Download presentation
Presentation is loading. Please wait.
Published byTracey Preston Modified over 8 years ago
2
940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour1
3
940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour2
4
The End!!!!!!! 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour3
5
منابع کتابخانه شیمی عمومی مورتیمر کتب ومنابع شیمی سایت های اینترنتی www.elearning.kmu.ac.ir 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour4
6
940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour5 Chemical Kinetics (Mark=2)
7
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] tt rate = [B] tt [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. 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour6
8
A B rate = - [A] tt rate = [B][B] tt time 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour7
9
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 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour8 Reaction Rates
10
Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) average rate = - [Br 2 ] tt = - [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 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour9
11
2NO2 2NO+O2 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour10
12
940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour11
13
Concentrations & Rates 0.3 M HCl6 M HCl Mg(s) + 2 HCl(aq) ---> MgCl 2 (aq) + H 2 (g) 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour12
14
940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour13
15
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? 1 +2.5 = 3.5 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour14
16
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 ________ 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour15
17
Deriving Rate Laws Expt. [CH 3 CHO] Disappear of CH 3 CHO (mol/L) (mol/Lsec) 10.100.020 20.200.081 30.300.182 40.400.318 940701.....56 slides http:\\academicstaff.kmu.ac.ir\aliasadipour16 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] 2 0.182 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.
18
Expression of R &K 940701.....56 slides http:\\academicstaff.kmu.ac.ir \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.
19
Expression of R 940701.....56 slides http:\\academicstaff.kmu.ac.ir \aliasadipourPresentation of Lecture Outlines, 14–18 Initial Concentrations (mol/L) ABH+H+ Initial Rate [mol/(L. s)] Exp. 10.010 0.000501.15 x 10 -6 Exp. 20.0200.0100.000502.30 x 10 -6 Exp. 30.0100.0200.000502.30 x 10 -6 Exp. 40.010 0.001001.15 x 10 -6 –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.
20
Expression of R 940701.....56 slides http:\\academicstaff.kmu.ac.ir \aliasadipourPresentation of Lecture Outlines, 14–19 Initial Concentrations (mol/L) ABH+H+ Initial Rate [mol/(L. s)] Exp. 10.010 0.000501.15 x 10 -6 Exp. 20.0200.0100.000502.30 x 10 -6 Exp. 30.0100.0200.000504.60 x 10 -6 Exp. 40.010 0.001001.15 x 10 -6 –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.
21
Expression of R 940701.....56 slides http:\\academicstaff.kmu.ac.ir \aliasadipourPresentation of Lecture Outlines, 14–20 Initial Concentrations (mol/L) ABH+H+ Initial Rate [mol/(L. s)] Exp. 10.010 0.000501.15 x 10 -6 Exp. 20.0200.0100.000502.30 x 10 -6 Exp. 30.0100.0200.000504.60 x 10 -6 Exp. 40.010 0.001001.15 x 10 -6 –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 +.
22
Expression of R 940701.....56 slides http:\\academicstaff.kmu.ac.ir \aliasadipourPresentation of Lecture Outlines, 14–21 Initial Concentrations (mol/L) ABH+H+ Initial Rate [mol/(L. s)] Exp. 10.010 0.000501.15 x 10 -6 Exp. 20.0200.0100.000502.30 x 10 -6 Exp. 30.0100.0200.000504.60 x 10 -6 Exp. 40.010 0.001001.15 x 10 -6
23
Expression of K 940701.....56 slides http:\\academicstaff.kmu.ac.ir \aliasadipourPresentation of Lecture Outlines, 14–22 Initial Concentrations (mol/L) ABH+H+ Initial Rate [mol/(L. s)] Exp. 10.010 0.000501.15 x 10 -6 Exp. 20.0200.0100.000502.30 x 10 -6 Exp. 30.0100.0200.000504.60 x 10 -6 Exp. 40.010 0.001001.15 x 10 -6 –You can now calculate the rate constant by substituting values from – any of the experiments. Using Experiment 1 you obtain:
24
940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour23
25
Zero-Order Reactions A product rate = rate = k [A] 0 = k k = rate [A] 0 =RATE= M/s [A] tt = 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 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour24
26
First-Order Reactions A product rate = k [A] k = rate [A] = 1/s or s -1 M/sM/s M = [A] tt = 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 = 0.693 k 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour25 rate = - [A] tt t ½ = t when [A] = [A] 0 /2 [A] tt = k - [A]
27
Second-Order Reactions 13.3 A productrate = k [A] 2 k = rate [A] 2 = 1/M s M/sM/s M2M2 = [A] tt = 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] tt 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour26 [A] tt = k - [A] 2
28
Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions OrderRate Law Concentration-Time Equation Half-Life 0 1 2 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] 0 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour27
29
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 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour28
30
Collision Theory O 3 (g) + NO(g) → O 2 (g) + NO 2 (g) 10 31 Collisin/Lit.S Reactions require (A) geometry (A) geometry (B) Activation energy (B) Activation energy 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour29
31
Three possible collision orientations-- a & b produce reactions, c does not. 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour30
32
Collision Theory TEMPERATURE 10 c 0 T 100-300% RATE 25 c 0 T 35 c 0 2% COLLISIN EFFECTIVE COLLISIONS 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour31
33
940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour32
34
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. 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour33
35
940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour34
36
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. 940701.....56 slides http:\\academicstaff.kmu.ac.ir\aliasadip our35 Exothermic ReactionEndothermic Reaction
37
940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour36
38
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 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour37 (Arrhenius equation)
39
Arrhenius Equation Can be arranged in the form of a straight line 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour38 Plot ln k vs. 1/T
40
k = A e ( -Ea/RT ) 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour39
41
940701.....56 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
42
940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour41
43
Mechanisms Proposed Mechanism Step 1 — slowHOOH + I - --> HOI + OH - Step 2 — fastHOI + I - --> I 2 + OH - Step 3 — fast2 OH - + 2 H + --> 2 H 2 O ------------------------------------------------------------------- 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 H 2 O 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour42 Most rxns. involve a sequence of elementary steps. 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 H 2 O 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 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 - ]
44
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. 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour43
45
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] 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour44
46
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-] 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour45 R= R1 of RDS
47
SN2 OH- +CH 3 Br CH 3 OH + Br- R=K[CH 3 Br][OH-] 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour46
48
940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour47
49
NO 2 + CO →NO (g) + CO 2(g) Rate = k[NO 2 ] 2 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour48 Single step Two possible mechanisms Two steps is rational mechanism Experimental Rate Laws and Mechanisms Proposed
50
940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour49
51
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 940701.....56 slides http:\\academicstaff.kmu.ac.ir\aliasadipour 50
52
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. 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour51 Br -
53
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 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour52 Cl 2 Catal.
54
C 2 H 4 + H 2 C 2 H 6 with a metal catalyst a. reactants b. adsorption c. migration/reaction d. desorption 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour53 2)Heterogeneous Catalysis
55
Enzyme Catalysis 13.6 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour54
56
Catalytic Converters 13.6 940701.....56 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
57
The End!!!!!!! 940701.....56 slideshttp:\\academicstaff.kmu.ac.ir\aliasadipour56
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.