Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration.

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]t

rate = [B]t

[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.

13.11GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023

A B

13.1

rate = -[A]

t

rate = [B]

t

time

2GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023

Reaction rate = change in Reaction rate = change in concentration of a reactant or concentration of a reactant or product with time.product with time.

Three “types” of ratesThree “types” of rates initial rateinitial rateaverage rateaverage rateinstantaneous rateinstantaneous rate

Reaction Rates Reaction Rates


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

average rate = -[Br2]

t= -

[Br2]final – [Br2]initial

tfinal - tinitial

slope oftangent

slope oftangent slope of

tangent

instantaneous rate = rate for specific instance in time13.14GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023

2NO22NO+O2



13.3

Concentrations & RatesConcentrations & Rates

0.3 M HCl 6 M HCl

Mg(s) + 2 HCl(aq) ---> MgClMg(s) + 2 HCl(aq) ---> MgCl22(aq) + H(aq) + H22(g)(g)


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]2.5

What is the reaction order?First in A, 2.5 in B

Overall reaction order?2.5 +1 = 3.5


Interpreting Rate Laws Interpreting Rate Laws Rate = k [A]Rate = k [A]mm[B][B]nn[C][C]pp

If m = 1, rxn. is 1st order in AIf m = 1, rxn. is 1st order in ARate = k [A]Rate = k [A]11

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]Rate = k [A]22

Doubling [A] increases rate by ________Doubling [A] increases rate by ________ If m = 0, rxn. is zero order.If m = 0, rxn. is zero order.

Rate = k [A]Rate = k [A]00

If [A] doubles, rate ________If [A] doubles, rate ________


Determine the rate law and calculate the rate constant for the following reaction from the following data:

S2O82- (aq) + 3I- (aq) 2SO4

2- (aq) + I3- (aq)

Experiment [S2O82-] [I-]

Initial Rate (M/s)

1 0.08 0.034 2.2 x 10-4

2 0.08 0.017 1.1 x 10-4

3 0.16 0.017 2.2 x 10-4

rate = k [S2O82-]x[I-]y

Double [I-], rate doubles (experiment 1 & 2)

y = 1

Double [S2O82-], rate doubles (experiment 2 & 3)

x = 1

k = rate

[S2O82-][I-]

=2.2 x 10-4 M/s

(0.08 M)(0.034 M)= 0.08/M•s

13.2

rate = k [S2O82-][I-]


Deriving Rate LawsDeriving Rate Laws

Expt. Expt. [CH[CH33CHO]CHO] Disappear of CHDisappear of CH33CHOCHO(mol/L)(mol/L) (mol/L•sec)(mol/L•sec)

11 0.100.10 0.0200.020

22 0.200.20 0.0810.081

33 0.300.30 0.1820.182

44 0.400.40 0.3180.318

Derive Derive rate law rate law and and kk for for CHCH33CHO(g) --> CHCHO(g) --> CH44(g) + CO(g)(g) + CO(g)

from experimental data for rate of disappearance from experimental data for rate of disappearance of CHof CH33CHOCHO

12GENERAL CHEMISTRY(KINETIKS)

Rate of rxn = k [CHRate of rxn = k [CH33CHO]CHO]22

Deriving Rate LawsDeriving Rate Laws

Rate of rxn = k [CHRate of rxn = k [CH33CHO]CHO]22

Here the rate goes up by ______ when initial conc. Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is doubles. Therefore, we say this reaction is _________________ order._________________ order.

Now determine the value of k. Use expt. #3 data—Now determine the value of k. Use expt. #3 data—0.182 mol/L•s = k (0.30 mol/L)0.182 mol/L•s = k (0.30 mol/L)22

k = 2.0 (L / mol•s)k = 2.0 (L / mol•s)Using k you can calc. rate at other values of [CHUsing k you can calc. rate at other values of [CH33CHO] CHO]

at same T.at same T.


Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 14–14

Expession of R &KIodide 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.

)(2)()(2)(3)( 2322 lOHaqIaqHaqIaqOH


A Problem to ConsiderInitial Concentrations (mol/L)

H2O2 I- H+ Initial Rate [mol/(L.s)]Exp. 1 0.010 0.010 0.00050 1.15 x 10-6

Exp. 2 0.020 0.010 0.00050 2.30 x 10-6

Exp. 3 0.010 0.020 0.00050 2.30 x 10-6

Exp. 4 0.010 0.010 0.00100 1.15 x 10-6

– 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.




Exp. 2 0.020 0.010 0.00050 2.30 x 10-6

Exp. 3 0.010 0.020 0.00050 2.30 x 10-6

Exp. 4 0.010 0.010 0.00100 1.15 x 10-6

– 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-.




Exp. 2 0.020 0.010 0.00050 2.30 x 10-6

Exp. 3 0.010 0.020 0.00050 2.30 x 10-6

Exp. 4 0.010 0.010 0.00100 1.15 x 10-6

– Comparing Experiment 1 and Experiment 4, you see that when – the H+ concentration doubles (with other concentrations constant),– the rate is unchanged.– This implies a zero-order dependence with respect to H+.


A Problem to Consider

– Because [H+]0 = 1, the rate law is:

]I][OH[kRate 22

– The reaction orders with respect to H2O2, I-, and H+, are 1, 1, and 0, respectively.

Initial Concentrations (mol/L)H2O2 I- H+ Initial Rate [mol/(L.s)]

Exp. 1 0.010 0.010 0.00050 1.15 x 10-6

Exp. 2 0.020 0.010 0.00050 2.30 x 10-6

Exp. 3 0.010 0.020 0.00050 2.30 x 10-6

Exp. 4 0.010 0.010 0.00100 1.15 x 10-6




Exp. 2 0.020 0.010 0.00050 2.30 x 10-6

Exp. 3 0.010 0.020 0.00050 2.30 x 10-6

Exp. 4 0.010 0.010 0.00100 1.15 x 10-6

– You can now calculate the rate constant by substituting values from– any of the experiments. Using Experiment 1 you obtain:

Lmol

010.0Lmol

010.0ksL

mol1015.1 6




Exp. 2 0.020 0.010 0.00050 2.30 x 10-6

Exp. 3 0.010 0.020 0.00050 2.30 x 10-6

Exp. 4 0.010 0.010 0.00100 1.15 x 10-6

– You can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 you obtain:

)/(102.1/010.0010.0

1015.1 216

smolLLmol

sk

Zero-Order Reactions

13.3

A product rate = -[A]

trate = k [A]0 = k

k = rate

[A]0=RATE= M/s

[A]

t= 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

t½ =[A]0

2k

[A] = [A]0 - kt

ZERO ORDER REACTION

0

2

4

6

8

10

12

0 2 4 6 8 10 12

TIME

[A]

First-Order Reactions

13.3

A product rate = k [A]

k = rate

[A]= 1/s or s-1

M/s

M=

[A]

t= 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

t½ =0.693

k


Second-Order Reactions

13.3

A product rate = -[A]

trate = k [A]2

k = rate

[A]2= 1/M•s

M/s

M2=

[A]

t= 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

+ kt

t½ = t when [A] = [A]0/2

t½ =1

k[A]0

SECOND ORDER REACTION

0

2

4

6

8

10

12

0 1 2 3 4 5

TIME

[A]

SECOND ORDER REACTION

0

0.1

0.2

0.3

0.4

0.5

0.6

0 1 2 3 4 5

TIME

1/[A

]

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

Summary of the Kinetics of Zero-Order, First-Orderand Second-Order Reactions

Order Rate 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

t½

ln2

k=

t½ =[A]0

2k

t½ =1

k[A]0

13.3

Collision TheoryCollision TheoryCollision TheoryCollision TheoryOO33(g) + NO(g) ---> O(g) + NO(g) ---> O22(g) + NO(g) + NO22(g)(g)

10 31 COLLISINS/LIT.SReactions require Reactions require (A)(A) geometry geometry(B) (B) Activation energy Activation energy


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

Collision TheoryCollision Theory

TEMPERATURE10 c0 T 100-300% RATE25 c0 T 35 c0 2% COLLISINEFFECTIVE COLLISIONS

Friday, April 21, 2023 GENERAL CHEMISTRY(KINETIKS) 28


① ① with sufficient energywith 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.


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.



A + B C + D

Exothermic Reaction Endothermic Reaction

The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction.


Temperature Dependence of the Rate Constant

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

(Arrhenius equation)


2.05

2.1

2.15

2.2

2.25

2.3

2.35

2.4

2.45

0 200 400 600 800 1000 1200

TEMPERATURE

K

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


0.74

0.76

0.78

0.8

0.82

0.84

0.86

0.88

0.9

0 0.0005 0.001 0.0015 0.002 0.0025

1/T

Ln

K

•k = A • e( -Ea/RT )

ln k = ln A-Ea/RT=

(-Ea/R)(1/T) + ln A

Ea/RTdecreases

- Ea/RTincreases

e- Ea/RT

increases kincreases

REACTI ONSPEEDS UP

I f Tincreases


300400C0 Rate20000 Times400500C0 Rate400 Times

Ea=60KJ/mol300310C0 Rate2 TimesEa=260KJ/mol300310C0 Rate25 Times


K in different T


Log(K2/K1)=-Ea/2.303R(1/T2-1/T1)

Log(K2/K1) =Ea/2.303R(T2-T1)/T1T2

Ln(K2/K1)=-Ea/R(1/T2-1/T1)

MechanismsMechanismsOO33 + NO reaction occurs in a single + NO reaction occurs in a single

ELEMENTARYELEMENTARY step. Most others involve a step. Most others involve a sequence of elementary steps.sequence of elementary steps.

Adding elementary steps gives NET reaction.Adding elementary steps gives NET reaction.


MechanismsMechanisms

Most rxns. involve a sequence of elementary Most rxns. involve a sequence of elementary steps.steps.

2 I2 I-- + H + H22OO22 + 2 H + 2 H++ ---> I ---> I22 + 2 H + 2 H22OO

Rate = k [IRate = k [I--] [H] [H22OO22]]NOTENOTE1.1. Rate law comes from experimentRate law comes from experiment2.2. Order and stoichiometric coefficients Order and stoichiometric coefficients

not necessarily the same!not necessarily the same!3.3. Rate law reflects all chemistry Rate law reflects all chemistry

including including the slowest step the slowest step in multistep in multistep reaction.reaction.


MechanismsMechanisms

Proposed MechanismProposed Mechanism

Step 1 — slowStep 1 — slow HOOH + IHOOH + I-- --> HOI + OH --> HOI + OH--

Step 2 — fastStep 2 — fast HOI + IHOI + I-- --> I --> I22 + OH + OH--

Step 3 — fastStep 3 — fast 2 OH2 OH- - + 2 H + 2 H++ --> 2 H --> 2 H22OO

Rate of the reaction controlled by slow step —Rate of the reaction controlled by slow step —

RATE DETERMINING STEPRATE DETERMINING STEP, rds., rds.

Rate can be no faster than rds!Rate can be no faster than rds!

The species HOI and OHThe species HOI and OH-- are are reaction reaction intermediates.intermediates.

Proposed MechanismProposed Mechanism

Step 1 — slowStep 1 — slow HOOH + IHOOH + I-- --> HOI + OH --> HOI + OH--

Step 2 — fastStep 2 — fast HOI + IHOI + I-- --> I --> I22 + OH + OH--

Step 3 — fastStep 3 — fast 2 OH2 OH- - + 2 H + 2 H++ --> 2 H --> 2 H22OO

Rate of the reaction controlled by slow step —Rate of the reaction controlled by slow step —

RATE DETERMINING STEPRATE DETERMINING STEP, rds., rds.

Rate can be no faster than rds!Rate can be no faster than rds!

The species HOI and OHThe species HOI and OH-- are are reaction reaction intermediates.intermediates.

Most rxns. involve a sequence of elementary steps.Most rxns. involve a sequence of elementary steps. 2 I2 I-- + H + H22OO22 + 2 H + 2 H++ ---> I ---> I22 + 2 H + 2 H22OO

Rate = k [IRate = k [I--] [H] [H22OO22]]


MechanismsMechanisms 2NO+F22NOF R=K[NO][F2]

1=NO+F2NOF+F R1=K1[NO][F2]

Rate limiting step(RDS)2=NO+F NOF R2=K2[NO][F]


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-]


SN2

OH- +CH3Br CH3OH + Br-

R=K[CH3Br][OH-]



Rate Laws Rate Laws and and MechanismsMechanisms

NO2 + CO reaction: Rate = k[NO2]2

Single step

Two possible Two possible mechanismsmechanisms Two steps:


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‘ 13.647GENERAL CHEMISTRY(KINETIKS)Friday, April 21, 2023

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.

Br-


Homogeneous CatalysisN2O (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) 2ClOCl2(g)+O2(g)Ea=140KJ/mol


C2H4 + H2 C2H6

with a metal catalyst

a. reactantsb. adsorptionc. migration/reactiond. desorption

Heterogeneous Catalysis


Enzyme Catalysis


Catalytic Converters

13.6

CO + Unburned Hydrocarbons + O2 CO2 + H2Ocatalytic

converter

2NO + 2NO2 2N2 + 3O2

catalyticconverter


The End!!!!!!!The End!!!!!!!


Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration.

Documents

rate doubles

rate expression

rate constant

rate of disappearance

increases rate

general chemistrykinetiks13

initial rate ms10

aa bb cc dd rate