## Presentation on theme: "911213.....56 slides http:\\aliasadipour.kmu.ac.ir."— Presentation transcript:

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The End!!!!!!! slides

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منابع کتابخانه شیمی عمومی مورتیمر کتب ومنابع شیمی سایت های اینترنتی
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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 = - D[A] Dt D[A] = change in concentration of A over time period Dt rate = D[B] Dt D[B] = change in concentration of B over time period Dt Because [A] decreases with time, D[A] is negative. slides

A B time D[A] rate = - Dt D[B] rate = Dt 911213.....56 slides

Reaction Rates Reaction rate = change in concentration of a reactant or product with time. Three “types” of rates initial rate average rate instantaneous rate slides

slope of tangent slope of tangent slope of tangent
Br2 (aq) + HCOOH (aq) Br- (aq) + 2H+ (aq) + CO2 (g) slope of tangent slope of tangent slope of tangent average rate = - D[Br2] Dt = - [Br2]final – [Br2]initial tfinal - tinitial instantaneous rate = rate for specific instance in time slides

2NO22NO+O2 slides

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Concentrations & Rates
0.3 M HCl 6 M HCl Mg(s) HCl(aq) ---> MgCl2(aq) + H2(g) slides

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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? = 3.5 slides

Interpreting Rate Laws
Rate = k [A]m[B]n[C]p If m = 1, rxn. is 1st order in A Rate = 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]2 Doubling [A] increases rate by ________ If m = 0, rxn. is zero order. Rate = k [A]0 If [A] doubles, rate ________ slides

Deriving Rate Laws Derive rate law and k for
CH3CHO(g) → CH4(g) + CO(g) from experimental data for rate of disappearance of CH3CHO Rate of rxn = k [CH3CHO]2 ???? Therefore, we say this reaction is _____ order. Here the rate goes up by _____ when initial conc. doubles. Expt. [CH3CHO] Disappear of CH3CHO (mol/L) (mol/L•sec) Now determine the value of k. Use expt. #3 data— 1,2,3 or 4 R= k [CH3CHO] mol/L•s = k (0.30 mol/L)2 k = 2.0 (L / mol•s) Using k you can calc. rate at other values of [CH3CHO] at same T. slides

Expression of R &K 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. slides

Expression of R &K Initial Concentrations (mol/L) A B H+
Initial Rate [mol/(L.s)] Exp. 1 0.010 1.15 x 10-6 Exp. 2 0.020 2.30 x 10-6 Exp. 3 Exp. 4 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. slides

Expression of R &K Initial Concentrations (mol/L) A B H+
Initial Rate [mol/(L.s)] Exp. 1 0.010 1.15 x 10-6 Exp. 2 0.020 2.30 x 10-6 Exp. 3 4.60 x 10-6 Exp. 4 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. slides

Expression of R &K Initial Concentrations (mol/L) A B H+
Initial Rate [mol/(L.s)] Exp. 1 0.010 1.15 x 10-6 Exp. 2 0.020 2.30 x 10-6 Exp. 3 4.60 x 10-6 Exp. 4 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+. slides

Expression of R &K Initial Concentrations (mol/L) A B H+
Initial Rate [mol/(L.s)] Exp. 1 0.010 1.15 x 10-6 Exp. 2 0.020 2.30 x 10-6 Exp. 3 4.60 x 10-6 Exp. 4 slides

Expression of R &K Initial Concentrations (mol/L) A B H+
Initial Rate [mol/(L.s)] Exp. 1 0.010 1.15 x 10-6 Exp. 2 0.020 2.30 x 10-6 Exp. 3 4.60 x 10-6 Exp. 4 You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain: slides

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[A] = [A]0 - kt Zero-Order Reactions
D[A] Dt = k - A product rate = rate = k [A]0 = k rate [A]0 k = =RATE= M/s [A] = [A]0 - kt [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 t½ = [A]0 2k slides

First-Order Reactions
rate = - D[A] Dt A product rate = k [A] rate [A] M/s M = k = = 1/s or s-1 D[A] Dt = k [A] - t½ = t when [A] = [A]0/2 [A] is the concentration of A at any time t t½ = 0.693 k [A]0 is the concentration of A at time t=0 [A] = [A]0exp(-kt) ln[A] = ln[A]0 - kt slides

Second-Order Reactions
rate = - D[A] Dt A product rate = k [A]2 rate [A]2 M/s M2 = D[A] Dt = k [A]2 - k = = 1/M•s [A] is the concentration of A at any time t 1 [A] = [A]0 + kt [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 t½ = 1 k[A]0 slides 13.3

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 Graphs of different rates of reaction copy the chapter from Langmuir for them! slides

Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions Order Rate Law Concentration-Time Equation Half-Life t½ = [A]0 2k rate = k [A] = [A]0 - kt ln2 k = 1 rate = k [A] ln[A] = ln[A]0 - kt 1 [A] = [A]0 + kt t½ = 1 k[A]0 2 rate = k [A]2 slides

Collision Theory O3(g) + NO(g) → O2(g) + NO2(g) 10 31 Collisin/Lit.S
Reactions require (A) geometry (B) Activation energy slides

Three possible collision orientations-- a & b produce reactions,
c does not. slides

Collision Theory TEMPERATURE 10 c0 T  100-300% RATE
25 c0 T  35 c  2%  COLLISIN EFFECTIVE COLLISIONS slides

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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. slides

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The Collision Theory of Chemical Kinetics
Activation 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. Exothermic Reaction Endothermic Reaction Exothermic. Products are more stable than reactants and after product formation, there is release of heat. endothermic. slides

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Temperature Dependence of the Rate Constant
(Arrhenius equation) k = A • exp( -Ea/RT ) Ea is the activation energy (J/mol) R is the gas constant (8.314 J/K•mol) T is the absolute temperature A is the frequency factor slides

Arrhenius Equation k = A • exp( -Ea/RT )
Can be arranged in the form of a straight line ln k = ln A-Ea/RT=(-Ea/R)(1/T) + ln A Plot ln k vs. 1/T  slope = -Ea/R slides

ln k = ln A-Ea/RT=(-Ea/R)(1/T) + ln A 300400C0 Rate20000 Times
k = A • e( -Ea/RT ) ln k = ln A-Ea/RT=(-Ea/R)(1/T) + ln A 300400C0 Rate20000 Times 400500C0 Rate400 Times Ea=60KJ/mol 300310C0  Rate2 Times Ea=260KJ/mol 300310C0  Rate25 Times slides

K in different T Log(K2/K1) =Ea/2.303R(T2-T1)/T1T2
Ln(K2/K1)=-Ea/R(1/T2-1/T1) Log(K2/K1)=-Ea/2.303R(1/T2-1/T1) Log(K2/K1) =Ea/2.303R(T2-T1)/T1T2 Log(K2/K1) =ΔH/2.303R(T2-T1)/T1T2 slides

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Most rxns. involve a sequence of elementary steps.
Mechanisms Rate = k [H2O2] [I-] Most rxns. involve a sequence of elementary steps. 2 I- + H2O H+ ---> I H2O Proposed Mechanism Step 1 — slow HOOH + I- --> HOI + OH- Step 2 — fast HOI + I- --> I2 + OH- Step 3 — fast 2 OH H+ --> 2 H2O 2 I- + H2O H+ ---> I H2O Rate = k [H2O2] [I-] Rate can be no faster than slow step RATE DETERMINING STEP,( rds.) The species HOI and OH- are intermediates. slides

Mechanisms NOTE Rate 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. slides

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] slides

SN1 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-] slides

SN2 OH- +CH3Br  CH3OH + Br- R=K[CH3Br][OH-] 911213.....56 slides

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Rate Laws andMechanisms
Proposed NO2 + CO →NO(g) + CO2(g) Rate = k[NO2]2 Experimental Two possible mechanisms Two steps is rational mechanism Single step slides

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ratecatalyzed > rateuncatalyzed Ea k
A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A • exp( -Ea/RT ) ratecatalyzed > rateuncatalyzed Ea k uncatalyzed catalyzed slides

Br - 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. slides

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) 2ClOCl2(g)+O2(g) Ea=140KJ/mol Cl2 Catal. slides

2)Heterogeneous Catalysis
C2H4 + H2  C2H6 with a metal catalyst a. reactants b. adsorption c. migration/reaction d. desorption slides

Enzyme Catalysis 13.6 911213.....56 slides
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2N2 + 3O2 Catalytic Converters CO2 + H2O catalytic converter catalytic
CO + Unburned Hydrocarbons + O2 catalytic converter 2N2 + 3O2 2NO + 2NO2 slides 13.6

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