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منابع کتابخانه شیمی عمومی مورتیمر کتب ومنابع شیمی سایت های اینترنتی slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour4

slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour5 Chemical Kinetics (Mark=2)

Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B rate = -  [A] tt rate =  [B] tt  [A] = change in concentration of A over time period  t  [B] = change in concentration of B over time period  t Because [A] decreases with time,  [A] is negative slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour6

A B rate = -  [A] tt rate = [B][B] tt time slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour7

Reaction rate = change in concentration of a reactant or product with time. Reaction rate = change in concentration of a reactant or product with time. Three “types” of rates Three “types” of rates initial rate initial rate average rate average rate instantaneous rate instantaneous rate slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour8 Reaction Rates

Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) average rate = -  [Br 2 ] tt = - [Br 2 ] final – [Br 2 ] initial t final - t initial slope of tangent slope of tangent slope of tangent instantaneous rate = rate for specific instance in time slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour9

2NO2  2NO+O slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour10

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Concentrations & Rates 0.3 M HCl6 M HCl Mg(s) + 2 HCl(aq) ---> MgCl 2 (aq) + H 2 (g) slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour12

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Reaction Order What is the rate expression for aA + bB → cC + dD Rate = k[A] x [B] y where x=1 and y=2.5? Rate = k[A][B] 2.5 What is the reaction order? First in A, 2.5 in B Overall reaction order? = slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour14

Interpreting Rate Laws Rate = k [A] m [B] n [C] p If m = 1, rxn. is 1st order in A If m = 1, rxn. is 1st order in A Rate = k [A] 1 If [A] doubles, then rate goes up by factor of __ If [A] doubles, then rate goes up by factor of __ If m = 2, rxn. is 2nd order in A. If m = 2, rxn. is 2nd order in A. Rate = k [A] 2 Rate = k [A] 2 Doubling [A] increases rate by ________ If m = 0, rxn. is zero order. If m = 0, rxn. is zero order. Rate = k [A] 0 Rate = k [A] 0 If [A] doubles, rate ________ slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour15

Deriving Rate Laws Expt. [CH 3 CHO] Disappear of CH 3 CHO (mol/L) (mol/Lsec) slides Derive rate law and k for CH 3 CHO(g) → CH 4 (g) + CO(g) from experimental data for rate of disappearance of CH 3 CHO Rate of rxn = k [CH 3 CHO] 2 ???? Therefore, we say this reaction is _____ order. Here the rate goes up by _____ when initial conc. doubles. Now determine the value of k. Use expt. #3 data— 1,2,3 or 4 R= k [CH 3 CHO] mol/Ls = k (0.30 mol/L) 2 k = 2.0 (L / mols) k = 2.0 (L / mols) Using k you can calc. rate at other values of [CH 3 CHO] at same T.

Expression of R &K slides \aliasadipourPresentation of Lecture Outlines, 14–17 –A series of four experiments was run at different concentrations, and the initial rates of X formation were determined. –From the following data, obtain the reaction orders with respect to A, B, and H +. –Calculate the numerical value of the rate constant.

Expression of R slides \aliasadipourPresentation of Lecture Outlines, 14–18 Initial Concentrations (mol/L) ABH+H+ Initial Rate [mol/(L. s)] Exp x Exp x Exp x Exp x –Comparing Experiment 1 and Experiment 2, you see that when the – A concentration doubles (with other concentrations constant), – the rate doubles. –This implies a first-order dependence with respect to A.

Expression of R slides \aliasadipourPresentation of Lecture Outlines, 14–19 Initial Concentrations (mol/L) ABH+H+ Initial Rate [mol/(L. s)] Exp x Exp x Exp x Exp x –Comparing Experiment 1 and Experiment 3, you see that when the – B concentration doubles (with other concentrations constant), – the rate 4 times. –This implies a second-order dependence with respect to B.

Expression of R slides \aliasadipourPresentation of Lecture Outlines, 14–20 Initial Concentrations (mol/L) ABH+H+ Initial Rate [mol/(L. s)] Exp x Exp x Exp x Exp x –Comparing Experiment 1 and Experiment 4, you see that when –the H + concentration doubles (with other concentrations constant), – the rate is the same. –This implies a zero-order dependence with respect to H +.

Expression of R slides \aliasadipourPresentation of Lecture Outlines, 14–21 Initial Concentrations (mol/L) ABH+H+ Initial Rate [mol/(L. s)] Exp x Exp x Exp x Exp x 10 -6

Expression of K slides \aliasadipourPresentation of Lecture Outlines, 14–22 Initial Concentrations (mol/L) ABH+H+ Initial Rate [mol/(L. s)] Exp x Exp x Exp x Exp x –You can now calculate the rate constant by substituting values from – any of the experiments. Using Experiment 1 you obtain:

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Zero-Order Reactions A product rate = rate = k [A] 0 = k k = rate [A] 0 =RATE= M/s  [A] tt = k - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 t ½ = t when [A] = [A] 0 /2 t ½ = [A] 0 2k2k [A] = [A] 0 - kt slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour24

First-Order Reactions A product rate = k [A] k = rate [A] = 1/s or s -1 M/sM/s M =  [A] tt = k [A] - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 [A] = [A] 0 exp(-kt) ln[A] = ln[A] 0 - kt t ½ = k slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour25 rate = -  [A] tt t ½ = t when [A] = [A] 0 /2  [A] tt = k - [A]

Second-Order Reactions 13.3 A productrate = k [A] 2 k = rate [A] 2 = 1/M s M/sM/s M2M2 =  [A] tt = k [A] 2 - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 1 [A] = 1 [A] 0 + kt t ½ = t when [A] = [A] 0 /2 t ½ = 1 k[A] 0 rate = -  [A] tt slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour26  [A] tt = k - [A] 2

Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions OrderRate Law Concentration-Time Equation Half-Life rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] 0 - kt 1 [A] = 1 [A] 0 + kt [A] = [A] 0 - kt t½t½ ln2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A] slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour27

First step in evaluating rate data is to graphically interpret the order of rxn Zeroth order: rate does not change with lower concentration First, second orders: Rate changes as a function of concentration slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour28

Collision Theory O 3 (g) + NO(g) → O 2 (g) + NO 2 (g) Collisin/Lit.S Reactions require (A) geometry (A) geometry (B) Activation energy (B) Activation energy slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour29

Three possible collision orientations-- a & b produce reactions, c does not slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour30

Collision Theory TEMPERATURE 10 c 0  T  %  RATE 25 c 0 T  35 c 0  2%  COLLISIN EFFECTIVE COLLISIONS slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour31

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① with sufficient energy The colliding molecules must have a total kinetic energy equal to or greater than the activation energy, Ea. Ea is the minimum energy of collision required for two molecules to initiate a chemical reaction. It can be thought of as the hill in below. Regardless of whether the elevation of the ground on the other side is lower than the original position, there must be enough energy imparted to the golf ball to get it over the hill slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour33

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The Collision Theory of Chemical Kinetics Activation Energy (E a )- the minimum amount of energy required to initiate a chemical reaction. Activated Complex (Transition State)- a temporary species formed by the reactant molecules as a result of the collision before they form the product slides our35 Exothermic ReactionEndothermic Reaction

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Temperature Dependence of the Rate Constant E a is the activation energy (J/mol) R is the gas constant (8.314 J/Kmol) T is the absolute temperature A is the frequency factor slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour37 (Arrhenius equation)

Arrhenius Equation Can be arranged in the form of a straight line slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour38 Plot ln k vs. 1/T 

k = A e ( -Ea/RT ) slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour39

slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour40 Ln(K2/K1)=-Ea/R(1/T2-1/T1) K in different T Log(K2/K1)=-Ea/2.303R(1/T2-1/T1) 300  400C 0   Rate20000 Times 400  500C 0   Rate400 Times Ea=60KJ/mol 300  310C 0   Rate2 Times Ea=260KJ/mol 300  310C 0   Rate25 Times

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Mechanisms Proposed Mechanism Step 1 — slowHOOH + I - --> HOI + OH - Step 2 — fastHOI + I - --> I 2 + OH - Step 3 — fast2 OH H + --> 2 H 2 O I - + H 2 O H + ---> I H 2 O slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour42 Most rxns. involve a sequence of elementary steps. 2 I - + H 2 O H + ---> I H 2 O 2 I - + H 2 O H + ---> I H 2 O Rate can be no faster than slow step RATE DETERMINING STEP,( rds.) The species HOI and OH - are intermediates. Rate = k [H 2 O 2 ] [I - ]

Mechanisms NOTES 1. Rate law comes from experiment 2. Order and stoichiometric coefficients not necessarily the same! 3.Rate law reflects all chemistry including the slowest step in multistep reaction slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour43

Mechanisms 2NO+F2  2NOF R=K[NO][F2] 1=NO+F2  NOF+F R1=K1[NO][F2] Rate limiting step(RDS) R=R1 2=NO+F  NOF R2=K2[NO][F] slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour44

SN1 OH- + (CH3)3CBr  (CH3)3COH + Br- R=K[(CH 3 ) 3 CBr] 1=(CH 3 ) 3 CBr  (CH 3 ) 3 C + +Br- R1=K1[(CH 3 ) 3 CBr ] 2= (CH 3 ) 3 C + +OH-  (CH 3 ) 3 COH R2=K2[(CH 3 ) 3 C+] +[OH-] slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour45 R= R1 of RDS

SN2 OH- +CH 3 Br  CH 3 OH + Br- R=K[CH 3 Br][OH-] slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour46

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NO 2 + CO →NO (g) + CO 2(g) Rate = k[NO 2 ] slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour48 Single step Two possible mechanisms Two steps is rational mechanism Experimental Rate Laws and Mechanisms Proposed

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A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A exp( -E a /RT ) EaEa k uncatalyzedcatalyzed rate catalyzed > rate uncatalyzed slides 50

For the decomposition of hydrogen peroxide: A catalyst(Br - ) speeds up a reaction by lowering the activation energy. It does this by providing a different mechanism by which the reaction can occur. The energy profiles for both the catalyzed and uncatalyzed reactions of decomposition of hydrogen peroxide are shown in Figure below slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour51 Br -

1)Homogeneous Catalysis N2O (g)  N 2 (g) + O 2 (g) 1) N2O (g)  N2 (g) + O (g) 2) O (g) + N2O (g)  N2(g)+ O 2 (g) Ea=240KJ/mol Cl2(g)  2Cl(g) N2O(g)+Cl(g)  N2(g)+ClO(g) 2ClO  Cl2(g)+O2(g) Ea=140KJ/mol slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour52 Cl 2 Catal.

C 2 H 4 + H 2  C 2 H 6 with a metal catalyst a. reactants b. adsorption c. migration/reaction d. desorption slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour53 2)Heterogeneous Catalysis

Enzyme Catalysis slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour54

Catalytic Converters slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour55 CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter catalytic converter 2NO + 2NO 2 2N 2 + 3O 2

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