The End!!!!!!!

Slide 1

911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir1

The End!!!!!!!911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir2


www.asadipour.kmu.ac.ir911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir4Chemical KineticsThermodynamics 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 Brate = -D[A]Dtrate = D[B]DtD[A] = change in concentration of A over time period DtD[B] = change in concentration of B over time period DtBecause [A] decreases with time, D[A] is negative. 911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir5A B

rate = -D[A]Dtrate = D[B]Dt

time911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir6Reaction rate = change in concentration of a reactant or product with time. Three types of rates initial rateaverage rateinstantaneous rate911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir7Reaction Rates

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

average rate = -D[Br2]Dt= -[Br2]final [Br2]initialtfinal - tinitialslope oftangentslope oftangentslope oftangentinstantaneous rate = rate for specific instance in time

911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir82NO22NO+O2




Concentrations & Rates

0.3 M HCl6 M HClMg(s) + 2 HCl(aq) ---> MgCl2(aq) + H2(g)

911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir12911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir13

Reaction OrderWhat 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]2.5What is the reaction order?First in A, 2.5 in BOverall reaction order?2.5 +1 = 3.5911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir14Interpreting Rate Laws Rate = k [A]m[B]n[C]pIf m = 1, rxn. is 1st order in ARate = k [A]1 If [A] doubles, then rate goes up by factor of __ If m = 2, rxn. is 2nd order in A. Rate = k [A]2Doubling [A] increases rate by ________If m = 0, rxn. is zero order. Rate = k [A]0If [A] doubles, rate ________911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir15Deriving Rate LawsExpt. [CH3CHO] Disappear of CH3CHO(mol/L) (mol/Lsec)10.100.02020.200.08130.300.18240.400.318911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir16Derive rate law and k for CH3CHO(g) CH4(g) + CO(g)from experimental data for rate of disappearance of CH3CHORate of rxn = k [CH3CHO]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 4R= k [CH3CHO]2 0.182 mol/Ls = k (0.30 mol/L)2 k = 2.0 (L / mols)Using k you can calc. rate at other values of [CH3CHO] at same T.Expression of R &K911213.....56 slideshttp:\\aliasadipour.kmu.ac.irPresentation of Lecture Outlines, 1417A 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 &K911213.....56 slideshttp:\\aliasadipour.kmu.ac.irPresentation of Lecture Outlines, 1418Initial Concentrations (mol/L)ABH+Initial Rate [mol/(L.s)]Exp. 10.0100.0100.000501.15 x 10-6Exp. 20.0200.0100.000502.30 x 10-6Exp. 30.0100.0200.000502.30 x 10-6Exp. 40.0100.0100.001001.15 x 10-6Comparing 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 &K911213.....56 slideshttp:\\aliasadipour.kmu.ac.irPresentation of Lecture Outlines, 1419Initial Concentrations (mol/L)ABH+Initial Rate [mol/(L.s)]Exp. 10.0100.0100.000501.15 x 10-6Exp. 20.0200.0100.000502.30 x 10-6Exp. 30.0100.0200.000504.60 x 10-6Exp. 40.0100.0100.001001.15 x 10-6Comparing 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 &K911213.....56 slideshttp:\\aliasadipour.kmu.ac.irPresentation of Lecture Outlines, 1420Initial Concentrations (mol/L)ABH+Initial Rate [mol/(L.s)]Exp. 10.0100.0100.000501.15 x 10-6Exp. 20.0200.0100.000502.30 x 10-6Exp. 30.0100.0200.000504.60 x 10-6Exp. 40.0100.0100.001001.15 x 10-6Comparing 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 &K911213.....56 slideshttp:\\aliasadipour.kmu.ac.irPresentation of Lecture Outlines, 1421

Initial Concentrations (mol/L)ABH+Initial Rate [mol/(L.s)]Exp. 10.0100.0100.000501.15 x 10-6Exp. 20.0200.0100.000502.30 x 10-6Exp. 30.0100.0200.000504.60 x 10-6Exp. 40.0100.0100.001001.15 x 10-6Expression of R &K911213.....56 slideshttp:\\aliasadipour.kmu.ac.irPresentation of Lecture Outlines, 1422Initial Concentrations (mol/L)ABH+Initial Rate [mol/(L.s)]Exp. 10.0100.0100.000501.15 x 10-6Exp. 20.0200.0100.000502.30 x 10-6Exp. 30.0100.0200.000504.60 x 10-6Exp. 40.0100.0100.001001.15 x 10-6You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain:


Zero-Order ReactionsA productrate = rate = k [A]0 = kk = rate[A]0=RATE= M/sD[A]Dt= k-[A] is the concentration of A at any time t[A]0 is the concentration of A at time t=0t = t when [A] = [A]0/2t =[A]02k[A] = [A]0 - kt911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir24

First-Order ReactionsA productrate = k [A]k = rate[A]= 1/s or s-1M/sM=D[A]Dt= 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]0exp(-kt)

ln[A] = ln[A]0 - kt

t =0.693k911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir25rate = -D[A]Dtt = t when [A] = [A]0/2Second-Order Reactions13.3A productrate = k [A]2k = rate[A]2= 1/MsM/sM2=D[A]Dt= k [A]2-[A] is the concentration of A at any time t[A]0 is the concentration of A at time t=01[A]=1[A]0+ ktt = t when [A] = [A]0/2t =1k[A]0rate = -D[A]Dt911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir26

First step in evaluating rate data is to graphically interpret the order of rxn

Zeroth order: rate does not change with lower concentrationFirst, second orders:Rate changes as a function of concentration911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir2727Graphs of different rates of reaction copy the chapter from Langmuir for them!

Summary of the Kinetics of Zero-Order, First-Orderand Second-Order ReactionsOrderRate LawConcentration-Time EquationHalf-Life012rate = krate = k [A]rate = k [A]2ln[A] = ln[A]0 - kt1[A]=1[A]0+ kt[A] = [A]0 - kttln2k=t =[A]02kt =1k[A]0911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir28Collision TheoryO3(g) + NO(g) O2(g) + NO2(g)

10 31 Collisin/Lit.SReactions require (A) geometry (B) Activation energy


Three possible collision orientations-- a & b produce reactions,c does not.911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir30Collision TheoryTEMPERATURE10 c0 T 100-300% RATE25 c0 T 35 c0 2% COLLISINEFFECTIVE COLLISIONS911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir31

911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir32 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.ir34The Collision Theory of Chemical KineticsActivation Energy (Ea)- 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.ir35Exothermic ReactionEndothermic Reaction35Exothermic. Products are more stable than reactants and after product formation, there is release of heat. endothermic.911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir36

Temperature Dependence of the Rate Constantk = A exp( -Ea/RT )Ea is the activation energy (J/mol)R is the gas constant (8.314 J/Kmol)T is the absolute temperatureA is the frequency factor911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir37(Arrhenius equation)Arrhenius Equation k = A exp( -Ea/RT ) Can be arranged in the form of a straight lineln k = ln A-Ea/RT=(-Ea/R)(1/T) + ln A Plot ln k vs. 1/T slope = -Ea/R

911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir38k = A e( -Ea/RT )

ln k = ln A-Ea/RT=(-Ea/R)(1/T) + ln A

911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir39300400C0 Rate20000 Times400500C0 Rate400 Times

Ea=60KJ/mol300310C0 Rate2 Times

Ea=260KJ/mol300310C0 Rate25 Times

911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir40Ln(K2/K1)=-Ea/R(1/T2-1/T1)Log(K2/K1) =Ea/2.303R(T2-T1)/T1T2K in different TLog(K2/K1)=-Ea/2.303R(1/T2-1/T1)Log(K2/K1) =H/2.303R(T2-T1)/T1T2911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir41

MechanismsProposed MechanismStep 1 slowHOOH + I- --> HOI + OH-Step 2 fastHOI + I- --> I2 + OH-Step 3 fast2 OH- + 2 H+ --> 2 H2O-------------------------------------------------------------------2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O

911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir42Most rxns. involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate can be no faster than slow stepRATE DETERMINING STEP,( rds.)The species HOI and OH- are intermediates.Rate = k [H2O2] [I-] Rate = k [H2O2] [I-] MechanismsNOTERate law comes from experiment

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.ir43Mechanisms 2NO+F22NOF R=K[NO][F2]

1=NO+F2NOF+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.ir44SN1

OH- + (CH3)3CBr(CH3)3COH + Br-R=K[(CH3)3CBr]1)(CH3)3CBr(CH3)3C+ +Br- R1=K1[(CH3)3CBr ]

2) (CH3)3C+ +OH- (CH3)3COH R2=K2[(CH3)3C+] +[OH-]911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir45SN2OH- +CH3Br CH3OH + Br-

R=K[CH3Br][OH-]


911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir47NO2 + CO NO(g) + CO2(g) Rate = k[NO2]2911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir48Single stepTwo possible mechanismsTwo steps

Two stepsis rational mechanism

ExperimentalRate Laws andMechanismsProposed911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir49

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.k = A exp( -Ea/RT )Eak

uncatalyzedcatalyzedratecatalyzed > rateuncatalyzed911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir50 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.ir51Br -1)Homogeneous Catalysis N2O (g) N2 (g) + O2 (g) 1) N2O (g) N2 (g) + O (g) O (g) + N2O (g) N2(g)+ O2 (g) Ea=240KJ/mol------------------------------------------------------- Cl2(g)2Cl(g) N2O(g)+Cl(g)N2(g)+ClO(g) 2ClOCl2(g)+O2(g) Ea=140KJ/mol

911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir52Cl2 Catal.C2H4 + H2 C2H6

with a metal catalyst

a. reactantsb. adsorptionc. migration/reactiond. desorption

911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir532)Heterogeneous CatalysisEnzyme Catalysis

13.6911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir54Catalytic Converters13.6

911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir55CO + Unburned Hydrocarbons + O2CO2 + H2Ocatalyticconvertercatalyticconverter2NO + 2NO22N2 + 3O2The End!!!!!!!911213.....56 slideshttp:\\aliasadipour.kmu.ac.ir56

The End!!!!!!!

Documents

Transcript of The End!!!!!!!