Chemical Kinetics in Situations Use of High- . . . Intermediate - PowerPoint PPT Presentation

Chemical Kinetics: . . . How Formulas of . . . Case of High . . . Example Chemical Kinetics in Situations Use of High- . . . Intermediate Between Usual and High Problem Concentrations: Fuzzy-Motivated Analysis of the . . . Derivation of the Formulas Resulting Formula for . . . Remaining Open . . . Olga Kosheleva and Vladik Kreinovich Home Page University of Texas at El Paso 500 W. University Title Page El Paso, TX 79968, USA ◭◭ ◮◮ olgak@utep.edu, vladik@utep.edu ◭ ◮ La´ ecio Carvalho Barros Instituto de Matem´ atica, Estat´ ıstica, Page 1 of 21 e Computa¸ c˜ ao Cient´ ıfica (IMECC Universidade Estadual de Campinas (UNICAMP) Go Back Campinas, SP, Brasil C.P. 6065 Full Screen laeciocb@ime.unicamp.br Close Quit

Chemical Kinetics: . . . How Formulas of . . . 1. Chemical Kinetics: Usual Formulas Case of High . . . • Chemical kinetics describes the rate of chemical reac- Example tions. Use of High- . . . Problem • For usual concentrations: Analysis of the . . . – the rate of a reaction between two substances Resulting Formula for . . . A and B Remaining Open . . . – is proportional to the product c A · c B of their con- Home Page centration. Title Page • Similarly, if we have a reaction ◭◭ ◮◮ A + . . . + B → . . . ◭ ◮ Page 2 of 21 with three or more substances: Go Back – the rate of this reaction Full Screen – is proportional to the products of the concentra- tions of all these substances c A · . . . · c B . Close Quit

Chemical Kinetics: . . . How Formulas of . . . 2. How Formulas of Chemical Kinetics Are Usu- Case of High . . . ally Derived Example • Molecules of both substances are randomly distributed Use of High- . . . in space. Problem Analysis of the . . . • So, for each molecule of the substance A: Resulting Formula for . . . – the probability that it meets a B-molecule Remaining Open . . . – is proportional to the concentration c B . Home Page • If the molecules meet, then (with a certain probability) Title Page they get into a reaction. ◭◭ ◮◮ • Thus, the mean number of reactions involving a given ◭ ◮ A-molecule is also proportional to c B . Page 3 of 21 • The total number of A-molecules in a given volume is Go Back proportional to c A . Full Screen • Thus, the total number of reactions per unit time is proportional to c A · c B . Close Quit

Chemical Kinetics: . . . How Formulas of . . . 3. Case of High Concentrations Case of High . . . • When the concentrations are very high, there is no need Example for the molecules to randomly bump into each other. Use of High- . . . Problem • Indeed, these molecules are everywhere. Analysis of the . . . • So, as soon as we have molecules of all needed type, Resulting Formula for . . . the reaction starts. Remaining Open . . . Home Page • In other words, in this case, the reaction rate is propor- tional to the concentration of the corresponding tuples. Title Page • Thus, the reaction rate is proportional to the minimum ◭◭ ◮◮ min( c A , . . . , c B ) of all the input concentrations ◭ ◮ c A , . . . , c B . Page 4 of 21 Go Back Full Screen Close Quit

Chemical Kinetics: . . . How Formulas of . . . 4. Example Case of High . . . • The formula min( c A , . . . , c B ) can be easily illustrated Example on a similar relation between predators and prey. Use of High- . . . Problem • Let us throw a bunch of rabbits into a zoo cage filled Analysis of the . . . with hungry wolves. Resulting Formula for . . . • Then each wolf will start eating its rabbit. Remaining Open . . . Home Page • This will continue as long as there are sufficiently many rabbits to feed all the wolves. Title Page • When c R ≥ c W , the number of eaten rabbits will be ◭◭ ◮◮ proportional to the number of wolves, i.e., to c W . ◭ ◮ • When there are few rabbits ( c R < c W ), the number of Page 5 of 21 eaten rabbits is sim the number of rabbits c R . Go Back • In both cases, the reaction rate is proportional to Full Screen min( c R , c W ). Close Quit

Chemical Kinetics: . . . How Formulas of . . . 5. Use of High-Concentration Reactions Case of High . . . 1) The high-concentration reaction rate indeed turned out Example to be very useful to describe biochemical processes. Use of High- . . . Problem 2) In many cases, difficult-to-solve computational prob- Analysis of the . . . lems can be reduced to problems of chemical kinetics. Resulting Formula for . . . • We can then solve the original problem by simulating Remaining Open . . . these reactions. Home Page • To make simulations as fast as possible, it is desirable Title Page to simulate reactions which are as fast as possible. ◭◭ ◮◮ • The reaction rate increases with the concentrations of ◭ ◮ the reagents. Page 6 of 21 • Thus, to speed up simulations, we should simulate high-concentration reactions. Go Back Full Screen • This simulation indeed speeds up the corresponding computations. Close Quit

Chemical Kinetics: . . . How Formulas of . . . 6. Problem Case of High . . . • We know the formulas for the usual and for the high Example concentrations. Use of High- . . . Problem • However, it is not clear how to compute the reaction Analysis of the . . . rate for concentrations between usual and high. Resulting Formula for . . . • Both r = c A · c B and r = min( c A , c B ) are particular Remaining Open . . . cases of t-norms – “and”-operations in fuzzy logic. Home Page • This is not a coincidence: Title Page – there is no reaction if one of the substances is miss- ◭◭ ◮◮ ing, so c A = 0 or c B imply that r = 0; ◭ ◮ – his is exactly the property of a t-norm. Page 7 of 21 • Fuzzy t-norms have indeed been effectively used to de- Go Back scribe chemical reactions. Full Screen • Problem: there are many possible “and”-operations, it is not clear which one to select. Close Quit

Chemical Kinetics: . . . How Formulas of . . . 7. Analysis of the Problem: General Case Case of High . . . • Let’s analyze the problem to find the most appropriate Example “and”-operation. Use of High- . . . Problem • Two molecules get into a reaction only when they are Analysis of the . . . close enough. Resulting Formula for . . . • When these molecules are close enough, then the reac- Remaining Open . . . tion rate is proportional to min( c A , c B ). Home Page Title Page ◭◭ ◮◮ ◭ ◮ Page 8 of 21 Go Back Full Screen Close Quit

Chemical Kinetics: . . . How Formulas of . . . 8. Case of Low Concentrations Case of High . . . • When concentrations are low , then, within each region, Example we have either zero or one molecule. Use of High- . . . Problem • The probability to have two molecules is very small Analysis of the . . . (proportional to the square of these concentrations). Resulting Formula for . . . • This probability can thus be safely ignored. Remaining Open . . . • In this case, for each region, the reaction occurs if we Home Page have both an A-molecule and a B-molecule. Title Page • The probability to have an A-molecule is ∼ c A . ◭◭ ◮◮ • The probability to have a B-molecule is ∼ c B . ◭ ◮ • The distributions for A and B are independent. Page 9 of 21 • Thus, the probability to have both A- and B-molecules Go Back in a region is equal to the product of these probabilities. Full Screen • This probability is thus proportional to the product of Close the concentrations c A · c B . Quit

Chemical Kinetics: . . . How Formulas of . . . 9. Case of High Concentrations Case of High . . . • When the concentrations are high, then each region Example has molecules of both types. Use of High- . . . Problem • The average number of A-molecules in a region is pro- Analysis of the . . . portional to c A : equal to k · c A for some k . Resulting Formula for . . . • Similarly, the average number of B-molecules in a re- Remaining Open . . . gion is equal to k · c B . Home Page • So the average reaction rate is proportional to Title Page min( k · c A , k · c B ) = k · min( c A , c B ) . ◭◭ ◮◮ ◭ ◮ • The rate is thus proportional to min( c A , c B ). Page 10 of 21 • This analysis leads us to the following reformulation of Go Back our problem. Full Screen Close Quit

Chemical Kinetics: . . . How Formulas of . . . 10. Resulting Formulation of the Problem in Pre- Case of High . . . cise Terms Example • Within a unit volume, we have a certain number r of Use of High- . . . “small regions”. Problem Analysis of the . . . • Small means that only molecules within the same re- Resulting Formula for . . . gion can interact with each other. Remaining Open . . . • We have a total of N A = N · c A A-molecules and N B = Home Page N · c B B-molecules. Title Page • Each of these molecules is randomly distributed among ◭◭ ◮◮ the regions. ◭ ◮ • So, it can be located in any of the r regions with equal probability. Page 11 of 21 • Within each region, the reaction rate is ∼ min( n A , n B ). Go Back Full Screen • The overall reaction rate is the average over all the regions. Close Quit