v6: chemical kinetics & transition states

6. Lecture SS 2005

Optimization, Energy Landscapes, Protein Folding 1

V6: Chemical Kinetics & Transition States

see chapter 19 in book of K. Dill

Aim: describe kinetics of processes on energy landscapes

(e.g. chemical reactions).

- temperature effect - detailed balance- mass action law- Arrhenius plots- concept of transition state/activation barrier transition state theory- -value analysis revisited- effect of catalysts

6. Lecture SS 2005


Reaction rates are proportional to concentrations

B

k

k

A

r

f

Lets consider a simple kinetic process, the interconversion between 2 states,

kf and kr : forward and reverse rate coefficients.

How do the amounts of A and B change with time t, given the initial amounts at

time t = 0 ? tBktAkdt

tBd

tBktAkdt

tAd

rf

rf

The two equations are coupled.

One can solve them by matrix algebra …

6. Lecture SS 2005


Excursion: coupled differential equations

yxy

yxx

2

4

y

x

y

x

12

14

We can use the method of diagonalisation to solve coupled ordinary differential equations.

For example, let x(t) and y(t) be differentiable functions and x' and y' their derivatives.

The differential equations are relatively difficult to solve:

By diagonalizing the square matrix, we get

but u' = ku for a constant k is easy to solve.

It has the solution u = Aekx where A is a constant

Remembering this fact, we translate the ODEs into matrix form

y

x

y

x1

21

11

20

03

21

11

www.algebra.com

6. Lecture SS 2005


Excursion: coupled differential equations

y

x

y

x1

21

11

20

03

21

11

y

x

v

u1

21

11

y

x

v

u1

21

11

By diagonalizing the square matrix, we get

We then put It follows that

Thus

v

u

v

u

20

03

The solutions of this system are found easily: t

t

Dev

Ceu2

3

with some constants C and D.

y

x

v

u1

21

11

y

x

v

u

21

11With

tt

tt

DeCey

DeCex23

23

2

6. Lecture SS 2005


Reaction rates are proportional to concentrations

tAkdt

tAdf

tkA

tAdtk

A

dAf

t

f 0ln

0

tk feAtA 0

With this technique, we could solve our system of coupled diff. equations.

If kr << kf, the first equation simplies to

If [A(t)] + [B(t)] = constant, then

tk feAtB 0constant

6. Lecture SS 2005


[A]eq and [B]eq : equilibrium concentrations.

To see that this is a condition of equilibrium follows from inserting

into

resulting in

At equilibrium, rates obey detailed balance

eqreqf BkAk

eqreqf BkAk

tBktAkdt

tBd

tBktAkdt

tAd

rf

rf

The principle of detailed balance says that the forward and reverse rates must be

identical for an elementary reaction at equilibrium:

Taken from Dill book

0,0

dt

tBd

dt

tAd

6. Lecture SS 2005



r

f

eq

eq

k

k

A

BK

The detailed balance condition relates the rate coefficients kf and kr to the

equilibrium constant K:


For more complex systems, the principle of detailed balance gives more information beyond

the statement of equilibrium. For a system having more than one elementary reaction, the

forward and reverse rates must be equal for every elementary reaction.

For this system:

AB

BA

eq

eq

BI

IB

eq

eq

IA

AI

eq

eq

k

k

B

A

k

k

I

B

k

k

A

I ,,

Let‘s consider a 3-state mechanism with

kIA 0, kBI 0, kAB 0.

6. Lecture SS 2005


These are two independent equations for 3

unknown concentrations

the system has an infinite number of

solutions.

In mechanism (b), all rates of the

Denominator in

are zero mechanism (b) is impossible.


BAIB

IBAI

kBkI

kIkA

This results in the mechanism shown right.

The only conditions for equilibrium are:

BIABIA

IBBAAI

IBBAAI

BIABIA

BI

IB

eqAI

IA

BA

ABeq

BA

ABeqeqeq

IA

AIeq

AB

BA

eq

eq

BI

IB

eq

eq

IA

AI

eq

eq

kkk

kkk

kkk

kkk

k

k

Ak

k

k

kA

k

kABA

k

kI

k

k

B

A

k

k

I

B

k

k

A

I

1or,1

or)2(

)3(,)1(

,,


6. Lecture SS 2005



The principle of detailed balance says that forward and backward reactions at

equilibrium cannot have different intermediate states.

That is, if the forward reaction is A I B,

the backward reaction cannot be B A.

The principle of detailed balance can be derived from microscopic statistical

mechanics.

6. Lecture SS 2005


The mass action laws describe mechanisms in chemical kinetics

Suppose the following reaction leading from reactants A, B, and C to product P:

PcCbBaA In general, the initial reaction rate depends on

- the concentrations of the reactants

- the temperature and pressure

- and on the coefficients a, b, and c.

Kinetic law of mass action (CM Guldberg & P Waage, 1864):

„the reactants should depend on on stoichiometry in the

same way that equilibrium constants do“.

cbaf CBAk

dt

Pd

Although mass action is in agreement with many experiments,

there are exceptions. These require a quantum mechanical

description.

6. Lecture SS 2005


Reaction rates depend on temperature

Consider a binary reaction in the gas phase:

PBAk2

Suppose that

By definition, the rate coefficient k2 is independent of [A] and [B].

But k2 can depend strongly on temperature.

BAkdt

Pd2

The observed dependence of the reaction rate on the temperature is much

greater than one would expect from just the enhanced thermal motions of the

molecules.

6. Lecture SS 2005


Arrhenius equation

1889, S. Arrhenius started from the

van‘t Hoff equation for the strong

dependence of the equilibrium constant

K on temperature:

2

ln

kT

h

dT

Kd

and proposed that kf and kr also have van‘t Hoff form

22

lnand

ln

kT

E

dT

kd

kT

E

dT

kdaraf

where Ea and E‘a have units of energy that are chosen to fit exp. data.

Ea and E‘a are called activation energies.

6. Lecture SS 2005


Activation energy diagram

According to Arrhenius, it is not the average energy of the reactants that

determines the reaction rates but only the high energies of the ‚activated‘

molecules.


There are two plateaus, one for the reactants and one for the products.

In between lies an energy maximum (also: transition state or activation barrier)

which is the energy that activated molecules must have to proceed from

reactants to products.

Measuring kf as a function of temperature, and using eq. (1) gives Ea.

Measuring the reverse rate gives E‘a.

Measuring the equilibrium constant versus temperature gives h°.

6. Lecture SS 2005


Population at different temperatures

From


r

f

k

kK it follows

aa EEh '

The figure shows how activation is interpreted according to the Boltzmann

distribution law: a small increase in temperature can lead to a relatively

large increase in the population of high-energy molecules.

6. Lecture SS 2005


Arrhenius plots

Integrating over

temperature T


2

ln

kT

E

dT

kdaf

gives:

BkTE

f

BkTE

f

BkTE

f

af

eAAek

eek

ek

BkT

Ek

a

a

a

with

ln

H2 + I2 2HI (open circles)

2HI H2 + I2 (full circles)

Diffusion of carbon in iron

The figures show examples of

chemical systems showing

Arrhenius behavior.

6. Lecture SS 2005


Activated processes

Arrhenius kinetics applies to many physical and chemical processes.

When should one treat a process as activated?

If a small increase in temperature gives a large increase in rate,

a good first step is to try the Arrhenius model.

E.g. breaking of bonds.

Counter example: highly reactive radicals. CNHHHCNH 223

These can be much faster than typical activated processes and they slow

down with increasing temperature.

We now describe a more microscopic approach to reaction rates,

called transition state theory.

6. Lecture SS 2005


The energy landscape of a reactionAn energy landscape defines how the

energy of a reacting system depends

on its degrees of freedom.

E.g. A + BC AB + C

Each reaction trajectory would involve

some excursions up the walls of the

valleys.

When averaged over multiple trajectories,

the reaction process can be described as

following the lowest energy route, along

the entrance valley over the saddle point

and out of the exit valley, because the

Boltzmann populations are highest along

that average route. Taken from Dill book

Energy surface for

D + H2 HD + H

The transition (saddle) point isdenoted by the symbol ‡.It is unstable: a ball placed on thesaddle point will roll downhillalong the reaction coordinate.

6. Lecture SS 2005


Calculating rate coefficients from TST

Let us consider the reaction

by transition state theory:

Divide the reaction process into two stages:

(1) the equilibrium between the reactants ant the transition state (AB)‡ with

‚equilibrium constant‘ K‡

(2) a direct step downhill from the TS to the product with rate coefficient k‡:

PBAk2

PABBA kK ‡‡ ‡

Key assumption of TST: step (1) can be expressed as an equilibrium between

the reactants A and B and the transition state (AB)‡ , with

BAB

KA

‡‡

even though (AB)‡ is not a true equilibrium state.

6. Lecture SS 2005



The overall rate is expressed as the number of molecules in the TS, [(AB)‡],

multiplied by the rate coefficient k‡ for the second product-forming step

BAKkABkdt

Pd ‡‡‡‡

Because the quantitiy K‡ is regarded as an equilibrium constant, it can be

expressed in terms of the molar partition functions:

kTD

BA

AB eqq

qK

‡‡

‡

where D‡ is the dissociation energy of the TS minus the dissociation energy of

the reactants.

q(AB)‡ is the partition function of the transition state.

6. Lecture SS 2005


The transition state

(left) Contour plot of a reaction pathway (- - -) on an energy landscape

for the reaction A + BC AB + C. The broken line shows the lowest-energy

path between reactants and products

(right) The transition state is an unstable point along the reaction pathway

(indicated by the arrow) and a stable point in all other directions that are

normal to the reaction coordinate.


6. Lecture SS 2005



Provided that the unstable degrees of freedom are independent of the stable

degrees of freedom, q(AB)‡ is factorable into two components

‡‡

‡ where qqqqAB

represents the partition function for all of the ordinary thermodynamic degrees of

freedom of the TS structure, and q represents the one nonequilibrium vibrational

degree of freedom of the bond along the reaction coordinate.

The partition function of a vibration is

hv

kT

eq kThv

1

1

for a weak vibration with low frequency .

Once the system has reached the transition state, it is assumed to proceed

as quickly to the product state as the system permits, namely at the frequency

of the reaction coordinate vibration,

‡k

6. Lecture SS 2005


deviations from TST

A factor , the transmission coefficient, is often introduced (k‡ = ) to

account for observed deviations from the simple rate theory.

In condensed-phase medium, or in complex systems, < 1.

This gives the relation between the rate coefficient k2 and the partition functions:

‡

‡‡

2

Kh

kT

eqq

q

h

kTk kTD

BA

6. Lecture SS 2005


relation between value analysis and TST

In V5, we characterized the effect of a protein mutant by its -value

0

ln

G

kkRT

wt

mut

G0 reflects whether the mutant stabilizes the folded state F over the unfolded

state U stronger or weaker than wild-type protein.

According to TST, both wild-type and mutant folding proceed via transition

states with activation free energies G‡wt

and G‡mut.

wtFU

mutFU GGG 0

‡‡‡;‡

‡‡

‡

‡

‡

‡

wtmutkTG

kTGG

kTG

kTG

wt

mut

kTGmut

kTGwt

GGGe

ee

e

k

k

ek

ek

wtmut

wt

mut

mut

wt

A -value of 1 meansthat G0 = G‡ forthis mutant the mutant has the sameeffect on the TS structureas on the folded state this part of the TS structure is folded asin the folded state F.

6. Lecture SS 2005


Catalysts speed up chemical reactions


Free energy barrier G‡ is reduced by a catalyst C.

Catalysts affect the rates of chemical reactions; e.g. enzymes accelerate

biochemical reactions.

Enzymes can achieve remarkable accelerations, e.g. by a factor of

2 x 1023 for orotine 5‘-phosphate decarboxylase.

Linus Pauling proposed in 1946 that catalysts work by stabilizing the

transition state.

PBA k 0

ABC

kc

Linus Pauling 1935

http://osulibrary.oregonstate.edu/specialcollections/coll/pauling/blood/pictures/large-1935i.005.html

6. Lecture SS 2005


Catalysts speed up chemical reactions

From transition theory we obtain for the catalyzed reaction rate kc (normalized to

the uncatalyzed reaction rate k0)

CAB

ABC

k

kc‡

‡

0

This ration represents the ‚binding constant‘ of the catalyst to the transition

state the rate enhancement by the catalyst is proportional to the binding affinity

of the catalyst for the transition state.

This has two important implications:

(1) to accelerate a reaction, Pauling‘s principle says to design a catalyst that

binds tightly to the transition state (and not the reactants or product, e.g.).

(2) a catalyst that reduces the transition state free energy for the forward reaction

is also a catalyst for the backward reaction.

6. Lecture SS 2005


Speeding up reactions by intramolecular localization or solvent preorganization


Reactants polarize, so water reorganizes.

Two neutral reactants become charged in the transition state.

Creating this charge separation costs free energy because it orients the solvent

dipoles.

6. Lecture SS 2005


Speeding up reactions by intramolecular localization or solvent preorganization

Enzymes can reduce the activation barrier by having a site with pre-organized

dipoles.


6. Lecture SS 2005


Funnel landscape describe diffusion and polymer folding

All the processes described sofar involve well-defined reactants and products, and

a well-defined reaction coordinate.

But diffusional processes and polymer conformational changes often cannot be

described in this way. The starting point of protein folding is not a single point on

an energy landscape but a broad distribution.

A bumpy energy landscape, such as occursin diffusion processes, polymer conformationalchanges, and biomolecule folding.A single minimum in the center may represent‚product‘, but there can be many different‚reactants‘, such as the many openconfigurations of a denatured protein.

http://www.dillgroup.ucsf.edu/

http://www.dillgroup.ucsf.edu/energy.htm

6. Lecture SS 2005


Summary

Chemical reactions and diffusion processes usually speed up with temperature.

This can be explained in terms of a transition state or activation barrier and an

equilibrium between reactants and a transient, unstable transition state.

For chemical reactions, the transition state involves an unstable weak vibration

along the reaction coordinate, and an equilibrium between all other degrees of

freedom.

Catalysts act by binding to the transition state structure.

They can speed up reactions by forcing the reactants into transition-state-like

configurations.

v6: chemical kinetics & transition states

Documents

equilibrium concentrations

condition of equilibrium

equilibrium constant

reverse rates

statement of equilibrium

forward reaction

detailed balanceaeq

detailed balancethese