Reaction kinetics: 1st order reactions•

•••

• • • • • • •

[A]

t

k1A B (+ C)

Decay reactions, like radio-activity;SN1 reactions

[A]k=dt

d[A]1Rate: -

Rewriting: - dtk=[A]

d[A]1

Integration gives: t

0

t

0

ktdtd[A][A]

1

So: ln[A]t – ln[A]0 = -kt or: =]A[

]A[ln

0

t -kt

Since [A]0 is constant we can also write: ln[A]t = -k1t + C

Plotting ln[A]t against t gives a straight line with slope -k:

slope = - kx

x

xx

x

t

ln[A]t

tln[A]t

The halflife, t1/2, is defined as the time that is needed to reduce the concentration of the reactant to 50% of its original value. In formula:

0

0

]A[

]A[2/1ln =-kt1/2 so

k

693.0=

k

2ln=t 2/1

=]A[

]A[ln

0

t -kt

Reaction kinetics: 2nd order reactionsk2

A + B Pdt

]A[dSo: - = - ]B][A[k=

dt

]P[d+=

dt

]B[d2

When [A] [B], this equation is mathematically rather complicated.A simplification reads as follows:take [P] = x, then [A] = [A]0 – x and [B] = [B]0 – x

The rate then becomes: = =]B][A[k=

dt

dx=

dt

]P[d2 k2([A]0-x)([B]0-x)

so dtk=dxx)-x)([B]-([A]

12

00

Integration gives: tk=]B][A[

]B[]A[ln

[A]-[B]

12

0

0

00

Plotting]A[

]B[ln against t gives a straight line with slope k2([B]0-[A]0)

Special cases:

• [A]0>>[B]0 (pseudo-first order kinetics)

k2CH3I + H2O CH3OH + HI

H2OExample:

- ]ICH['k=]OH][ICH[k=dt

]ICH[d3232

3 in which k'=k2[H2O]

This is a pseudo-first order reaction, since [H2O] is constant. The second-order rate constant k2 can be calculated from k' and [H2O]. In a dilute aqueous solution, [H2O]=55 M.

Special cases:

• [A]0 = [B]0

k2A + B P

222 ]A[k=]B][A[k=

dt

]P[d=

dt

]A[d-

Integration gives: tk=[A]

1-

[A]

12

0t

Plotting of 0t [A]

1-

]A[

1against t gives k2 as the slope.

Reversible reactions

A Bk1

k-1Take the simplest possibility:

On t = 0: [A] = [A]0 [B] = 0 t = t: [A] = [A]0-x [B] = x

= k1[A] – k-1[B] = k1([A]0 – x) – k-1x = k1[A]0 – (k1 + k-1)x

Integration gives: (1)

At equilibrium, the net reaction rate = 0, so [B]t is constant(=[B]e = xe), so: k1[A]e = k-1[B]e = k-1xe

dt

dx

t)k+-(k=]A[k

x)k+(k-]A[kln 1-1

01

1-101

[A]0

xe

t

[A]t

[B]t

There is an equilibrium constant:

so: (2)

e0

e

e

e

1-

1

x-]A[

x=

]A[

]B[=

k

k=K

Combining eq (1) with (2) gives: t)k+-(k=x

x-xln 1-1

e

e

This is the rate equation for a first order process!

Determination of (k1 + k-1) by plotting against t e

e

x

x-xln

Eq (2) gives 1

1

k

k 2 equations, 2 unknowns

Individual values of k1 and k-1 can be determined

1

1e0 k

k1x[A]

Preequilibria

A + B A·Bk1

k-1

Ck2

Very complicated kinetics, unless you assume that [A·B] is constant during a large part of the reaction (steady state approach)

0=dt

d[A·B]k1[A][B] = k-1[A·B] + k2[A·B] = (k-1 + k2)[A·B]

So the rate equation now becomes:

= k2[A·B] = 21

21

kk

[A][B]kk

Two possibilities:

- rapid breakdown of A·B, k2>>k-1, so = k1[A][B]

- slow breakdown of the complex: k2<<k1,k-1, so:

= k2[A·B] = =

k2K[A][B]

[A·B]

[C]

A0

t

[A·B]

[C]

A0

t

xe

= k2[A·B] =21

21

kk

[A][B]kk

]B][A[k

kk

1-

12

A + B A·Bk1

k-1

Ck2

Interpretation of rate constants:the Arrhenius equation

Eact

ES

P

Reaction coordinate

X‡ (TS)

Every reaction has to overcome an energy barrier: the transition state (TS, X‡).At higher temperature, more particles are able to overcome the energy barrier.

Arrhenius equation:

RT

E-A·expk a

obs

Ea can be determined by measuring kobs at two different temperatures:

21

a

2

1

T

1

T

1

R

Eexp

k

k

Idem, from statistical mechanics (collision theory)

RT

E-P·Z·expk a

obs

P = probability factor (not every collision is effective)Z = collision number (number of collisions per second)

Arrhenius:

RT

E-A·expk a

obs

Idem, from transition state theory:

k‡

A + B productsK‡

X‡

[A][B]

][X=K

‡‡ or [X‡] = K‡[A][B]

= k‡[X‡] = k‡K‡[A][B] = k[A][B], so k = k‡K‡

Statistical mechanics gives us the following relation:

h

Tk=k B‡ so ‡B K

h

Tk=k

kB = Boltzmann’s constant;h = Planck’s constant

Eact

ES

P

Reaction coordinate

X‡ (TS)

For all equilibria we can write: G0 = - RT ln K, so for our case we get: G‡ = - RT ln K‡

Expressing K‡ in terms of G‡ and RT gives the following equation for k:

RT

ΔGexp

h

Tkk

‡B

Since G‡ = H‡ - TS‡, we can also write:

R

ΔSexp

RT

ΔHexp

h

Tkk

‡‡B

(1)

(2)

Eq (1) and (2) are called the Eyring equations

‡B Kh

Tk=k

The Eyring and Arrhenius equations resemble each other:

Arrhenius:

RT

E-A·expk a

obs

so:R

E

d(1/T)

klnd a

Eyring:R

ΔHT

d(1/T)

klnd ‡

so Ea = H‡ + RT

In order to determine H‡ and S‡ it is easier to differentiate ln (k/T) to 1/T:

R

ΔH

d(1/T)

ln(k/T)d ‡

ln k

1/T

ln A

slope = R

Ea

1/T

slope =

T

kln

R

ΔH‡

So, the procedure to determine activation parameters is:

- determine k at different temperatures- plotting ln(k/T) against 1/T gives H‡

-

R

ΔSexp

RT

ΔHexp

h

Tkk

‡‡B then gives S‡

and when you have H‡ and S‡, you also have G‡ sinceG = H-TS

Interpretation of activation parameters

• G‡, the Gibbs free energy of activation, determines at which rate a certain reaction will run at a given temperature

• H‡ is a measure for the amount of binding energy that is lost in the transition state relative to the ground state (including solvent effects)

• S‡ is a measure for the difference in (dis)order between the transition state and the ground state– for monomolecular reactions: S‡ 0 J/mol.K

– for a bimolecular reaction: S‡ << 0 J/mol.K(two particles have to come together in the transition state to form one particle, demanding a much greater order)

Example:

N

CH2Ph

O

NH2

HH

N

S N+Me2Me2N

N

CH2Ph

O

NH2

H

N

S NMe2Me2N

H

MBH

+

BNAH(NADH model)

methylene blue(MB+)

+

BNA+

(NAD+ model)

ln(At-A)

t (sec)

xx

xx

xx

xx

xx

x

G‡ = 62.8 kJ/mol (very fast rx)H‡ = 33.0 kJ/mol (rel. low, compensation of C-H bond cleavage by hydration TS)S‡ = -100 J/mol.K (bimolecular rx)

Another example:

N

CH2Ph

O

NH2

CNH

N

CH2Ph

O

NH2

H

CH3CN

1

+ CN

H‡ = 85 kJ/mol (relatively high: no new bonds to beformed, no compensation for the partial cleavageof the C-C bond in the transition state; acetonitrileis aprotic, compensation of H‡ by solvation willbe less than in water

S‡ = 0 J/mol.K (monomolecular reaction)

Application of activation parameters for the elucidationof reaction mechanisms:

H3CO

O P

O

O

O

H3CO

O+ H2O + H2PO4

A S‡ of +12 J/mol.K was found monomolecular process

H3CO

O P

O

O

O

PhH3C

O

OHHO P

O

O

O

Ph+ H2O +

A S‡ of -117 J/mol.K was found bimolecular process; rate determining step in this case is the attack of water on the carbonyl group.

Look in your course book for the exact reaction mechanisms!

Solvation (solvent effects)

Influence of solvation on the reaction rate:

CH3I + Cl CH3Cl + I

k(H2O) = 10-7 l.mol-1.s-1; k(DMF) = 10-1 l.mol-1.s-1

so G‡ ~ 30 kJ/mol

H

O

NCH3

CH3

= DMF

H2O

DMF

G‡H2O

G‡DMF

E

reaction progressG‡

DMF < G‡H2O

What is the background of this strong solvent effect?

In H2O there is more solvation than in DMF, due to hydrogen bonds.

Note the changes in entropy: loss of S‡ because of orientation of the substrates, gain of S‡ because of the liberation of water (less solvated transition state). The balance is not easy to predict!In general, in case of ions, the ground state is more solvated than the transition state:

O2N ON

OO

CN

OO2N

O C O

O2N ON

OO

‡

+

TS (‡) is hardly solvated due to the spreading of charge.Again a strong solvent effect here: k(H2O)= 7.4x10-6 s-1; k(DMF) = 37 s-1

Cl

H

O H

HO H

HO HH

O HH

OH

HOH

HOH

HO

H

H

OH

HO H

HO

H

Cl

HO

H

HOH

HOH

HO

H

C I

H H

H

H

OH

HO H

HO

H

IH3C

HO H

+ x H2O

Solvation effects in (bio)polymers

Polymers or enzymes may have apolar pockets, which leads to:- less solvation and therefore higher reaction rates;- changes in pKa’s of acidic/basic groups:

N

NRH3C

N

NRH3C

H

+ H+ R = CH3 or compound with polymer

R = CH3: pKa = 9.7R = polymer: pKa = 7.7

Ka = [PyN][H+][PyNH+]

E.g. lysine, R-NH2 + H+ R-NH3+

pKa (H2O) = 10.4, in some enzymes pKa = 7 !

The energy diagram

Consider the gas phase chlorination of methane:

CH3CH4 + Cl + H-Cl H3C H Cl H3C H Cl

pote

ntia

len

ergy

re

interatomic distance

translation

Reaction course is via the route of lowest energy (“mountain pass”)

CH4 + Cl•

H3C···H···Cl

CH3• + HCl

rH-Cl

rC-H

A cross-section of this “mountain landscape” gives the well-known energy diagram:

pote

ntia

len

ergy

reaction course

H3C H Cl

H3C H Cl

‡ transition state

H3C···H···Cl

What does the transition state look like?

Hammond postulate:The TS closely resembles the species with the highest energy content

E

reaction course

Figure a

E

reaction course

Figure b

groundstate

product

transition state

ground state

transition state

product

Exothermic reaction (a) has a low Ea, TS resembles ground state;Endothermic reaction (b) has a high Ea, TS resembles the product

Kinetic isotope effects

H

D

D

H Cl2

H

D

D

Cl

H

Cl

D

H

+ HCl 89%

+ DCl 11%

Difference in effectivity of C-H and C-D bond cleavage:primary isotope effect

The background is the difference in bond strength, caused by the difference in mass between H and D:E0 of covalent bond is given by: 1/2h = 1/2hc(1/);1/(C-H) 3000 cm-1, E0(C-H) 18 kJ/mol;1/(C-D) 2200 cm-1, E0(C-D) 13 kJ/mol

E

C-H(D) bond distance

E0C-D

E0C-HDCD

DCH

kobs =

RT

EA·exp a

RT

5000exp

RT13.000-x

-A·exp

RT18.000-x

-A·exp

k

k

D

H 7.5

This is the maximum primary kinetic isotope effect at ~25ºC.

The isotope effect tells us something about the transition state. For this, we have to look at the stretching vibrations of theC-H(D) bond:H3C H Cl

H3C H Cl

Antisymmetrical stretching vibration: leads to reaction

Symmetrical stretching vibration: involvement of H(D) depends on the structure of the transition state

In this case:Ea = D

x

H3C H Cl Symmetrical stretching vibration: involvement of H(D) depends on the structure of the transition state:

H exactly in the middle between C and Cl: symmetrical transition state, kinetic isotope effect is maximum (~7.5). H not in the middle: isotope effect < 7.5

E

reactants products

E0‡

EaD Ea

H

E0C-D

E0C-H

H3C · · · · ·H · · · · ·Cl

a

H3C · · H · · · · · · · ·Cl

E0C-D

E0C-H

E0‡

EaH

E0H-Cl

E0D-Cl

EaD

b

E0‡

E0C-D

E0C-H

EaD

EaH

c

H3C · · · · · · · · H · ·Cl

symmetrical TS early TS late TS

Rule of thumb:kH/kD 4-7: bond cleavage, symmetrical transition statekH/kD 1-4: bond cleavage, asymmetrical transition state;

or no bond cleavage (secondary isotope effect)

Example:

C2H5OCH2

O2N

O

BC2H5O

CHO2N

O

+ + HB

Maximum isotope effect at symmetrical TS, so whenpKa(acid) = pKa(HB)

5

x

x

x

x x

x

x

pKa of HB

kH/kD

5

10

0 10

N

X

:B = OH , H2O, HCOO , CH3COO , XC6H5O ,

, etc.

CN

CN

H

H

COOH

baseCN

CNH

CN

CN

D

Dbase

CN

CND

+ base-H+

+ base-D+

kH

kD= 6.0

+ H2Ok1

k -1COO + H3O+

COOD + D2Ok1

k -1COO + D3O+

k1

k -1= Ka = 6.09x10-5

k1

k -1= Ka = 1.95x10-5

Some more examples: