CHAPTER 19:  CHEMICAL KINETICS

aA bB cC P+ + →

law of mass action

rates depend on stoichiometry in the same way as equilibrium constants

[ ] [ ] [ ]fa b cdP k A B C

dt=

rate constant for forward reaction

However, kinetic mechanisms do not necessarily follow stoichiometry

They do for “elementary reactions”

unimolecular decaybimolecular reactions

A P→A B P+ →

Many reactions proceed in 2 or more steps

Detailed Balance

A Bfk

A Brk

At equilibrium:   [ ] [ ]f req eqk A k B= forward and reverse  rates are equal

[ ] [ ] [ ]

[ ]f r

d Ak A k B

dtd B

= − +

[ ] [ ] [ ]

[ ] [ ] at equilibrium 0

f r

d Bk A k B

dtd A d B

dt dt

= −

⇒ = =

[ ][ ]

f

r

dt dtkB

KA k

⇒ = =

A, B, I

steady state ⇒ [ ] [ ] [ ] 0d A d B d I

d d d= = =y

dt dt dt

detailed balance is a stronger statementeach forward rate must = its corresponding backward ratebackward rate

[ ][ ]

[ ][ ]

[ ][ ]

, , eq eq eqAI IB BA

IA BI ABeq eq eq

I B Ak k kA k I k B k

= = =

A BB A→→

via intermediate I

directly

In scenario (b)

1AI IB BA

IA BI AB

k k kk k k

=

detailed balance

at equilibrium forward and reverse reactions must havethe same intermediates

Temperature dependence

A B P+ →2k

“2” ⇒ bimolecularA B P+ →

[ ] [ ][ ]2

d Pk A B

dt=

rate

2 22HI H I→ +

> T from 700 to 800 Kreaction rate coefficient > 40x

Recall van’t Hoff Equation

2

d nK hΔ= 2dT RT

( )2

f ad n k E

dT RT=

Arrhenius proposed

( )2

2

'r a

dT RTd n k E

dT RT=

Arrhenius:  It is not the average energy but thehigh energy tail of the distributionthat is important Rates for forward and reverse 

ti f H + I > 2HImeasure , ', o

f a r ak E k E K h→ → →Δ

'oa ah E EΔ = −

/aE RTfk Ae−=

reactions for H2 + I2 ‐> 2HI

f

activated vs. non‐active reactionsMany reactions are activatedHowever, radical reactions tend to be non‐activated,

Transition state theory

minimum energy pathway: saddle point = transition state (TS)

Now consider 

A B P+ →2k

K ≠ k ≠

( )A B AB P≠+ → →K ≠ k ≠

equilibrium direct downhill

( )[ ][ ]

#AB

KA B

≠⎡ ⎤⎣ ⎦=

[ ] ( ) [ ][ ]#

#2

d Pk AB k K A B

dtk k K

≠≠ ≠

≠

⎡ ⎤= =⎣ ⎦

=

Diffusion coefficient of C in Fe.

2

/D kTAB

A B

qK eq q

≠≠

≠ Δ= D≠Δ = Dissoc. energy of TS – dissoc. energy of reactants

ABq q qξ≠ ≠=

Nonequil reaction coord1

2

1

k

h

ξξ

ξ

υπ μ

υ

=

<<

weak vibration

Nonequil. reaction coord.

ordinary degreesof freedom

/kT

1

11 h

kTkTqhe ξ

ξ

ξ υξυ

−

<<

= ≈−

k ξυ≠ = = rate coefficient for downhill step

/2

D kT

A B

kT qk eh q q

≠≠

Δ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ “‐” unstable degree of 

A Bh q q

kT Kh

≠

⎝ ⎠⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

gfreedom has been factored out

kT nK G H T S≠ ≠ ≠Δ Δ Δ

/ / /2

G kT H kT S k

kT nK G H T SkT kTk e e eh h

≠ ≠ ≠−Δ −Δ Δ

− = Δ = Δ − Δ

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Arrhenius/

/ /

aE RTf

S k H kT

k Ae

kTk e e≠ ≠

−

Δ −Δ

=

⎛ ⎞⎜ ⎟

TSTfk e e

h= ⎜ ⎟⎝ ⎠

TST

/S kTkTA eh

≠Δ=h

fastest possible rate  ( )when 0G≠Δ =

126 24 10kT x s= = (one per 0 16 ps) at T = 300 K6.24 10x sh

= = (one per 0.16 ps) at T = 300 K

k ≠ = max speed

throttled by bottlenecky

Only the fraction ( ) [ ][ ]/ 1K AB A B≠≠ ⎡ ⎤= <<⎣ ⎦ reach the bottleneck

1RT nk G≠ ≠− = Δ > ⇒ free energy barrier

enthalpic – e.g., must break a bondentropic – e.g., need a certain orientation

Primary Kinetic Isotope effect

isotopic substitution most important when it is at a reacting bond

( )CH CH C H≠→ → +

( )CD CD C D≠→ → +

vs.

( )CD CD C D→ → +

( )/

/

CH

CH CD

D kTCH

D D kTCHH

qe

qk ek

≠≠

≠ ≠

Δ

Δ −Δ

⎛ ⎞⎜ ⎟⎝ ⎠= ≈⎛ ⎞ /CDD kTD CD

CD

qke

q≠≠ Δ⎛ ⎞

⎜ ⎟⎝ ⎠

( )1D D h≠ ≠Δ Δ ( )2

1 12

CH CD CD CH

CD CH CH

D D h υ υ

υ υ υ

≠ ≠−Δ Δ = − −

⎛ ⎞− = −⎜ ⎟⎝ ⎠

using reduced masses

1exp 12 2

CHH

D

hkk kT

υ⎡ ⎤⎛ ⎞= − →⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

7.78 at T = 300K

Catalysis

3 2H N NH+ →

Catalysts generally work by stabilizing TS

E g High barrier due to cost of breaking triple bond2 2 33 2H N NH+ →Fe catalyst (Haber‐Bosch process)

A B P+ →0k≠⎡ ⎤

E.g.,  High barrier due to cost of breaking triple bond

[ ][ ]0 0

ABkT kTk Kh h A B

ABCK

≠

≠

≠

≠

⎡ ⎤⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎣ ⎦= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎡ ⎤⎢ ⎥⎣ ⎦

Without catalyst

With catalyst[ ][ ][ ]C

C C

KA B C

kTk Kh

≠

≠

⎢ ⎥⎣ ⎦=

⎛ ⎞= ⎜ ⎟⎝ ⎠

t h t

With catalyst

rate enhancement

( ) [ ]0 0

C CB

ABCk K Kk K AB C

≠≠

≠ ≠

⎡ ⎤⎢ ⎥⎣ ⎦= = =

⎡ ⎤⎢ ⎥⎣ ⎦( ) [ ]⎢ ⎥⎣ ⎦

BK = binding constant of catalyst to TS

Pauling:  enhancement BK∝

Thus, you want catalyst to bind tightly to TS

If catalyst lowers TS in energy → speeds up bothforward and reverse processes

slow step in A + B could be their diffusion through the solventto proper orientation

Solve by pre‐organized solvent cavity

3 3 3Cl CH Br ClCH Br− −+ → +

Solventgas 1

relative rate (cm3 molecule‐1 s‐1)

acetone 1010−

1510−

1610−methanolwater

Acid/Base catalysis

AH H A+ −→ +aK

AH H A→ +

R P→k

H +

[ ]a

H AK

AH

+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦=

ak[ ] [ ][ ]a

d Rk AH R

dt= −

Bronsted law: the stronger the acid, the faster the R→ P reactionBronsted law:  the stronger the acid, the faster the R → P reaction

log log , 0a a ak K cα α= + >

loga apk K= −

log a a bk pK cα= − +

Base catalysis

log logb b bk K cβ= +log logb b b

a

k K cE a G b

β += Δ +

for acid dissoc.