11.1 reaction rate 3

11.1 11.1 REACTION RATEREACTION RATE

CH EM I S T RY U N I TCH EM I S T RY U N I TS K 0 2 7S K 0 2 7

OBJECTIVES:OBJECTIVES:

CHAPTER 11 : REACT ION K INET ICSCHAPTER 11 : REACT ION K INET ICS

11.1 REACTION RATE11.1 REACTION RATE

1. Write the integrated rate laws for zero order,

1st order and 2nd order reaction

2. Define half-life.

S K 0 2 7 C H EM I S T RYS K 0 2 7 C H EM I S T RY

2. Define half-life.

3. Draw the respective graphs for the different

order reactions.

4. Solve quantitative problems.



Integrated Rate Laws

o The integrated rate law is an equation that

describes the concentration of a reactant as a

function of time.function of time.

It can use to determine rate constant, half life

and concentration of reactant at specific time




Integrated Rate Equation

(a) For Zero-order reaction

For reaction, A → product

Rate = k[A]oRate = d[A]−Rate =

dt−

= kdt

d[A]−The differential rate equation;

[A]o = initial concentration of A

[A]t = concentration of A at time t

k = rate constant

t = time t

The integrated rate equation; [A]o – [A]t = kt





o Half life(t1/2) of a zero order reactiono Half life (t1/2) is defined as the length of time required

for the concentration of a reactant to decrease to half

of its initial value.of its initial value.

Substituting t = t1/2 ,

2

[A]o[A]t =

[A]o – [A]t = ktSince,

1/2o

o kt2

[A][A] =−Thus,

2k

[A]t o1/2 =




Characteristic graphs for zero order reaction

RateRate = k[A]o

[A]

[A]o – [A]t = kt

[A]t = [A]o – kt

y = C + mx

[A]

t

[A]o

Gradient = k






(b) For first-order reaction

Rate = k[A]1Rate = d[A]− Rate = k[A]1Rate =

dt

d[A]−

= k[A]dt


The integrated rate law; ln[A]o – ln[A]t = kt





Half life(t1/2) of a first-order reaction

kt[A]

[A]ln

t

o=Since,

Substituting t = t1/2 ,

2

[A]o[A]t =

k

2ln t1/2 =

[A]t

1/2

o

o kt2[A]

[A]ln =




Characteristic graphs for first-order reaction

ln[A]

ln[A]o

Rate = k [A]

rate

ln[A]o – ln[A] t = kt

ln[A]o[A]

t

[A]

t

t[A]



k

2ln t1/2 =

Graph of half life(t1/2) for first-order reaction

[A]o[A]

t1/2 is independent of

t1 = t2[A]o/2

[A]o/4

t1 t2time

t1/2 is independent of

initial concentration



The reaction 2A → B is first order in A with

a rate constant of 2.8 x 10-2 s-1 at 800oC.

How long will it take for A to decrease from 0.88

M to 0.14 M ?

Exercise:

[A]0 = 0.88 M

[A]t = 0.14 M

Solution:

kt[A]

[A]ln

t

o=

2108.2

)14.088.0(lnt

−×

= t = 66 s




Exercise:

2. For the first order decomposition of H2O2(aq)

given that k = 3.66 x 10-3 s-1 and [H2O2 ]o = 0.882 M,

determine;

a) the time at which [H2O2] = 0.600 M

b) the [H O ] after 225 s.

(105.26 s)

2 2

b) the [H2O2 ] after 225 s. ([H2O2] = 0.387 M)

3. The decomposition of ethane, C2H6 to methyl

radicals is a 1st order reaction with a rate constant

of 5.36 x 10-4 s-1 at 700o C.

C2H6(g) → 2CH3(g)

Calculate the half life of the reaction in minutes.(t1/2 = 21.5 min)




Exercise:

The decomposition of nitrogen pentoxide is as below;

time, t/min 0 10 20 30 40 50 60

N2O5(g) → 2 NO2(g) + ½ O2(g)

time, t/min 0 10 20 30 40 50 60

[N2O5] x 10-4 M 176 124 93 71 53 39 29

The decomposition is first order reaction.

a)Plot a linear graph to prove it.

b)From the plot determine rate constant, k




time, t/min 0 10 20 30 40 50 60

[N2O5] x 10-4 M 176 124 93 71 53 39 29

ln[N2O5] -4.04 -4.39 -4.68 -4.95 -5.24 -5.55 -5.84

Solution:

Plot ln [N O ] vs t gives a

ln [N2O5] vs t

y = -0.0296x - 4.0686

-7

-6

-5

-4

-3

-2

-1

0

0 20 40 60 80

t (s)

ln [N2O5]

Plot ln [N2O5] vs t gives a

linear plot & -ve gradient as

the linear equation for

The integrated rate law

for first order rxn is

ln[N2O5]t = ln [N2O5]o-kt

From the plot gradient

= k = 0.0296 s-1



The following results were obtained from an

experimental investigation on dissociation of

dinitrogen pentokside at 45oC

N2O5(g) → 2 NO2(g) + ½ O2(g)

Exercise:

N2O5(g) → 2 NO2(g) + ½ O2(g)

time, t/min 0 10 20 30 40 50 60

[N2O5] x 10-4 M 176 124 93 71 53 39 29

Plot graph of [N2O5] vs time, determine

i) The order of the reaction

ii) the rate constant k



Solution:

M

180

160

140 t1/2 for the reaction is a constant and

does not depend on the initial [N O ]

2nd t1/2 = 20 min (88 x 10-4 M → 44 x 10-4 M)

1st t1/2 = 20 min (176 x 10-4 M → 88 x 10-4 M)

t (min)

[N2O5] x 10-4M

140

80

120

100

80

60

40

20

10 20 30 40 50 60 70

1/2

does not depend on the initial [N2O5]

Thus, the above reaction is first order

k = 1min035.0

20

2ln −=





(b) For second-order reaction

Rate = k[A]2Rate = d[A]

− Rate = k[A]2Rate = dt

d[A]−

= k[A]2dt


The integrated rate law; kt[A]

1

[A]

1

ot

=−





Half life(t1/2) of a second-order reaction

Since, kt[A]

1

[A]

1

ot

=−

[A]Substituting t = t1/2 ,

k[A]

1t

o

1/2 =

1/2

oo

kt[A]

1

2[A]

1=−

2

[A]o[A]t =




Characteristic graphs for second-order reaction

rate

Rate = k [A]2

1/[A]

kt[A]

1

[A]

1

ot

=−

[A]

t

[A]

1/[A]o

t

1/[A] – 1/[A]o

t



[A]o

Graph of half life(t1/2) for second-order reaction

k[A]

1t

o

1/2 =

time

[A]o/2

[A]o/4

t1

t1/2 is dependent on

initial concentration

t2

t2 = 2t1



Exercise:

Iodine atoms combine to form molecular iodine in

the gaseous phase,

I(g) + I(g) → I2(g)

This reaction follows second order and has a high This reaction follows second order and has a high

rate constant 7.0 x 109 M-1s-1

If the initial concentration of iodine was 0.086 M,

i) calculate it’s concentration after 2 min.

ii) calculate the half life of the reaction if the

initial concentration of iodine is 0.06 M and

0.42 M respectively.

[I] = 1.190 x 10-12 M , 3.4 x 10-10 s 2.4 x 10-9sS K 0 2 7 C H EM I S T RYS K 0 2 7 C H EM I S T RY



Exercise:

1) The decomposition of HI is second order, at 500oC, the half-

life of HI is 2.11 min when the initial HI concentration is

0.10 M. What will be the half-life (in minutes) when the

initial HI concentration is 0.010 M? (21 min)

2) The rate constant for the first-order decomposition of 2) The rate constant for the first-order decomposition of

N2O5(g) at 100oC is 1.46 ×10-1s-1.

a) If the initial concentration of N2O5 in a reaction vessel is

4.5 ×10-3 mol/L, what will the concentration be 20.0 s after

the decomposition begins? (2.4 ×10-4 M)

b) What is the half-life (in s) of N2O5 at 100oC? (4.75 s)

c) If the initial concentration of N2O5 is 4.50 ×10-3 M, what

will be the concentration be after three half-lives? (5.62

×10-4 M) S K 0 2 7 C H EM I S T RYS K 0 2 7 C H EM I S T RY



Summary

Order Rate LawIntegrated Rate

Law Half-Life

[A]

Linear Plot

[A] vs t0

1

2

rate = k

rate = k [A]

rate = k [A]2

ln[A]o - ln[A]t = kt

[A]o-[A]t = kt

t½ln2

k=

t½ =[A]02k

t½ =1

k[A]0kt

[A]

1

[A]

1

ot

=− 1/[A]t vs t

ln[A]t vs t

[A]t vs t




Problem-solving Tip

1. Which equation to use:o The rate-law expression relates rate and conc.

o The integrated rate equation relates time and conc.conc.

2. To know the Order of reactiono Order is stated

o Rate law expression is given

o Unit of rate constant, k.

order 0 M time-1

order 1 time-1

order 2 M-1 time-1


11.1 reaction rate 3

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