Chemistry 142 Chapter 13: Chemical Kinetics

Outline I.

Kinetics II. Rate A. Integrated Rate Law B. Reaction Rate and Temperature III. Reaction Mechanisms IV. Catalysis

at t = 0 [A] = 8 [B] = 8 [C] = 0 at t = 16 [A] = 4 [B] = 4 [C] = 4

Chapter 13 Examples – Rate

Tro, Chemistry: A Molecular Approach at t = 0 [X] = 8 [Y] = 8 [Z] = 0 at t = 16 [X] = 7 [Y] = 7 [Z] = 1 2

Concentration as a function of time 2 N

2

O

5 (g) 

4 NO

2 (g)

+ O

2 (g)

Chapter 13 – Examples: Rate 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) a) Calculate dinitrogen 200-300 s.

the average rate of decomposition of pentaoxide during the time interval b) Calculate the average rate of formation same interval.

of oxygen gas during the c) What is instantaneous rate the of nitrogen dioxide formation at 250 s?

0,025 0,02 0,015 0,01 0,005 0 0 Time (s) 0 100 200 300 400 500 600 700 [N 2 O 5 ] (M) 0.0200

0.0169

0.0142

0.0120

0.0101

0.0086

0.0072

0.0061

[N 2 O 5 ] vs. time (s)

100 200 300 400

time (s)

500 600 700 800

Continuous Monitoring Sampling

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Tro, Chemistry: A Molecular Approach 8

Energy Profile for the Isomerization of Methyl Isonitrile

9

Effective Collisions Orientation Effect

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Rate Laws of Elementary Steps

Tro, Chemistry: A Molecular Approach 11

Chapter 13 – Examples Mechanism Validation The balanced equation for the reaction of nitrogen dioxide and fluorine gases is: 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g) the experimentally determined rate law is rate = k[NO 2 ][F 2 ] A suggested mechanism is k 1 NO 2 (g) + F 2 (g)  NO 2 F (g) + F (g) F (g) + NO 2 (g) k 2  NO 2 F (g) Is this an acceptable mechanism?

slow fast

Chapter 13 – Examples Mechanism Validation The balanced equation for the reaction of nitrogen dioxide and fluorine gases is: 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g) the experimentally determined rate law is rate = k[NO 2 ][F 2 ] Another suggested mechanism is NO 2 (g) + F 2 (g)  NO 2 (g) + O (g)  NO 3 (g) NOF 2 (g) + NO 2 (g)  NO 2 F NO 3 (g) + NOF (g) NOF 2 (g)  NO 2 F + O (g) (g) (g) + NOF + NO 2 (g) (g) Is this an acceptable mechanism?

slow fast fast fast

Chapter 13 – Examples Mechanism Validation The balanced equation for the reaction of nitrogen monoxide and chlorine gases is: 2 NO (g) + Cl 2 (g)  2 NOCl the experimentally determined rate law is (g) rate = k[NO] 2 [Cl 2 ] A suggested mechanism is NO (g) k 1 + Cl 2 (g)  k -1 NOCl 2 (g) NOCl 2 (g) + NO (g) k 2  2 NOCl (g) fast slow Is this an acceptable mechanism?

Temperature’s Effect on Reaction Rates

Tro, Chemistry: A Molecular Approach 17

Arrhenius Plot

Chapter 13 – Examples Arrhenius Plot For the reaction: 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) Several values of k were obtained at different temperatures. Calculate the activation energy for the reaction.

T (°C) 20 30 40 50 60 k (s -1 ) 2.0 x 10 -5 7.3 x 10 -5 2.7 x 10 -4 9.1 x 10 -4 2.9 x 10 -3

T (°C) 20 30 40 50 60 T (K) 293 303 313 323 333 1/T (1/K) 3.41 x 10 -3 3.30 x 10 -3 3.19 x 10 -3 3.10 x 10 -3 3.00 x10 -3 Chapter 13 – Examples Arrhenius Plot 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) k (s -1 ) 2.0 x 10 -5 7.3 x 10 -5 2.7 x 10 -4 9.1 x 10 -4 2.9 x 10 -3 ln k -10.82

-9.53

-8.22

-7.00

-5.84

Calculate the activation energy for the reaction.

Arrhenius Plot ln k versus 1/T

-6 -8 -10 -12 0 2,90E-03 -2 -4 3,00E-03 3,10E-03 3,20E-03 3,30E-03 3,40E-03 3,50E-03 ln(k) = (-12174 K)/T + 30.694

R² = 0.9998

1/T (1/K)

Chapter 13 – Examples Arrhenius Equation For the gas phase reaction: CH 4 (g) + 2 S 2 (g)  CS 2 (g) + 2 H 2 S (g) The rate constant at 550 °C is 1.1 L/mol s and at 625 °C the value of k is 6.4 L/mol s. Calculate the activation energy for the reaction.

 Energetics in the Presence of a Catalyst

2 O 3 3 O 2 The catalyst, a chlorine atom, is not consumed during the chemical reaction.

Catalysis

Enzyme-Substrate Binding Lock and Key Mechanism

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Enzymatic Hydrolysis of Sucrose

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Types of Catalysts

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Catalytic Hydrogenation H

2

C=CH

2

+ H

2

→ CH

3

CH

3 Tro, Chemistry: A Molecular Approach 27

Molecular View of Catalytic Converters

2 NO

(g) 

N

2 (g)

+ O

2 (g)

First Order Reactions

ln[A] 0 ln[A] Tro, Chemistry: A Molecular Approach time 31

C 4 H 9 Cl

Rate Data for

(aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Tro, Chemistry: A Molecular Approach

Time (sec)

0.0

[C 4 H 9 Cl], M

0.1000

50.0

100.0

150.0

200.0

0.0905

0.0820

0.0741

0.0671

300.0

400.0

500.0

800.0

10000.0

0.0549

0.0448

0.0368

0.0200

0.0000

32

C

4

H

9

Cl

(aq)

+ H

2

O

(l) 

C

4

H

9

OH

(aq)

+ 2 HCl

(aq)

Concentration vs. Time for the Hydrolysis of C 4 H 9 Cl

0.12

0.1

0.08

0.06

0.04

0.02

0 0 Tro, Chemistry: A Molecular Approach 200 400 33

time, (s)

600 800 1000

C

4

H

9

Cl

(aq)

+ H

2

O

(l) 

C

4

H

9

OH

(aq)

+ 2 HCl

(aq)

Rate vs. Time for Hydrolysis of C 4 H 9 Cl

2.5E-04 2.0E-04 1.5E-04 1.0E-04 5.0E-05 0.0E+00 0 Tro, Chemistry: A Molecular Approach 100 200 300 34 400

time, (s)

500 600 700 800

C

4

H

9

Cl

(aq)

+ H

2

O

(l) 

C

4

H

9

OH

(aq)

+ 2 HCl

(aq) LN([C 4 H 9 Cl]) vs. Time for Hydrolysis of C 4 H 9 Cl 0 -0.5

-1 -1.5

-2 -2.5

-3 -3.5

-4 -4.5

0 100 200 Tro, Chemistry: A Molecular Approach slope = -2.01 x 10 -3

k =

2.01 x 10 -3 s -1 300 400 time, (s) 500 y = -2.01E-03x - 2.30E+00 600 700 800 t 1 2  0 .

693

k

 0 .

693 2 .

01  10  3 s 1  345 s 35

Chapter 13 – Examples Reaction Order Using the data below verify that the decomposition of dinitrogen pentaoxide is first order and calculate the value of the rate constant, where rate = -Δ[N 2 O 5 ]/Δt 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) Time (s) 0 50 100 200 300 400 [N 2 O 5 ] (mol/L) 0.100

0.0707

0.0500

0.0250

0.0125

0.00625

Chapter 13 – Examples Reaction Order Using the data below verify that the decomposition of dinitrogen pentaoxide is first order and calculate the value of the rate constant, where rate = -Δ[N 2 O 5 ]/Δt 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) 0,120 0,100

First Order Reaction Concentration [N 2 O 5 ] versus Time

0,080 Time (s) 0 50 100 200 300 400 [N 2 O 5 ] (mol/L) 0.100

0.0707

0.0500

0.0250

0.0125

0.00625

0,060 0,040 0,020 0,000 0 100 200

time (s)

300 400 500

Chapter 13 – Examples Reaction Order Using the data below verify that the decomposition of dinitrogen pentaoxide is first order and calculate the value of the rate constant, where rate = -Δ[N 2 O 5 ]/Δt 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) 0,000 0 -1,000 -2,000 Time (s) 0 50 100 200 300 400 [N 2 O 5 ] (mol/L) 0.100

0.0707

0.0500

0.0250

0.0125

0.00625

ln[N 2 O 5 ] -2.303

-2.649

-2.996

-3.689

-4.382

-5.075

-3,000 -4,000 -5,000 -6,000

First Order Reaction ln[N 2 O 5 ] versus Time

100 200 300 400 500 ln[N 2 O 5 ] = -0.0069(t) - 2.3026

time (s)

Half-Life of a First-Order Reaction Is Constant

Tro, Chemistry: A Molecular Approach 39

Chapter 13 – Examples Half-life A certain first order reaction has a half-life of 20.0 minutes.

a. Calculate the rate constant.

b. How much time is required for this reaction to be 75% complete?

Second Order Reactions

1/[A] l/[A] 0 time Tro, Chemistry: A Molecular Approach 41

Butadiene reacts to form its dimer according to: 2 C 4 H 6 (g)  C 8 H 12 (g) a) Is this reaction first order or second order?

Chapter 13 – Examples Rate Laws b) What is the value of the rate constant function of the reaction?

c) What is the half-life for the reaction under the conditions of this experiment?

time (s) 0 1000 1800 2800 3600 4400 5200 6200 [C 4 H 6 ] (mol/L) 0.0100

0.00625

0.00476

0.00370

0.00313

0.00270

0.00241

0.00208

Butadiene reacts to form its dimer according to: 2 C 4 H 6 (g)  C 8 H 12 (g) a) Is this reaction first order or second order?

b) What is the value of the rate constant function of the reaction?

c) What is the half-life for the reaction conditions experiment?

under of the this

First Order Reaction Plot ln[C 4 H 6 ] versus time

-4,500 0 -5,000 1000 2000 3000 4000 5000 6000 time (s) 0 1000 1800 2800 3600 4400 5200 6200 7000 -5,500 -6,000 -6,500

time (s)

Chapter 13 – Examples Rate Laws 600 500 400 300 200 100 0 0 [C 4 H 6 ] (mol/L) 0.0100

0.00625

0.00476

0.00370

0.00313

0.00270

0.00241

0.00208

Second Order Reaction Plot 1/[C 4 H 6 ] versus time

1000 2000 ln[C 4 H 6 ] -4.605

-5.075

-5.348

-5.599

-5.767

-5.915

-6.028

-6.175

3000 4000

time (s)

1/[C 4 H 6 ] (L/mol) 100 160 210 270 319 370 415 481 1/[C 4 H 6 ] = (0.0612L/mol s)t + 99.36 L/mol 5000 6000 7000

Half-life for a second order reaction

Zero Order Reactions

[A] 0 [A] time Tro, Chemistry: A Molecular Approach 45

Tro, Chemistry: A Molecular Approach 46

Chapter 13 – Examples: Rate a) 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) Calculate the average rate of decomposition dinitrogen of pentaoxide during the time interval 200-300 s.

b) Calculate the average rate of formation of oxygen gas during the same interval.

c) d) What is the instantaneous rate of nitrogen dioxide formation at 250 s?

Is this reaction first order, second order, or zero order?

Time (s) 0 100 200 300 400 500 600 700 [N 2 O 5 ] (M) 0.0200

0.0169

0.0142

0.0120

0.0101

0.0086

0.0072

0.0061

Chapter 13 – Examples: Rate 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) d) Is this reaction first order, Time (s) 0 second order, or zero order?

[N 2 O 5 ] vs. time (s)

100 200 300 0,025 400 0,02 500 0,015 600 0,01 700 0,005 0 0 200 400

time (s) ln[N 2 O 5 ] vs. time (s)

600 800 200,0 150,0 200 400 600 800 -4,050 0 100,0 -4,250 -4,450 50,0 -4,650 -4,850 -5,050

time (s)

0,0 0 [N 2 O 5 ] (M) ln[N 2 O 5 ] 1/[N 2 O 5 ] 0.0200

-3.912

50.0

0.0169

-4.080

59.2

0.0142

-4.255

70.4

0.0120

-4.423

83.3

0.0101

-4.595

99.0

0.0086

-4.76

116 0.0072

-4.93

139 0.0061

-5.10

164

1/[N 2 O 5 ] vs. time (s)

100 200 300 400

time (s)

500 600 700 800

Chapter 13 – Examples Initial rate method

Describe the general form of the rate law for the reaction given the results of four experiments below: NH 4 + (aq) + NO 2 (aq)  N 2 (g) + 2 H 2 O (l) Experiment 1 2 3 [NH 4 + ] 0 0.100 M 0.100 M 0.200 M [NO 2 ] 0 0.0050 M 0.010 M 0.010 M Initial rate (mol/L s) 1.35 x 10 -7 2.70 x 10 -7 5.40 x 10 -7

Chapter 13 – Examples Initial rate method

Describe the general form of the rate law for the reaction given the results of three experiments below: CH 3 COOCH 3 (aq) + OH (aq)  CH 3 COO (aq) + CH 3 OH (aq) Experiment 1 2 3 [CH 3 COOCH 3 ] 0 0.040 M 0.040 M 0.080 M [OH ] 0 0.040 M 0.080 M 0.080 M Initial rate (mol/L s) 0.00022

0.00045

0.00090

Chapter 13 – Examples Initial rate method

Describe the general form of the rate law for the reaction given the results of three experiments below: CH 3 COOH (aq) + OH (aq)  CH 3 COO (aq) + CH 3 OH (aq) Experiment 1 2 3 [CH 3 COOH ] 0 0.040 M 0.040 M 0.080 M [OH ] 0 0.040 M 0.080 M 0.080 M Initial rate (mol/L s) 0.00022

0.00045

0.00090

k (L/mol s) 0.0138

0.0141

0.0141

Chapter 13 – Examples Initial Rate Method

The reaction between bromate ions and bromide ions in an acidic solution is given by the equation: BrO 3 (aq) + 5 Br (aq) + 6 H + (aq)  3 Br 2 (l) + 3 H 2 O (l) Describe the general form of the rate law for this reaction given the results of four experiments below.

Experiment 1 2 3 4 [BrO 3 ] 0 0.10 M 0.20 M 0.20 M 0.10 M [Br ] 0 0.10 M 0.10 M 0.20 M 0.10 M [H + ] 0 0.10 M 0.10 M 0.10 M 0.20 M Initial rate (mol/L s) 8.0 x 10 -4 1.6 x 10 -3 3.2 x 10 -3 3.2 x 10 -3