Materials Chemistry II - Kinetics Lecture 1 PDF

Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz  Statistics, combinatorics, and probability 3 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Throwing the dice What is the p...

Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz  Statistics, combinatorics, and probability 3 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Throwing the dice What is the probability of throwing “6” ? 1 The answer is easy … = 16.7 % 6 How did we calculated ? The probability is given by the ratio of the success cases divided by the number of all possibilities. Number of all possibilities 6 {1, 2, 3, 4, 5, and 6} Number of the success cases 1 4 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Throwing the dice - 2 dice What is the probability of a certain number of points on 2 dice together? 2 1x 2.8 % 3 2x 5.6 % 2 3 4 5 6 7 4 3x 8.3 % 5 4x 11.1 % 3 4 5 6 7 8 6 5x 13.9 % 7 6x 16.7 % 4 5 6 7 8 9 8 5x 13.9 % 9 4x 11.1 % 5 6 7 8 9 10 10 3x 8.3 % 6 7 8 9 10 11 11 2x 5.6 % 12 1x 2.8 % 7 8 9 10 11 12 36 x 5 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 What is the probability to become a millionaire? Lottery - 6 from 49 disclaimer: No responsibility can be taken for hazardous behavior of the attendees of the MCh 2 lecture. 6 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Lotto 6 from 49 What is the probability of getting all 6 numbers right? Is the order of the numbers important? {8, 25, 32, 11, 49, and 35} {11, 8, 27, 32, 35, and 49} No, it is not. Then, all possibilities is the number of all combinations of 6 from 49. 𝑛𝑛 𝑛𝑛 𝑛𝑛! n – the number of elements in a set 𝐶𝐶 𝑘𝑘 = = 𝑘𝑘 𝑛𝑛 − 𝑘𝑘 ! 𝑘𝑘! k – the number of elements in a subset 49 49 49 ! 49 ! 49 𝑥𝑥 48 𝑥𝑥 47 𝑥𝑥 46 𝑥𝑥 45 𝑥𝑥 44 𝑥𝑥 43! 𝐶𝐶 6 = = = = 6 49 − 6 ! 6! 43! 6! 43! 𝑥𝑥 6 𝑥𝑥 5 𝑥𝑥 4 𝑥𝑥 3 𝑥𝑥 2 𝑥𝑥 1 𝐶𝐶 649 = 13 983 816 1 The probability to get all 6 numbers right is = 0.000 007 15 % 13 983 816 7 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Olympic games 8 runners compete in the 100 m sprint final. How many possibilities are there to win the gold, silver, and bronze medals? Is the order of the medal winners important ? Yes, it is. Then, all possibilities are the number of all variations / k-permutations of 3 from 8. 𝑛𝑛! n – the number of elements in a set 𝑃𝑃 𝑛𝑛, 𝑘𝑘 = 𝑛𝑛 − 𝑘𝑘 ! k – the number of elements in a subset 8! 8 ! 8 𝑥𝑥 7 𝑥𝑥 6 𝑥𝑥 5! 𝑃𝑃 8, 3 = = = = 8 𝑥𝑥 7 𝑥𝑥 6 8 − 3 ! 5! 5! 𝑃𝑃 8, 3 = 336 8 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Olympic games 8 runners compete in the 100 m sprint final. How many possibilities are there for the runners to reach the finish line? Is the order important ? Yes, it is. Then, all possibilities are the number of all permutations of 8. 𝑃𝑃 𝑛𝑛 = 𝑛𝑛! n – the number of elements in a set 𝑃𝑃 8 = 8! = 8 𝑥𝑥 7 𝑥𝑥 6 𝑥𝑥 5 𝑥𝑥 4 𝑥𝑥 3 𝑥𝑥 2 𝑥𝑥 1 𝑃𝑃 8 = 40 320 9 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 MCh 2 exam 69 students write the MCh 2 exam. How many possibilities are there to order to students based on the achieved number of points ? Is the order important ? Yes, it is. Can more than one student reach the same number of points ? Yes, it can. Then, all possibilities are the number of all permutations with repetition of 69. 𝑛𝑛! n – the number of elements in a set 𝑃𝑃 𝑛𝑛, 𝑘𝑘𝑖𝑖 = 𝑘𝑘1 ! 𝑘𝑘2 ! 𝑘𝑘3 ! … 𝑘𝑘𝑚𝑚 ! ki – the number of elements in a subset 69! 𝑃𝑃 69, 3, 4 = 3! 4! 𝑃𝑃 69, 3, 4 = 1.188 𝑥𝑥 1096 10 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Topics and questions 1) Statistics and average values 2) What is a system? What is a configuration of a system? What is weight of a configuration? What is a statistical ensemble? 3) Large numbers and Stirling’s approximation 4) Method of undetermined multipliers (Lagrange) 5) Boltzmann distribution Partition functions - q and Q 11 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Statistics Statistics and its methods can help us to discuss the behavior of atoms and molecules since there are just too many of them. Avogadro’s number 6.023 x 1023 molecules mol-1 Statistical thermodynamics represents a link between the molecular properties and bulk thermodynamic properties. 12 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Average values For example, The pressure of a gas depends on average force exerted by its molecules. There is no need to specify, the molecules’ velocities and directions. slow fast different angle Any macroscopic physical quantity can be expressed as an average value over all molecules. 13 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25  What is a system? What is a configuration of a system? What is weight of a configuration? What is a statistical ensemble? 14 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 What is a system ? A system can be anything a star a table a student a stone a molecule We will focus on molecules, start with a few of molecules (1, 2, 3, …), and generalize the results for many molecules 1023 15 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 What is a configuration of a system ? A configuration of a system is a state this system can be in. Any individual molecule can be in states with various energies 𝜀𝜀0 , 𝜀𝜀1 , 𝜀𝜀2 , 𝜀𝜀3 , … The lowest energy level ε0 is often taken as 0 𝜀𝜀0 = 0 and all other energies will be measured relative to ε0 In order to calculate the actual internal energy of a system U, we may have to add a constant U0 different configurations of a system (with different energies) energy states of a molecule 𝐸𝐸 = 𝜀𝜀0 𝐸𝐸 = 𝜀𝜀1 𝐸𝐸 = 𝜀𝜀4 𝜀𝜀4 𝜀𝜀4 𝜀𝜀4 𝜀𝜀4 𝜀𝜀3 𝜀𝜀3 𝜀𝜀3 𝜀𝜀3 𝜀𝜀2 𝜀𝜀2 𝜀𝜀2 𝜀𝜀2 𝜀𝜀1 𝜀𝜀1 𝜀𝜀1 𝜀𝜀1 𝜀𝜀0 = 0 𝜀𝜀0 𝜀𝜀0 𝜀𝜀0 16 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 2 molecules and 2-level energy system Let’s discuss a system with 2 molecules (distinguishable particles) assume 2 energy levels 𝜀𝜀0 𝜀𝜀1 sketch all possibilities 0 𝜀𝜀 How many configurations do we have for different system energies E ? 𝜀𝜀 𝐸𝐸 = 0 𝑊𝑊 = 1 0 𝜀𝜀 𝐸𝐸 = 𝜀𝜀 𝑊𝑊 = 2 W is weight of a configuration 0 𝜀𝜀 𝐸𝐸 = 2𝜀𝜀 𝑊𝑊 = 1 0 17 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 3 molecules and 3-level energy system Let’s discuss a system with 3 molecules (distinguishable particles) assume 3 energy levels 𝜀𝜀 2𝜀𝜀 0 𝐸𝐸 = 0 𝑊𝑊 = 1 𝐸𝐸 = 𝜀𝜀 𝑊𝑊 = 3 𝐸𝐸 = 2𝜀𝜀 𝑊𝑊 = 3 𝑊𝑊 = 3 𝑊𝑊 = 6 𝑊𝑊 = 3𝑥𝑥𝑥𝑥𝑥𝑥 𝐸𝐸 = 3𝜀𝜀 𝑊𝑊 = 1 𝑊𝑊 = 3! 𝐸𝐸 = 4𝜀𝜀 𝑊𝑊 = 3 𝑊𝑊 = 3 3! 𝐸𝐸 = 5𝜀𝜀 𝑊𝑊 = 3 𝑊𝑊 = 2! 𝐸𝐸 = 6𝜀𝜀 𝑊𝑊 = 1 18 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 3 molecules and 3-level energy system Let’s discuss a system with 3 molecules (distinguishable particles) assume 3 energy levels 𝜀𝜀 2𝜀𝜀 0 𝐸𝐸 = 0 𝑊𝑊 = 1 𝐸𝐸 = 𝜀𝜀 𝑊𝑊 = 3 𝐸𝐸 = 2𝜀𝜀 𝑊𝑊 = 3 𝑊𝑊 = 3 𝑊𝑊 = 6 𝑊𝑊 = 3𝑥𝑥𝑥𝑥𝑥𝑥 𝐸𝐸 = 3𝜀𝜀 𝑊𝑊 = 1 𝑊𝑊 = 3! N – number of molecules ni – population of the i-th energy levels 3! 𝑊𝑊 = 2! (For a given energy), 𝑁𝑁! a system will be found in the configuration with the largest weight 𝑊𝑊 = 𝑛𝑛0 ! 𝑛𝑛1 ! 𝑛𝑛2 ! … 19 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 What is a statistical ensemble ? A statistical ensemble is a collection of configurations a system can be in. microcanonical ensemble thermally isolated ensemble constant quantities N - number of particles V - volume E - energy characteristic thermodynamic state function S - entropy canonical ensemble an ensemble in contact with a reservoir with temperature T constant quantities N - number of particles V - volume T - temperature characteristic thermodynamic state function A - Helmholtz energy 20 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 System in a dominating configuration (for a given energy E) The weight of a configuration is given by N – number of particles 𝑁𝑁! ni – population of the i-th energy level 𝑊𝑊 = εi – energy of the i-th energy level 𝑛𝑛0 ! 𝑛𝑛1 ! 𝑛𝑛2 ! … The (constant) total energy of a system is given by 𝐸𝐸 = 𝑛𝑛𝑖𝑖 𝜀𝜀𝑖𝑖 𝑖𝑖 The (constant) total number molecules of a system is given by 𝑁𝑁 = 𝑛𝑛𝑖𝑖 𝑖𝑖 We look for a distribution of populations of the molecules with 𝑛𝑛0 , 𝑛𝑛1 , 𝑛𝑛2 , 𝑛𝑛3 ,.. with maximum of the weight W ! Weight W can be a large number 𝑑𝑑𝑑𝑑 = 0 Populations ni are not independent variables 21 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Large numbers numbers normal 1; 1000 large 1023 Avogadro’s number 6.023 x 1023 molecules mol-1 𝑛𝑛! 𝑛𝑛! = 𝑛𝑛 𝑛𝑛 − 1 𝑛𝑛 − 2 𝑛𝑛 − 3 … 1 factorial (Fakultät) 100! 9.3 𝑥𝑥 10157 𝑛𝑛 - natural number 23 huge 1010 It is practical to calculate with 𝑙𝑙𝑙𝑙 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 large 1023 → normal 𝑙𝑙𝑙𝑙𝑙𝑙 1023 = 23 𝑙𝑙𝑙𝑙𝑙𝑙 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 23 10 23 = 1023 huge 1010 → large 𝑙𝑙𝑙𝑙𝑙𝑙 10 22 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Stirling’s approximation 𝑙𝑙𝑙𝑙 𝑛𝑛! ≈ 𝑛𝑛 𝑙𝑙𝑙𝑙 𝑛𝑛 − 𝑛𝑛 The weight of a configuration is given by 𝑁𝑁! 𝑊𝑊 = 𝑛𝑛0 ! 𝑛𝑛1 ! 𝑛𝑛2 ! … 𝑁𝑁! 𝑙𝑙𝑙𝑙 𝑊𝑊 = 𝑙𝑙𝑙𝑙 = 𝑙𝑙𝑙𝑙 𝑁𝑁! − 𝑙𝑙𝑙𝑙 𝑛𝑛0 ! 𝑛𝑛1 ! 𝑛𝑛2 ! … 𝑛𝑛0 ! 𝑛𝑛1 ! 𝑛𝑛2 ! … = 𝑙𝑙𝑙𝑙 𝑁𝑁! − 𝑙𝑙𝑙𝑙 𝑛𝑛0 ! + 𝑙𝑙𝑙𝑙 𝑛𝑛1 ! + 𝑙𝑙𝑙𝑙 𝑛𝑛2 ! + ⋯ = 𝑙𝑙𝑙𝑙 𝑁𝑁! − 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 ! 𝑖𝑖 ≈ 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 − 𝑁𝑁 − 𝑛𝑛𝑖𝑖 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 − 𝑛𝑛𝑖𝑖 𝑖𝑖 ≈ 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 − 𝑁𝑁 − 𝑛𝑛𝑖𝑖 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 + 𝑛𝑛𝑖𝑖 𝑁𝑁 = 𝑛𝑛𝑖𝑖 𝑖𝑖 𝑖𝑖 𝑖𝑖 𝑙𝑙𝑙𝑙 𝑊𝑊 ≈ 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 − 𝑛𝑛𝑖𝑖 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 𝑖𝑖 23 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Method of undetermined multipliers (Lagrange) We have a weight function 𝑙𝑙𝑙𝑙 𝑊𝑊 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 and look for a maximum 𝑑𝑑 𝑙𝑙𝑙𝑙 𝑊𝑊 = 𝑑𝑑𝑛𝑛𝑖𝑖 = 0 𝜕𝜕 𝑛𝑛𝑖𝑖 𝑖𝑖 with constrains constant number of molecules 𝑁𝑁 = 𝑛𝑛𝑖𝑖 0= 𝑑𝑑𝑛𝑛𝑖𝑖 𝑖𝑖 𝑖𝑖 constant energy 𝐸𝐸 = 𝑛𝑛𝑖𝑖 𝜀𝜀𝑖𝑖 0 = 𝜀𝜀𝑖𝑖 𝑑𝑑𝑛𝑛𝑖𝑖 𝑖𝑖 𝑖𝑖 Build another function f(ni) with a help of Lagrange multipliers 𝛼𝛼 and − 𝛽𝛽 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 𝑑𝑑 𝑓𝑓 𝑛𝑛𝑖𝑖 = 𝑑𝑑𝑛𝑛𝑖𝑖 + 𝛼𝛼 𝑑𝑑𝑛𝑛𝑖𝑖 − 𝛽𝛽 𝜀𝜀𝑖𝑖 𝑑𝑑𝑛𝑛𝑖𝑖 = 0 𝜕𝜕 𝑛𝑛𝑖𝑖 𝑖𝑖 𝑖𝑖 𝑖𝑖 24 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Method of undetermined multipliers (Lagrange) We have a weight function 𝑙𝑙𝑙𝑙 𝑊𝑊 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 and look for a maximum 𝑑𝑑 𝑙𝑙𝑙𝑙 𝑊𝑊 = 𝑑𝑑𝑛𝑛𝑖𝑖 = 0 𝜕𝜕 𝑛𝑛𝑖𝑖 𝑖𝑖 with constrains constant number of molecules 𝑁𝑁 = 𝑛𝑛𝑖𝑖 0= 𝑑𝑑𝑛𝑛𝑖𝑖 𝑖𝑖 𝑖𝑖 constant energy 𝐸𝐸 = 𝑛𝑛𝑖𝑖 𝜀𝜀𝑖𝑖 0 = 𝜀𝜀𝑖𝑖 𝑑𝑑𝑛𝑛𝑖𝑖 𝑖𝑖 𝑖𝑖 Build another function f(ni) with a help of Lagrange multipliers 𝛼𝛼 and − 𝛽𝛽 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 𝑑𝑑 𝑓𝑓 𝑛𝑛𝑖𝑖 = 𝑑𝑑𝑛𝑛𝑖𝑖 + 𝛼𝛼 𝑑𝑑𝑛𝑛𝑖𝑖 − 𝛽𝛽 𝜀𝜀𝑖𝑖 𝑑𝑑𝑛𝑛𝑖𝑖 = 0 𝜕𝜕 𝑛𝑛𝑖𝑖 𝑖𝑖 𝑖𝑖 𝑖𝑖 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 𝑑𝑑 𝑓𝑓 𝑛𝑛𝑖𝑖 = + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 𝑑𝑑𝑛𝑛𝑖𝑖 = 0 𝜕𝜕 𝑛𝑛𝑖𝑖 𝑖𝑖 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 =0 for all 𝑑𝑑𝑛𝑛𝑖𝑖 𝜕𝜕 𝑛𝑛𝑖𝑖 25 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Method of undetermined multipliers (Lagrange) … continues 1 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 we want to solve a set of equations + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 = 0 for all 𝑑𝑑𝑛𝑛𝑖𝑖 𝜕𝜕 𝑛𝑛𝑖𝑖 according to Stirling’s approximation 𝑙𝑙𝑙𝑙 𝑊𝑊 ≈ 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 − 𝑛𝑛𝑖𝑖 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 𝑖𝑖 𝜕𝜕𝜕𝜕𝜕𝜕 𝑊𝑊 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕 𝑛𝑛𝑗𝑗 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 then = − 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕𝜕𝜕 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑁𝑁 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 𝑁𝑁 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝜕𝜕 1 𝜕𝜕𝜕𝜕 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 𝑁𝑁 𝜕𝜕𝑛𝑛𝑖𝑖 𝑁𝑁 𝜕𝜕𝑛𝑛𝑖𝑖 26 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Method of undetermined multipliers (Lagrange) … continues 1 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 we want to solve a set of equations + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 = 0 for all 𝑑𝑑𝑛𝑛𝑖𝑖 𝜕𝜕 𝑛𝑛𝑖𝑖 according to Stirling’s approximation 𝑙𝑙𝑙𝑙 𝑊𝑊 ≈ 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 − 𝑛𝑛𝑖𝑖 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 𝑖𝑖 𝜕𝜕𝜕𝜕𝜕𝜕 𝑊𝑊 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕 𝑛𝑛𝑗𝑗 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 then = − 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕𝜕𝜕 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑁𝑁 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 𝑁𝑁 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 = 1 𝑙𝑙𝑙𝑙 𝑁𝑁 + 1 𝜕𝜕𝜕𝜕 𝜕𝜕 𝑛𝑛0 + 𝑛𝑛1 + 𝑛𝑛2 +.. = =1 for all i 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 27 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Method of undetermined multipliers (Lagrange) … continues 1 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 we want to solve a set of equations + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 = 0 for all 𝑑𝑑𝑛𝑛𝑖𝑖 𝜕𝜕 𝑛𝑛𝑖𝑖 according to Stirling’s approximation 𝑙𝑙𝑙𝑙 𝑊𝑊 ≈ 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 − 𝑛𝑛𝑖𝑖 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 𝑖𝑖 𝜕𝜕𝜕𝜕𝜕𝜕 𝑊𝑊 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕 𝑛𝑛𝑗𝑗 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 then = − 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕𝜕𝜕 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑁𝑁 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 𝑁𝑁 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 1 𝜕𝜕 𝑛𝑛𝑗𝑗 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑗𝑗 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 = 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 + 𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 𝑗𝑗 𝜕𝜕𝑛𝑛𝑗𝑗 1 𝜕𝜕𝑛𝑛𝑗𝑗 = 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 + 𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑖𝑖 𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 28 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Method of undetermined multipliers (Lagrange) … continues 1 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 we want to solve a set of equations + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 = 0 for all 𝑑𝑑𝑛𝑛𝑖𝑖 𝜕𝜕 𝑛𝑛𝑖𝑖 according to Stirling’s approximation 𝑙𝑙𝑙𝑙 𝑊𝑊 ≈ 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 − 𝑛𝑛𝑖𝑖 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 𝑖𝑖 𝜕𝜕𝜕𝜕𝜕𝜕 𝑊𝑊 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕 𝑛𝑛𝑗𝑗 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 then = − 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕𝜕𝜕 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑁𝑁 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 𝑁𝑁 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 1 𝜕𝜕 𝑛𝑛𝑗𝑗 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑗𝑗 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 = 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 + 𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 𝑗𝑗 𝜕𝜕𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑗𝑗 = 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 + = 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 + 1 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 𝑗𝑗 29 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Method of undetermined multipliers (Lagrange) … continues 1 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 we want to solve a set of equations + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 = 0 for all 𝑑𝑑𝑛𝑛𝑖𝑖 𝜕𝜕 𝑛𝑛𝑖𝑖 according to Stirling’s approximation 𝑙𝑙𝑙𝑙 𝑊𝑊 ≈ 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 − 𝑛𝑛𝑖𝑖 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 𝑖𝑖 𝜕𝜕𝜕𝜕𝜕𝜕 𝑊𝑊 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕 𝑛𝑛𝑗𝑗 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 then = − 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕𝜕𝜕 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑁𝑁 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 𝑁𝑁 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 1 𝜕𝜕 𝑛𝑛𝑗𝑗 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑗𝑗 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 = 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 + 𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 𝑗𝑗 𝜕𝜕𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑗𝑗 = 1 𝑓𝑓𝑓𝑓𝑓𝑓 𝑛𝑛𝑗𝑗 = 𝑛𝑛𝑖𝑖 = 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 + 1 = 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 + 1 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 30 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Method of undetermined multipliers (Lagrange) … continues 1 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 we want to solve a set of equations + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 = 0 for all 𝑑𝑑𝑛𝑛𝑖𝑖 𝜕𝜕 𝑛𝑛𝑖𝑖 according to Stirling’s approximation 𝑙𝑙𝑙𝑙 𝑊𝑊 ≈ 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 − 𝑛𝑛𝑖𝑖 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 𝑖𝑖 𝜕𝜕𝜕𝜕𝜕𝜕 𝑊𝑊 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕 𝑛𝑛𝑗𝑗 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 then = − 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕𝜕𝜕 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑁𝑁 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 𝑁𝑁 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 1 𝜕𝜕 𝑛𝑛𝑗𝑗 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑗𝑗 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 = 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 + 𝑛𝑛𝑗𝑗 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 𝑗𝑗 𝜕𝜕𝑛𝑛𝑗𝑗 = 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 + 1 = 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 + 1 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 31 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Method of undetermined multipliers (Lagrange) … continues 2 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 we want to solve a set of equations + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 = 0 for all 𝑑𝑑𝑛𝑛𝑖𝑖 𝜕𝜕 𝑛𝑛𝑖𝑖 according to Stirling’s approximation 𝑙𝑙𝑙𝑙 𝑊𝑊 ≈ 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 − 𝑛𝑛𝑖𝑖 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 𝑖𝑖 𝜕𝜕𝜕𝜕𝜕𝜕 𝑊𝑊 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕 𝑛𝑛𝑗𝑗 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 then = − 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 1 − 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 + 1 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 1 = 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 + 1 32 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Method of undetermined multipliers (Lagrange) … continues 2 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 we want to solve a set of equations + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 = 0 for all 𝑑𝑑𝑛𝑛𝑖𝑖 𝜕𝜕 𝑛𝑛𝑖𝑖 according to Stirling’s approximation 𝑙𝑙𝑙𝑙 𝑊𝑊 ≈ 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 − 𝑛𝑛𝑖𝑖 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 𝑖𝑖 𝜕𝜕𝜕𝜕𝜕𝜕 𝑊𝑊 𝜕𝜕 𝑁𝑁 𝑙𝑙𝑙𝑙 𝑁𝑁 𝜕𝜕 𝑛𝑛𝑗𝑗 𝑙𝑙𝑙𝑙 𝑛𝑛𝑗𝑗 then = − 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝜕𝜕𝑛𝑛𝑖𝑖 𝑗𝑗 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 1 − 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 + 1 = 𝑙𝑙𝑙𝑙 𝑁𝑁 + 1 − 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 − 1 = 𝑙𝑙𝑙𝑙 𝑁𝑁 − 𝑙𝑙𝑙𝑙 𝑛𝑛𝑖𝑖 𝑛𝑛𝑖𝑖 𝑛𝑛𝑖𝑖 = − 𝑙𝑙𝑙𝑙 −𝑙𝑙𝑙𝑙 + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 =0 𝑁𝑁 𝑁𝑁 𝑛𝑛𝑖𝑖 𝑙𝑙𝑙𝑙 = 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 𝑁𝑁 𝑛𝑛𝑖𝑖 = 𝑒𝑒 𝛼𝛼−𝛽𝛽𝜀𝜀𝑖𝑖 𝑁𝑁 𝑛𝑛𝑖𝑖 = 𝑒𝑒 𝛼𝛼 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 𝑁𝑁 33 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Method of undetermined multipliers (Lagrange) … end 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 we want to solve a set of equations + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 = 0 for all 𝑑𝑑𝑛𝑛𝑖𝑖 𝜕𝜕 𝑛𝑛𝑖𝑖 𝑛𝑛𝑖𝑖 = 𝑒𝑒 𝛼𝛼 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 N – number of particles 𝑁𝑁 ni – population of the i-th energy level εi – energy of the i-th energy level 𝑛𝑛𝑖𝑖 = 𝑒𝑒 𝛼𝛼 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 𝑁𝑁 34 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Method of undetermined multipliers (Lagrange) … end 𝜕𝜕 𝑙𝑙𝑙𝑙 𝑊𝑊 we want to solve a set of equations + 𝛼𝛼 − 𝛽𝛽𝜀𝜀𝑖𝑖 = 0 for all 𝑑𝑑𝑛𝑛𝑖𝑖 𝜕𝜕 𝑛𝑛𝑖𝑖 𝑛𝑛𝑖𝑖 = 𝑒𝑒 𝛼𝛼 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 N – number of particles 𝑁𝑁 ni – population of the i-th energy level εi – energy of the i-th energy level 𝑁𝑁 = 𝑛𝑛𝑗𝑗 = 𝑁𝑁𝑒𝑒 𝛼𝛼 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑗𝑗 = 𝑁𝑁𝑒𝑒 𝛼𝛼 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑗𝑗 𝑗𝑗 𝑗𝑗 𝑗𝑗 1 = 𝑒𝑒 𝛼𝛼 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑗𝑗 𝑗𝑗 1 −𝛽𝛽𝜀𝜀𝑗𝑗 = 𝑒𝑒 𝛼𝛼 ∑𝑗𝑗 𝑒𝑒 𝑛𝑛𝑖𝑖 𝑝𝑝𝑖𝑖 = = 𝑒𝑒 𝛼𝛼 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 pi – probability of the occupation of the i-th energy level 𝑁𝑁 pi – Boltzmann distribution 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 𝑝𝑝𝑖𝑖 = ∑𝑗𝑗 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑗𝑗 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 𝑞𝑞 = 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑗𝑗 𝑝𝑝𝑖𝑖 = molecular partition function (Zustandssumme) 𝑞𝑞 𝑗𝑗 35 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 System in a dominating configuration (for a given energy E) The weight of a configuration is given by N – number of particles 𝑁𝑁! ni – population of the i-th energy level 𝑊𝑊 = εi – energy of the i-th energy level 𝑛𝑛0 ! 𝑛𝑛1 ! 𝑛𝑛2 ! … The (constant) total energy of a system is given by 𝐸𝐸 = 𝑛𝑛𝑖𝑖 𝜀𝜀𝑖𝑖 𝑖𝑖 The (constant) total number molecules of a system is given by 𝑁𝑁 = 𝑛𝑛𝑖𝑖 𝑖𝑖 We look for a distribution of populations of the molecules with 𝑛𝑛0 , 𝑛𝑛1 , 𝑛𝑛2 , 𝑛𝑛3 ,.. with maximum of the weight W ! Weight W can be a large number 𝑑𝑑𝑑𝑑 = 0 Populations ni are not independent variables 36 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 System in a dominating configuration (for a given energy E) The weight of a configuration is given by N – number of particles 𝑁𝑁! ni – population of the i-th energy level 𝑊𝑊 = εi – energy of the i-th energy level 𝑛𝑛0 ! 𝑛𝑛1 ! 𝑛𝑛2 ! … The (constant) total energy of a system is given by 𝐸𝐸 = 𝑛𝑛𝑖𝑖 𝜀𝜀𝑖𝑖 𝑖𝑖 The (constant) total number molecules of a system is given by 𝑁𝑁 = 𝑛𝑛𝑖𝑖 𝑖𝑖 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 −𝛽𝛽𝜀𝜀𝑗𝑗 1 (*) (* 3rd lecture) Boltzmann distribution 𝑝𝑝𝑖𝑖 = 𝑞𝑞 = 𝑒𝑒 𝛽𝛽 = 𝑞𝑞 𝑗𝑗 𝑘𝑘𝑘𝑘 q - molecular partition function k – Boltzmann constant 1.38 x 10-23 J K-1 T – temperature (K) 37 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Boltzmann distribution for a 2-level system 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 1 𝑝𝑝𝑖𝑖 = 𝑞𝑞 = 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 𝛽𝛽 = 𝑞𝑞 = 𝑒𝑒 −𝛽𝛽𝛽 + 𝑒𝑒 −𝛽𝛽𝜀𝜀 = 1 + 𝑒𝑒 −𝛽𝛽𝜀𝜀 𝑞𝑞 𝑘𝑘𝑘𝑘 𝑖𝑖 𝑒𝑒 −𝛽𝛽0 1 𝜀𝜀0 = 0 𝑝𝑝0 = = 𝑞𝑞 1 + 𝑒𝑒 −𝛽𝛽𝛽𝛽 𝑒𝑒 −𝛽𝛽𝜀𝜀 𝑒𝑒 −𝛽𝛽𝛽𝛽 𝜀𝜀1 = 𝜀𝜀 𝑝𝑝1 = = 𝑞𝑞 1 + 𝑒𝑒 −𝛽𝛽𝛽𝛽 at T → 0, only state ε0 is possible at T → ∞, both states are equally possible 50% (Please pay attention: not all molecules are in state ε1) 38 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Molecular (microcanonical) partition function q (in general) 1 𝑞𝑞 = 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 𝛽𝛽 = summation over the molecules within the system 𝑖𝑖 𝑘𝑘𝑘𝑘 𝜀𝜀2 𝑔𝑔2 𝑞𝑞 = 𝑔𝑔𝑗𝑗 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑗𝑗 gj – degeneracy of the j-th energy level summation over the energy levels 𝜀𝜀1 𝑔𝑔1 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝜀𝜀0 𝑔𝑔0 How can the partition function be interpreted? 1 T → 0, 𝛽𝛽 = lim 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑗𝑗 → 0, 𝜀𝜀0 = → ∞, 𝛽𝛽→∞ 0 ≡0 lim 𝑞𝑞 = 𝑔𝑔0 q is equal to the degeneracy 𝑘𝑘𝑘𝑘 𝑇𝑇→0 𝑘𝑘𝑘𝑘 except for ε0 0 of the ground state 1 lim 𝑞𝑞 = ∞ q is equal to the number T → ∞, 𝛽𝛽 = → 0, lim 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑗𝑗 → 1, 1 → ∞ 𝑘𝑘𝑘𝑘 𝛽𝛽→0 𝑗𝑗 𝑇𝑇→∞ of energy states, up to ∞ q gives an indication of the average number of states that are thermally accessible to a molecule at the temperature T of the system 39 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Boltzmann distribution as a function of temperature 1 𝑞𝑞 = 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 𝛽𝛽 = 𝑘𝑘𝑘𝑘 𝑖𝑖 𝜀𝜀𝑖𝑖 𝜀𝜀 −𝑘𝑘𝑘𝑘 − 𝑖𝑖 𝑒𝑒 𝑛𝑛𝑖𝑖 𝑒𝑒 𝑘𝑘𝑘𝑘 𝑝𝑝𝑖𝑖 = 𝜀𝜀𝑗𝑗 = 𝜀𝜀𝑗𝑗 −𝑘𝑘𝑘𝑘 𝑁𝑁 −𝑘𝑘𝑘𝑘 ∑𝑗𝑗 𝑒𝑒 ∑𝑗𝑗 𝑒𝑒 𝜀𝜀𝑖𝑖 −𝑘𝑘𝑘𝑘 𝑒𝑒 𝑛𝑛𝑖𝑖 = 𝑁𝑁 𝜀𝜀𝑗𝑗 −𝑘𝑘𝑘𝑘 ∑𝑗𝑗 𝑒𝑒 Figure: Atkins, Physical Chemistry 40 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Canonical partition function molecular (microcanonical) partition function εi – energy of the i-th energy level 1 𝑞𝑞 = 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 𝛽𝛽 = 𝑖𝑖 𝑘𝑘𝑘𝑘 describes an microcanonical ensemble at constant energy E 𝐸𝐸 = 𝑛𝑛𝑖𝑖 𝜀𝜀𝑖𝑖 𝑖𝑖 canonical partition function 𝑄𝑄 = 𝑒𝑒 −𝛽𝛽𝐸𝐸𝑖𝑖 Ei – energy of the state i of the system 𝑖𝑖 describes an canonical ensemble at constant temperature T for independent molecules 𝐸𝐸𝑖𝑖 = 𝜀𝜀𝑖𝑖 1 + 𝜀𝜀𝑖𝑖 2 + … + 𝜀𝜀𝑖𝑖 𝑁𝑁 𝑄𝑄 = 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 1 −𝛽𝛽𝜀𝜀𝑖𝑖 2 −⋯−𝛽𝛽𝜀𝜀𝑖𝑖 𝑁𝑁 = 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 1 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 2 … 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 𝑁𝑁 𝑖𝑖 𝑖𝑖 𝑖𝑖 𝑖𝑖 𝑁𝑁 𝑄𝑄 = 𝑒𝑒 −𝛽𝛽𝜀𝜀𝑖𝑖 = 𝑞𝑞 𝑁𝑁 𝑄𝑄 = 𝑞𝑞𝑁𝑁 𝑖𝑖 41 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Distinguishable and indistinguishable molecules distinguishable chemical nature 𝐸𝐸 = 0 𝜀𝜀 𝑊𝑊 = 1 0 Ar vs. Ne 𝐸𝐸 = 𝜀𝜀 𝜀𝜀 𝑊𝑊 = 2 0 Ti in a solid state with a set of coordinates 𝐸𝐸 = 2𝜀𝜀 𝜀𝜀 0 𝑊𝑊 = 1 𝑄𝑄 = 𝑒𝑒 −𝛽𝛽(0+0) + 𝑒𝑒 −𝛽𝛽(𝜀𝜀+0) + 𝑒𝑒 −𝛽𝛽(0+𝜀𝜀) + 𝑒𝑒 −𝛽𝛽(𝜀𝜀+𝜀𝜀) 𝑄𝑄 = 1 + 2𝑒𝑒 −𝛽𝛽𝛽𝛽 + 𝑒𝑒 −2𝛽𝛽𝛽𝛽 𝑄𝑄 = 1 + 2𝑒𝑒 −𝛽𝛽𝛽𝛽 +𝑒𝑒 −2𝛽𝛽𝛽𝛽 𝜀𝜀 indistinguishable particles free to move 𝐸𝐸 = 0 0 𝑊𝑊 ′ = 1 Ar gas 𝑊𝑊 𝐸𝐸 = 𝜀𝜀 𝜀𝜀 𝑊𝑊 ′ = 1 𝑊𝑊 ′ = 0 𝑁𝑁! Ti vapor (gas phase) 𝜀𝜀 𝐸𝐸 = 2𝜀𝜀 0 𝑊𝑊 ′ = 1 𝑄𝑄 = 𝑒𝑒 −𝛽𝛽(0+0) + 𝑒𝑒 −𝛽𝛽(0+𝜀𝜀) + 𝑒𝑒 −𝛽𝛽(𝜀𝜀+𝜀𝜀) 𝑄𝑄 = 1 + 𝑒𝑒 −𝛽𝛽𝛽𝛽 + 𝑒𝑒 −2𝛽𝛽𝛽𝛽 𝑄𝑄 = 1 + 𝑒𝑒 −𝛽𝛽𝛽𝛽 + 𝑒𝑒 −2𝛽𝛽𝛽𝛽 42 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Throwing the dice - 2 dice What is the probability of a certain number of points on 2 dice together? 2 1x 2.8 % 3 2x 5.6 % 2 3 4 5 6 7 4 3x 8.3 % 5 4x 11.1 % 3 4 5 6 7 8 6 5x 13.9 % 7 6x 16.7 % 4 5 6 7 8 9 8 5x 13.9 % 9 4x 11.1 % 5 6 7 8 9 10 10 3x 8.3 % 6 7 8 9 10 11 11 2x 5.6 % 12 1x 2.8 % 7 8 9 10 11 12 36 x 43 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Throwing the dice - 2 dice What is the probability of a certain number of points on 2 dice together? 2 1x 4.8 % 3 1x 4.8 % 2 3 4 5 6 7 4 2x 9.5 % 5 2x 9.5 % 3 4 5 6 7 8 6 3x 14.3 % 7 3x 14.3 % 4 5 6 7 8 9 8 3x 14.3 % 9 2x 9.5 % 5 6 7 8 9 10 10 2x 9.5 % 6 7 8 9 10 11 11 1x 4.8 % 12 1x 4.8 % 7 8 9 10 11 12 21 x 44 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Distinguishable and indistinguishable molecules distinguishable chemical nature 𝐸𝐸 = 0 𝜀𝜀 𝑊𝑊 = 1 0 Ar vs. Ne 𝐸𝐸 = 𝜀𝜀 𝜀𝜀 𝑊𝑊 = 2 0 Ti in a solid state with a set of coordinates 𝐸𝐸 = 2𝜀𝜀 𝜀𝜀 0 𝑊𝑊 = 1 𝑄𝑄 = 𝑒𝑒 −𝛽𝛽(0+0) + 𝑒𝑒 −𝛽𝛽(𝜀𝜀+0) + 𝑒𝑒 −𝛽𝛽(0+𝜀𝜀) + 𝑒𝑒 −𝛽𝛽(𝜀𝜀+𝜀𝜀) 𝑄𝑄 = 1 + 2𝑒𝑒 −𝛽𝛽𝛽𝛽 + 𝑒𝑒 −2𝛽𝛽𝛽𝛽 𝑄𝑄 = 1 + 2𝑒𝑒 −𝛽𝛽𝛽𝛽 +𝑒𝑒 −2𝛽𝛽𝛽𝛽 𝜀𝜀 indistinguishable particles free to move 𝐸𝐸 = 0 0 𝑊𝑊 ′ = 1 Ar gas 𝑊𝑊 𝐸𝐸 = 𝜀𝜀 𝜀𝜀 𝑊𝑊 ′ = 1 𝑊𝑊 ′ = 0 𝑁𝑁! Ti vapor (gas phase) 𝜀𝜀 𝐸𝐸 = 2𝜀𝜀 0 𝑊𝑊 ′ = 1 𝑄𝑄 = 𝑒𝑒 −𝛽𝛽(0+0) + 𝑒𝑒 −𝛽𝛽(0+𝜀𝜀) + 𝑒𝑒 −𝛽𝛽(𝜀𝜀+𝜀𝜀) 𝑄𝑄 = 1 + 𝑒𝑒 −𝛽𝛽𝛽𝛽 + 𝑒𝑒 −2𝛽𝛽𝛽𝛽 𝑄𝑄 = 1 + 𝑒𝑒 −𝛽𝛽𝛽𝛽 + 𝑒𝑒 −2𝛽𝛽𝛽𝛽 45 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 The Gibbs paradox Canonical partition function describes a canonical ensemble, an ensemble at constant temperature T Ei – energy of the state i of the system 𝑄𝑄 = 𝑒𝑒 −𝛽𝛽𝐸𝐸𝑖𝑖 1 𝛽𝛽 = 𝑖𝑖 𝑘𝑘𝑘𝑘 distinguishable - different in chemical nature – Ar vs. Ne - Ti in a solid state with a set of coordinates 𝑄𝑄 = 𝑞𝑞 𝑁𝑁 indistinguishable - particles free to move – Ar gas 𝑞𝑞 𝑁𝑁 - Ti vapor (gas phase) 𝑄𝑄 = 𝑁𝑁! 46 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Harmonic oscillator Let’s assume a harmonic oscillator with infinite number of energy levels 𝜀𝜀0 = 0 ℎ 𝜀𝜀1 = ℏ𝜔𝜔 ℏ= ℎ Planck’s constant, 6.63 x 10-34 Js 2𝜋𝜋 𝜀𝜀2 = 2 ℏ𝜔𝜔 … Let’s calculate the canonical partition function 𝑄𝑄 = 𝑒𝑒 −𝛽𝛽𝐸𝐸𝑖𝑖 𝑖𝑖 𝑄𝑄 = 𝑒𝑒 −𝛽𝛽0 + 𝑒𝑒 −𝛽𝛽ℏ𝜔𝜔 + 𝑒𝑒 −2𝛽𝛽𝛽𝛽𝛽 + 𝑒𝑒 −3𝛽𝛽𝛽𝛽𝛽 + … 1 + 𝑒𝑒 −𝛽𝛽ℏ𝜔𝜔 + 𝑒𝑒 −2𝛽𝛽𝛽𝛽𝛽 + 𝑒𝑒 −3𝛽𝛽𝛽𝛽𝛽 + … geometric series 𝑎𝑎0 = 1 𝑥𝑥 = 𝑒𝑒 −𝛽𝛽𝛽𝛽𝛽 ∞ 𝑎𝑎0 1 𝑎𝑎𝑛𝑛 = 𝑄𝑄 = 1 − 𝑥𝑥 1 − 𝑒𝑒 −𝛽𝛽𝛽𝛽𝛽 𝑛𝑛=0 47 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Comprehension What is a system ? What is a configuration ? What is a statistical ensemble ? What is a microcanonical ensemble and what it its characteristic state function ? What is a canonical ensemble and what it its characteristic state function ? What are the major steps in derivation of the Boltzmann distribution ? What is the Stirling’s approximation ? What is the method of undetermined multipliers ? What is a microcanonical partition function and how it can be interpreted ? What is a canonical partition function ? What is the difference between distinguishable and indistinguishable molecules ? What is the Gibbs paradox ? 48 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Self-testing An isolated and equally spaced three-level system contains two molecules. Assume that the ground state energy is 0. The energy difference between the energy levels is ε > 0. The total energy of this system is 2ε. What configurations are possible if the particles are (a) distinguishable ? (b) indistinguishable ? 49 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Self-testing (solution) (a) distinguishable ? (b) indistinguishable ? 𝐸𝐸 = 0 𝑊𝑊 = 1 𝐸𝐸 = 0 𝑊𝑊 = 1 𝐸𝐸 = 𝜀𝜀 𝑊𝑊 = 2 𝐸𝐸 = 𝜀𝜀 𝑊𝑊 = 1 𝑊𝑊 = 2 𝑊𝑊 = 1 𝐸𝐸 = 2𝜀𝜀 𝐸𝐸 = 2𝜀𝜀 𝑊𝑊 = 1 𝑊𝑊 = 1 𝐸𝐸 = 3𝜀𝜀 𝑊𝑊 = 2 𝐸𝐸 = 3𝜀𝜀 𝑊𝑊 = 1 𝐸𝐸 = 4𝜀𝜀 𝑊𝑊 = 1 𝐸𝐸 = 4𝜀𝜀 𝑊𝑊 = 1 50 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25 Literature P. Atkins, Physical Chemistry (Oxford University Press) D. Music, compendium, RWTH Aachen University 51 Materials Chemistry II – Kinetics – Lecture 1 Statistical methods Stanislav Mráz | Winter semester 2024 / 25