard reaction and

reorganization Gibbs energy,

respectively.

Harmonic oscillations of displacement between qReat and qProd.

(3) The factor “κv” is a measure of probability that the

system will convert from reactants (DA) to products (D+A-).

(4) To understand this, let’s see effect of nuclear coordinate

rearrangement on electronic energy levels (Fig. 22.29).

(5) Transmission probability exponentially decreases with

the thickness of the barrier (Ch 8.3). The electron

transfer probability exponentially decreases with the

distance between D and A.

ket e-r (eq. 22.60)

Where r corresponds to the edge to edge distance of energy

barrier and depends on the material property through

which the electron travels. In vacuum, 28nm-1 < < 35 nm-1.

Fig. 22. 29. Correspondence between electronic energy level (left) and

nuclear energy level (right) during electron transfer. (a) Initially (q0R), an

electron is in HOMO of DA. LUMO of D+A- is too high to accept electron.

(b) AT q*, DA and D+A- become degenerate. Electron transfer by

“tunneling” through energy barrier of height V. (c) The system relaxes to

the equilibrium nuclear configuration (q0P), in which HOMO of DA is

higher than LUMO of D+A- (Stable).

Harmonic oscillations

Harmonic oscillations

(mathematical model for tunneling)

(b) The expression for the rate of electron transfer

(1) DA complex and medium surrounding DA should rearrange spatially as charge

is redistributed for D+A-.

(2) Rearrangement of D and A molecules and solvent molecules contribute to the

Gibbs energy of activation, Δ‡G

(3) Δ‡G = (ΔrG+)2/4 (Further Info. 22.1, We will accept this as it is..)

where ΔrG is standard reaction Gibbs energy for the electron transfer, is the

reorganization energy.

Now we know the mathematical expressions for ket.

(4) By combining eq. 22.59(ket κv‡ exp(-Δ‡G/RT)) and 22.60(ket e-r):

Rate constant of electron transfer: ket e-re- Δ‡G/RT (eq. 22.62)

(5) Two limitation of eq. 22.62

- It describes only systems where the electroactive species are far and wavefunctions

of D and A do not overlap.

- It applies at only high temperatures, under which thermal fluctuation can bring

reactants to the transition state.

(mathematical model for Gibbs energy)

Exercise (#4) For an electron donor-acceptor pair, ket=2.02 105 s-1 when

the doner-acceptor distance r=1.11nm and ket=4.51 105 s-1 when the

distance r=1.23nm. Assuming that ΔrG° and λ are the same in both

experiments, estimate the value of β.

Since ket e-re- Δ‡G/RT, for the same donor and acceptor at different distances,

eqn. 22.63 can be applied.

𝒍𝒏𝒌𝒆𝒕 = −𝜷𝒓 + 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 (at fixed temperature)

The slope of a plot of 𝑙𝑛𝑘𝑒𝑡 versus r is – 𝛽. The slope of a line defined by two

point is :

𝒔𝒍𝒐𝒑𝒆 =∆𝒙

∆𝒚=𝒍𝒏𝒌𝒆𝒕,𝟐 − 𝒍𝒏𝒌𝒆𝒕,𝟏

𝒓𝟐 − 𝒓𝟏=–𝜷

so 𝜷 = −𝟔. 𝟕𝒏𝒎−𝟏

We are assuming that ΔrG° and λ are the same

22.10 Electron Transfer Processes at electrodes*Electrode-solute/solution interaction is different from a discrete complex in the bulk of the solution.

~ In homogeneous system, electron transfers through tunneling (Ch. 22.9)

~ In heterogeneous system (with electrode), “specific interaction between electrode surface and solute/solvent” changes electron transfer conditions from bulk solution.

(a) The electrode–solution interface (5 different models)(1) Electrical double layer model: a boundary model of liquid and

solid surface (a sheet of positive charge at surface of electrode and

a sheet of negative charge next to it, from which Galvani potential

difference is created, between bulk electrode and bulk solution.

This model describes an abrupt change between two extremes.)

(2) Helmholtz layer model: solvated ions arrange along the surface

of the electrode but are held away by hydration spheres. (22.31 )

(This describes gradual changes of electrical potential between two

extremes. This ignores the disrupting effect of thermal motion.)

Fig. 22. 31. model that treats the system as two rigid planes of charge. One is outer Helmholtz plane

(OHP), which is due to ions with solvating molecules, and the other is inner Helmholtz plane (IHP),

which is due to the electrode. From IHP to OHP, potential changes linearly from M to S.

IHP

(3) Gouy-Chapman model (diffuse double layer): takes into

account the disordering effect of thermal motion (no longer a

linear decrease from M to S). (Fig 22.32)

(4) Stern model: Neither Helmholtz and G-C model is good, in

that H. model overemphasize the rigidity of local solution and G-

C model underemphasize the structure. Stern model is

combination of these two, in which ions close to the electrode are

constrained in rigid Helmholtz plane while the ions are dispersed

outside of the plane. (OHP + diffuse double layer = linear

potential decrease + gradual potential decrease) (Fig 22.33)

(5) Grahame model: New plane is added to Stern model, i.e.

Inner Helmholtz Plane (IHP)by metal electrode + Stern model (OHPby

confined ions+diffuse double layerby thermal distrubance)

→ Fig 22.31+Fig 22.33 →

Fig. 22. 32. G-C model of the electrical double layer. The plot of electrical potential versus

distance shows the meaning of diffuse double layer.

Fig. 22. 33. Stern model of the electrode-solution interface.

Fig. 22. 32

Fig. 22. 33

* Galvani potential difference :

potential difference between metal

electrode and bulk solution, i.e.

overall potential difference

(b) The Butler – Volmer equation

Explain electrochemical reactions at the electrode (electron transfer between electrode and reactants) using mathematical models: for example, reduction of ion at electrodes by the transfer of a single electron (electron transfer as a rate determining step).

(1) “Butler – Volmer” equation: j=j0{e(1-)f - e-f} (j is the current density, the current flowing through the unit area of electrode), where f=F/RT (F: Faraday’s constant),

=E-E (overpotential; E is electrode potential at equilibrium, when there is no net current, and E is the electrode potential, when the current is drawn from the cell (This shows how much extra-potential can be made to generate current in an electrode.)

is the transfer coefficient, which shows if the transition state of electroactive species is reactant-like (=0) or product-like (=1).

j0 is the exchange current density (the magnitude of the equal but opposite value of the current density at equilibrium)

See Fig. 22.34, which shows

“how Butler-Volmer eq. predicts current density, j”

Fig. 22. 34. The dependence of the current density, j, on the overpotential,

, for different values of the transfer coefficient, .

Anode

(oxidation)

Cathode

(reduction)

linear

when

< 0.01V

exponential

(mathematical model for the electron transfer in

the heterogeneous system, i.e. electrode-solution)

(Exchange current densities reflect intrinsic rates of electron transfer between an analyte and the

electrode. Such rates provide insights into the structure and bonding in the analyte and the electrode.)

(2) When the overpotential is small (η < 0.01 V), fη << 1, exponential can be expanded.

j = j0 { 1 + (1 – α) fη + …… … - (1-αfη + ….)} ≈ j0fη ⇒ η = RTj/Fjo (since, f=F/RT)

∴ j∝ η: linear dependency of current density, j, to overpotential, . (Fig. 22.34)

Brief Illustration

Electrode (cathode, reduction, electrons transfer from electrode to reactants)

: Pt (s) | H2 (g) | H+ (aq), Exchange current density, jo = 0.79 mA/cm2 at RT

when overpotential, η = 5.0mV ( η < 0.01 V), current density, j=?

At given temperature, f = F/RT = 1/(25.69mV)

j = j0fη = (0.79 mA/cm2) 1/(25.69 mV) (5.0 mV) = 0.15 mA/cm2

(3) When the overpotential is very high and positive, (η > 0.12V)

e(1-α)fη >> e-αfη therefore j>0, electrode (surface) at which current flows out (electrons flow in), thereby oxidation happens. This electrode corresponds to anode in electrolysis

j = j0e(1-α)fη ln j = ln j0 + (1-α)fη

(4) When the overpotential is very large but negative, (η < -0.12V)

e(1-α)fη << e-αfη therefore j<0, electrode (surface) at which current flows in (electrons flow out), thereby reduction happens. This electrode corresponds to cathode in electrolysis

j = -j0e-αfη ln (-j) = ln j0 - αfη

(Some exchange current density j0 and transfer coefficient are given in Table 22.3.)

j=j0{e(1-)f - e-f}

j=j0{e(1-)f - e-f}

j=j0{e(1-)f - e-f}

cathode

If the overpotential is large, we cannot expand the exponential but divide conditions.

(c) The Electrolysis(1) Definitions

- Electrolysis: a method of using a direct electric current (DC) to drive an otherwise non-spontaneous chemical reaction.

- The applied potential difference = zero current potential (at equilibrium) + cell overpotential (when current is flowing)

- Cell overpotential: sum of overpotentials at two electrodes + ohmic drop (IRs, where Rsis internal resistance of the cell) due to the current through the electrolyte

(2) Relative rate of gas evolution or metal deposition during electrolysis can be estimated from Butler-volmer eq. and Table 22.3, assuming equal transfer coefficient,

At cathode (η < -0.12V), j=j0{e(1-)f - e-f} reduces to j -j0e-αfη

ratio of cathodic currents (reduction of gas and metal): j/j = j0/j0e(-)f

where j and j are current density of electrodeposition and gas evolution, respectively,

and j0 and j0 are corresponding exchange current densities (Table 22.3).

(3) Above equation shows that metal deposition is favored (j>j) by a large exchange current density (j0>j0) and high overpotential (i.e. - is positive and large).

*Pt is better electrode (cathode) material than

other metals (it has higher exchange current

density and thereby requires less

overpotential).

*Also Fe reduction is more favorable than gas

evolution (i.e. j0 of Fe reduction is larger).

- example: H2 evolution at various metal electrodes (See Table 22.3 and Data Section)

~ for Pb electrode, j0 ~ 5 pA/cm2 The exchange current density is too small.

Therefore, a high overpotential is needed to induce significant H2 evolution.

~ for Pt electrode, j0 ~ 1 mA/cm2 Since the exchange current density is large, the

gas evolution occurs at a much lower overpotential

cf) exchange current density, j0, depends on crystal orientation

~ Cu on Cu electrode, (100) face j0 = 1 mA/cm2

(111) face j0 = 0.4 mA/cm2

j = -j0e-αfη (note η is negative in cathode, η < -0.12V, reduction, i.e. –αfη > 0)

(d) Working galvanic cells

- In working galvanic cells, the overpotential results in a smaller potential than under

zero-current conditions. Also the cell potential decreases as current is generated (due to

the irreversible cell reaction).

Cell: M│M+ (aq) ║ M + (aq)│M potential of cell E = ∆R –L

Potential difference at electrodes = zero-current potential + overpotentials (when current is flowing)

∆X = EX+η X (X is Left or Right electrode, E is potential at zero current, E when current flowing)

Working Cell Potential=E=∆R-∆L=E+(ηR–ηL) (where E=ER-EL is zero current cell potential)

Since η R and η L is negative (cathode) and positive (anode), respectively

E = E – │ η R │ – │η L │ (working cell potential is smaller than the zero current cell potential)

We need to consider the Ohmic potential difference, IRs, showing cell’s irreversibility.

E = E - │η R│-│η L│ - IRs

We can calculate overpotentials using Butler-Volmer eq. for a given current I.

Assumption: high overpotential limit, same area of electrodes (A), both transfer coefficients = 1/2

j = - j0 e- α fη η R = j = j0 e(1-α)fη

η L = )ln(1

0R

R

j

j

f

)ln(

)1(

1

0L

L

j

j

f

Cathode Anode

= E – IRs -

LR

LR

jj

jj

F

RT

00

))((ln

2

RTFf /&2

1

= E – IRs -)(

)/(ln

2

00

2

LR jj

AI

F

RT

(- jR) = jL = I/A

= E – IRs -)(

)(ln

4

jA

I

F

RTwhere 2

1

00 )( RLjjj

E = E - │η R│-│η L│ - IRs )ln()1(

1)ln(

1

00 L

L

R

R

j

j

fj

j

f = E – IRs -

Brief Illustration

Suppose, A=10cm2, exchange current densities, j0L=j0R=5Acm-2, Rs=10.

At T=298K, RT/F=25.7mV. Zero current cell potential E=1.5V. I=10mA.

Working potential E ?

E = E – IRs - = 1.5V – 10mA10 - 425.7mVln(10mA/(10cm25Acm-2)

= 0.9V

)(

)(ln

4

jA

I

F

RT

Exercise (#5) The transfer coefficient of a certain electrode in contact with M3+

and M4+ in aqueous solution at 25°C is 0.39. The current density is found to be

55.0mA cm-2 when the overpotential is 125mV. What is the overpotential reqired

for a current density of 75 mA cm-2?

The conditions are in the limit of large (>0.12V), positive overpotentials (anode), so equation

22.70 applies : j = j0 e(1-α)fη

𝒍𝒏 𝒋 = 𝒍𝒏 𝒋𝟎 + 𝟏 − 𝜶 𝒇𝜼

Where, at 25C, 𝒇 =𝑭

𝑹𝑻= 𝟑𝟖. 𝟗𝑽−𝟏

Subtracting this equation from the same relationship between another set of currents and

overpotentials (as if we compared gas evolution and metal deposition), we have

𝒍𝒏𝒋′

𝒋= 𝟏 − 𝜶 𝒇(𝜼′ − 𝜼)

which rearranges to (since we are interested in the overpotential)

𝜼′ = 𝜼 +𝒍𝒏

𝒋′

𝒋

𝟏 − 𝜶 𝒇= 𝟎. 𝟏𝟐𝟓 +

𝒍𝒏𝟕𝟓𝟓𝟓

𝟏 − 𝟎. 𝟑𝟗 ∗ 𝟑𝟖. 𝟗= 𝟎. 𝟏𝟑𝟖𝑽

(exchange current density is a constant)

Exercise (#6) To a first approximation, significant evolution or deposition occurs in

electrolysis only if the overpotential exceeds about 0.6V. To illustrate this criterion,

determine the effect that increasing the overpotential from 0.40V to 0.60V has on the

current density in the electrolysis of a certain electrolyte solution, which is 1.0mA cm-2

at 0.4V and 25°C. Take α=0.5.

Since overpotential is 0.6V, it is for anode. And thus O2 g is produced at the anode in this

electrolysis and H2 g at the cathode. The net reaction is

𝟐𝑯𝟐𝑶 𝒍 → 𝟐𝑯𝟐 𝒈 + 𝑶𝟐 𝒈

for a large positive overpotential (>0.12V) we use eqn 22.70 (same as the previous exercise):

𝒍𝒏 𝒋 = 𝒍𝒏 𝒋𝟎 + 𝟏 − 𝜶 𝒇𝜼

which implies 𝒍𝒏𝒋′

𝒋= 𝟏 − 𝜶 𝒇 𝜼′ − 𝜼 = 𝟏 − 𝟎. 𝟓 ∗ 𝟑𝟖. 𝟗 ∗ (𝟎. 𝟔 − 𝟎. 𝟒) = 𝟑. 𝟗

and 𝒋′ = 𝒋𝒆𝟑.𝟗 = 𝟏𝒎𝑨𝒄𝒎− 𝟐 ∗ 𝒆𝟑

. 𝟗 = 𝟒𝟗𝒎𝑨 𝒄𝒎−𝟐

Hence, the anodic current density increases by roughly a factor of 50 with a corresponding

increase of overpotential (from 0.4V to 0.6V) in O2 g evolution.