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Chemical Kinetics Thermodynamics – does a reaction take place?

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1 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. Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS) 13.1

2 13.1 A B time D[A] rate = - Dt D[B] rate = Dt
Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS) 13.1

3 Reaction Rates Reaction rate = change in concentration of a reactant or product with time. Three “types” of rates initial rate average rate instantaneous rate Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

4 slope of tangent slope of tangent slope of tangent 13.1
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 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS) 13.1

5 2NO22NO+O2 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

6 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

7 13.3

8 Concentrations & Rates
0.3 M HCl 6 M HCl Mg(s) HCl(aq) ---> MgCl2(aq) + H2(g) Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

9 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 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

10 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 ________ Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

11 [S2O82-] [I-] Initial Rate (M/s) 1 0.08 0.034 2.2 x 10-4 2 0.017
Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O82- (aq) + 3I- (aq) SO42- (aq) + I3- (aq) Experiment [S2O82-] [I-] Initial Rate (M/s) 1 0.08 0.034 2.2 x 10-4 2 0.017 1.1 x 10-4 3 0.16 rate = k [S2O82-]x[I-]y y = 1 x = 1 rate = k [S2O82-][I-] Double [I-], rate doubles (experiment 1 & 2) Double [S2O82-], rate doubles (experiment 2 & 3) k = rate [S2O82-][I-] = 2.2 x 10-4 M/s (0.08 M)(0.034 M) = 0.08/M•s Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS) 13.2

12 Deriving Rate Laws Derive rate law and k for CH3CHO(g) --> CH4(g) + CO(g) from experimental data for rate of disappearance of CH3CHO Expt. [CH3CHO] Disappear of CH3CHO (mol/L) (mol/L•sec) Rate of rxn = k [CH3CHO]2 GENERAL CHEMISTRY(KINETIKS)

13 Deriving Rate Laws Rate of rxn = k [CH3CHO]2 Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is _________________ order. Now determine the value of k. Use expt. #3 data— 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. Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

14 Expession of R &K Iodide ion is oxidized in acidic solution to triiodide ion, I3- , by hydrogen peroxide. A series of four experiments was run at different concentrations, and the initial rates of I3- formation were determined. From the following data, obtain the reaction orders with respect to H2O2, I-, and H+. Calculate the numerical value of the rate constant. Copyright © Houghton Mifflin Company.All rights reserved.

15 A Problem to Consider Initial Concentrations (mol/L) H2O2 I- 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 H2O2 concentration doubles (with other concentrations constant), the rate doubles. This implies a first-order dependence with respect to H2O2. Copyright © Houghton Mifflin Company.All rights reserved.

16 A Problem to Consider Initial Concentrations (mol/L) H2O2 I- 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 3, you see that when the I- concentration doubles (with other concentrations constant), the rate doubles. This implies a first-order dependence with respect to I-. Copyright © Houghton Mifflin Company.All rights reserved.

17 A Problem to Consider Initial Concentrations (mol/L) H2O2 I- 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 4, you see that when the H+ concentration doubles (with other concentrations constant), the rate is unchanged. This implies a zero-order dependence with respect to H+. Copyright © Houghton Mifflin Company.All rights reserved.

18 A Problem to Consider Initial Concentrations (mol/L) H2O2 I- 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 Because [H+]0 = 1, the rate law is: The reaction orders with respect to H2O2, I-, and H+, are 1, 1, and 0, respectively. Copyright © Houghton Mifflin Company.All rights reserved.

19 A Problem to Consider Initial Concentrations (mol/L) H2O2 I- 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 You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain: Copyright © Houghton Mifflin Company.All rights reserved.

20 A Problem to Consider Initial Concentrations (mol/L) H2O2 I- 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 You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain: Copyright © Houghton Mifflin Company.All rights reserved.

21 Zero-Order Reactions [A] is the concentration of A at any time t
rate = - D[A] Dt A product rate = k [A]0 = k rate [A]0 D[A] Dt = k - k = =RATE= M/s [A] is the concentration of A at any time t [A] = [A]0 - kt [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 t½ = [A]0 2k 13.3

22 First-Order Reactions
A product rate = k [A] D[A] Dt = k [A] - rate [A] M/s M = k = = 1/s or s-1 [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 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS) 13.3

23 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 13.3

24 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!

25 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 13.3

26 Collision Theory O3(g) + NO(g) ---> O2(g) + NO2(g)
10 31 COLLISINS/LIT.S Reactions require (A) geometry (B) Activation energy Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

27 Three possible collision orientations-- a) & b) produce reactions,
while c) does not.

28 Collision Theory TEMPERATURE 10 c0 T  100-300% RATE
25 c0 T  35 c  2%  COLLISIN EFFECTIVE COLLISIONS Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

29 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

30 ① with sufficient energy
The colliding molecules must have a total kinetic energy equal to or greater than the activation energy, Ea. The activation energy is the mini- mum 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(平面) 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. Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

31 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. Products are more stable than reactants and after product formation, there is release of heat. endothermic. Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

32 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

33 13.4 A + B C + D Exothermic Reaction Endothermic Reaction
The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction. Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS) 13.4

34 Temperature Dependence of the Rate Constant
k = A • exp( -Ea/RT ) (Arrhenius equation) 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 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS) 13.4

35 Arrhenius Equation k = A • exp( -Ea/RT )
Can be arranged in the form of a straight line ln k = ln A-Ea/RT Plot ln k vs. 1/T  slope = -Ea/R Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

36 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 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

37 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

38 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 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

39 Mechanisms O3 + NO reaction occurs in a single ELEMENTARY step. Most others involve a sequence of elementary steps. Adding elementary steps gives NET reaction. Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

40 Mechanisms Most rxns. involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] 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. Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

41 Most rxns. involve a sequence of elementary steps.
Mechanisms Most rxns. involve a sequence of elementary steps. 2 I- + H2O H+ ---> I H2O Rate = k [I-] [H2O2] 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 Rate of the reaction controlled by slow step — RATE DETERMINING STEP, rds. Rate can be no faster than rds! The species HOI and OH- are reaction intermediates. Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

42 Mechanisms 2NO+F22NOF R=K[NO][F2] 1=NO+F2NOF+F R1=K1[NO][F2]
Rate limiting step(RDS) 2=NO+F NOF R2=K2[NO][F] Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

43 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-] Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

44 SN2 OH- +CH3Br  CH3OH + Br- R=K[CH3Br][OH-] Wednesday, April 26, 2017
GENERAL CHEMISTRY(KINETIKS)

45 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

46 Rate Laws and Mechanisms
NO2 + CO reaction: Rate = k[NO2]2 Two possible mechanisms Two steps: Single step Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

47 A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A • exp( -Ea/RT ) Ea k uncatalyzed catalyzed ratecatalyzed > rateuncatalyzed Ea < Ea Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS) 13.6

48 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 (机制) by which the reac- tion can occur. The energy profiles for both the catalyzed and uncatalyzed reactions of decomposition of hydrogen peroxide are shown in Figure below. Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

49 Homogeneous Catalysis
N2O (g) (g)  N2 (g) + O2 (g) 1== N2O (g)  N2 (g) + O (g) 2==O (g) + N2O (g)  N2(g)+ O2 (g) Ea=240KJ/mol Cl2 Cl2(g)2Cl(g) N2O(g)+Cl(g)N2(g)+ClO(g) 2ClOCl2(g)+O2(g) Ea=140KJ/mol Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

50 Heterogeneous Catalysis
C2H4 + H2  C2H6 with a metal catalyst a. reactants b. adsorption c. migration/reaction d. desorption Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)

51 Enzyme Catalysis 13.6 Wednesday, April 26, 2017
GENERAL CHEMISTRY(KINETIKS) 13.6

52 Catalytic Converters catalytic converter catalytic converter 13.6
CO + Unburned Hydrocarbons + O2 CO2 + H2O catalytic converter 2NO + 2NO2 2N2 + 3O2 Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS) 13.6

53 The End!!!!!!! Wednesday, April 26, 2017 GENERAL CHEMISTRY(KINETIKS)


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