Slide 2 911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir1 Slide
3 The End!!!!!!! 911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir2
Slide 4 911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir3 Slide 5
www.asadipour.kmu.ac.ir 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir4 Slide 6 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. 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir5 Slide 7 A B rate = - [A] tt
rate = [B][B] tt time 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir6 Slide 8 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
911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir7 Reaction Rates
Slide 9 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
911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir8 Slide 10 2NO2
2NO+O2 911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir9 Slide 11
911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir10 Slide 12
911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir11 Slide 13
Concentrations & Rates 0.3 M HCl6 M HCl Mg(s) + 2 HCl(aq)
---> MgCl 2 (aq) + H 2 (g) 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir12 Slide 14 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir13 Slide 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? 2.5 +1 = 3.5
911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir14 Slide 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 ________ 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir15 Slide 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 911213.....56 slides
http:\\aliasadipour.kmu.ac.ir16 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. Slide 18
Expression of R &K 911213.....56
slideshttp:\\aliasadipour.kmu.ac.irPresentation of Lecture
Outlines, 1417 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. Slide 19 Expression of R &K 911213.....56
slideshttp:\\aliasadipour.kmu.ac.irPresentation of Lecture
Outlines, 1418 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. Slide 20 Expression of R
&K 911213.....56
slideshttp:\\aliasadipour.kmu.ac.irPresentation of Lecture
Outlines, 1419 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. Slide 21 Expression of R
&K 911213.....56
slideshttp:\\aliasadipour.kmu.ac.irPresentation of Lecture
Outlines, 1420 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 +. Slide 22 Expression of
R &K 911213.....56
slideshttp:\\aliasadipour.kmu.ac.irPresentation of Lecture
Outlines, 1421 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 Slide 23 Expression of R &K
911213.....56 slideshttp:\\aliasadipour.kmu.ac.irPresentation of
Lecture Outlines, 1422 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: Slide 24 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir23 Slide 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 = t when [A] = [A] 0 /2 t = [A] 0
2k2k [A] = [A] 0 - kt 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir24 Slide 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 = 0.693 k 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir25 rate = - [A] tt t = t when
[A] = [A] 0 /2 Slide 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 = t when [A] = [A] 0 /2 t =
1 k[A] 0 rate = - [A] tt 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir26 Slide 28 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
911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir27 Slide 29
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 tt ln2 k = t = [A] 0 2k2k t = 1 k[A] 0
911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir28 Slide 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 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir29 Slide 31 Three possible
collision orientations-- a & b produce reactions, c does not.
911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir30 Slide 32
Collision Theory TEMPERATURE 10 c 0 T 100-300% RATE 25 c 0 T 35 c 0
2% COLLISIN EFFECTIVE COLLISIONS 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir31 Slide 33 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir32 Slide 34 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. 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir33 Slide 35 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir34 Slide 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. 911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir35
Exothermic ReactionEndothermic Reaction Slide 37 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir36 Slide 38 Temperature
Dependence of the Rate Constant k = A exp( -E a /RT ) 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 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir37 (Arrhenius equation) Slide 39
Arrhenius Equation k = A exp( -E a /RT ) Can be arranged in the
form of a straight line ln k = ln A-Ea/RT=(-E a /R)(1/T) + ln A
Plot ln k vs. 1/T slope = -E a /R 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir38 Slide 40 k = A e ( -Ea/RT )
ln k = ln A-Ea/RT=(-E a /R)(1/T) + ln A 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir39 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 Slide 41 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir40 Ln(K2/K1)=-Ea/R(1/T2-1/T1)
Log(K2/K1) = Ea/2.303R(T2-T1)/T1T2 K in different T
Log(K2/K1)=-Ea/2.303R(1/T2-1/T1) Log(K2/K1) = H /2.303R(T2-T1)/T1T2
Slide 42 911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir41 Slide
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 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir42 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 - ] Slide 44 Mechanisms
NOTE 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.
911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir43 Slide 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] 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir44 Slide 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-] 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir45 Slide 47 SN2 OH- +CH 3 Br CH
3 OH + Br- R=K[CH 3 Br][OH-] 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir46 Slide 48 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir47 Slide 49 NO 2 + CO NO (g) +
CO 2(g) Rate = k[NO 2 ] 2 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir48 Single step Two possible
mechanisms Two steps is rational mechanism Experimental Rate Laws
andMechanisms Proposed Slide 50 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir49 Slide 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
911213.....56 slides http:\\aliasadipour.kmu.ac.ir 50 Slide 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. 911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir51 Br -
Slide 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
911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir52 Cl 2 Catal.
Slide 54 C 2 H 4 + H 2 C 2 H 6 with a metal catalyst a. reactants
b. adsorption c. migration/reaction d. desorption 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir53 2)Heterogeneous Catalysis
Slide 55 Enzyme Catalysis 13.6 911213.....56
slideshttp:\\aliasadipour.kmu.ac.ir54 Slide 56 Catalytic Converters
13.6 911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir55 CO +
Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter
catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 Slide 57 The End!!!!!!!
911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir56
