# ( 1) j=1. 1 Computer Experiment 5: Computational Thermochemistry. 1.1 Background:

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1 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 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: N E j Q = g j!e " k B!T # ( 1) j=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):! =! i trans +! j rot +! k vib +! l el. ( 2) Thus the molecular partition can be factorised according to: q = q trans!q rot!q vib!q el. ( 3)

2 Computational Thermochemistry 2 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: q vib = e! h! 2k B T 1!e! h! k B T. ( 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: N! l el q el = g l!e " k B!T # ( 5) l=1 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 ' ( = k T 2 ) # "lnq & B $ % "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

3 Computational Thermochemistry 3 partition function (some authors even say that the partition function is the wave function of modern thermodynamics ) 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: q tot = q trans!q rot!q vib!q elec à H tot = 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 = C V T dt. 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 = T " 0 1 T # $ #T k T #lnq ' 2 dt. ( 8) B %& #T () V Evaluation of the integral and reintroduction of the inner energy leads to: # S!S 0 = k B T "lnq & + k $ % "T '( B lnq! k B lnq V ( ) T =0. ( 9) The temperature- independent term S0 can thus be identified as S 0 = ( k B lnq), which T =0 is referred to as the zero point entropy, the entropy at 0 K. The remaining terms equal the temperature- dependent entropy according to: 10) " S = k B T!lnQ % + k # $!T &' B lnq. ( V 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.

4 Computational Thermochemistry 4 Table 1: Thermodynamic State Functions. Function Inner energy U Entropy S Enthalpy H Gibbs free enthalpy G Statistical expression "!lnq % U = k B T 2 # $!T & ' V " S = k B T!lnQ % + k # $!T &' B lnq V "!lnq % H = k B T 2 # $!T & ' + k TV "!lnq % B # $!V & V " G = k B TV!lnQ % (k # $!V &' B T lnq T ' T 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 + h! i 1 & )!. ( 2k B k B e h! i /k B i=1 $ % T "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

5 Computational Thermochemistry 5 energy. The analogue expressions for the entropic contributions are derived in the same way: 12) " 3 S trans = 5 2 R + R! ln V! 2!Mk T % " % 2 B N A # $ h 2 &' # $ &' ( " 3% + S rot = 1 2 R 3 + ln! " "! I I I! 8!2 k B T % # $ h 2 &' # $ &' )*,- 3.6(7) " h# S vib = R i k B T! 1 e h# i /k B T.1. ln % /kbt / ( ) # $ 1.e.h#i i=1 &' 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 Bond Dissociation Energy and Atomization Energy The strength of intramolecular bonds in chemical terms is defined by the binding energy E bind, which in case of diatomic molecules is identical to the dissociation energy E diss. For larger molecules ABn, the binding energy is the arithmetic mean of the sum over all n possible A- B bond dissociation energies: E bind AB = 1 n n diss! E i. ( 13) i=1 ( 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 E atom is defined as the reaction energy of the atomization reaction, which conveys the gaseous species AB n into one atom of type A and n

6 Computational Thermochemistry 6 atoms of type B in the gas- phase. A possible approach to the determination of the atomization energy of a molecular compound AB n lies in the addition of the enthalpy of formation for the gas- phase formation reaction (A + n B à AB n ) and the atomization reaction for the transfer of the most stable elemental modifications A x, B y to the gas- phase according to 1/x A x à A and 1/y B y 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 D [kj/mol]

7 Computational Thermochemistry 7 H-CH H-CH H-CH H-C 6 H H-CCH H-CH 2 CHO H-COCH Table 3: Gibbs free reaction enthalpy of some reactions Reaction ΔG0 [kj/mol] H 2 + O 2 2H 2O (l) N 2 + 3H 2 2NH H 2O (g) + CO H 2 + CO 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

8 Computational Thermochemistry 8 2 Computer Experiment 6: Computational Chemical Kinetics 2.1 Background 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: ( 14) A k k 1!!" B!! 2 " C 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.

9 Computational Thermochemistry Concentration (mm) A(t) B(t) C(t) 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 k 1 and k 2. 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 data 2 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.

10 Computational Thermochemistry A B C Absorbance (OD) ε (M -1 cm -1 ) Wavelength (nm) wavelength (nm) time (sec) 10 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: ( 15) With the rate law: ( 16) A!! k " B d! A$ "# %& dt = ' d! "# B $ %& dt = 'k! A$ "# %& We are interested in the time courses! A$ t "# %& ( ),! B$ t "# %& ( ). Since matter is conserved there is a conservation law, which states that: ( 17) And thus: ( 18)! A $ t "# %& ( )+! B$ t "# %&! B "# $ t %& ( ) = const ( ) = const ' A$ ( )! "# %& t 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:

11 Computational Thermochemistry 11 ( 19)! A $ t "# %& ( ) = A 0 e 'kt where A0 is the initial concentration of [A]. Thus! B "# irreversible bimolecular reaction of the form: ( 20) The solution is: ( 21) unless A 0 = B 0, then: ( 22) A + B k!!" C + D ( ) ( ) ( 1 kt = ln B " A% t + 0 #$ &' A 0!B 0 A " 0 B % )* t #$ &',-! A $ t "# %& ( ) = 1 kt + 1 A 0 ( ) = A 0 ( 1'e 'kt ). For an 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: C 1 ( t i+1 )!C 1 ( t i ) = f t i+1!t 1 C 1 ( t i ),...,C N t i i ( 23) ( ( )) $ t %& 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.

12 Computational Thermochemistry 12 C N ( t i+1 )!C N ( t i ) = f t i+1!t N C 1 ( t i ),...,C N t i i ( 24) ( ( )) Which can be solved for the unknown C X ( t i+1 ) (X=1 N): C X ( t i+1 ) =C X ( t i )+ t i+1!t i ( ( )) ( 25) ( )f X C 1 ( t i ),...,C N t i 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 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: ( 26) " k = Aexp! E % a # $ RT &' Where A is the pre- exponential factor, R is the gas constant, T the temperature and the key quantity is the activation energy E a. The larger E a, 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.

13 Computational Thermochemistry 13 ( 27) A + B! # k 1 %## # $ AB k "1 { } * k * ## $ C + D If one assumes the pseudo- steady state for { AB} *, one comes to the conclusion that 5 one can write the rate constant for the bimolecular decay as: ( 28) k = K * k * where K * = k 1 / k!1 (the equilibrium constant for the TS) and quantum mechanics leads to the conclusion that: ( 29) k * =! k B T h Where k B = 1.38x10!38 J / 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: ( 31) k = k B T h # & "G* exp! $ % RT '( = k T B h # * "H & # * exp "S &! exp $ % RT '( $ % R '( 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.

14 Computational Thermochemistry 14 A = k B T h " *!S % exp # $ R &' ( 32) E a =!H * ( 33) The quantity!h * is the one that one wants to calculate while!s * is slightly more difficult to predict 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: Energy (kcal/mol) H 1 -H 2 Distance (A) H 2 -H 3 Distance (A) H 2 -H 3 Distance (Angström) H 1 -H 2 Distance (Angström) Figure 3: Two- dimensional potential energy surface for the system H- H- H in a linear arrangement. A transition state is observed around the H 1 - H 2 and H 2 - H 3 distances being both 1.0 Angström.

15 Computational Thermochemistry 15 Once such a point was found, the electronic energy difference E TS!E educt is known and is the most important contributor to the activation enthalpy; The vibrational contribution to the activation enthalpy is given by: *!H vib ( 34) = R # & k B $ % exp ("h! TS n / k B T )"1'( 3N "6 educt # & h! "R n 2 ) 1 + 2k n=0 B $ % exp ("h! educt n / k B T )"1'( 3N "7 TS h! n ) n=0 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)!H vib 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!h vib 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 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 expression 7 ( 35)!( T ) = ! h" TS $ 24 "# k B T %& 7 Wigner, EP Z. Phys. Chem. Abt B 1932, 19, 203

16 Computational Thermochemistry 16 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 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.

17 Computational Thermochemistry 17 The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. 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: ( 36) k # H = exp! "H % k D $ % * ( H )! "H * D Where H and D have been written for the two isotopes since hydrogen to k B T ( ) deuterium is the most frequently studied isotope effect. The difference!h * ( H )"!H * ( D) is entirely determined by the different ZPE contributions to!h * vib. & '( 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.

18 Computational Thermochemistry Description of the Experiment 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 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 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.

19 Computational Thermochemistry 19 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 kj/mol. 9 9 Weiss, S and Leroi, GE, J. Chem. Phys., 1968, 48, 962