12 Kinetics 6.pdf

(1)

CHAPTER 12:

CHEMICAL KINETICS

The study of how fast reactions take place.

Some happen almost instantaneously, while

others take millions of years.

Chemical Kinetics

Kinetic

Resulting from motion

Chemical Kinetics

Concerns the “motion” or

progress of a reaction

How long does a reaction take?

Reaction rate

What path does it take?

Mechanism

According to COLLISION THEORY,

the

RATE

of a chemical reaction

is proportional to the number of collisions per second

The

activation

energy

(E

_a

) is the

minimum amount

of energy

required to initiate

a chemical

reaction

number of collisions

rate

s



Molecules must collide with the proper orientation in

order for a reaction to occur.

Cl + NOCl → Cl

2

+ NO

An effective collision

results in reaction.

Correct orientation to

facilitate reaction

An ineffective collision

results in no reaction.

Incorrect orientation does

not favor reaction

When molecules collide in an effective collision,

they form an

activated complex

also known as the

transition state

.

The

rate

of a reaction

increases

at higher temperatures

.

Can we determine the rate or velocity of a car,

knowing the distance traveled and the time taken?

0 10 20 30 40 50 60

0.00 0.20 0.40 0.60 0.80 1.00

D

is

ta

nc

e

(m

i)

Time (hr)

velocity

=

distance change

time change

=

D

d

D

t

=

33

mi

-

22

mi

0.60

hr

-

0.40

hr

=

55

mi hr

The rate is the slope of the line!

D

d

(2)

Reaction Rate Analogy

rate

=

concentration change

time change

=

D

[ ]

A

D

t

velocity

=

distance change

time change

=

D

d

D

t

Average Reaction Rate

2 N

2

O

5

(

g

)

4 NO

2

(

g

) + O

2

(

g

)

0.0000 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300

0 100 200 300 400 500 600 700 800

C

on

ce

ntr

ati

on

(M

)

Time (s)

N

2

O

5

NO

2

O

2

Average Reaction Rate

2 N

2

O

5

(

g

)

4 NO

2

(

g

) + O

2

(

g

)

0.0000 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300

0 100 200 300 400 500 600 700 800

C

on

ce

ntr

ati

on

(M

)

Time (s)

N

2

O

5

NO

2

O

2

Average Reaction rate

=D

[ ]

O2

Dt =

0.0049M-0.0040M

400s-300s =9´10

-6_M

sec

Rate of formation of O2:

=D

[

NO2

]

Dt =

0.0197M-0.0160M

400s-300s =3.7´10

-5_M

sec

Rate of formation of NO2:

=D

[

N2O5

]

Dt =

0.0101M-0.0120M

(

)

400s-300s =-1.9´10

-5_M

sec

Rate of decomposition of N2O5:

Initially, only reactant molecules are present.

As time progresses, there are more and more

product molecules.

A → B

[A] decreases

[A] rate =

t D 

D

[B] rate =

t D

D

The rate of a reaction can be represented as either a

decrease of reactive molecules with time or as an

increase of product molecules with time.

(3)

Br

2

(

aq

) + HCOOH(

aq

) → 2Br

–

(

aq

) + 2H

+

(

aq

) + CO

2

(

g

)

average rate = - D[Br] Dt

=-[Br₂]_final-[Br₂]_initial

t_final-t_initial

Measuring Reaction Progress

and Expressing Reaction Rate

Br

2

(

aq

) + HCOOH(

aq

) → 2Br

–

(

aq

) + 2H

+

(

aq

) + CO

2

(

g

)

Using time and concentration data, we can calculate

the average rate over a time interval.

Br

2

(

aq

) + HCOOH(

aq

) → 2Br

–

(

aq

) + 2H

+

(

aq

) + CO

2

(

g

)

=-0.0101-0.0120

(

)

M

50.0-0.0

(

)

s =3.80x10

-5_M/_s

=-0.00846-0.0120

(

)

M

100.0-0.0

(

)

s =3.54x10

-5_M/_s

First 50 seconds:

First 100 seconds:

General Rate of Reaction

1

2

D

[

N

2

O

5

]

D

t

æ

è

çç

ö

ø

÷÷=

1

4

D

[

NO

2

]

D

t

æ

è

çç

ö

ø

÷÷=

D

[ ]

O

2

D

t

General rate of reaction = rate of consumption of a reactant or formation of a product divided

by its coefficient in the balanced equation

2 N

₂

O

₅

(

g

)

4 NO

₂

(

g

) + O

₂

(

g

)

a

A +

b

B →

c

C + d

D

1 [A] 1 [B] 1 [C] 1 [D] rate =

a t b t c t d t

D D D D

    

D D D D

Write rate expressions for the reaction:

CO

2

(

g

) +

2

H

2

O(

g

)



CH

4

(

g

) +

2

O

2

(

g

)

1 [A]

1 [B] 1 [C] 1 [D]

rate =

a t

b t

c

t

d

t

D



 



D

2 2 4 2

[CO ]

1

[H O]

[CH ]

1

[O ]

rate =

2

t

D



 



D

For the indicated reaction:

a.

Write the rate of reaction in terms of [I

–

] and [I

3–

].

a.

If

, what is the value of

?

3 I–₍_aq_{) + H}

3AsO4(aq) + 2 H+(aq) I3–(aq) + H3AsO3(aq) + H2O(l)

D_{éë ùû}I -Dt =-4.8´10

-4M s

Déë ùûI3

-Dt

D

éë ùû

I

3

-D

t

=

1

3

D

éë ùû

I

-D

t

æ

è

ç

ö

ø

÷

=

1

3

(

-

4.8

´

10

-4

)

=

1.6

´

10

-4

M

s

Rate

=

1

3

D

éë ùû

I

-D

t

æ

è

ç

ö

ø

÷

=

D

éë ùû

I

3

(4)

General Rate of Reaction

0.0000 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300

0 200 400 600 800

Co

nc

en

tr

ati

on

(M

)

Time (s)

Instantaneous rate of reaction = slope of the tangent to a concentration-versus-time curve at time t

Initial rate of reaction = the instantaneous rate at the beginning of a reaction ( t = 0)

The instantaneous rate is the rate for a specific

instant in time and is equal to the slope of a tangent

to the curve at any particular point in time.

The rate of a reaction is directly proportional to the

concentration of reactant.

Time (s) [Br2](M) Rate (M/s)

0.0 0.0120 4.20 x 10–5

50.0 0.0101 3.52 x 10–5

250.0 0.00500 1.75 x 10–5

Br

2

(

aq

) + HCOOH(

aq

) → 2Br

–

(

aq

) + 2H

+

(

aq

) + CO

2

(

g

)

 

2 250 s2 50 s

Br 2 Br 

5

rate at 50.0 s 3.52 10 ₂ rate at 250.0 s 1.75 10

  

 



2

rate [Br ] rate [Br ]k 2

k

is called the rate constant.

We can use concentration and rate data for any

value of t to calculate the value of k for a reaction.

Time (s) [Br2](M) Rate (M/s)

0.0 0.0120 4.20 x 10–5

50.0 0.0101 3.52 x 10–5

250.0 0.00500 1.75 x 10–5

Br

2

(

aq

) + HCOOH(

aq

) → 2Br

–

(

aq

) + 2H

+

(

aq

) + CO

2

(

g

)

2 rate [Br ]k

at

t

= 50.0 s

2

rate

[Br ]

k 

5

3 1

3.52 10 s

3.49 10 s

0.0101 / M k

M 

  

  

Strategy For reactions containing gaseous species, progress is generally monitored by measuring pressure. Pressures are converted to molar concentrations using the ideal gas equation, and rate expressions are written in terms of molar concentrations.

Write the rate expressions for each of the following reactions: (a) I-₍aq) + OCl-₍aq) → Cl-₍aq) + OI-₍aq)

(b) 2O3(g) → 3O2(g)

(c) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Solution (a) All the coefficients in this equation are 1. Therefore,

rate = − = − = = Δ[I_Δt-] Δ[OCl_Δt-] Δ[Cl_Δt-] Δ[OI_Δt-]

Solution (b) rate = − =

(c) rate = − = − = = Δ[O3]

Δt Δ[OΔt 2] 1

2 1 3

Δ[NH3]

Δt 1

4 1 5 Δ[OΔt 2] 4 1 Δ[NO] Δt 1 6 Δ[HΔt 2O]

(5)

Strategy Determine the rate of reaction and, using the stoichiometry of the reaction, convert to rates of change for the specified individual species. Consider the reaction

4NO2(g) + O2(g) → 2N2O5(g)

At a particular time during the reaction, nitrogen dioxide is being consumed at the rate 0.00130 M/s. (a) At what rate is molecular oxygen being consumed? (b) At

what rate is dinitrogen pentoxide being produced?

Solution

rate = − = − =

We are given

= −0.00130 M/s

where the minus sign indicates that the concentration of NO2 is decreasing the

time.

Δ[NO2]

Δt 1

4 Δ[OΔt 2] 1 2 Δ[NΔt 2O5]

Δ[NO2]

Δt

Solution The rate of reaction, therefore, is rate = − = − (−0.00130 M/s)

= 3.25×10-4_M_/s

(a) 3.25×10-4_M_{/s = −}

= −3.25×10-4_M_/s

Molecular oxygen is being consumed at a rate of 3.25×10-4M/s.

(b) 3.25×10-4_M_{/s =}

2(3.25×10-4_M_{/s) =}

= 6.50×10-4_M_/s

Dinitrogen pentoxide is being produced at a rate of 6.50×10-4_M_/s.

Δ[NO2]

Δt 1

4 1 4

Δ[O2]

Δt Δ[O2]

Δt

Δ[N2O5]

Δt 1 2 Δ[N2O5]

Δt Δ[N2O5]

Δt

Think About It Remember that the negative sign in a rate expression indicates that a species is being consumed rather than produced. Rates are always expressed as positive quantities.

The rate law is an equation that relates the

rate of reaction to the concentrations of reactants.

The sum of the exponents, (m + n), is the overall reaction order.

a

A +

b

B Products

Rate

=

-

D

[ ]

A

D

t

=

k

[ ]

A

m

B

[ ]

n

Rate constant

Order with respect to B

Order with respect to A

Dependence of Reaction Rate

on Reactant Concentrations

The concentrations of the

PRODUCTS

do

not appear

in the

rate law.

The value of the

exponents

,

m

and

n

,

must be determined

experimentally

; they cannot

be written from the balanced

equation.

a

A +

b

B Products

Rate

=

-

D

[ ]

A

D

t

=

k

[ ]

A

m

B

[ ]

n

Reaction Order

The sum of the exponents, (m + n), is the overall reaction order.

a

A +

b

B Products

Rate

=

D

[ ]

A

D

t

=

k

[ ]

A

m

B

[ ]

n

Rate Law Overall Reaction Order

Rate = k Zero order

Rate = k[A] First order

Rate = k[A][B] or Rate = k[A]2 _{Second order}

Rate = k[A][B]2_{or Rate =}_k_[A]2_[B] _{Third order}

Reaction Rate Units

rate

=

concentration change

time change

rate

=

M

s

=

Ms

(6)

The initial rate is the rate

at the beginning of a reaction.

F

2

(

g

) + 2 ClO

2

(

g

)  2 FClO

2

(

g

)

Experiment [F2](M) [ClO2](M) Initial Rate (M/s)

1 0.10 0.010 1.2 x 10–3

2 0.10 0.040 4.8 x 10–3

3 0.20 0.010 2.4 x 10–3

Initial Rate Data for the Reaction between F

2

and ClO

2

1: Experimental Determination of Rate Law

from Initial Rates

F

2

(

g

) + 2ClO

2

(

g

) → 2FClO

2

(

g

)

Experiment [F2](M) [ClO2](M) Initial Rate (M/s)

1 0.10 0.010 1.2 x 10–3

2 0.10 0.040 4.8 x 10–3

3 0.20 0.010 2.4 x 10–3

Initial Rate Data for the Reaction between F

2

and ClO

2

[F2]

doubles constant [ClO2] doubles Rate

 

2 32 1



F

0.20

2

F

0.10

M

 







3 3 3 1

rate

2.4 10 / s 2

rate

1.2 10 / s

M

The reaction is first order in F

2

;

x

= 1

rate =

k

[F

2

][ClO

2

]

y

rate =

k

[F

2

]

x

[ClO

2

]

y

from Initial Rates

F

2

(

g

) + 2ClO

2

(

g

) → 2FClO

2

(

g

)

Experiment [F2](M) [ClO2](M) Initial Rate (M/s)

1 0.10 0.010 1.2 x 10–3

2 0.10 0.040 4.8 x 10–3

3 0.20 0.010 2.4 x 10–3

Initial Rate Data for the Reaction between F

2

and ClO

2

[F2]

constant [ClOx 4 2] Rate x 4







2 22 1





ClO

0.040

4

ClO

0.010

M

 







3 2 3 1

rate

4.8 10 / s 4

rate

1.2 10 / s

M

The reaction is first order in ClO

2

;

y

= 1

rate =

k

[F

2

][ClO

2

]

y

rate =

k

[F

2

][ClO

2

]

from Initial Rates

Experiment [A](M) [B](M) Initial Rate (M/s)

1 0.10 0.015 2.1 x 10–4

2 0.20 0.015 4.2 x 10–4

3 0.10 0.030 8.4 x 10–4

[A]

x 2 constant [B] Rate x 2

a

A +

b

B →

c

C +

d

D

Initial Rate Data for the Reaction between A and B

 

12



A

0.10

0.5

A

0.20

M

 







4 1 4 2

rate

2.1 10 / s 0.5

rate

4.2 10 / s

M

The reaction is first order in A;

x

= 1

rate =

k

[A]

x

[B]

y

rate =

k

[A][B]

y

from Initial Rates

Experiment [A](M) [B](M) Initial Rate (M/s)

1 0.10 0.015 2.1 x 10–4

2 0.20 0.015 4.2 x 10–4

3 0.10 0.030 8.4 x 10–4

[A]

constant x 2 [B] Rate x 4

a

A +

b

B →

c

C +

d

D

Initial Rate Data for the Reaction between A and B

 

31



B

0.030

2

B

0.015

M

 







4 3 4 1

rate

8.4 10 / s 4

rate

2.1 10 / s

M

The reaction is second order in B;

y

= 2

rate =

k

[A][B]

y

rate =

k

[A][B]

2

Three important things to remember

about the Rate Law:

1.

The exponents in a rate law must be determined

from a table of experimental data.

2.

Comparing changes in individual reactant

concentrations with changes in rate, shows how

the rate depends on each reactant concentration.

(7)

Use Initial Rate Data to Determine a Rate Law

Determine the rate law and calculate the rate

constant, including its units, for the reaction:

S

2

O

82–

(aq) + 3 I

–

(aq)  2 SO

42–

(aq) + I

3–

(aq)

Experiment [S2O82–](M) [I–](M) Initial Rate (M/s)

1 0.080 0.034 2.2 x 10–4

2 0.080 0.017 1.1 x 10–4

3 0.16 0.017 2.2 x 10–4

Use Initial Rate Data to Determine a Rate Law

(14.3A)

Solution:

Step 1:

In experiments 1 and 2, [S

2

O

82–

] is constant. The [I

–

] is

doubled, and rate doubles.

Experiment [S2O82–](M) [I–](M) Initial Rate (M/s)

1 0.080 0.034 2.2 x 10–4

2 0.080 0.017 1.1 x 10–4

3 0.16 0.017 2.2 x 10–4

Initial Rate Data for the Reaction between S

2

O

82–

and I

–

            1 2

I 0.034 ₂

0.017 I M M 4 1 4 2

rate _{2.2 10 / s 2} rate 1.1 10 / s

M M     



The reaction is first

order in I

–

(14.3A)

Solution:

Step 2:

In experiments 2 and 3, [S

2

O

82–

] is doubled, [I

–

] is constant,

and rate doubles.

Experiment [S2O82–](M) [I–](M) Initial Rate (M/s)

1 0.080 0.034 2.2 x 10–4

2 0.080 0.017 1.1 x 10–4

3 0.16 0.017 2.2 x 10–4

Initial Rate Data for the Reaction between S

2

O

82–

and I

–

     _{ } _     2 2 8 3

2 2 8 ₂

S O _0.16

2 0.080 S O M M       4 3 4 2

rate _{2.2 10 / s 2} rate 1.1 10 / s

M

The reaction is first

order in S

2

O

82–

The rate law is: rate =

k

[S

2

O

82–

] [I

–

]

(14.3A)

Solution:

Step 3:

Use the data from any experiment to calculate

k.

Experiment [S2O82–](M) [I–](M) Initial Rate (M/s)

1 0.080 0.034 2.2 x 10–4

2 0.080 0.017 1.1 x 10–4

3 0.16 0.017 2.2 x 10–4

Initial Rate Data for the Reaction between S

2

O

82–

and I

–





4

1 1 3

2 2 8 ₃ ₃

rate 2.2 10 / s _0.081 _s

(0.16 )(0.017 )

S O I

M k M M M                  

Use Initial Rate Data to Determine a Rate Law: Dependence

of Reaction Rate on Reactant Concentration

The gas-phase reaction of nitric oxide with hydrogen at 1280

°

C

is

2NO(

g

) + 2H

2

(

g

) → N

2

(

g

) +2H

2

O(

g

)

From the following data collected at 1280

°

C, determine

(a)

rate law

(b)

rate constant, including units

(c)

rate of the reaction when [NO] = 4.8

×

10

-3

M

and [H

₂

] = 6.2

×

10

-3

M

.

Experiment [NO] (

M

) [H

2

] (

M

) Initial rate (

M

/s)

1

5.0

×

10

-3

2.0

×

10

-3

1.3

×

10

-5

2

1.0

×

10

-2

2.0

×

10

-3

5.0

×

10

-5

3

1.0

×

10

-2

4.0

×

10

-3

1.0

×

10

-4

Solution The rate of reaction, therefore, is = ≈ 4 =

Canceling identical terms in the numerator and denominator gives

Therefore, x = 2. The reaction is second order in NO.

Dividing the rate from experiment 3 by the rate from experiment 2, we get

= = 2 =

Canceling identical terms in the numerator and denominator gives

Therefore, y = 1. The reaction is first order in H2. The overall rate law is

rate = k[NO]2_[H₂_]

5.0×10-5M/s

1.3×10-5_M_/s

rate2

rate1

k(1.0×10-2M)x_(2.0×10-3M)y

k(5.0×10-3_M₎x_(2.0_×₁₀-3_M₎y

(1.0×10-2_M₎x (5.0×10-3_M₎x= 2x = 4

1.0×10-4_M_/s

5.0×10-5_M_/s

rate3

rate2

k(1.0×10-2M)x_(4.0×10-3_M₎y

k(1.0×10-2_M₎x_(2.0_×₁₀-3_M₎y

(8)

Solution We can use data from any of the experiments to calculate the value and units of k. Using the data from experiment 1 gives

k = =

(c) Using the rate constant determined in part (b) and the concentrations of NO and H2 given in the problem statement, we can determine the reaction rate as

follows:

rate = (2.6×102M-2_∙s-1_)(4.8×10-3M)2_(6.2×10-3M)

= 3.7×10-5M∙s-1

1.3×10-5_M_/s

(5.0×10-3_M₎2_(2.0×10-3_M₎

rate [NO]2_[H

2] = 2.6×10

2M-2_∙s-1

Think About It The exponent for the concentration of H2 in the rate law is 1,

whereas the coefficient for H2 in the balanced equation is 2. It is a common error

to try to write a rate law using the stoichiometric coefficients as the exponents. Remember that, in general, the exponents in the rate law are not related to the coefficients in the balanced equation. Rate laws must be determined by examining a table of experimental data.

A rate law can be used to determine

the concentration of a reactant

at a specific time during a reaction

rate =

k

[A]

x

[B]

y

rate rate constant

rate law

A

first order reaction

is a reaction whose rate

depends on the concentration of one of the reactants

raised to the first power.

C

2

H

6

→ 2

·

CH

3

rate =

k

[C

2

H

6

]

2N

2

O

5

(

g

) → 2NO

2

(

g

) + O

2

(

g

)

rate =

k

[N

2

O

5

]

Dependence of Reactant Concentration on Time:

First Order Reactions

In a first-order reaction of the type

A → products

The rate can be expressed as the rate of change in reactant

concentration,

as well as in the form of the rate law:

rate =

k

[A]

Setting the two expressions equal to each other yields:

 

A

rate =

t

D



D

 

A

 

A

k

t

D





D

First Order Reactions

Using calculus, it is possible to show that:

ln is the natural logarithm

[A]

0

and [A]

t

refer to the concentration of A at times 0 and

t

The equation above is sometimes called the integrated rate law

for a

first order reaction.

 

0 A ln_At_{ }_kt

The rate constant for the reaction

2A



B

is 0.075 s

–1

at 110°C. The reaction is first order in A.

How long (in seconds) will it take for [A] to decrease from

1.25

M

to 0.71

M

?

Solution:

Step 1:

Use the equation below to calculate time in seconds.

t

= 75 seconds









0



3 1



0.71

ln_1.25t_{ }7.5 10 s_   _t

 

0

(9)

The decomposition of hydrogen peroxide is first

order in H

2

O

2

.

2 H

2

O

2

(

aq

)



2 H

2

O(

l

) + O

2

(

g

)

The rate constant for this reaction

at 20°C is 1.8 x 10

–5

s

–1

.

If the starting concentration of H

2

O

2

is 0.75

M

, determine

a) the concentration of H

2

O

2

remaining after 3 h

b) how long it will take for the H

2

O

2

concentration to

decrease to 0.10

M

Strategy Use ln ([A]t/[A]0) = –kt to find [H2O2]t where t = 3 h, and then solve

for t to determine how much time must pass for [H2O2]t to equal 0.10 M. [H2O2]0

= 0.75 M; time t for part (a) is (3 h)(60 min/h)(60 s/min) = 10,800 s.

The decomposition of hydrogen peroxide is first order in H2O2.

2H2O2(aq) → 2H2O(l) + O2(g)

The rate constant for this reaction at 20°C is 1.8×10-5_s-1_{. If the start}

concentration of H2O2 is 0.75 M, determine (a) the concentration of H2O2

remaining after 3 h and (b) how long it will take for the H2O2 concentration to

drop to 0.10 M.

Solution (a) ln

(b) ln [H2O2]t

[H2O2]0 = –kt

[H2O2]t

0.75 M= –(1.8×10-5 s-1)(10,800 s) = –0.1944

Solution Take the inverse natural logarithm of both sides of the equation to get

[H2O2]t = (0.823)(0.75 M) = 0.62 M

The concentration of H2O2 after 3 h is 0.62 M.

(b) ln

The time required for the peroxide concentration to drop to 0.10 M is 1.1×105_s

or about 31 h. [H2O2]t

0.75 M= e–0.1944 = 0.823

0.10 M

0.75 M = –2.015 = –(1.8×10-5 s-1)t

2.015

1.8×10-5 _s-1₌t = 1.12×105 s

Think About It Don’t forget the minus sign. If you calculate a concentration at time t that is greater than the concentration at time 0 (or if you get a negative time

required for the concentration to drop to a specified level), check your solution for this common error.

Dependence of Reactant Concentration on Time:

First Order Reactions

Rearrangement of the first-order integrated rate law gives:

Rearrangement in this way has the form

of the linear equation

y = mx

+

b.

 

0 A ln_At_{ }_kt

ln[A]

t

= –

kt

+ ln[A]

0

ln[A]

t

= –

kt

+ ln[A]

0

Slope = –

k

Intercept = ln[A]

0

The rate of decomposition of azomethane is studied by monitoring

the partial pressure of the reactant as a function of time.

CH

3

—N=N—CH

3

(

g

) → N

2

(

g

) + C

2

H

6

(

g

)

The data obtained at 300

°

C are listed in the following table:

Time (s) Pazomethane (mmHg)

0 284

100 220

150 193

200 170

250 150

300 132

Solution The table expressed in ln P is

Plotting these data gives a straight line, indicating that the reaction is indeed first order. Thus, we can use ln ([A]t/[A]0) = –kt in terms of pressure.

ln = –kt

Pt and P0 can be pressures at any two times during the experiment. P0 need not be

the pressure at 0 s–it need only be at the earlier of the two times.

Pt

P0

Time (s) ln P

0 5.649

100 5.394

150 5.263

200 5.136

250 5.011

(10)

Dependence of Reactant Concentration on Time:

Plotting the data gives a straight line, indicating the reaction is first

order.

ln[A]t = –kt + ln[A]0

Slope = –2.55 x 10

–3

s

–1

Intercept = 5.65

4.80 4.90 5.00 5.10 5.20 5.30 5.40 5.50 5.60 5.70

0 100 200 300 400

ln

P

Time (s)

Worked Example 14.5 (cont.)

Solution Using data from times 100 s and 250 s of the original table (Pazomethane

versus t), we get

ln = –kt

ln 0.682 = –k(150 s)

k = 2.55×10-3_s-1

150 mmHg 220 mmHg

Think About It We could equally well have determined the rate constant by calculating the slope of the plot of ln P versus t. Using the two points labeled on

the plot, we get

slope =

= 2.55×10-3_s-1

Remember that slope = –k, so k = 2.55×10-3_s-1_.

5.011 – 5.394 250 – 100

Dependence of Reactant Concentration on Time:

Ethyl iodide (C

2

H

5

I) decomposes at a certain temperature in the gas

phase as follows:

C

2

H

5

I(

g

) → C

2

H

4

(

g

) + HI(

g

)

Determine the rate of the reaction, after verifying that the reaction is

first order.

Time (s) [C2H5I] (M)

0 0.36

15 0.30

30 0.35

48 0.19

75 0.13

Dependence of Reactant Concentration on Time:

Solution:

Plot ln[C

2

H

5

I] vs time. If a straight line results, the reaction is first

order. The slope is equal to

k

.

Slope = –1.3 x 10

–2

s

–1

;

k

= 1.3 x 10

–2

s

–1

Time (s) [C2H5I] (M) ln[C2H5I]

0 0.36

-1.02

15 0.30

-1.20

30 0.35

-1.39

48 0.19

-1.66

75 0.13

-2.04

-2.50

-2.00 -1.50 -1.00 -0.50 0.00

0 20 40 60 80

ln

[C2 H5

I]

Time (s)

The

half-life

(

t

1/2

) is the time required for the reactant

concentration to drop to half its original value.

A



Products

The

half-life

(

t

1/2

) is the time required for the reactant

concentration to drop to half its original value.

t

=

t

1/2

when [A]

t

= ½[A]

0

.

 

0 A

lnAt kt

 

A0 1 ln

At t

k 

 

0 1/2

0 A 1 ln

1 A 2 t

k

 t1/20.693_k

rearranges

(11)

The decomposition of ethane (C

2

H

6

) to methyl

radicals (CH

3

) is a first order reaction with a rate

constant of 5.36 x 10

–4

s

–1

at 700°C.

Calculate the half-life of the reaction in minutes.

1/2

0.693

t

k



Strategy Use t½ = 0.693/k to calculate t½ in seconds, and then convert to

minutes.

The decomposition of ethane (C2H6) to methyl radicals (CH3) is a first-order

reaction with a rate constant of 5.36×10-4_s-1_{at 700}°C. C2H6 → CH3

Calculate the half-life of the reaction in minutes.

Solution t½ = = = 1293 s

1293 s × = 21.5 min

The half-life of ethane decomposition at 700°C is 21.5 min. 0.693

k

0.693 5.36×10-4_s-1

1 min 60 s

Think About It Half-lives and rate constants can be expressed using any units of time and reciprocal time, respectively. Track units carefully when you convert from one unit of time to another.

Dependence of Reactant Concentration on Time

Calculate the half-life of the decomposition of azomethane,

k

= 2.55

×

10

–3

s

-1

.

Solution:

Step 1:

Use the equation below to calculate half-life:

1/2 _{2.55 x 10 s}0.6933 1 272 s

t    

1/2 0.693 t

k 

A

Second–Order Reaction

is a reaction whose rate

depends on the concentration of one reactant raised to the

second power or on the product of the concentrations of two

different reactants that are first order in each.

 

0

1 ₊ 1

At A

kt 

 

1/2

0

1 A t

k 

Second-order integrated rate law:

Second-order half-life:

Iodine atoms combine to form molecular iodine in the

gas phase:

I(

g

) + I(

g

)



I

₂

(

g

)

This reaction is second order and has a rate constant

7.0 x 10

9

M

–1

s

–1

at 23°C.

a) if the initial concentration of I is 0.086

M

, calculate

the concentration after 2.0 minutes.

b) calculate the half-life of the reaction when the initial

concentration of I is 0.60

M

and when the initial

concentration of I is 0.42

M

.

Strategy Use 1/[A]t = kt + 1/[A]0 to determine [I]t at t = 2.0 min; use t½ = 1/k[A]0 to determine t½ when [I]0 = 0.60 M and when [I]0 = 0.42 M.

Iodine atoms combine to form molecular iodine in the gas phase: I(g) + I(g) → I2(g)

This reaction is second order and has a rate constant of 7.0×109_M-1_∙s-1_{at 23}°C. (a) If the initial concentration of I is 0.086 M, calculate the concentration after

2.0 min. (b) Calculate the half-life of the reaction when the initial concentration of I is 0.60 M and when the initial concentration of I is 0.42 M.

Solution t = (2.0 min)(60 s/min) = 120 s

(a) = kt +

= (7.0×109_M-1_∙s-1_{)(120 s) +}

1 [A]t

1 [A]0

(12)

Solution = 8.4×1011M-1

[A]t = = 1.2×10-12_M

The concentration of atomic iodine after 2 min is 1.2×10-12_M_.

(b) When [I]0 = 0.60 M,

t½ = = = 2.4×10-10 s

When [I]0 = 0.42 M,

t½ = = = 3.4×10-10 s

1 8.4×1011_M-1

1

k[A]0

1 (7.0×109M-1_∙s-1_)(0.60M)

1

k[A]0

1 (7.0×109_M-1_∙s-1_)(0.42_M₎

Think About It (a) Iodine, like the other halogens, exists as diatomic molecules at room temperature. It makes sense, therefore, that atomic iodine would react quickly, and essentially completely, to form I2 at room temperature. The very low remaining concentration

of I after 2 min makes sense. (b) As expected, the half-life of this second-order reaction is not constant. (A constant half-life is a characteristic of first-order reactions.)

The rate of a zero order reaction is a constant,

independent of reactant concentration.

Units for

k

:

Zero-Order

Reaction

Rate

=

M

s

=

Ms

-1

=

k

Rate

=

k

[ ]

A

m

In a zero order reaction,

Rate

=

k

[ ]

A

0

=

k

Units for

k

:

First-Order

Reaction

Rate

=

M

s

=

k M

( )

Rate

=

k

[ ]

A

m

In a first order reaction,

Rate

=

k

[ ]

A

1

k

=

1

s

=

s

-1

Units for

k

:

Second-Order

Reaction

Rate

=

M

s

=

k M

( )

2

Rate

=

k

[ ]

A

m

In a second order reaction,

Rate

=

k

[ ]

A

2

k

=

1

M

×

s

=

M

-1

s

-1

Units for

k

:

Third-Order

Reaction

Rate

=

M

s

=

k M

( )

3

Rate

=

k

[ ]

A

m

In a third order reaction,

Rate

=

k

[ ]

A

3

k

=

1

M

2

×

s

=

M

(13)

Reaction Order Summary

The sum of the exponents, (m + n), is the overall reaction order.

a

A +

b

B Products

Rate Law Overall Reaction Order Units for k Rate = k Zero order M/s or Ms–1

Rate = k[A] First order 1/s or s–1

Rate = k[A][B] Second order 1/(M•s) or M–1_s–1

Rate = k[A][B]2 _{Third order} _1/(M_2•_{s) or M}–2_s–1

Rate

=

D

[ ]

A

D

t

=

k

[ ]

A

m

B

[ ]

n

Types of Rate Laws

Rate

=

k

[ ]

A

m

B

[ ]

n Differential rate law

Shows how the rate depends on concentrations

ln

[ ]

A

t

A

[ ]

₀

=

-

kt

Integrated rate law

Shows how the concentrations depend on time

The exact form of the both rate laws depends on the overall order of the reaction.

Zero Order

Reaction

Rate

=

-

d

[ ]

A

dt

=

k

[ ]

A

0

=

k

Differential rate law

A

[ ]

=

-

kt

+

[ ]

A

₀ Integrated rate law

t

₁ 2

=

A

[ ]

0

2

k

Half-life

Zero order reactions are relatively uncommon.

Graph of a Zero-Order Reaction

A

[ ]

=

-

kt

+

[ ]

A

₀

y = mx + b

0 5 10 15 20 25

0 2 4 6 8 10 12

[

A

]

t

Half-Life of a Zero-Order Reaction

Half-life

is the time required for a reactant to

reach half of its original concentration.

The half-life of a zero-order reaction depends on

the concentration. Since the concentration

decreases as the reaction proceeds, the half-life

decreases as the reaction proceeds.

t

₁ 2

=

A

[ ]

0

2

k

Half-Life of a Zero-Order Reaction

0 5 10 15 20 25

0 2 4 6 8 10 12

[

A

]

t

₁

2

=

A

[ ]

0

2

k

t1/2 = 5

(14)

Zero-Order Reaction Example

a) Determine the rate constant,

k

, of a zero order reaction if

the initial concentration of substance A is 1.5 M and after

120 seconds the concentration of substance A is 0.75 M.

A

[ ]

=

-

kt

+

[ ]

A

₀

0.75

=

-

k

( )

120

+

1.5

k

=

(

1.5

-

0.75

)

120

=

0.0063

M

s

Zero-Order Reaction Example

b) What is the half-life of substance A if its original concentration

is 1.2 M?

k

=

0.0063

M

s

t

1

2

=

A

[ ]

0

2

k

t

₁ 2

=

1.2

2

( )

(

0.0063

)

=

95

s

First Order

Reaction

Rate

=

-

d

[ ]

A

dt

=

k

[ ]

A

1

=

k

[ ]

A

Differential rate law

ln

[ ]

A

=

-

kt

+

ln

[ ]

A

₀ Integrated rate law

t

₁ 2

=

ln2

k

=

0.693

k

Half-life

A

[ ]

=

[ ]

A

₀

e

-kt

simplifies to…

Graph of a First Order Reaction

A

[ ]

=

[ ]

A

0

e

-kt

0 5 10 15 20 25

0 2 4 6 8 10 12

[

A

]

t

Graph of a First Order Reaction

ln

[ ]

A

=

-

kt

+

ln

[ ]

A

₀

-4 -3 -2 -1 0 1 2 3 4

0 2 4 6 8 10 12

ln[

A

]

t

y

= mx + b

Half-Life of a First-Order Reaction

Half-life

reach half of its original concentration.

The half-life of a first-order reaction doesn’t

depend on the concentration.

The half-life for a first-order reaction is a

constant value.

t

1 2

=

ln2

k

=

0.693

(15)

Half-Life of a First-Order Reaction

0 5 10 15 20 25

0 2 4 6 8 10 12

[

A

]

t

t1/2 = 1.2

t

1 2

=

ln2

k

=

0.693

k

t1/2 = 1.2 t1/2 = 1.2

First-Order Reaction Example

At 500

°

C, cyclopropane (C

3

H

6

) rearranges to propene (CH

3

–

CH=CH

2

). The reaction is first order, and the rate constant is

6.7 x 10

–4

s

–1

. If the initial concentration of C

3

H

6

is 0.0500 M:

a) What is the molarity of C

3

H

6

after 30 min?

A

[ ]

=

[ ]

A

₀

e

-kt 30 min = 1800 s

A

[ ]

=

(

0.0500

)

e

-

(

6.7´10-4

)

(

1800

)

=

0.015

M

First-Order Reaction Example

At 500°C, cyclopropane (C3H6) rearranges to propene (CH3–CH=CH2). The reaction is first order, and the rate constant is 6.7 x 10–4_s–1_{. If the initial}

concentration of C3H6 is 0.0500 M:

b) How many minutes does it take for the C

3

H

6

concentration

to drop to 0.0100 M?

ln

[ ]

A

=

-

kt

+

ln

[ ]

A

₀

kt

=

ln

[ ]

A

0

-

ln

[ ]

A

=

ln

A

[ ]

0

A

[ ]

æ

è

çç

ö

ø

÷÷

t

=

ln

[ ]

A

0

A

[ ]

æ

è

çç

ö

ø

÷÷

k

=

ln 0.0500

0.0100

æ

è

ç

ö

ø

÷

6.7

´

10

-4

=

2402

s

=

40min

Second-Order

Reaction

Rate

=

-

d

[ ]

A

dt

=

k

[ ]

A

2 Differential rate law

1

A

[ ]

=

kt

+

[ ]

A

1

0 Integrated rate law

t

₁ 2

=

1

k

[ ]

A

₀ Half-life

Graph of a Second-Order Reaction

0 5 10 15 20 25

0 2 4 6 8 10 12

[

A

]

t

Graph of a Second-Order Reaction

1

A

[ ]

=

kt

+

[ ]

A

1

₀

y = mx + b

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16

0 2 4 6 8 10 12

1

—

[

A

]

(16)

Half-Life of a Second-Order Reaction

t

₁ 2

=

1

k

[ ]

A

₀

Half-life

reach half of its original concentration.

The half-life of a second-order reaction depends

on the inverse concentration. Since the

concentration decreases as the reaction

proceeds, the half-life of a second-order

reaction increases as the reaction proceeds.

Half-Life of a Second-Order Reaction

0 5 10 15 20 25

0 5 10 15 20

[

A

]

t

₁

2

=

1

k

[ ]

A

0

t1/2 = 5

t1/2 = 10

Second-Order Reaction Example

Iodine atoms combine to form molecular iodine (I

2

) in the gas

phase. The reaction is second order and has a rate constant of

7.0 x 10

9

M

–1

s

–1

at 23

°

C.

a) If the initial concentration of iodine is 0.086 M, what is the

concentration of iodine after 2.0 min?

2 min = 120 s

1

A

[ ]

=

kt

+

[ ]

A

1

0

1

A

[ ]

=

(

7.0

´

10

9

)

( )

120

+

0.086

1

=

8.4

´

10

11

A

[ ]

=

1

8.4

´

10

11

=

1.2

´

10

-12

M

Second-Order Reaction Example

Iodine atoms combine to form molecular iodine (I2) in the gas phase. The reaction is second order and has a rate constant of 7.0 x 109_M–1_s–1_{at 23}°C.

b) Calculate the half-life of the reaction when the initial

concentration of iodine is 0.60 M.

t

₁ 2

=

1

k

[ ]

A

₀

t

₁ 2

=

1

7.0

´

10

9

(

)

(

0.60

)

=

2.4

´

10

-10

s

Reaction Order Example

Nitrosyl bromide decomposes at 10°C:

a) Given the following data, determine whether the reaction is zero, first, or second order.

Time (s) 0 10 20 30 40

[NOBr] 0.0400 0.0303 0.0244 0.0204 0.0175

The order of the reaction cannot be determined from the balanced equation.

2 NOBr(g)

2 NO(g) + Br

2

(g)

Reaction Order Example

In a zero order reaction, a plot of [NOBr] vs. t would give a straight line. This is obviously not a zero order reaction.

0.0150 0.0200 0.0250 0.0300 0.0350 0.0400 0.0450

0 10 20 30 40 50

[N

O

Br

]

(17)

Reaction Order Example

In a first order reaction, a plot of ln [NOBr] vs. t would give sort of a straight line. This is not a first order reaction.

-4.20 -4.00 -3.80 -3.60 -3.40 -3.20 -3.00

0 10 20 30 40 50

ln

[N

O

Br

]

t (sec)

Reaction Order Example

In a second order reaction, a plot of 1/[NOBr] vs. t would give a straight line. This is a second order reaction.

20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0 60.0

0 10 20 30 40 50

1/

[N

O

Br

]

t (sec)

Reaction Order Example

b) Determine the rate constant for the reaction.

2 NOBr(g)

2 NO(g) + Br

2

(g)

1

A

[ ]

=

kt

+

[ ]

A

1

0

y = mx + b

In a second order reaction, the slope of the line is the rate constant.

Reaction Order Example

b) Determine the rate constant for the reaction.

2 NOBr(g)

2 NO(g) + Br

2

(g)

Microsoft Excel gives the slope (value of the rate constant): 0.80

Rate Law Overall Reaction Order Units for k Rate = k Zero order M/s or Ms–1

Rate = k[A] First order 1/s or s–1

Rate = k[A][B] Second order 1/(M•s) or M–1_s–1 Rate = k[A][B]2 _{Third order} _1/(M_2•_{s) or M}–2_s–1

The rate constant k = 0.80 M–1_s–1_.

Dependence of Reaction Rate on Temperature

The dependence of the rate constant on temperature can be

expressed by the

Arrhenius equation

.

A

represents the collision frequency and is called the frequency

factor.

E

a

is the activation energy (in kJ/mol).

R

is the gas constant (8.314 J/mol K).

T

is the absolute temperature.

e

is the base of the natural logarithm.

a/ E RT

k



Ae



Dependence of Reaction Rate on Temperature

Taking the natural log of both sides, the Arrhenius equation may be

written as:

Rearrangement gives the linear form of the Arrhenius equation:

a

ln

k

ln

A

E

RT





a

1

ln

k

E

ln

A

R

T



 

 





 



 

(18)

Graphical Method for Determining the Activation

Energy for a Reaction

Rate constants for the reaction

CO(

g

) + NO

2

(

g

) → CO

2

(

g

) + NO(

g

)

were measured at four different temperatures. The data are shown in

the table. Determine the activation energy for the reaction.

k

(

M

–1

·

s

–1

)

T

(K)

0.0521

288

0.101

298

0.184

308

0.332

318

Solution Taking the natural log of each value of k and the inverse of each value

of T gives

A plot of these data yields the following graph: ln k 1/T (K-1₎

–2.95 3.47×10-3

–2.29 3.36×10-3

–1.69 3.25×10-3

–1.10 3.14×10-3

Solution The slope is determined using the x and y coordinates of any two

points on the line. Using the points that are labeled on the graph gives

slope = = –5.5×103_K

The value of the slope is –5.5×103_{K. Because the slope = –}_E a/R,

Ea = –(slope)(R)

= –(–5.5×103_{K)(8.314 J/K∙mol)}

= 4.6×104_{J/mol or 46 kJ/mol}

–1.4 – (–2.5) 3.2×10-3_K-1_{– 3.4×10}-3_K-1

Think About It Note that while k has units M-1∙s-1, ln k has no units.

Graphical Method for Determining the Activation

Energy for a Reaction

Solution:

Plot ln

k

versus 1/

T

and determine the slope of the line;

slope = –

E

a

/

R

.

slope = –5.6 x 10

3

K = –

E

a

/

R

E

a

= –(–5.6 x 10

3

K )(8.314 J/mol·K) = 46 kJ/mol

k (M–1_·_s–1₎ _T_(K) _ln_k _1/_T_(K–1₎

0.0521 288 0.00347 -2.95 0.101 298 0.00336 -2.29 0.184 308 0.00325 -1.69 0.332 318 0.00314 -1.10 -3.50

-3.00 -2.50 -2.00 -1.50 -1.00 -0.50 0.00

ln

k

1/T (K–1)

a

1

ln

k

E

ln

A

R

T



 

 





 



 





Two point Form of the Arrhenius Equation

1.

If the rate constants at two different temperatures are

known, it is possible to calculate the activation

energy.

2.

If the activation energy and the rate constant at one

temperature are known, it is possible to determine

the rate constant at any other temperature.

a 1

2 2 1

1

ln

k

E

k

R

T

















Arrhenius Equation Example

Temp (°C) k (M–1s–1) 283 3.52 x 10–7

356 3.02 x 10–5

393 2.19 x 10–4

427 1.16 x 10–3

508 3.95 x 10–2

Rate constants for the gas phase decomposition of hydrogen

iodide are listed in the following table:

a) What is the activation energy of the reaction?

(19)

Temp (°C) Temp (K) k (M–1s–1) 1/T (1/K) ln k 283 556 3.52 x 10–7 0.00180 –14.860

356 629 3.02 x 10–5 _0.00159 _–10.408

393 666 2.19 x 10–4 _0.00150 _–8.426

427 700 1.16 x 10–3 _0.00143 _–6.759

508 781 3.95 x 10–2 _0.00128 _–3.231

ln

k

=

-

E

a

R

æ

è

ç

ö

ø

÷

1

T

æ

è

ç

ö

ø

÷+

ln

A

slope

=

-

E

a

R

E

a

=

-

slope

´

R

-15.00 -13.00 -11.00 -9.00 -7.00 -5.00 -3.00

0.0012 0.0013 0.0014 0.0015 0.0016 0.0017 0.0018

ln

k

1/T

E

a

=

-

slope

´

R

=

- -

(

2.24

´

10

4

)

(

8.314

)

=

1.86

´

10

5

J

mol

=

186

kJ

mol

ln

k

₁

=

-

E

a

R

æ

è

ç

ö

ø

÷

1

T

₁

æ

è

ç

ö

ø

÷+

ln

A

ln

k

₂

=

-

E

a

R

æ

è

ç

ö

ø

÷

1

T

₂

æ

è

ç

ö

ø

÷+

ln

A

ln

k

2

k

₁

æ

è

ç

ö

ø

÷=

E

a

R

æ

è

ç

ö

ø

÷

1

T

₁

-1

T

₂

æ

è

ç

ö

ø

÷

b) Calculate the activation energy of the reaction from the rate

constants at 283

°

C and 508

°

C.

ln

k

2

k

₁

æ

è

ç

ö

ø

÷=

E

a

R

æ

è

ç

ö

ø

÷

1

T

₁

-1

T

₂

æ

è

ç

ö

ø

÷

b) Calculate the activation energy of the reaction from the rate

constants at 283

°

C and 508

°

C.

ln 3.95

´

10

-2

3.52

´

10

-7

æ

è

ç

ö

ø

÷=

E

a

8.314

æ

è

ç

ö

ø

÷

1

556

-1

781

æ

è

ç

ö

ø

÷

E

a

=

1.87

´

10

5

J

mol

=

187

kJ

mol

Arrhenius Equation Example

ln

k

2

k

1

æ

è

ç

ö

ø

÷=

E

a

R

æ

è

ç

ö

ø

÷

1

T

1

-

1

T

2

æ

è

ç

ö

ø

÷

c) What is the rate constant at 618

°

C?

Hard way

ln

k

=

-

E

a

R

æ

è

ç

ö

ø

÷

1

T

æ

è

ç

ö

ø

÷+

ln

A

y =

m

x

+ b

Easy way

Arrhenius Equation Example

c) What is the rate constant at 618

°

C?

ln

k

=

-

E

a

R

æ

è

ç

ö

ø

÷

1

T

æ

è

ç

ö

ø

÷+

ln

A

y =

m

x

+ b

Use Excel’s SLOPE and INTERCEPT functions to calculate m and b.

ln

k

=

slope

´

1

891

æ

è

(20)

Strategy Rearrange and solve for Ea using the following

The rate constant for a particular first-order reaction is given for three different temperatures:

Using the data, calculate the activation energy of the reaction. T (K) k (s-1₎

400 2.9×10-3

450 6.1×10-2

500 7.0×10-1

Ea = R k1

k2

ln 1

T2 1

T2 –

Solution

The activation energy of the reaction is 91 kJ/mol.

Think About It A good way to check your work is to use the value of Ea that

you calculated (and Equation 14.11) to determine the rate constant at 500 K. Make sure it agrees with the value in the table.

E_a = 8.314 J/K∙mol

2.9×10 -3

6.1×10 -2

ln 1 450 K– 400 K1

= 91,173 J/mol = 91 kJ/mol

Strategy Rearrange and solve for k2 using the following

Ea = 8.3×104 J/mol, T1 = 423 K, T2 = 573 K, R = 8.314 J/K∙mol, and k1 = 2.1×10-2 s-1.

A certain first-order reaction has an activation energy of 83 kJ/mol. If the rate constant for this reaction is 2.1×10-2_s-1_{at 150}°C, what is the rate constant at 300°C?

k2 =

k1

1

T2

1

T2

–

Ea

R e

Solution

The rate constant of 300°C is 10 s-1_.

k2 = 2.1

×10-2_s-1

1 573 K–423 K1 8.3×104_J/mol

8.314 J/K∙mol e

= 1.0×101_s-1

Think About It Make sure that the rate constant you calculate at a higher temperature is in fact higher than the original rate constant. According to the Arrhenius equation, the rate constant always increases with increasing temperature. If you get a smaller k at a

higher temperature, check your solution for mathematical errors.

The sequence of steps that sum to give the overall

reaction is called the

reaction mechanism

.

A balanced chemical equation does not indicate

how a reaction actually takes place.

Step 1:

A + B → C

Step 2:

C + B → D

Overall reaction:

A + 2B → D

Chemical species that appear in the reaction

mechanism, but not in the overall chemical

equation, are called

intermediates

.

Step 1:

Step 2:

NO + NO

N

2

O

2

+ O

2

N

2

O

2

2NO

2

Overall reaction:

2NO + O

₂

2NO

₂

Each step in a reaction mechanism represents an

elementary reaction

, one that occurs in a single

collision of the reactant molecules.

The

molecularity

of an elementary reaction is essentially the

number of reactant molecules involved in the collision.

unimolecular

(

one

reactant molecule)

A → products rate =

k

[A] first order

bimolecular

(

two

reactant molecules)

A + B → products rate =

k

[A][B] second order

(21)

Reaction Mechanism

H

2

+ ICl

HCl + HI

HI + ICl

HCl + I

2

Elementary step

Elementary step

H

2

+ 2ICl

2HCl + I

2 Overall reaction Step 1:

Step 2:

Sum:

Reaction Mechanism

N

2

O

N

2

+ O

N

2

O + O

N

2

+ O

2

Unimolecular step

Bimolecular step

2N

2

O

2N

2

+ O

Step 2:

Sum:

rate = k[N2O]

rate = k[N2O][O]

In a reaction mechanism consisting of more than one

elementary step, the rate law for the overall process

is given by the

rate–determining step

.

The rate determining step is the

slowest

step in the

sequence.

A proposed mechanism must satisfy two requirements:

1.

The sum of the elementary reactions must be the

overall balanced equation for the reaction.

2.

The rate determining step must have the same rate

law as that determined from the experimental data.

Deducing a Plausible Reaction Mechanism from the

Experimentally Determined Rate Law

The decomposition of hydrogen peroxide can be facilitated by iodide

ions:

Step 1:

(slow)

Step 2:

H

2

O

2

+ I

–

H

2

O

2

+ IO

–

H

2

O + IO

–

H

2

O + O

2

+ I

–

Overall reaction:

2H

2

O

2

2H

2

O + O

2

Rate =

k

1

[H

2

O

2

][I

–

]

Reaction Mechanism Example

ICl + ICl

I

2

+ Cl

2

Cl

2

+ H

2

2HCl

(slow)

H

2

+ 2ICl

2HCl + I

Step 2:

Sum:

The experimentally determined rate law is rate = k[H2][ICl].

H

2

+ 2ICl

2HCl + I

2

rate = k[ICl]2

Mechanism 1

Reaction Mechanism Example

H

2

+ ICl

HI + HCl

ICl + HCl

HI + Cl

2

(slow)

H

2

+ 2ICl

2HI + Cl

Step 2:

Sum:

The experimentally determined rate law is rate = k[H2][ICl].

H

2

+ 2ICl

2HCl + I

2

rate = k[H2][ICl]

(22)

Reaction Mechanism Example

H

2

2H

ICl + H

HCl + I

(slow)

H