chemical kinetics - students

Copyright©2000 by Houghton Mifflin Company. All rights

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ChemistryFIFTH EDITION

Chapter 12Chemical Kinetics


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Chemical Kinetics

• The area of chemistry that concerns reaction rates.


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Reaction Rate

•Change in concentration (conc) of a reactant or product per unit time.

Rate = conc of A at time conc of A at time 2 1

2 1

t tt t

At

Concentration of reactants and products as a function of time for the reaction at 300˚C

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

Time (+ 1 s) Concentration (mol/L)

NO2 NO O2

0 0.0100 0 0

50 0.0079 0.0021 0.0011

100 0.0065 0.0035 0.0018

150 0.0055 0.0045 0.0023

200 0.0048 0.0052 0.0026

250 0.0043 0.0057 0.0029

300 0.0038 0.0062 0.0031

350 0.0034 0.0066 0.0033

400 0.0031 0.0069 0.0035


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Figure 12.1Definition of Rate

Starting with a flask of nitrogen dioxide at 300 ºC, the concentration of nitrogen dioxide, nitric oxide, and oxygen are plotted vs. time.

Rate of consumption of NO2 = Rate of consumption of NO = 2 times the rate of consumption of O2


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Figure 12.2Representation of the Reaction of 2NO2(g) g) + O2(g)

The figure shows the initial concentration, and the concentration as time passes. NO2 is converted to NO and O2


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Rate Laws

• Rate = k[NO2]n

• k = rate constant

• n = rate order


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Types of Rate Laws

• Differential Rate Law: expresses how rate depends on concentration.

• Integrated Rate Law: expresses how concentration depends on time.


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Figure 12.3A Plot of the Concentration of N2O5 as a Function of

Time for the Reaction

Note that the reaction rate at [N2O5]=0.90 M is twice the rate at [N2O5]=0.45 M

Rate Laws


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]ON[]ON[]O[N

- Rate 521

5252 kk

t

The reaction is first orderNote that the order is not the same as the coefficient from the reaction


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Method of Initial Rates

•Initial Rate: the “instantaneous rate” just after the reaction begins.

•The initial rate is determined in several experiments using different initial concentrations.


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Overall Reaction Order

•Sum of the order of each component in the rate law.

•rate = k[H2SeO3][H+]2[I]3

•The overall reaction order is 1 + 2 + 3 = 6.

Example 12.1

The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation

BrO3- (aq) + 5Br-(aq) + 6H+(aq) 3Br2(l) +3H2O(l)

The following table gives the results from four experiments. Using this data, determine the orders for all three reactions, the overall reaction order, and the value of the rate constant.


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Example 12.1

Experiment Initial Concentration of BrO3

- (mol/L)

Initial Concentration of Br- (mol/L)

Initial Concentration of H+ (mol/L)

Measured Initial Rate (mol/L·s)

1 0.10 0.10 0.10 8.0x10-4

2 0.20 0.10 0.10 1.6x10-3

3 0.20 0.20 0.10 3.2x10-3

4 0.10 0.10 0.20 3.2x10-3


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The results from four experiments to study the reactionBrO3

- (aq) + 5Br-(aq) + 6H+(aq) 3Br2(l) +3H2O(l)

Example 12.1

Rate = k[BrO3-]n[Br-]m[H+]p

Determine the values of n, m, and p by comparing rates from the various experiments.

Compare exp 1 to exp 2 where only [BrO3-] changes.

n = 1


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Example 12.1

Compare exp 3 to exp 2 where only [Br-] changes.

m = 1

Compare exp 4 to exp 1 where only [H+] changes.

p = 2


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Problem 12.23The reaction

2NO (g) + Cl2 (g) 2NOCl (g)

Was studied at -10ºC. The following results were obtained where

a.What is the rate law?

b.What is the value of the rate constant?


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t

Cl

][

-Rate 2

[NO]0 (mol/L) [Cl2]0 (mol/L) Initial Rate (mol/L·min)

0.10 0.10 0.18

0.10 0.20 0.36

0.20 0.20 1.45


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First-Order Rate Law

• Integrated first-order rate law is

• ln[A] = kt + ln[A]o

• The reaction is first order in A if a plot of ln[A] versus t is a straight line.

Rate = A

A

t

k

For For aaA A Products in a 1st-order reaction, Products in a 1st-order reaction,

Example 12.2The decomposition of N2O5 in the gas phase was studied at constant temperature.

2N2O5(g) 4NO2(g) + O2(g)

The following results were collected:

Using this data verify that

The reaction is first order in

[N2O5], and calculate the

Value of the rate constant,

Where the rate = -∆[N2O5]/∆t


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[N2O5] (mol/L) Time (s)

0.1000 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400

Example 12.3Using the data given in example 12.2, calculate [N2O5] at 150 s after the start of the reaction.

From example 12.2, [N2O5] = 0.0500 mol/L at 100 s, and [N2O5] = 0.0250 mol/L at 200 s.

ln[N2O5] = -kt + ln[N2O5]0

Where t = 150 s, k = 6.93x10-3 s-1, [N2O5]0=0.10000 mol/L


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Problem 12.29The decomposition of hydrogen peroxide was studied, and the following data was obtained at a particular temperature.

Assuming that

Determine the rate law, the integrated rate law. What is the value of the rate constant.

Calculate [H2O2] exactly 4000 s after the start of the reaction.

Time (s) [H2O2] (mol/L)

0 1.00

120 ±1 0.91

300 ±1 0.78

600 ±1 0.59

1200 ±1 0.37

1800 ±1 0.22

2400 ±1 0.13

3000 ±1 0.082

3600 ±1 0.050


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t

OHRate

][ 22


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Half-Life of a First-Order Reaction

The time required for a reactant to reach half its original concentration is called the half-life.

Calculate the half-life of the reaction in example 12.2


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Figure 12.5A Plot of (N2O5) Versus Time for the Decomposition Reaction of N2O5

Consider the following numbers

[N2O5] (mol/L) t(s)

0.100 0

∆t=100 s [N2O5]t=100 = 0.050 = 1

0.0500 100 [N2O5]t=0 0.100 2

∆t=100 s [N2O5]t=200 = 0.025 = 1

0.0250 200 [N2O5]t=100 0.050 2

∆t=100 s [N2O5]t=300 = 0.0125 = 1

0.0125 300 [N2O5]t=200 0.0250 2


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Note that it always takes 100 seconds for [N2O5] to decrease by 1/2


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If the reaction is first order in [A],

ln([A]0/[A]) = kt

By definition, when t = t1/2

[A] = [A]0/2


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• t1/2 = half-life of the reaction

• k = rate constant

•For a first-order reaction, the half-life does not depend on concentration.

tk1/2

0 693.

Example 12.4

A certain first-order reaction was a half-life of 20.0 minutes.

a.Calculate the rate constant for this reaction.

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


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Problem 12.37

A certain first order reaction is 45.0% complete in 65 s. What is the rate constant for this process? What is the half-life for this process?


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Problem 12.39

A first order reaction is 38.5% complete in 480 s.

a.Calculate the rate constant.

b.What is the value of the half-life.

c.How long will it take for the reaction to go to 25% completion?

d.How long will it take for the reaction to go to 75% completion?

e.How long will it take for the reaction to go to 95% completion?


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Second-Order Rate Law• For aA products in a second-order

reaction,

• Integrated rate law is

Rate = A

A

t

k 2

1A

+ 1

A o

kt


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Half-Life of a Second-Order Reaction

•t1/2 = half-life of the reaction

•k = rate constant

•Ao = initial concentration of A

•The half-life is dependent upon the initial concentration.

tk1/2

oA

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Example 12.5Butadiene reacts to form its dimer according to the equation

2C4H6 (g) C8H12 (g)

The following data were collected for this reaction at a given temperature:

[C4H6] (mol/L) Time (+ 1 s)

0.08000 0

0.00625 1000

0.00476 1800

0.00370 2800

0.00313 3600

0.00270 4400

0.00241 5200

0.00209 6200


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a. Is this reaction first or second order?b. What is the value of the rate constant for the reaction?c. What is the half-life for the reaction under the conditions

of this experiment?

Problem 12.31

The rate of the reaction

NO2(g) + CO(g) NO(g) + CO2(g)

depends only on the concentration of nitrogen dioxide below 225˚C. At a temperature below 225˚C, the following data were collected

Time (s) [NO2] (mol/L)

0 0.500

1.20 x 103 0.444

3.00 x 103 0.381

4.50 x 103 0.340

9.00 x 103 0.250

1.80 x 104 0.174


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Determine the rate law and the integrated rate law. What is the value of the rate constant? Calculate the [NO2] at 2.70 x 104 s after the start of the reaction.

Problem 12.33The decomposition of ethanol (C2H5OH) on an alumina (Al2O3) surface

C2H5OH (g) C2H4 (g) + H2O (g)

Was studied at 600K. Concentration versus time data were collected for this reaction, and a plot of [A] versus time resulted in a straight line with a slope of -4.00x10-5 mol/L·s.

a.Determine the rate law and the integrated rate law. What is the value of the rate constant for this reaction?

b.If the initial concentration of C2H5OH was 1.25 x 10-2 M, calculate the half life for this reaction.

c.How much time is required for all the 1.25 x 10-2 M C2H5OH to decompose?


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Zero-Order Rate Law

• For aA products in a zero-order reaction,

Rate = k[A]0 = k(1) = k

• Integrated rate law is

[A] = -kt + [A]0


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Half-Life of a Zero-Order Reaction

•t1/2 = half-life of the reaction

•k = rate constant

•Ao = initial concentration of A

•The half-life is dependent upon the initial concentration.

k

At

2

][ 0

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Figure 12.7A Plot of (A) Versus t for a Zero-Order Reaction

A first order reaction has a constant rate.


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A Summary

1. Simplification: Conditions are set such that only forward reaction is important.

2. Two types: differential rate law integrated rate law

3. Which type? Depends on the type of data collected - differential and integrated forms can be interconverted.


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A Summary (continued)

4. Most common: method of initial rates.

5. Concentration v. time: used to determine integrated rate law, often graphically.

6. For several reactants: choose conditions under which only one reactant varies significantly (pseudo first-order conditions).

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

Order Rate LawConcentration-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

t½ln 2k

=

t½ =[A]0

2k

t½ =1

k[A]0

Order Units

zero

first

second

third

molV-1s-1

s-1

Vmol-1s-1

(Vmol-1)2s-1


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Arrhenius Equation

• Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy).

• Orientation of reactants must allow formation of new bonds.


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Arrhenius Equation (continued)

• k = rate constant• A = frequency factor

• Ea = activation energy

• T = temperature• R = gas constant

k Ae E RT a /

Example 12.7The reaction

2N2O5 (g) 4NO2 (g) + O2 (g)

Was studied at several temperatures, and the following values of k were obtained:

k (s-) T (˚C)

2.0 x 10-5 20

7.3 x 10-5 30

2.7 x 10-4 40

9.1 x 10-4 50

2.9 x 10-3 60


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Calculate the value of Ea for this reaction.

Problem 12.53Experimental values for the temperature dependence of the rate constant for the gas phase reaction

NO + O3 NO2 + O2

Are as follows:

T (K) k (L/mol·s)

195 1.08 x 109

230 2.95 x 109

260 5.42 x 109

298 12.0 x 109

369 35.5 x 109


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Make the appropriate graph using this data and determine the activation energy for this reaction.

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Alternate Form of the Arrhenius Equation

At two temperatures, T1 and T2

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