Reaction Rate Theory D E reaction coordinate + k A B AB

The Arrhenius Equation k + A B AB d [AB] r = = k [A] [B] k = v e d t - Eact / RT E Svante Arrhenius 1859 - 1927 Nobel Prize 1903 Eact + Empirical ! reaction parameter

Transition State Theory To determine the rate we must know the concentration on top of the barrier. The relative concentration between a reactant and product in a Chemical reaction is given by the Chemical Equilibrium The Chemical Equilibrium is given by the chemical potential of the reactant and the product. That we know how to calculate.

The Chemical Equilibrium The chemical potential for the reactant and the product can be determined if we know their Partition Functions Q. Here Qi is the partition function for the gas i and qi the partition function for the gas molecule i Let us assume that we know qi then

The Chemical Equilibrium If we assume an ideal gas and normalize the pressure with p0=1 bar We obtain the important result that the Equilibrium Constant K(T) is given by the Partitions Functions of the reactants and products Thus we can determine the concentration of a product on top of the barrier if we know the relevant Partion Functions

Partition Functions Obviously are Partition Functions relevant. We shall here deal with the Canonical Partition Function in which N, V, and T are fixed. Remember, that although we talk of a partition function for an individual molecule we always should keep in mind that this only applicable for a large ensample of molecules, i.e. statistics Consider a system with i energy levels with energy ei and degeneration gi Where Pi is the probability for finding the system in state i

å e = P Ludwig Boltzmann (1844-1906) Boltzmann Statistics: RT = - ¥ å / Boltzmann Statistics: The high temperature/diluted limit of Real statistical thermodynamics There is some really interesting Physics here!! S = k ln (W)

Partition Functions Why does the partition function look like this? Lets see if we can rationalize the expression: Let us consider a system of N particles, which can be distributed on i states with each the energy ei and Ni particles. It is assumed the system is very dilute. I.e. many more available states than particles. Constraint 1 Constraint 2 Requirement: The Entropy should be maximized (Ludwig Boltzmann)

Partition Functions Where we have utilized Stirling approximation: Only valid for huge N Problem: Optimize the entropy and fulfill the two constraints at the same time. USE LAGRANGE UNDERTERMINED MULTIPLIERS

Partition Functions Result: The Entropy Maximized when If we now utilize the first constraints: Which reminds us of q the partition function

Partition Functions The second constraint: We have to relate the average energy to some thermodynamical data This can be found by considering a particle confined in a box Now if we wants to perform the sum above we need to have an analytical expression for the energy in state i

Partition Functions By inserting this in the result of constraint 1 and assuming close lying states

Partition Functions Utilizing this in constraint 2 Thus I.e. temperature is just a Lagrange multiplyer

Partition Functions Since constraint 1 gave Since constraint 2 gave and the entropy is max for Thus the form of the partition function comes as a result of maximizing the entropy with 2 constraints

Translational Partition Functions As we have assumed the system to be a particle capable of moving in one dimension we have determined the one-dimensional partition function for translational motion in a box of length l Now what happens when we have several degrees of freedom? If the different degrees of freedom are independent the Hamiltonian can be written as a sum of Hamiltonians for each degree of freedom Htot=H1+H2+…. Discuss the validity of this: When does this not work? Give examples

Translational Partition Functions If the hamiltonian can be written as a sum the different coordinates are indrependant and Thus for translational motion in 3. Dimensions. qtrans3D = qtransx qtransy qtransz=

Partition Functions It is now possible to understand we the Maxwell-Boltzman distribution comes from

Average: Maxwell-Boltzmann distribution of velocities 500 – 1500 m/s at 300 K

Partition Functions Similarly can we separate the internal motions of a molecule in Part involving vibrations, rotation and nuclei motion, and electronic motion i.e. for a molecule we have Now we create a system of many molecules N that are in principle independent and as they are indistinguishable we get an overall partition function Q What if they were distinguishable ???

Partition Functions What was the advantage of having the Partition Function?

Partition Functions Similarly can we separate the internal motions of a molecule in Part involving vibrations, rotation and nuclei motion, and electronic motion i.e. for a mulecule we have Now we create a system of many molecules N that are in principle independent and as they are indistinguishable we get an overall partition function Q

The Vibrational Partition Function Consider a harmonic potential If there are several normal modes:

The Rotational (Nuclear) Partition Function Notice: Is not valid for H2 WHY? TRH2=85K, TRCO=3K

The Rotational (Nuclear) Partition Function Molecular symmetry  Types of molecules C1, Ci, and Cs 1 CO, CHFClBr, meso-tartraric acid, and CH3OH C2, C2v, and C2h 2 H2, H2O2, H2O, and trans-dichloroethylene C3v and C3h 3 NH3, and planar B(OH)3 The Symmetri factor: This has strong impact on the rotational energy levels. Results in fx Ortho- and para-hydrogen For a non-linear molecule:

Effect of bosons and fermions If two fermions (half intergral spin) are interchanges the total wave function must be anti symmetric i.e. change sign. Consider Hydrogen each nuclei spin is I=1/2 From two spin particles we can form 2 nuclear wave function: and which are (I+1)(2I+1)=3 and I(I+1)=1 degenerate respectively Since the rotation wave function has the symmetry is it easily seen that if the nuclear function is even must j be odd and visa versa

Ortho and Para Hydrogen This means that our hydrogen comes in two forms: Ortho Hydrogen Which has odd J and Para Hydrogen which has even J incl. 0 Notice there is 3 times as much Ortho than Para, but Para has the lowest energy a low temperature. If liquid Hydrogen should ever be a fuel we shall see advertisements Hydroprod Inc. Hydroprod Inc. Buy your liq. H2 here!!! Absolute Ortho free Hydrogen for longer mileages

Liquid Hydrogen This has severe consequences for manufacturing Liq H2 !! The ortho-para exchange is slow but will eventually happen so if we have made liq. hydrogen without this exchange being in equilibrium we have build a heating source into our liq. H2 as ¾ of the H2 will End in J=1 instead of 0. i.e. 11% loss due to the internal conversion of Ortho into Para hydrogen

The Electronic Partition Function Does usually not contribute exceptions are NO and fx. H atoms which will be twice degenerate due to spin What about He, Ne, Ar etc??

Partition Functions Summary

Partition Functions Example Knowing the degrees of internal coordinates and their energy distribution calculate the amount of molecules dissociated into atoms a different temperatures. T(K) KH2(T) pH/p0 KN2(T) pN/p0 KO2(T) pO/p0 298 5.81*10-72 2.41*10-36 6.35*10-160 2.52*10-80 6.13*10-81 7.83*10-41 1000 5.24*10-18   2.29 *10-9 2.55*10-43 5.05*10-22 4.12*10-19 6.42*10-10 2000 3.13*10-6 1.76*10-3 2.23*10-18 1.80*10-9 1.22*10-5 3.49*10-3 3000 1.77*10-3 1.72*10-1 1.01*10-9 3.18*10-5 5.04*10-1 5.01*10-1   We see why we cannot make ammonia in the gas phase but O radicals may make NO at elevated temperatures