PETER PAZMANY 

SEMMELWEIS UNIVERSITYCATHOLIC UNIVERSITY 



Development of Complex Curricula for Molecular Bionics and Infobionics Programs within a consortial* framework** 
Consortium leader 



PETER PAZMANY CATHOLIC UNIVERSITY 

Consortium members 

SEMMELWEIS UNIVERSITY, DIALOG CAMPUS PUBLISHER 

The Project has been realised with the support of the European Union and has been co-financed by the European Social Fund *** 
**Molekuláris bionika és Infobionika Szakok tananyagának komplex fejlesztése konzorciumi keretben 
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INTRODUCTION TO BIOPHYSICS 

(Bevezetés a biofizikába) 

DERIVATION OF THE RATE CONSTANT 

(A sebességi együttható származtatása) 
GYÖRFFY DÁNIEL, ZÁVODSZKY PÉTER 



Introduction 

. 
The temperature dependence of the rate
constant is described by the empirical
Arrhenius equation 

. 
Several theories were established for calculating the value of rate constant andexplaining the temperature dependence of the rate constant 
. 
The collision theory considers reactions
between atoms or molecules as collisions 
between rigid spheres 


. 
Only collisions with sufficiently large kinetic energy along the straight line connecting thecentres of colliding molecules or atoms lead to reaction 
. 
Results from collision theory are in onlyqualitative agreement with the experimentalresults 
. 
Transition state theory is based on free
energies of different states 

. 
The state with the highest free energy along
the reaction path is called the transition state 



. 
The difference in free energy between the
initial state and the transition state is the 
activation energy 

. 
Some improvements of transition state theorywere carried out by Eyring 
. 
The crucial assumption of Eyring's theory isthat transition state itself is also in a free energy valley 
. 
The transition state theory provides some
quantitatively useful predictions 



Arrhenius equation 

. 
It was known that the rate of reaction dependson the temperature: 
– 
Warming speeds up and 

– 
Cooling slows down the reactions 


. 
Arrhenius describes this relation in his famous equation 


/ RT 


k=Ae-EA
where k is the rate constant, EA is the activation energy, R=8.314 J·mole-1·K-1 the gas constant, T is the temperature and A is the so called pre-exponential factor which can be different for different reactions and its value can be measured 


The temperature dependence of the rate 
constant 




(1859-1927) 




Determination of activation energy 

. 
We can calculate the activation energy from the temperature dependence of the rate constant 
. 
To make this calculation simpler, we derive a
linear relationship from the Arrhenius equation 



. 
Let us set out from the original form of the Arrheniusequation 
k=Ae-EA/ RT 
. 
Let us take its logarithm 
EA
ln k=ln A- 
RT 


. 
Now we would like to determine the activation energy of a particular reaction 
. 
Let us measure the velocity of the reaction at different temperatures 
. 
Let us plot ln k against 1/T to get a linear 
relationship 



Linear form of Arrhenius plot 



Introduction to biophysics: Derivation of rate constant 




15 
Introduction to biophysics: Derivation of rate constant 


16 
Introduction to biophysics: Derivation of rate constant 



Introduction to biophysics: Derivation of rate constant 

The Maxwell-Boltzmann 
distribution 

. 
Collision theory derives the rate constant fromthe number of collisions 
. 
We can count collisions only if we know thevelocities of atoms 
. 
We do not know the velocities of all atoms but we know their probability distribution 
. 
Velocities of atoms follow the Maxwell-
Boltzmann distribution 



. 
We already know that energies of particles
follow the Boltzmann distribution 
- E/ k BT 
e 
p. E .=.. E . 
- Ej /k BT
. .. Ej . e
j 

where p(E) is the probability that a particle has energy E or in other words, the ratio of particles with energy E; .(E) is the density of states of energy E, i.e. the number of states with energy E and kB=1.38·10-23 J·K-1 is the Boltzmann constant 


. 
To get the distribution of particle velocities, wehave to determine a relationship betweenvelocity and energy 
. 
The kinetic energy of particles depends on
velocity of those particles 
2 
Ek = mv

2 
where Ek is the kinetic energy, v is the velocity and m is the mass of the particle 


. 
First of all, let us reduce the problem to onedimension and consider only the component ofvelocity in the x direction 
. 
According to the Boltzmann distribution 
-..v x ./ k BT -mv 2 x /2 k BT 
p. v x .= . e=. e
-..v x ./ k BT -mv 2 x /2 k BT

. e. e

-. -. 

where p(v x) is the probability that the x component of the velocity of a particle is v x, .(v x) is the kinetic energy corresponding to the x 
component of velocity 


. 
Considering the integral 
-ax2

.. edx =.. 
-. a 

the probability is 

m -m v2 x/ 2 k BT
p. v .=. e
x 2 . kBT 
. 

This expression is the Maxwell-Boltzmann distribution for one component of the velocity 

distribution 


. 

Based on the results for one direction, let us build the expression for the total velocity vector 
. 

The three dimensional velocity is 
v2 =v2 .v2 .v2 
xyz 
. 

In an ideal gas, the one-dimensionalcomponents of velocity are independent ofeach other so the probability that the velocityvector is v is 
p. v .= p .v x . p.v y. p . vz . 



. 

The probability that the particle has velocitycharacterized by the vector v is 
33/ 2 m -m.v x 2. v2 y. v2 z ./ 2 kBT m -mv 2/ 2 k BT
p. v .=. e=. .e
2 . k BT 2 . kBT 
. 

The above expression only tells us what theprobability of a particle with the velocity vector v(vx,vy,vz) is but we are interested in the 
distribution of absolute values i.e. the lengthsof vectors 



. 

The end points of vectors of identical lengths and beginning at the same point lie on thesurface of a sphere with radius v 
. so 
Nv .4 . v2 

where Nv is the number of vectors with length v 
. 

Thus the probability distribution for v is 
.2. kBT .3/ 2 


m -mv2 /2 k BT 
p. v.=4 .v2 e 



Number of vectors of length v 



. 

Based on the distribution, we can obtain the 
average speed i.e. the expected value of v 

3 / 2.. 3 m -mv2 / 2 kB
v=.v·p .v. dv=.4 . ve Tdv 
.2 . k BT .
00 

. 3 -ax21 
. 

Making use of the integral . xe= 02 a2 
the average velocity is 
8 kBT 
v=
..
. m 




(1831-1879) 






The collision theory 

. 

Collision theory in pure form as described hereapplies only to gases 
. 

Atoms or molecules are modelled by
Newtonian rigid spheres 

– 
They are not compressible 


– 
Interaction between them occurs only when theytouch each other 


. 

They do not lose any kinetic energy during thecollision 



. 

Let us consider two atoms (spheres) A and Bwith radius rA and rB 
. 	Let r  denote the effective radius r =rA+rB
eff	eff 
. 	Let vA and vB denote the velocity of the A and the B sphere, respectively 
. 

To simplify calculations, let us consider the B sphere immobile and use the relative velocity v=vA-vB rather than vA and vB 



Relative velocity 





. 

Let c denote the collision parameter defined as the distance between straight paths of centresof spheres  before collision 
– In the case of immobile sphere this  is not a real pathbut a line which is parallel to the other path and goacross the centre of the immobile sphere 
. 

Collision occurs only if the collision parameter is smaller than the effective radius i.e. 
c.reff 

. 

In the case of c=0, the collision is frontal and if  a collision does not occur 
c>reff




. 

After collision the A atom is diverted by a . angle which is a function of the collision parameter 
. 	In the case of frontal collision .=. and if c>reff then .=0 



Reaction cross section 





Direction of collisions 





Collision cylinder 





. 

Let us consider an A molecule traveling withvelocity v in a unit volume within which there are NB B molecules 
. 

Collision occurs if the centre of a B molecule is in a circular .·reff 2 area around the centre of A 
. 

In unit time, an A molecule covers a distance v so it moves through a .·reff2·v collision volume 
. In a unit volume, there are NA A molecules, so 
the number of collisions in unit volume and in unit time: 
Z =NAN B . reff 2 v 




. 

We are interested, however, not in collisions of one molecule but collisions of an ensemble of molecules 
. 

Velocities of molecules are not the same but they follow the Maxwell-Boltzmann distributionas discussed earlier 
. 

In the expression describing the number ofcollisions, we should substitute the velocity v of one molecule by the average velocity v of the ensemble of molecules 



. 

According to the Maxwell-Boltzmann
distribution, the average velocity v is 

=.8 k BT vm. 
. 

Since we consider relative motions of molecules we should derive the average ofrelative velocities from the expression above 


Introduction to biophysics: Derivation of rate constant 




Introduction to biophysics: Derivation of rate constant 




Introduction to biophysics: Derivation of rate constant 





Introduction to biophysics: Derivation of rate constant 
www.itk.ppke.hu 
Introduction to biophysics: Derivation of rate constant 


45 


. 

Using the expression for the average relative
velocity, we obtain that the total number of
collisions in a unit volume and in unit time is 

2 .8 kBT
Z =N AN B . reff 

.. 



. 
For a reaction to occur, some rearrangementof valence electrons is required which is energetically expensive 
. 
Thus, for a reaction to occur, a simple collisionis not enough but it requires a collision withenough energy along the straight lineconnecting the centres of atoms 
. 
We are interested in the proportion of
collisions with enough energy 




. 

According to the Maxwell Boltzmanndistribution, the fraction of particles withrelative velocity v is 
.2 . kBT .3/ 2

. 2 -. v2/ 2 kB
p. v. dv=4 .ve Tdv 



. 
The distribution of kinetic energies from the
relative velocities is 

. .
3/ 2

. 2. 1 -./ kB
p... d .=4 .eTd .

2. kBT ..2 .. 
taking into account that 

22 . d . 

v = and dv = 
..2 .. 



Probability density function of collision
energies 





. 

In the case of a collision, only the kineticenergy due to the velocity component alongthe line connecting the centres of particlesgets utilized 
. 

We should determine this velocity based onthe relative velocity 
. 

The figure below helps us to understand it 




Velocity component between centres 





. 
Since 

22
v

c =vrel . reff -d2 ./reff 
the component of the kinetic energy we areinterested in is 
.c=..reff 2 -d 2 ./reff 2 



. 

Only those collisions lead to reaction wherethis component of kinetic energy is higher thana given limit energy 

.c..0 

. 

Given the kinetic energy . of relative velocity, 
we can define a maximum value of d where . 

c 
is exactly .0 

22 2
.0=.c=..reff -d max ./reff 
thus 
22

dmax =reff .1-.0 /.. 



. 

Since reaction occurs only when 
d .d max 


we can define a modified effective reaction cross section 
2
Aeff =. d max 





. 

The total number of collisions in unit time having enough energy for the reaction to occur, which is the rate of the reaction, is the integral over the distribution of relative kineticenergies from .0 to infinity 
v=.. vrel p... Aeff ... d . N AN B 

.0 

where v is the rate of the reaction and vrel is the relative velocity of molecules 



. 

After substitution the equation is 
. 

42 .0 -./ kB
v=..2 . ... reff .1-.eTNANB

3 T 3 .

.0 . .. kB 
. 

Integrating the equation we get 
=.8 k BT 2 -.0 / kBTN AN B
ve
. reff 
.. 




. 
Since 
v=kN AN B 


based on the equation above, the rate constant is 

2 -.0 / k BT 
=.8 kBT
ke
.reff 
.. 




. 

The relation between the gas constant and theBoltzmann constant is 
R=kBAN 


where AN=6.022·1023 mol-1 is the Avogadro constant which is the number of atoms or molecules in a mole 
. 
Thus 

E0 / R=.0 / kB 
where E0 relates to one mole material 



Amedeo Avogadro
(1776-1856) 





. 

Based on the collision theory, we obtain amolecular description of both the exponentialand the preexponential factor in the Arrheniusequation 
2 
=.8 kBT

Ath =.c .reff 
.. 

where A is the theoretically calculatedpreexponential factor and .c is the collision frequency 



. 

Based on the theory, predictions can be madeand results calculated from the theory can becompared with the experimental data 
. 

Unfortunately, most of the theoretical resultsare at most in a weak agreement with theexperimental data 



Comparison of theoretically and
experimentally obtained reaction rates 





. 

To improve the agreement between the resultsobtained by theory and experiments, a steric factor can be introduced which reflects the fact that the assumption of spherical particlescauses serious inaccuracy and that theorientation of particles during the collision hasa significant influence of whether a reaction occurs 
. 

A considerable insufficiency of the collisiontheory is that we cannot calculate the stericfactor in advance so the collision theory is unsuitable for predicting the rate constant 



. 

This more accurate model of the reaction rate is the transition state theory proposed by Henry Eyring and Michael Polanyi 
. 

In order to understand the transition state theory, we require some quantum mechanicalintroduction 



. 

In the early 1900s it became apparent thatexperimental results can only be explained byassuming that energy is not continuous but itcan adopt only discrete values 
. 

These energy levels are predictable by theSchrödinger equation 
H .= Ei . 

where H is the Hamiltonian operator, . is the wave function and Ei is the energy of a given energy level 



(1805-1865) 






. 

The wave function .(x,y,z) does not have a easy-to-grasp meaning but its square .2 is the probability density function of the location ofthe particle 
. 

Hamiltonian operator describes the relevantforces acting on the particle studied 
. 

To obtain the Hamiltonian operator for our problem, we can set out from two basic operators: the operator of momentum and theposition coordinate 



. 

The operator of the momentum is 
. d
.p= 
i dx 

where ppj is the momentum operator, i is the imaginary unit and 
h
.= 
2. 

where h=6.626·10-34 is the Planck constant 



. 

The operator of the position coordinate is 
x.=x× 

Where xxj is the operator of position and x. represents the multiplication by x 
. 

Based on these operators, we can define theoperator of kinetic energy 
.p2 .2 d 2 .
E = =- 

2 m 2 m dx2 
where E is the operator of the kinetic energy andm is the mass of the particle 



Solution of the Schrödinger equation 

. 

To obtain the energies and the wave function, we can solve the Schrödinger equation 
. 

For different problems, the Hamiltonianoperator can adopt different forms, butgenerally it contains two terms correspondingto the kinetic and potential energy 
H = E..V.. x. 




Erwin Schrödinger
(1887-1961) 





Werner Heisenberg
(1901-1976) 





(1858-1947) 






Translational motion 
. 

Let us consider a particle in a box allowingonly one-dimensional motion 
. 

The walls of the box are represented
mathematically by 

V .0.=. and V .l.=. 
where l is the length of the box and at any other 0<x<l position 
V . x.=0 




One-dimensional translational motion 





. 

Thus the Hamiltonian operator for the one-dimensional translation is 
.2 d2 
H =- 

2 m dx2 
and the Schrödinger equation is 
.2 
d .

- E .
2 m dx2 =




. 

After rearranging the Schrödinger equation weobtain the 
d2 .

.k2 .=0 
dx2 

second order differential equation where 
2 mE k2 
= .2 




. 

The solution of the Schrödinger equation is 

.. x.=A sin kx. B cos kx 
where A and B are constant 

. 
To get the value of A and B, we can utilize theconstraint imposed by the potential energy at the walls, namely 
V .0.=. and V .l.=. 



. 

Because the potential energy at the walls isinfinity, the probability that a particle staysthere is zero, so 
.2 .0.=0 
and so 
.. 0.=0 




. 

If x=0 then 
A sin kx=0 and B cos kx=1 

. 

Since .(0) must be zero, B also must be zero 
. 
So for .(x) we obtain that 

.. x .=A sin kx 



. 

The second boundary condition will help us todetermine the value of A 
.. l.= A sin kl=0 
. 

Disregarding the trivial but uninterestingsolution A=0 the equation above is satisfiedonly if 
kl=n. 
where 
n=1, 2,3,... 
is a positive integer 

82 


. 

Setting out from the equation above we get 

.
2 mE

nl=n . 
.2 
. 

And the energy values of different levels are 
2 .2 .22 h2 En = n= n
2 ml28 ml2 



Energy levels for a particle in a box 






. 

The wave function can be obtained by usingthe property of probability density functions that 
l 

... x.2 dx=1 
0 
. 

Substituting the expression we got for .(x) 

lA2 l
A2 .sin2 kx dx==1 
02 




. 

Thus the normalized wave function is 
.n . x.=.2 sin n. x 
ll 



Wave functions and density functions for 
a particle in a box 








. 

Let us generalize the problem to three
dimensions 

. 

The Schrödinger equation for a particleconfined within a three dimensional box is 
.2 .2 .2 .2 - .. .. x,y,z.= E .. x,y,z.
2 22
2 m .. x. y. z.




. 

The equation can be separated and we get 

.. x,y,z.=.. x ... y ... z. 
and 

22 2
h2 nx ny nz
E =E .E .E = ..
nx y z 22 2
..
8 mll l
xyz 
where lx, ly and lz are the length of the box in the x, y and z dimensions, respectively 


Introduction to biophysics: Derivation of rate constant 

Partition function for translation 

. 

The general form of Boltzmann partition
function is 

- En / kT 
Z =. e 
n 
. 

Substituting into the equation the energies weobtained by solving the Schrödinger equation, we get the partition function Zt (t referring to 
translation) 

222
h2 ny nz
- ..
8 mkT .nlx 2 xl 2 y lz 2 . 
zt =. e 
n 



Introduction to biophysics: Derivation of rate constant 


. 

If the energy levels are sufficiently close toeach other that so many energy levels arefilled then the sum can be replaced by anintegral 
222
h2 nx ny nz
.- ..
222
8 mkT .lx lylz .
Zt =.e dn 
0 
so 

3 / 23 /2
2 . mkT 2 . mkT 
Zt =l xl yl z =V 
h2




.h2 ...
where V is the volume 



. 
For us, the most important motion is vibration
which can be modelled as a harmonic oscillator 

. 
Let us imagine our two-atom system as two
bodies with mass m connected by a spring 

. 
There is an equilibrium distance between theatoms where the potential energy isconsidered zero and which is at the x=0 place 



. 
The potential energy as a function of the
deviation from the equilibrium point is 

kx2 V . x.= s 
2 
where V(x) is the potential energy and ks is the spring constant which is characteristic of thegiven spring (or of the bond between the atomsin our case) 



A model of harmonic oscillator 





. 
Thus the total Hamiltonian operator for the vibration of a two-atom system is 
.2 d2 kx2 H =- . s 
2 . dx 22 
where µ is the reduced mass and the Schrödinger equation is 
.2 d 2 .. x. ksx2 
- ... x.=E .. x.

2 . dx22 



. 
After rearranging the equation we obtain asecond-order differential equation for . 
d 2 . 2 . .. x2 
- E- .=0 

.2 ..
dx22 
where 
.=2. .=.ks 
. 
is the angular frequency 




. 

To obtain the wave functions, we apply a trick, namely we substitute 
2 E 

.=... 
k= and x 

.. . 
into the equation above to get the simple form that 
d2 ... k-.2 ..=0 dx2 



. 
The solution can be written as 
-.2 
.=e 2 f ... 

where f(.) is some unknown function of . which we would like to determine 
. 

Substituting this expression into the
differential equation above we can write 

d2 f df 

-2 . .. k-1. f =0 
d .2 d . 




. 
Let us write f as a polynomial 
n f ...=. ci .i 
i= 0 
. 

It can be shown (derivation omitted) that sincethe polynomial has to be of finite length, 
k=2n .1 





. 
Polynomials f(x) satisfying the 
d 2 f df 

-2 x .2 nf =0 
dx2 dx 

differential equation are called Hermite polynomials and denoted by Hn(x) 
. 

Polynomials corresponding to the first seven
(0-6) degrees are listed in the following table 




Hermite polynomials 





(1822-1901) 






. 

Based on the equation k=2n+1, we obtain the energy levels: 
11
En =...n. .=h ..n..
22 
where n=0, 1, 2 … is a non-negative integer 
. 

It can be observed that even in the groundstate (n=0) there is an oscillation with energy En=h./2 which is called zero-point energy 



Wave function and energy levels for a
harmonic oscillator 





Partition function of vibration 

. 

Repeating the argument used in the case oftranslation, we get for the partition function ofvibrational motion that 
. 

-.n.1 /2 .h./ kT -h ./2 kT .1.e-h ./ kT .e
zv =. e=e-2 h./ kT ..... n =0 



. 

Since the expression in the brackets in theequation above has the form 
1. x2. x3.... 
which is the series expansion of 
.1- x.-1 

the simple form of the partition function 
-h ./2 kT 
e
zv = 
-h ./kT 
1-e




. 

The third form of movement which contributes to the energy of a given particle is rotation 
. 

The rotation of a molecule consisting of two

atoms of masses m1 and m2 separated by a distance r can be modelled as a single particle with mass 
m1 m2
.= 
m1.m2 

orbiting at a distance r around a centre point 



Rotation of a two-atomic molecule 





Spherical polar coordinates of a particle 





. 

The particle performing rotation has onlykinetic energy so the Hamiltonian operator for it is 
.2 d 2 
H =- .2 .. dx2 

and thus the Schrödinger equation is 
.2 d 2 .. x.

-=E .. x.

2 . dx2 




. 

To make the calculations simpler, let usconsider . as a function of r, . and ., where r is the radius of the orbit and . and . are polar angles 
. 

Since r is constant we consider . as a function of only the two angles 



. 

Thus the Hamiltonian operator expressed by .

and . is  
H =-  .2 2 .  .  
where . is  
.=  1 r2  . .2 . .2  .cot .  . . .  .  1 sin2 .  .2 ..2  .  

the Laplace operator 





Pierre Simon Laplace
(1749-1827) 





. 

After substitutions and rearrangement, the

Schrödinger equation is 

.2 . .. 1 .2 . 2 . r2 
.cot .. =- E . ..2 .. sin2 . ..2 .2 
where the differential equation to be solved is 
.2 . .. 1 .2 . 2 I
.cot .. . E .=0 
.2..2 .. sin2 . ..2 

where I=µr2 is the moment of inertia 



. 

Omitting derivation again,  the wave function is obtained as: 
.l ,m.. , ..=sin|m|. Pl ,m cos. eim. 

which are called spherical harmonics and where 
is a Legendre polynomial, l and m are 

Pl,m 
quantum numbers 

l=0,1 ,2... and m=-l ,-l.1...0 ,1... l-1, l 




Spherical harmonics 

l  m  Yl,m  
0  0  (1/4.)1/2  
1  0  (3/4.)1/2 cos .  
±1  (3/8.)1/2 sin . e±i.  
0  (5/16.)1/2 (3 cos2 . -1)  
2  ±1  (15/8.)1/2 cos . sin . e±i.  
±2  (15/32.)1/2 sin . e±2i.  
0  (7/16.)1/2 (5 cos3 . -3 cos .)  
3  ±1  (21/8.)1/2 (5 cos2 .-1) sin . e±i.  
±2  (105/32.)1/2 sin2 . cos . e±2i.  
±3  (35/64.)1/2 sin3 . e±3i.  




Adrien-Marie Legendre
(1752-1833) 





. 
The energy levels are 

l .l.1..2 
El = 
2 I 
. 

It can be seen that the energy is the functionof only the l quantum number so statescharacterized by different m values but by the same l have the same energy and this energylevel is said to be degenerate and it has a degeneracy of 2l+1 



Partition function of rotation 
. 

The corresponding partition function is 

. 

-El / kT 
zr =. .2 l.1. e

l=0 

which can be approximated by an integral if 

2 Ik 
T » 

.2 




. 
Then 
2 I kT 
zr = 
..2 

where . is a symmetry factor which is .=1 for heteronuclear and .=2 for homonuclear diatomic molecules 



. 

For nonlinear molecules, there are three moments of inertia and the corresponding partition function is 
.. IxI yIz 2 kT 3/ 2 zr ,nonlinear = 
. ..2 .
where Ix, Iy and Iz are the moments of inertia corresponding to the rotational degrees offreedom 


Introduction to biophysics: Derivation of rate constant 

Gibbs free energy as a function of
the partition function 

. 

We already know that the Helmholtz free energy expressed by the partition function is 
F =-kBT ln Z 
. 
Based on this the pressure is 

. F . ln Z 
p=-=

.. V .T kBT .. V .T 



G=F . pV 
and so 
. ln Z

G-G .0.=-k BT ln Z .kBTV ..V .T 
where G(0) is the Gibbs free energy at absolute zero temperature 



. 
For an ideal gas where 
pV =Nk BT 
the expression above becomes 

G-G .0.=-k BT ln Z .N kBT 



. 
The partition function of the whole system forindistinguishable particles can be expressed by thepartition functions of the individual particles 
N
zZ = N! 

which can be approximated based on the Stirlingformula 
z

ln Z = N ln .N
N 




. 
Thus 

G-G .0.=-Nk BT ln Nz -Nk BT .Nk BT 
and after simplification 

G-G . 0.=-Nk BT ln z 
N 



Introduction to biophysics: Derivation of rate constant 

Relation of the equilibrium constant
and the partition functions 

. 

Now that we know the partition functions wecan express the equilibrium constant K as a function of them 
. 

Let us set out from the standard reaction free energy 
.rG0 =-RT ln K 




. 

Let us consider a simple bimolecular reaction 
aA.bB . cC .dD 
where a, b, c and d are the stoichiometric 
coefficients of the A, B, C and D substances 

. 
Let 

°° °°
G° 

.r =cGC ,m .dGD ,m -aGA ,m -bGB ,m 
be the standard reaction free energy of thereaction above 



. 
Since 

°° °
GX ,m =GX ,m . 0.-kBT ln zX 
where z0X is the standard partition function of an X particle 



. 

Thus, the standard reaction free energy, whichregards to single molecules rather than a moleof molecules, is 
°°° °° 

.rGm =cGC ,m .0 ..dGD ,m .0.-aG A,m .0 .-bG B ,m .0. 

°° ° 
-k BT .c ln zC .d ln zD-a ln z° A-b ln zB . 




. 

Since G(0)=U(0), the first part of the right
hand side of the equation above is 
°° 
.r .0 =cU C ,m . 0..dU D,m . 0. 
°° 

-aU A,m .0.-bU B,m . 0. 
the standard reaction internal energy per molecule, i.e. the difference between the molar ground state energy of the products and the reactants 



. 
Since 

.rG°=-RT ln K = AN kBT 
thus 
G° 
.r

ln K=- ANk BT 
and thus 
°° 
.d

. zC ,m .c . zD ,m -.r .0 / RT 
K =° °.be
. zA,m .a . zB,m 




. 

The other approach to get the value of the rateconstant of a reaction is the transition state or activated complex theory 
. 

Transition state is the state with the highestenergy along a given reaction coordinate 
. 

The main advantage of transition state theoryin contrast to the collision theory is that it doesnot require any intuitive, non-measurable andincalculable parameter such as the steric factor 



. 

For simplicity and for easy understanding, letus investigate a simple bimolecularcombination reaction with A and B as reactants 
. 

We assume an activated complex which is
close to or at the transition state 

k‡
k1 C‡ .
A.B . P 


where A and B are reactants, P is the product andC‡ is the activated complex and k1 and k‡ are the corresponding rate constants 



. 

Transition state theory assumes that theconversion of the activated complex to productis the slow step of the reaction so itdetermines the velocity 
v=k‡[C‡ ] 


where v is the velocity and k‡ is the rate constant of the unimolecular transformation of the activated complex to the product, respectively 



. 
The velocity of the reaction expressed by the rateconstant of the whole reaction is 
v=k2 [ A][ B ] 

where k2 is the rate constant of the whole 
reaction, [A] and [B] are the concentrations of 
reactants A and B, respectively 




. Since k1 is fast, an equilibrium is established 
between the reactants and the activated 
complex 

[C ‡]=K ‡[ A][ B] 

where [C‡], [A] and [B] are the concentration ofthe activated complex, the A and the B reactant,respectively and K‡ is the equilibrium constant 
. 
Thus 
k2 =k‡ K ‡ 




. 

Now the task is for us to determine the value of k‡ and K‡ 
. 

To be able to continue we need the evaluation of the reaction coordinate and some skew to the area of statistical mechanics and quantummechanics 



Reaction coordinates 

. 

Let us consider a concrete problem to get apicturesque image on the problem of thereaction coordinate 
. 

Let the reaction be that an “A-B molecule and C atom” system converts to a “A atom B-Cmolecule” system 
. 

The energy of the system can be described asa function of three degrees of freedom 



Reaction coordinates 





. 	The three reaction coordinates are: rAB the distance between atoms A and B, rBC the distance between B and C, and . the angle
between the vector connecting A and B andthe vector connecting B and C atoms 
. 

If we want a picturesque image we have to reduce the number of reaction coordinates to two 
. 

Let us set the value of . to 180 degrees andthus the potential energy will be the functionof only two independent variables 



. 

Making use of these two reaction coordinates, we can depict a potential energy surface as afunction of the distances between atoms 
. 

Three locations on the potential energy surfaceare important: 
– 
A-B molecule and separated C atom 

– 
B-C molecule and separated A atom 

– 
Transition state 



. 

There is a valley connecting the first and thesecond states (with broken yellow line in thefigure below) 



. 

The line running on the bottom of the valleycorresponds to the reaction coordinate 
. 

The highest point of the valley mentioned
above is considered as the transition state 

. 

In our case, transition state is when B is about halfway between A and C 
. 

In a direction perpendicular to the linecorresponding to the reaction coordinate, thetransition state is at the bottom of a potentialenergy well, so it is a saddle point 



Potential energy landscape 





Potential energy surface in two 
dimensions 





. 

Let us get back to the reaction of the A and B atoms as reactants 
. 

In this reaction, the potential energy can beconsidered as a function of the distance between A and B, so this distance may bechosen as reaction coordinate 
. 

A reaction profile plots the potential energyagainst the reaction coordinate 
. 

Transition state is at the top of the curve 



Reaction profile 






. 

Although products are lower in potential
energy, a barrier has to be passed for the
reaction to occur 

. 

In the absence of significant entropy changeduring the reaction, potential energy can besubstituted by the free energy 
. 

The difference between free energy ofproducts and reactants is the reaction freeenergy which is always negative if the reactiontakes place spontaneously 
.Gr =G products-Greactants 




The value of the rate constant k‡ 

. 

To transform to products, the system first hasto pass the transition state 
. 

We assume that around the transition state there is a shallow valley along the reactioncoordinate in which the behaviour of the activated complex can be treated as aharmonic oscillator with frequency . 
. 

So the frequency by which the activatedcomplex gets to and pass the transition stateis also . 



Reaction profile proposed by Eyring 





. 

Since not every oscillation leads to passing thetransition state by the activated complex, it isonly stated that k‡ is proportional to thefrequency of vibration along the reactioncoordinate 
k‡=. . 


where . is the frequency and . is the transmission factor which often approaches 1 



. 

Since the activated complex occurs onlytransiently, practically its concentration is not measurable so the value of the equilibriumconstant cannot be calculated based on concentrations 
. 

Instead of using concentrations we can use thepartition functions of the substances 
. 
Let us get back to the reaction 
A.B . C‡ . P 





. 

Combining transition state theory and thestatistical mechanical derivation of the equilibrium constant, the rate constant for thisreaction is obtained as: 
z° C‡ ,m -.r .0 / RT 
k2 =k‡ K ‡ =k‡ e
°° 
zA ,m ·zB ,m 

where k2 is the rate constant of the whole reaction, k‡ is the rate constant of the transformation of the activated complex toproduct and K‡ is the equilibrium constant of it 



. 

Let us continue with the partition function ofthe activated complex 
. 

Since we assume that the activated complexpasses the transition state toward products bya vibrational motion, it is natural to separatethe part of the partition function describingthis vibrational mode from others 



. 

The partition function of this oscillation –

normalized by the zero point vibration – is 

1
z' = 
-h ./ kT 
1-e

where . is the same frequency as we havealready seen in the expression describing the rate constant k‡ of the activated complex-product transformation 



. 

The expression can be approximated by 

k BT
z'. 

h. 

since h./kT << 1 and the exponential can beexpanded as a series 
. 

Thus the total partition function is 
kBT 
zC ‡. zC ‡
h. 




. 

The equilibrium constant K‡ is then 
kT 
K ‡ = K

h . 

where K is the function of only z and independent of the vibration responsible for passing thetransition state 



. 

Thus we get the Eyring equation for the rate constant of the whole reaction 
kBT kBT
k2=. . K =. K
h . h 
. 

The partition functions can be obtained byspectroscopic measurements so we cancalculate K and thus the rate constant of the total reaction k2 


Introduction to biophysics: Derivation of rate constant 

Thermodynamic description of the 
transition state theory 

. 

Having described the theory with the help ofstatistical and quantum mechanics, let us look at a purely thermodynamic analysis 
. 

We can use the same expression as a startingpoint as in the statistical mechanicaldescription, namely 
[C ‡]
K ‡ 
= 
[ A][ B ] 




. 

We can talk about the reaction free energy .G‡ for the formation of the activated complex C‡ at quasi-equilibrium 
.G‡ fwd =-RT ln c ° K 

where the c° is the standard concentration which ensures that the free energy has the proper unit, and following from this 
1 -.G fwd / RT 
K =° e ‡ 
c 




Modified reaction profile 





. 

Thus for the rate constant k‡ we obtain 
k BT 1 -. G fwd / RT 
k2=. ° e ‡ 
hc 

. 

Let us introduce the activation enthalpy .H‡ and activation entropy .S‡ 
. 

Expressing the activation free energy withthem we get 
.G‡ =. H ‡-T .S ‡ 



. 

Substituting the expression above by that for the rate constant 
k BT 1 . S ‡ / R -. H ‡/ RT 
k2=. ° ee
hc 




. 

To determine the relationship betweenactivation enthalpy and activation energy, letus set out from the Arrhenius equation 
k=Ae-EA/ RT 
. 

After rearrangements and derivation with
respect to the temperature we obtain 

EA =RT 2 .ln k 

. T 




. 
Since 

. ln k 2 . H ‡ 
=.

. T TRT 2 
the activation energy expressed with theactivation enthalpy is 
EA=. H ‡.2 RT 



. 

Substituting it to the equation expressing thetotal rate constant the equation will be 
2 kT RT .S ‡ / R -EA/ RT 
k2 =eee
hp0 
where 

2 kT RT . S ‡ / R
A=ee
hp0 

is the preexponential factor which appears in theArrhenius equation