16-1 copyright ©the mcgraw-hill companies, inc. permission required for reproduction or display....

16-1

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 14

Kinetics: Rates and Mechanisms of Chemical Reactions

16-2


Figure 16.1 Reaction rate: the central focus of chemical kinetics

16-3


Figure 16.2 The wide range of reaction rates.

16-4


Factors That Influence Reaction Rate

Under a specific set of conditions, every reaction has its owncharacteristic rate, which depends upon the chemical nature ofthe reactants.

Five factors can be controlled during the reaction:

1. Concentration - molecules must collide to react;

2. Physical state - molecules must mix to collide;

3. Surface area – if reactants are in different phases, reaction occurs

where these phases meet

4. Temperature - molecules must collide with enough energy to react;

5. The use of a catalyst.

16-5


Figure 16.3 The effect of surface area on reaction rate.

16-6


Figure 16.4 Collision energy and reaction rate.

16-7


Expressing the Reaction Rate

reaction rate - changes in the concentrations of reactants orproducts per unit time

reactant concentrations decrease while product concentrationsincrease

rate of reaction = -

for A B

change in concentration of A

change in time= -

conc A2-conc A1

t2-t1

(conc A)-

t

16-8


Table 16.1 Concentration of O3 at Various Time in its

Reaction with C2H4 at 303K

C2H4(g) + O3(g) C2H4O(g) + O2(g)

Time (s) Concentration of O3 (mol/L)

0.0

20.0

30.0

40.0

50.0

60.0

10.0

3.20x10-5

2.42x10-5

1.95x10-5

1.63x10-5

1.40x10-5

1.23x10-5

1.10x10-5

(conc A)-

t

16-9


Figure 16.5

The concentrations of O3 vs. time during its reaction with C2H4

C2H4(g) + O3(g) C2H4O(g) + O2(g)

- [C2H4]

t

rate =

- [O3]

t

=

16-10


Figure 16.6 Plots of [C2H4] and [O2] vs. time.

Tools of the Laboratory

16-11


Table 16.2 Initial Rates for a Series of Experiments in the Reaction Between O2 and NO

Experiment

Initial Reactant Concentrations (mol/L)

Initial Rate (mol/L*s)

1

2

3

4

5

O2 NO

1.10x10-2 1.30x10-2 3.21x10-3

1.10x10-2 3.90x10-2 28.8x10-3

2.20x10-2

1.10x10-2

3.30x10-2

1.30x10-2

2.60x10-2

1.30x10-2

6.40x10-3

12.8x10-3

9.60x10-3

2NO(g) + O2(g) 2NO2(g)

16-12


Determining Reaction Orders

Using initial rates -

Run a series of experiments, each of which starts with a different set of reactant concentrations, and from each obtain an initial rate.

See Table 16.2 for data on the reaction

O2(g) + 2NO(g) 2NO2(g) rate = k [O2]m[NO]n

Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant.

k [O2]2m[NO]2

n

k [O2]1m[NO]1

n=

rate2

rate1 =

[O2]2m

[O2]1m

=

6.40x10-3mol/L*s

3.21x10-3mol/L*s

[O2]2

[O2]1

m

=1.10x10-2mol/L

2.20x10-2mol/L m; 2 = 2m m = 1

Do a similar calculation for the other reactant(s).

16-13


Sample Problem 16.3

PLAN:

SOLUTION:

Determining Reaction Order from Initial Rate Data

PROBLEM: Many gaseous reactions occur in a car engine and exhaust system. One of these is

NO2(g) + CO(g) NO(g) + CO2(g) rate = k[NO2]m[CO]n

Use the following data to determine the individual and overall reaction orders.

Experiment Initial Rate(mol/L*s) Initial [NO2] (mol/L) Initial [CO] (mol/L)

1

2

3

0.0050

0.080

0.0050

0.10

0.10

0.40

0.10

0.10

0.20

Solve for each reactant using the general rate law using the method described previously.

rate = k [NO2]m[CO]n

First, choose two experiments in which [CO] remains constant and the [NO2] varies.

16-14


Sample Problem 16.3 Determining Reaction Order from Initial Rate Data

continued

0.080

0.0050

rate 2

rate 1

[NO2] 2

[NO2] 1

m=

k [NO2]m2[CO]n

2

k [NO2]m1 [CO]n

1

=

0.40

0.10=

m; 16 = 4m and m = 2

k [NO2]m3[CO]n

3

k [NO2]m1 [CO]n

1

[CO] 3

[CO] 1

n=

rate 3

rate 1=

0.0050

0.0050=

0.20

0.10

n; 1 = 2n and n = 0

The reaction is 2nd order in NO2.

The reaction is zero order in CO.

rate = k [NO2]2[CO]0 = k [NO2]2

16-15


Table 16.3 Units of the Rate Constant k for Several Overall Reaction Orders

Overall Reaction Order Units of k (t in seconds)

0 mol/L*s (or mol L-1 s-1)

1 1/s (or s-1)

2 L/mol*s (or L mol -1 s-1)

3 L2 / mol2 *s (or L2 mol-2 s-1)

16-16


Figure 16.14

The effect of temperature on the distribution of collision energies

16-17


Table 16.6 Rate Laws for General Elementary Steps

Elementary Step Molecularity Rate Law

A product

2A product

A + B product

2A + B product

Unimolecular

Bimolecular

Bimolecular

Termolecular

Rate = k [A]

Rate = k [A]2

Rate = k [A][B]

Rate = k [A]2[B]

REACTION MECHANISMS

16-18


Sample Problem 16.8

PLAN:

SOLUTION:

Determining Molecularity and Rate Laws for Elementary Steps

PROBLEM: The following two reactions are proposed as elementary steps in the mechanism of an overall reaction:

(1) NO2Cl(g) NO2(g) + Cl (g)

(2) NO2Cl(g) + Cl (g) NO2(g) + Cl2(g)

(a) Write the overall balanced equation.

(b) Determine the overall rxn order for each step.

(a) The overall equation is the sum of the steps.(b) The molecularity is the sum of the reactant particles in the step.

2NO2Cl(g) 2NO2(g) + Cl2(g)

(c) Write the rate law for each step.

rate2 = k2 [NO2Cl][Cl]

(1) NO2Cl(g) NO2(g) + Cl (g)

(2) NO2Cl(g) + Cl (g) NO2(g) + Cl2(g)

(a)

Step(1) is unimolecular.Step(2) is bimolecular.

(b)

rate1 = k1 [NO2Cl](c)

16-19


The Rate-Determining Step of a Reaction Mechanism

The overall rate of a reaction is related to the rate of the slowest, or rate-determining step.

Correlating the Mechanism with the Rate Law

The elementary steps must add up to the overall equation.

The elementary steps must be physically reasonable.

The mechanism must correlated with the rate law.

16-1 copyright ©the mcgraw-hill companies, inc. permission required for reproduction or display....

Documents