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The End!!!!!!!

The End!!!!!!!. منابع. کتابخانه شیمی عمومی مورتیمر کتب ومنابع شیمی سایت های اینترنتی www.asadipour.kmu.ac.ir. A B. rate =. D [A]. D [B]. rate = -. D t. D t. Chemical Kinetics. Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed?.

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The End!!!!!!!

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  1. http:\\aliasadipour.kmu.ac.ir

  2. The End!!!!!!! http:\\aliasadipour.kmu.ac.ir

  3. http:\\aliasadipour.kmu.ac.ir

  4. منابع • کتابخانه • شیمی عمومی مورتیمر • کتب ومنابع شیمی • سایت های اینترنتی • www.asadipour.kmu.ac.ir http:\\aliasadipour.kmu.ac.ir

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

  6. A B time rate = D[A] D[B] rate = - Dt Dt http:\\aliasadipour.kmu.ac.ir

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

  8. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) slope of tangent slope of tangent slope of tangent [Br2]final – [Br2]initial D[Br2] average rate = - = - Dt tfinal - tinitial instantaneous rate = rate for specific instance in time http:\\aliasadipour.kmu.ac.ir

  9. 2NO22NO+O2 http:\\aliasadipour.kmu.ac.ir

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  12. Concentrations & Rates 0.3 M HCl 6 M HCl Mg(s) + 2 HCl(aq) ---> MgCl2(aq) + H2(g) http:\\aliasadipour.kmu.ac.ir

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  14. 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 http:\\aliasadipour.kmu.ac.ir

  15. 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 ________ http:\\aliasadipour.kmu.ac.ir

  16. 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) 1 0.10 0.020 20.20 0.081 30.30 0.182 40.400.318 Now determine the value of k. Use expt. #3 data— 1,2,3 or 4 R= k [CH3CHO]2 0.182 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. http:\\aliasadipour.kmu.ac.ir

  17. 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. http:\\aliasadipour.kmu.ac.ir

  18. Expression of R &K • Comparing Experiment 1 and Experiment 2, you see that when the • Aconcentration doubles (with other concentrations constant), • the rate doubles. • This implies a first-order dependence with respect to A. http:\\aliasadipour.kmu.ac.ir

  19. Expression of R &K • Comparing Experiment 1 and Experiment 3, you see that when the • Bconcentration doubles (with other concentrations constant), • the rate 4 times. • This implies a second-orderdependence with respect to B. http:\\aliasadipour.kmu.ac.ir

  20. Expression of R &K • 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-orderdependence with respect to H+. http:\\aliasadipour.kmu.ac.ir

  21. Expression of R &K http:\\aliasadipour.kmu.ac.ir

  22. Expression of R &K • You can now calculate the rate constant by substituting values from • any of the experiments. Using Experiment 1 you obtain: http:\\aliasadipour.kmu.ac.ir

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  24. A product rate [A]0 D[A] - = k Dt [A]0 t½ = 2k Zero-Order Reactions rate = k [A]0 = k rate = =RATE= M/s k = [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 http:\\aliasadipour.kmu.ac.ir

  25. A product rate = [A] M/s D[A] - M = k [A] Dt ln[A] = ln[A]0 - kt [A] = [A]0exp(-kt) 0.693 t½ = D[A] k rate = - Dt First-Order Reactions rate = k[A] k = = 1/s or s-1 t½ = t when [A] = [A]0/2 [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 http:\\aliasadipour.kmu.ac.ir

  26. A product rate = [A]2 M/s D[A] 1 1 - M2 = k [A]2 = + kt Dt [A] [A]0 t½ = D[A] rate = - Dt 1 k[A]0 Second-Order Reactions rate = k[A]2 = 1/M•s k = [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 http:\\aliasadipour.kmu.ac.ir 13.3

  27. 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 http:\\aliasadipour.kmu.ac.ir

  28. Concentration-Time Equation Order Rate Law Half-Life 1 1 = + kt [A] [A]0 = [A]0 t½ = t½ t½ = ln2 2k k 1 k[A]0 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions [A] = [A]0 - kt rate = k 0 ln[A] = ln[A]0 - kt 1 rate = k [A] 2 rate = k [A]2 http:\\aliasadipour.kmu.ac.ir

  29. Collision Theory O3(g) + NO(g) → O2(g) + NO2(g) 10 31Collisin/Lit.S Reactions require (A) geometry (B) Activation energy http:\\aliasadipour.kmu.ac.ir

  30. Three possible collision orientations-- a & b produce reactions, c does not. http:\\aliasadipour.kmu.ac.ir

  31. Collision Theory • TEMPERATURE • 10 c0 T  100-300% RATE • 25 c0 T  35 c0  2%  COLLISIN • EFFECTIVE COLLISIONS http:\\aliasadipour.kmu.ac.ir

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  33. 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. http:\\aliasadipour.kmu.ac.ir

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  35. 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 http:\\aliasadipour.kmu.ac.ir

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  37. Temperature Dependence of the Rate Constant (Arrhenius equation) k = A •exp( -Ea/RT ) Eais 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 http:\\aliasadipour.kmu.ac.ir

  38. 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 http:\\aliasadipour.kmu.ac.ir

  39. 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  Rate2Times Ea=260KJ/mol 300310C0  Rate25 Times http:\\aliasadipour.kmu.ac.ir

  40. K in different T 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 http:\\aliasadipour.kmu.ac.ir

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

  43. 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. http:\\aliasadipour.kmu.ac.ir

  44. 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] http:\\aliasadipour.kmu.ac.ir

  45. 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-] http:\\aliasadipour.kmu.ac.ir

  46. SN2 • OH- +CH3Br  CH3OH + Br- • R=K[CH3Br][OH-] http:\\aliasadipour.kmu.ac.ir

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  48. Rate Laws andMechanisms Proposed NO2 + CO →NO(g) + CO2(g) Rate = k[NO2]2 Experimental Two possible mechanisms Two steps Two steps is rational mechanism Single step http:\\aliasadipour.kmu.ac.ir

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  50. uncatalyzed catalyzed 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 http:\\aliasadipour.kmu.ac.ir

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