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CHAPTER 21: Reaction Dynamics I. Microscopic Theories of the Rate Constant.

A. The Reaction Profile (Potential Energy diagram): Highly schematic and generalized. A---B-C

B. Collision Theory of Bimolecular Gas-Phase Reactions.

Consider the elementary reaction in gas phase:

A + B → P

(collision freq. between A + B) = ZAB From kinetic theory of gases:

ZAB = NANBσcrel f = collision density (# of collisions of A and B per volume per time)

NANB = number densities (# of A or B per unit volume)

Rate   Collision   frequency Steric  

requirement Energy  

requirement = x x

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σ=πd2 = collision "cross section" or target area d = 12dA+ d

B( )

crel

= average relative velocity = (8kBT/πµ)1/2

µ = mAmB/(mA + mB) = "reduced mass"

f = symmetry factor: = 1 if A ≠ B = 1/2 if A = B Special Case: A and B identical.

µ = mA/2; f = 1/2

ZAA = √2 NA2πdA2(8kBT/πmA)1/2 In terms of molar concentrations [A] and [B]:

ZAB = σcrel f Navo2[A][B]

The Energy requirement:

Now, what about fraction of collisions with KE > Ea:

fraction... ≈ e-Ea/kT when Ea is on the "tail" of distribution.

Turns out that the collision cross section can be viewed as a function of the relative kinetic energy of two colliding molecules, and that when you integrate over kinetic energy distribution, you get a simple exponential dependence on the activation energy.

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Steric requirement: multiplicative factor θ, such that σ θ gives the effective cross-sectional area.

θ = fraction of collisions in a productive geometric arrangement. Total rate expression: Rate =- d[A]/dt = θ σc

relfNavo e-Ea/kT [A][B]

= k2[A][B] By comparison with Arrhenius theory. A = θ σc

relfNavo

In problems, substitutions of numerical values for parameters gives k2 in units length3/mol-s, which should be converted to Liters/mol-s Consider rxn (Hydrogen exchange) at T = 300 K: H + H2 → H2 + H Calculate Arrhenius A. (i) estimate rH, rH2 ≈ 1.0 x 10-10 m = (1Å) (ii) µ = (mHmH2)/(mH + mH2) = {(1.0 * 2.0)/(1.0 + 2.0)}Navo-1 (iii) kB = 1.38 x 10-23 J/K ↑ J=kg m2/s2 (iv) A = 4.1 x 1014 cm3/mol sec * 1 liter/1000 cm3 = 4.1 x 1011 L/mol sec with steric P=1 (v) Experiment gives A = 5.4 x 1010 L/mol s, or Atheory/Aexpt = 7.6 Discrepancy corrected by steric (or geometric) factor θ.

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Example: angle of approach of reactants is important.

Multiply k2 by θ = steric factor = 1/7.6

C. Transition State Theory

1. A more general theory of rate constants, with no adjustable parameter like θ, the steric factor.

2. Potential energy surface of the rxn: H + HD → H2 + D

One-dimensionalization along path:

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Steric requirement: multiplicative factor P, such that "P gives the effective cross-sectional area. P = fraction of collisions in a productive geometric arrangement. Total rate expression:

Rate =- d[A]/dt = P"crel

fNavo e-Ea/kT [A][B]

= k2[A][B] By comparison with Arrhenius theory.

A = P"crel

fNavo

In problems, substitutions of numerical values for parameters gives k2 in units length3/mol-s, which should be converted to Liters/mol-s Consider rxn (Hydrogen exchange) at T = 300 K: H + H2 ! H2 + H Calculate Arrhenius A. (i) estimate rH, rH2 # 1.0 x 10

-10 m = (1Å) (ii) µ = (mHmH2)/(mH + mH2) = {(1.0 * 2.0)/(1.0 + 2.0)}Navo-1 (iii) kB = 1.38 x 10

-23 J/K $ J=kg m2/s2 (iv) A = 4.1 x 1014 cm3/mol sec * 1 liter/1000 cm3 = 4.1 x 1011 L/mol sec with steric P=1 (v) Experiment gives A = 5.4 x 1010 L/mol s, or Atheory/Aexpt = 7.6 Discrepancy corrected by steric (or geometric) considerations. Example: angle of approach of reactants is important.

Multiply k2 by P = steric factor = 1/7.6

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3. TST theory says, H + HD

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↔ H--H--D → H2 + D ↑ in quasi-equilibrium

A +B→K≠

AB≠ →products Reactants are in pre-equilibrium with AB≠, the activated

complex

with equilibrium constant K≠.

4. Define: K≠=pAB≠po

pApB

p° is the standard pressure to keep K≠ unitless (=1 bar) 5. Now, we may convert to a concentration basis using px = RT[X]

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[AB≠] =RT

poK≠[A][B]

The forward rate = k≠[AB≠] with k≠= unimolecular decomposition rate constant of activated complex. Forward rate = k≠K≠[A][B](RT/p°) = k2[A][B] And so finally the TST second order rate constant is: k2 = k≠K≠(RT/p°) 6. Our next task is to find K≠ (= e =−ΔG≠°/RT according to Chapter 6) where ΔG≠° = molar free energy of activation (standard) = G°(AB≠) - G°(A + B) Remember that in general an equilibrium constant is a ratio of

partition functions of products over reactants, with the “product” being in this case the activated complex.

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K≠ =NavoqAB≠o

qAoqBoe−ΔEo /RT

where: ΔEo = Eo(AB≠) – Eo(A) – Eo(B)

is the difference in zero-point electronic energies. 7. The A and B partition functions are given as usual (Chap 15), but the

one for the activated complex is a bit tricky. The activated complex species has a vibration which leads to its

decomposition. The partition function for this vibration is

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q =1

1 − e−hν / kT≈

11 − (1 − hν /kT + ...)

≈ hν /kT

The activated complex also has other modes of vibration, possibly, and these contribute to the remainder of its partition function

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q AB≠o , so

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qAB≠o ≈ (hν /kT)q AB≠

o

8. So finally:

K≠ =kThν

NavoqAB≠o

qAoqBoe−ΔEo /RT ≡ kT

hνK≠

9. Now putting it all together we get the Eyring rate constant:

k2 = k≠ kThν

#

$%%

&

'((K

≠(RT / po)

10. But here we make one more connection: the unimolecular decomp

rate constant of the activated complex is the same as the vibration frequency ν assuming every vibration causes a decomposition, or if not, is related by a barrier-crossing transmission coefficient κ as:

k≠ = κν

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So a cancelation takes place and, voila, this unknown frequency goes

away:

k2 = κkTh

"

#$$

%

&''K

≠(RT / po)

The above equilibrium constant is a pressure-based one, which can

be replaced by a concentration-based one like Eyring used:

k2 = κkTh

"

#$$

%

&''Kc

≠

11. Thermodynamic version of TST: The above rate constant is couched in terms of statistical mechanical

partition functions, which is greatly useful when you have structural knowledge of the activated complex.

Without this knowledge one can still make progress using

thermodynamic description. Here we use the activation free energy:

K≠ = e−ΔG≠ /RT

k2 = κkTh

"

#$$

%

&''(RT / p

o)e−ΔG≠/RT

Activation free energy can be further decomposed into enthalpy and

entropy of activation: ΔG≠ = ΔH≠ - TΔS≠ 12. Relation of TST quantities to Arrhenius form: Need to find apparent Ea in terms of ΔH≠° and TST. E

a= RT2(∂lnk / ∂T) = ΔH≠ +2RT for gas phase bimolecular reaction.

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A = e2 kThRTpoeΔS

≠ /R

Note: entropy part is the only term unknown, and that part can be

identified as the steric factor of collision theory:

θ = eΔS≠ /R

Determine A by expt - find ΔS≠, infer something about activated

complex complexity. 13. Example bimolecular reaction

H + H2 → (H-H-H)* → H2 + H

From experiment: A = 5.4 x 1013 cm3/mole sec = 5.4 x 1010 L/mole sec Ea = 8.22 kcal/mole

Find ΔS≠ at 300°K.

ΔS≠ = -6.75 R (fairly large and negative)

Indicating H3* state is much more orderly than H2(g) + H(g) state.

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II. Reactions in Solution.

A. Two limiting types of behavior. 1. Low activation energies: Rxn occurs every time there's a collision; rate then depends on how

fast reactants can diffuse through solvent and encounter each other. Diffusion-Controlled Reaction:

Smoluchowski expression:

k = 4πdDNavof

relative diff coeff D = DA + DB (depends on solvent viscosity, size of A + B)

d = sum of radii

Example: H+(aq) + Ac-(aq) → HAc

f = electrostatic factor = 1 if non ionic 2. High activation energies: (Activation-controlled limit)

• Rxn rate depends on collision event itself, probability of success. • Solvent plays a role by stabilizing effect on A + B independent

species relative to AB≠ complex. • Will focus on case 2 for remainder of discussion.

B. Reaction between non-polar molecules:

• solvent forces are negligible • solvent acts simply as space filler • treat same as gas-phase (use collision theory, e.g.)

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C. Reaction between ions or polar molecules when solvent is polar:

• electrostatic effects important. Effect of dielectric constant (ε) of solvent:

Find ΔG≠ = G≠ - G (∞ separation) Interparticle force f = ZAZBe2/4πεoεx2 (x = distance separ.) f(x=∞) = 0 Calculate ΔGes≠ (electrostatic portion of ΔG≠) by computing work w done

on ions bringing them togther.

w = − f x( )∞

d

∫ dx

= − ZAZ

Be2 / 4πε

oεx2( )

∞

d

∫ dx

= + ZAZ

Be2 / 4πε

oεx( )|

∞

d

= + ZAZ

Be2 / 4πε

oεd( )

= electrostatic contrib to ΔG≠ per molecular pair

molar ΔG≠ = ΔGnes≠ + Z

AZBe2 / 4πε

oεd( )Navo

↑ ↑ molar non-electrostatic to make per mole contribution

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Smoluchowski expression: k = 4#dDNavof relative diff coeff = DA + DB (depends on solvent viscosity, size of A + B) d = sum of radii

Example: H+(aq) + Ac+-(aq) ! HAc f = electrostatic factor = 1 if non ionic 2. High activation energies: (Activation-controlled limit)

• Rxn rate depends on collision event itself, probability of success. • Solvent plays a role by stabilizing effect on A + B independent species relative to

AB! complex. • Will focus on case 2 for remainder of discussion.

B. Reaction between non-polar molecules:

• solvent forces are negligible • solvent acts simply as space filler • treat same as gas-phase (use collision theory, e.g.)

C. Reaction between ions or polar molecules when solvent is polar:

• electrostatic effects important. Effect of dielectric constant (() of solvent:

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Plug into TST:

k2= constants × e

− ΔGnes≠ + ZAZBe

2 /4πεoεd( )Navo( )/RT

k2= k

2oe

− ZAZBe2 /4πεoεd( )Navo( )/RT

ln k = ln k

o−Z

AZBe2 / 4πε

oεdk

BT

Theory says: ln k vs. 1/ε gives straight line How do we change ε? Change solvents, change temperature.

k↑ as ↓ε, why? D. Ionic strength (salt) effects on reactions between ions in solution Consider rxn in solution: A + B → AB≠ → products In TST, we used K≠ ~ [AB≠]/[A][B]. With equilib constant of rxn in non-

ideal solution, should use activities. K≠ = a≠/aAaB = ([AB≠]/[A][B])( γ≠/γAγB) ↑ act. coeff [AB≠] = K≠ [A][B]γAγB/γ≠ TST: Rate = k≠[AB≠] (However, rate still prop to [AB≠], not activites) ~ K≠[A][B]γAγB/γ≠ Rate = ko[A][B]γAγB/γ≠ ↑ ↑ old rate const correction without activities factor considered

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Find %G! = G! - G () separation) Interparticle force f = ZAZBe

2/4#(o(x2 (x = distance separ.) f(x=)) = 0 Calculate %Ges! (electrostatic portion of %G!) by computing work w done on ions

bringing them togther.

w = ! f x( )"

d

# dx

= ! ZAZ

Be2 / 4$%

o%x2( )

"

d

# dx

= + ZAZ

Be2 / 4$%

o%x( )|

"

d

= + ZAZ

Be2 / 4$%

o%d( )

= electrostatic contrib to &G' per molecular pair

molar !G" = !Gnes" + Z

AZBe2 / 4#$

o$d( )Navo

$ $ molar non-electrostatic to make per mole contribution Plug into TST:

k2= constants ! e

" #Gnes$ + ZAZBe

2 /4%&o&d( )NavoRT( )

k2= k

2oe

" ZAZBe2 /4%&o&d( )NavoRT( )

ln k = ln ko!Z

AZBe2 / 4"#

o#dk

BT

Theory says: ln k vs. 1/( gives straight line How do we change (? Change solvents, change temperature.

k$ as *(, why?

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k = koγAγB/γ≠ log k = log ko + log γAγB/γ≠ Use D-H limiting law log y = -BZ2√I ↑ ionic strength log k = log ko + 1.02 ZAZB√I

Origin of effect, non-independent behavior of ions at moderate conc:

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Notes: