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Computational Thermochemistry 1

1 Computer Experiment 5: Computational Thermochemistry

1.1 Background: Within the frame of this experiment you will employ methods of statistical

thermodynamics in order to derive thermochemical reaction quantities and energies of

chemical bonds, which both are well known subjects to the experimental chemist. It will

become evident that frequency analysis is an essential tool in theoretical

thermochemistry.

1.1.1 Basics of Statistical Thermodynamics The link between (microscopic) molecular quantities and the (macroscopic) observables

of a chemical system is provided by the system’s partition function Q, which is based on

the Boltzmann distribution over the available microstates and in the general formulation

reads:

Q = g

j!e"

Ej

kB !T

j=1

N

# ( 1)

Here N denotes the total number of accessible states, Ej the energy of the jth state, kB the

Boltzmann constant and T the Kelvin temperature (the term kBT can be interpreted as

the available thermal energy). The degeneracy factor gj is directly given by the

degeneracy of the jth state and equals one if the corresponding state is not degenerated.

Consider a homogenous chemical system consisting of molecules, e.g. a monomolecular

gas. From the macroscopic perspective, the system’s components are the molecules and

the system’s energy states Ej depend on the arrangement of the components and their

energies. The corresponding partition function is referred to as the canonical or system

partition function Q, whereas the microcanonical or molecular partition function q treats

the single molecule as the superordinated system, thus being a sum over the molecular

energy states. In first approximation, the total energy of a molecule can be split up into

contributions from each of the molecule’s degree of freedom (which is a direct result

from the Born-‐Oppenheimer approximation):

! = !

itrans + !

jrot + !

kvib + !

lel . ( 2)

Thus the molecular partition can be factorised according to:

q = qtrans!q

rot!q

vib!q

el . ( 3)


With this result the actual calculation of the molecular partition function is possible by

applying the quantum chemical solutions of appropriate problems, as for example the

particle in a box or the harmonic oscillator, to the degree-‐of-‐freedom dependent

expressions of the molecular partition function. For example, the application of the

harmonic oscillator energy eigenvalues to the vibrational partition function of a

diatomic molecule yields the expression:

qvib

=e!

h!2kBT

1!e!

h!kBT

. ( 4)

Thus the only quantity which has to be known is the frequency of the system’s normal

mode. Similar expressions for the other degree-‐of-‐freedom dependent partition

functions can be derived, depending on molecular quantities like the moment of inertia

or the volume, which in first approximation can be determined according to the ideal gas

law. As an exception to this, the electronic partition function includes a summation over

the system’s quantum states, which are given by solving the electronic Schrödinger

equation:

q

el= g

l!e"!lel

kB !T

l=1

N

# ( 5)

In most cases, the excited electronic states cannot be populated by the available

thermal energy and thus are left out of the summation. A further simplification is

realized by setting the zero point of energy to the electronic energy of the reactant’s

ground state, in which case the partition function is reduced to the degeneracy factor g1.

As laid out above, the partition function contains all necessary information for the

determination of the system’s thermochemical quantities. For example, the general

expression of the mean energy is proportional to the derivation of the partition

function’s natural logarithm with respect to the temperature:

E =!

"lnQ"ß

#

$%%%%

&

'((((

= kBT 2 )

"lnQ"T

#

$%%%%

&

'(((( . ( 6)

In the thermodynamic equilibrium, this formulation equals the inner energy U of the

system if the volume remains constant and the inner energy at 0 K is zero. Since the

inner energy is the basic quantity in equilibrium thermodynamics, expressions for all

other thermochemical state functions can be derived on the basis of the system’s


partition function (some authors even say that the partition function is the “wave

function of modern thermodynamics”).

1.1.2 Thermochemical Quantities After the system’s partition function is calculated, it may be used for the determination

of thermochemical quantities as shown above for the example of the inner energy. Since

partition function factorisation is the basis of these calculations, it is common to

distinguish between the contributions from the different molecular degrees of freedom

to the quantity under investigation, e.g. in case of the enthalpy:

qtot= q

trans!q

rot!q

vib!q

elec à

Htot= H

trans+ H

rot+ H

vib+ H

elec . ( 7)

For example, in classical thermodynamics the entropy (of a reversible process) is

defined according to Clausius: dS =

CV

TdT . The heat capacity CV is given as the first

derivative of the inner energy U with respect to the temperature at constant volume,

thus yielding:

S !S

0=

1T

0

T

"##T

kBT 2 #lnQ#T

$

%&&&&

'

())))V

dT . ( 8)

Evaluation of the integral and reintroduction of the inner energy leads to:

S !S

0= k

BT"lnQ"T

#

$%%%%

&

'((((V

+ kB

lnQ! kB

lnQ( )T=0

. ( 9)

The temperature-‐independent term S0 can thus be identified as S

0= k

BlnQ( )

T=0, which

is referred to as the zero point entropy, the entropy at 0 K. The remaining terms equal

the temperature-‐dependent entropy according to:

S = k

BT!lnQ!T

"

#$$$$

%

&''''V

+ kB

lnQ . (

10)

The remaining thermochemical quantities can be derived in an analogue way by starting

at the classical expression and substituting a known function by its statistical

counterpart (most often the inner energy U).

The following table summarizes the most common thermochemical functions and their

expression in terms of the partition function.


Table 1: Thermodynamic State Functions.

Function Statistical expression Inner energy U

U = k

BT 2 !lnQ

!T

"

#$$$$

%

&''''V

Entropy S

S = k

BT!lnQ!T

"

#$$$$

%

&''''V

+ kB

lnQ

Enthalpy H

H = k

BT

2 !lnQ

!T

"

#$$$$

%

&''''V

+ kBTV

!lnQ

!V

"

#$$$$

%

&''''T

Gibbs free enthalpy G

G = k

BTV

!lnQ!V

"

#$$$$

%

&''''T

(kBT lnQ

Most common quantum chemical program packages use the ideal gas law and the

quantum models of the rigid rotator and the harmonic oscillator as the basis for the

calculation of the partition function. Carrying out the differentiations according to the

statistical expressions (see table) yield constant enthalpy contributions of Htrans = Hrot =

3/2 RT for the rotational and translational degrees of freedom (as it is predicted by

classical thermodynamics), whereas the vibrational part is given by

H

vib= R

h!i

2kB

+h!

i

kB

!1

eh!

i/k

BT "1

#

$%%%%

&

'(((((

i=1

3N"6(7)

) . (

11)

The summation is carried out over all vibrational degrees of freedom, which in the case

of a non-‐linear molecule equal 3N-‐6. If the structure under investigation is a transition

state, i.e. a maximum on the potential energy surface along one direction, the

corresponding vibrational degree of freedom is imaginary and left out in the summation,

thus reducing the number of real vibrational degrees of freedom to 3N-‐7. The first part

of the vibrational enthalpy is a sum of temperature-‐independent terms (hυi/2kB), which

is referred to as the zero point energy (ZPE), whereas the second part depends on the

temperature and considers those molecules which are not in the vibrational ground

state. If the electronic ground state is not degenerated, Helec and Selec reduce to zero and

reaction enthalpies are directly given by the difference of the reactant’s electronic


energy. The analogue expressions for the entropic contributions are derived in the same

way:

Strans

=52R + R ! ln

VN

A

!2!Mk

BT

h2

"

#$$$$

%

&'''''

32

"

#

$$$$$$$$

%

&

'''''''''

Srot

=12R 3 + ln

!"! I

1I

2I

3!

8!2kBT

h2

"

#

$$$$

%

&

'''''

32

"

#

$$$$$$$$

%

&

'''''''''

(

)

*****

+

,

-----

Svib

= Rh#

i

kBT!

1

eh#i/kBT .1. ln 1.e.h#i/kBT( )

"

#$$$$

%

&'''''i=1

3.6(7)

/

(

12)

Here V and M denote the system’s volume according to the ideal gas law and the

molecular mass, respectively. The symmetry number σ is given by the order of the

rotational subgroup in the molecule’s point group and can be understood as the number

of sub-‐turns which transfer the molecule into it’s own starting structure during a full

360° turn. The principle moments of inertia I1, I2, I3 are the eigenvalues of the

diagonalized inertia matrix, which are included in the results of the frequency

calculation as well as the vibrational frequencies υi. Knowledge of these basic

thermochemical quantities allows the calculation of the free Gibbs enthalpy according to

G = H !T "S . As already mentioned above, the output of frequency calculations

contains information about the partition functions as well as most thermochemical

quantities.

1.1.3 Bond Dissociation Energy and Atomization Energy The strength of intramolecular bonds in chemical terms is defined by the binding energy

Ebind, which in case of diatomic molecules is identical to the dissociation energy Ediss. For

larger molecules ABn, the binding energy is the arithmetic mean of the sum over all n

possible A-‐B bond dissociation energies:

E

ABbind =

1n

Eidiss

i=1

n

! . ( 13)

Even for medium-‐sized molecules, there is often a significant difference between the dissociation

energy and the binding energy.

In contrast to the binding energy, the atomization energy Eatom is defined as the reaction energy of

the atomization reaction, which conveys the gaseous species ABn into one atom of type A and n


atoms of type B in the gas-‐phase. A possible approach to the determination of the atomization

energy of a molecular compound ABn lies in the addition of the enthalpy of formation for the gas-‐

phase formation reaction (A + n B à ABn) and the atomization reaction for the transfer of the most

stable elemental modifications Ax, By to the gas-‐phase according to 1/x Ax à A and 1/y By à B,

respectively.

1.2 Description of the Experiment To understand the concepts, you will investigate simple chemical systems and

determine the change in Gibbs free enthalpy ΔG0 as well as bond dissociation energies

and atomization energies. The reactions include the Haber-‐Bosch reaction, the

“Knallgas” reaction of hydrogen and oxygen to water, and the gas-‐phase reaction

between water and carbon monoxide:

o H2 + O2 → 2H2O

o N2 + 3H2 → 2NH3

o H2O + CO → H2 + CO2

1. Optimize the geometry of all participating reactants as a first step. Perform a DFT structure

optimization starting with the B3LYP functional and the SVP basis set.

2. Run a frequency analysis for the optimized geometries and look for the relevant

thermochemical data in the output file

3. Calculate the change in Gibb free enthalpy for all investigated reactions and compare the

results to experimental data.

Next, the strength of different C-‐H bonds will be examined on simple organic molecules and the

atomization energy of these compounds will be determined. As sample systems you may compare

the following molecules: methane, acetylene, benzene, and acetic aldehyde. Alternatively, you could

choose your own set of organic molecules and compare the results of your calculation.

4. Optimize the structures of all molecules. Again start with the B3LYP/SVP combination.

5. Determine the bond dissociation energies of all C-‐H bonds and the C-‐H binding energy in

methane as well as the atomization energies for all molecules and compare your results

to experimental data.

6. Repeat your calculations for the enlarged basis sets TZVP and TZVPP. Do you observe any

basis set effects ?

Table 2: Bond dissociation energies of some C-‐H bonds

Bond D0298 [kJ/mol]


H-CH3 438.9

H-CH2 462.0

H-CH 424.0

H-C6H5 473.1

H-CCH 556.1

H-CH2CHO 394.6

H-COCH3 373.8

Table 3: Gibbs free reaction enthalpy of some reactions

Reaction ΔG0

[kJ/mol] H2 + O2 → 2H2O(l) -572.0

N2 + 3H2 → 2NH3 -92.3

H2O(g) + CO → H2 + CO2 -41.2

1.3 Literature • McQuarry, D. A.; Simon, J. D. Physical Chemistry – A molecular approach;

University Science Books: Sausalito, 1997

• Wedler, G. Lehrbuch der physikalischen Chemie; Wiley-‐VCH: Weinheim, 1997

• Jensen, F. Introduction to computational chemistry; Wiley-‐VCH: Chichester, 1999

• Holleman, A. F.; Wiberg, N. Lehrbuch der anorganischen Chemie; Walter de Gruyter: Berlin, 1995

• Lide, D. Handbook of chemistry and physics; CRC Press: Boca Raton, 2000


2 Computer Experiment 6: Computational Chemical Kinetics

2.1 Background

2.1.1 Chemical Kinetics Reacting chemical systems are mathematically described by sets of coupled first

order differential equations. The determination of the rate law and the solution to

the associated differential equation system is the subject of descriptive kinetics. The

result of the analysis is the concentration of each species involved in the reaction as

a function of time. The rate at which different species are produced or consumed is

proportional to the rate constants k with is a characteristic quantity of the reacting

partners and is a function of temperature (vide infra). The evolution of a chemical

reaction system is fully determined from the initial concentrations of each species

and the set of k’s.1

Consider for example the comparatively simple reaction:

Ak1! "! B k2! "! C

( 14)

In which a substance A is transformed to B which then decays to the final product C.

The time course of the concentrations of A, B and C is shown in Figure 1

1 This is only “half true” – there are systems which display “deterministic chaos”, e.g. the evolution of the system is still fully determined by the rate laws but the trajectories become “infinitely sensitive” to the initial conditions. In this case, it is not possible to predict the outcome of the chemical evolution since the initial conditions are never precisely enough known. While in most chemical reaction systems, the time course of the various concentrations can be reasonably well fitted to sets of exponential curves, it is not necessarily true that the solutions of the differential equations are always exponentials. For example, there are oscillating chemical reactions and other oddities which are of no concern in the framework of this course.


0 2 4 6 8 100.0

0.2

0.4

0.6

0.8

1.0

A(t) B(t) C(t)

Con

cent

ratio

n (m

M)

Time (Sec) Figure 1: The time course of the reaction A → B → C. The intermediate B is formed from the decay of A and its concentration peaks at around 1.6 sec in this example. The extent to which B accumulates depends on the ratio of the rate constants k1 and k2. The intermediate B slowly decays towards the final product C.

In chemistry and biochemistry such reactions are usually followed by recording

some type of spectra during the course of the reaction. For example, consider the

absorption spectra in Figure 2. Let species A have an absorption spectrum with two

dominant peaks at ~350 and 430 nm, species B with a prominent absorption at 450

nm and 600 nm and species C a spectrum with peaks around 310 nm and 700 nm.

The absorbance as a function of time that results from this reaction system is then

shown on the right hand side of Figure 2. It has a fairly complicated appearance

since it is determined from the overlapping contributions of several species the

relative contributions of which evolve in time. Nevertheless, the successful

deconvolution of such data2 through analysis of the experimental data yields the

absorption spectra of the individual species as well as the rate constants k1 and k2

which may then be subjected to theoretical analysis.

2 In order to obtain the spectra of the individual species from the convoluted absorption envelope, one may use techniques like singular value decomposition (SVD). The details are outside the scope of this course.


300 400 500 600 7000

500

1000

1500

2000

2500

3000

A B C

ε (M

-1 c

m-1)

Wavelength (nm)

300400

500600

700

0

500

1000

1500

2000

2500

0

24

68

10

Abso

rban

ce (O

D)

time (sec)wavelength (nm) Figure 2: Hypothetical absorption spectra of species A, B and C (left). Three dimensional plot of absorbance versus time and wavelength. Note the appearance and disappearance of the peak around 600 nm caused by the intermediate B.

We give two examples of rate laws where the associated differential equation

systems have closed form solutions. The simplest chemical reaction is a

unimolecular decay of the form:

Ak! "! B

( 15)

With the rate law:

d A!"#$%&

dt='

d B!"#$%&

dt='k A!"#

$%&

( 16)

We are interested in the time courses A!"#$%& t( ), B!"#

$%& t( ) . Since matter is conserved there

is a conservation law, which states that:

A!"#$%& t( )+ B!"#

$%& t( ) = const

( 17)

And thus:

B!"#$%& t( ) = const' A!"#

$%& t( )

( 18)

which solves half of the problem. Such conservation laws always exist in chemical

reactions and serve to reduce the dimensionality of the associated differential

equation system. The remaining equation is not difficult to solve and the solution is:


A!"#$%& t( ) = A

0e'kt

( 19)

where A0 is the initial concentration of [A]. Thus B!"#$%& t( ) = A

01'e'kt( ) . For an

irreversible bimolecular reaction of the form:

A + B k! "! C + D

( 20)

The solution is:

kt =1

A0!B

0

lnB

0

A0

A"#$%&' t( )

B"#$%&' t( )

(

)

*****

+

,

------

( 21)

unless A0= B

0, then:

A!"#$%& t( ) =

1kt + 1

A0

( 22)

In the general case, the differential equation systems will be far too complex to

allow a closed form mathematical solution. In this case, one has to resort to

numerical techniques. While this is a rather specialized area of numerical

mathematics, we briefly illustrate the principle that underlies such simulations:

assume that you have N species with concentrations C1(t),…,CN(t). The rate laws are

of the form dC1/dt=f1(C1,…,CN), dC2/dt=f2(C1,…,CN) and so on.3 Now, we can try to

replace the differential on the left hand side by a finite difference to obtain an

iterative equation set:

C1

ti+1( )!C

1ti( )

ti+1! t

i

= f1

C1

ti( ),...,CN

ti( )( )

( 23)

…

3 Each function f contains the concentrations of potentially all species as well as the collection of rate constants k that describe the chemical network.


CN

ti+1( )!C

Nti( )

ti+1! t

i

= fN

C1

ti( ),...,CN

ti( )( )

( 24)

Which can be solved for the unknown C

Xti+1( ) (X=1…N):

C

Xti+1( ) =C

Xti( )+ t

i+1! t

i( ) fX

C1

ti( ),...,CN

ti( )( ) ( 25)

Thus, given the concentrations at time t0=0, it is straightforward to obtain the

concentrations at times t1, t2,… and therefore to simulate the entire evolution of the

chemical system. In order for the simple finite difference approximation to work,

the time-‐step must be chosen small enough.4

2.1.2 Transition State Theory From a microscopic point of view, one is interested to understand the value of k

from first physical principles. From phenomenological considerations, one already

knows that k is temperature dependent and obeys an Arrhenius type of equation:

k = Aexp !

Ea

RT

"

#$$$$

%

&'''''

( 26)

Where A is the “pre-‐exponential factor”, R is the gas constant, T the temperature and

the key quantity is the “activation energy” Ea. The larger Ea

, the slower the

reaction. More detailed insight into the reaction scenario is obtained from Eyring’s

transition state theory. In this case, one assumes that the two reacting partners A

and B have to pass through a special geometrical arrangement AB{ }*

(the

transition state, TS) before decaying to the products C and D. The kinetic scheme

that one writes is:

4 “Small enough“ is roughly such that it is sufficiently smaller than the fastest event in the chemical reaction. The simple equations given above are still numerically rather unstable. Much more sophisticated methods such as the Runge-‐Kutta procedure or ultimately, the “stiff-‐stable” Gear algorithm are used in practice for the integration of differential equation systems.


A + B !

k1

k"1

# $##% ### AB{ }* k*

# $# C + D

( 27)

If one assumes the “pseudo-‐steady state” for AB{ }*

, one comes to the conclusion

that5 one can write the rate constant for the bimolecular decay as:

k = K *k*

( 28)

where K* = k

1/k!1 (the equilibrium constant for the TS) and quantum mechanics

leads to the conclusion that:

k* = !

kBT

h

( 29)

Where kB= 1.38x10!38J /K is Boltzmann’s constant, h is Planck’s constant and κ is

a “transmission coefficient” which is usually close to unity.6 The “equilibrium

constant” K * is related to the free energy of the transition state over the initial state

by:

K * = exp !

"G*

RT

#

$%%%%

&

'(((((

( 30)

And hence:

k =

kBT

hexp !

"G*

RT

#

$%%%%

&

'(((((

=k

BT

hexp !

"H*

RT

#

$%%%%

&

'(((((exp

"S*

R

#

$%%%%

&

'(((((

( 31)

Where !H* and !S * are the enthalpy and the entropy of the transition state.

Thus, one identifies the parameters of the Arrhenius equation with:

5 This is explained in detail in any textbook on physical chemistry, e.g. P.W. Atkins, Physikalische Chemie, VCH, Weinheim, in the latest edition. 6 It also accounts for tunneling through the barrier.


A =

kBT

hexp

!S *

R

"

#$$$$

%

&'''''

( 32)

Ea=!H

*

( 33)

The quantity !H* is the one that one wants to calculate while !S * is slightly more

difficult to predict.

2.1.3 Quantum Chemical Calculation of Transition States The procedure to find transition states from quantum chemical calculations is

analogous to that for finding minimum structures. The first step is to search for

stationary points on the potential energy surface; therefore the gradients of the

energy with respect to the nuclear coordinates have to vanish. Next, the Hessian

matrix has to be determined in order to characterize the stationary point as a

minimum, maximum or saddle point. Only saddle points with a single negative

frequency correspond to transition states. An example is shown below which shows

the two-‐dimensional potential energy surface of the system H-‐H-‐H in a linear

arrangement:

Figure 3: Two-‐dimensional potential energy surface for the system H-‐H-‐H in a linear arrangement. A transition state is observed around the H1-‐H2 and H2-‐H3 distances being both 1.0 Angström.

0.60.8

1.01.2

1.41.6

1.82.0

2.22.4

0

20

40

60

80

100

120

0.6

0.9

1.2

1.5

1.8

2.1

2.4

Energy (kcal/mol)

H 2-H

3 D

istan

ce (A

)

H1 -H

2 Distance (A)

78 51

423324

24

3342

5160

69

15

15

12

12

78

9.0

9.0

6.0

6.0

3.0

3.0

87

0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.40.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

2.2

2.4

H2-H

3 Dist

ance

(Ang

ström

)

H1-H

2 Distance (Angström)


Once such a point was found, the electronic energy difference ETS !Eeduct is known

and is the most important contributor to the activation enthalpy; The vibrational

contribution to the activation enthalpy is given by:

!Hvib

* = Rh!

n

TS

2kB

1+2

exp "h!n

TS /kBT( )"1

#

$

%%%%%%

&

'

(((((((n=0

3N"7

)

"Rh!

n

educt

2kB

1+2

exp "h!n

educt /kBT( )"1

#

$

%%%%%%

&

'

(((((((n=0

3N"6

)

( 34)

Where the sum is performed over all vibrational degrees of freedom of the TS and

the reactants respectively. For unimolecular reactions the TS has one vibrational

degree of freedom less than the products (one mode corresponds to the transition

state and does not contribute to the ZPE of the TS) !Hvib

* is usually a few kcal/mol

negative and therefore reduces the barrier.

For bimolecular reactions, there is a constant contribution from the rotations and

translations to !H* of -‐4RT. In this case, the TS has more vibrational degrees of

freedom than the separated reactants which makes !Hvib

* a few kcal/mol positive.

Likewise, there is a substantial contribution to the activation entropy for

bimolecular reactions which raises the free-‐energy barrier by 10-‐15 kcal/mol.

Nevertheless, everything that is required in order to calculate the free energy of

activation can be approximately deduced from the outcome of a frequency

calculation at the geometries of the reactants and the TS respectively.

The tunnelling correction ! T( ) may be estimated from Wigner's expression7

! T( ) = 1+

124

h"TS

kBT

!

"####

$

%&&&&&

2

( 35)

7 Wigner, EP Z. Phys. Chem. Abt B 1932, 19, 203


where !TS is the absolute value of the imaginary frequency corresponding to the

transition state.

A word of caution: to find a transition state with quantum chemical programs is not

easy! Since the programs have no way to “guess” a transition state from a stable

structure alone, it is important to “guide” the programs to the desired transition

states by providing structures that are close to the final TS. However, this means

necessarily, that the outcome of the calculations depend on the skill of the computer

chemist to guess a reasonable TS – there is no guarantee that the TS the program

may find based on the guessed structure is also the one of lowest energy. Do not be

disappointed, if the program does not find the desired TS – just provide a better

starting structure. In calculating chemical reactions, the chemical intuition and

knowledge of the theoretician is crucial for success! Please refer to section Error!

Reference source not found. for appropriate input into the ORCA program.

2.1.4 Kinetic Isotope Effects There are two origins of the kinetic isotope effect. The first, quantum mechanical

tunnelling through the reaction potential energy barrier is usually only important at

very low temperatures and for reactions involving very light atoms. It may be

estimated from the change in the transmission coefficient caused by the change in

the imaginary frequency that leads to the transition state (see Wigner’s expression

above).

More importantly however, kinetic isotope effects are caused by differences in the

activation energy for reactions involving different isotopes since the reactants and

the TS have different zero point vibrational energies. The effect on the activation

barrier for a reaction involving an H/D atom is shown below.


Figure 4: Definition of energetic quantities in the calculation of chemical reactions on the basis of qualitative potential energy surfaces. On the x-‐axis, the “reaction path” and on the y-‐axis the total energy is plotted.

The net effect is that the activation energy is higher for the heavier isotope, and

therefore the reaction will be slower (a 'normal' isotope effect). The maximum

isotope effect is obtained when the bond involving the isotope is completely broken

in the transition state, in which case the difference in activation energies is simply

the difference in zero point energies of the stretching frequencies for the bond being

broken.8

From transition state theory, one readily deduces:

kH

kD

= exp !"H

*H( )!"H

*D( )

kBT

#

$

%%%%%

&

'

((((((

( 36)

Where “H” and “D” have been written for the two isotopes since hydrogen to

deuterium is the most frequently studied isotope effect. The difference

!H

*H( )"!H

*D( ) is entirely determined by the different ZPE contributions to

!Hvib

* .

8 In some reactions it is the zero point energy difference between the transition states which governs the kinetic isotope effect. In this case the activation energy is larger for the lighter isotope and an inverse isotope effect is observed, in which the heavier isotope containing reactant undergoes faster reaction.

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2.2 Description of the Experiment

2.2.1 Transition State of Glyoxal • Take the optimized geometry of the glyoxal isomer which has been calculated in

the previous section and rotate one carbonyl group Error! Bookmark not

defined. by 90°, so that it is arranged perpendicular to the opposite carbonyl

group.

• Calculate the transition state using B3LYP/SVP.

• Find out how quickly glyoxal interconverts at room temperature by calculating

the activation energy and the rate constant.

2.2.2 H/D Kinetic Isotope Effect • Calculate the H→D effect on the reaction rate of the reaction CH4 + OH• → H2O +

CH3. Build all molecules using MOLDEN and export the Z-‐matrix to the Gaussian

format. Then run a geometry optimization for all compounds using B3LYP/SVP.

(HINT: in the transition state one of the C-‐H bonds must be significantly

stretched and the H-‐O bond should already be partially formed. You should try to

start with such a structure. A rule of thumb is to stretch the bond to ~1.5 times

its equilibrium value).

• Think of a possible transition state structure and try to guide the optimization to

this transition state.

• Calculate the activation enthalpy and activation entropy using the zero-‐point

corrected energies. Calculate the reaction rates and predict and the resulting

isotope effect.

2.2.3 Rotational Barrier of Ethane • Study the rotational barrier of ethane. Determine the energies of the eclipsed

and staggered conformations. Perform a relaxed and an unrelaxed potential

energy surface scan and plot the results.

Refer to section Error! Reference source not found. for a description of how to

generate relaxed surface scans with the ORCA program.


• Build the ethane molecule using MOLDEN and export the Z-‐matrix.

• Perform an unrelaxed surface scan using the Scan keyword in the command line

and enter the parameters to change (See sample input below).

• Perform a relaxed surface scan using the Opt keyword and enter the parameters

to change. (See sample input below).

• What happens generally if one does not relax the rest of the geometry during the

surface scan? Will you over-‐ or understimate the barriers calculated by

unrelaxed surface scans ?

Compare to the available experimental data. The rotational barrier of ethane is about 1024 cm-‐1 ≈12.25 kJ/mol.9

9Weiss, S and Leroi, GE, J. Chem. Phys., 1968, 48, 962

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