C8 REACTION KINETICS

CHEMISTRY UNIT
KOLEJ MATRIKULASI MELAKA

SHARED BY MISS DALINA BINTI DAUD

CHAPTER 8
REACTION KINETICS

8.1 Reaction rate
8.2 Collision Theory
8.3 Factors affecting reaction rate

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Learning Outcomes:

8.1 Reaction Rate

(a) Define reaction rate.
(b) Write differential rate equation.
(c) Determine reaction rate based on differential

rate equation of a reaction.
(d) Define rate law, order of reaction and half-life.
(e) Write rate law with respect to the order of

reaction.
(f) Write the integrated rate equation for zero, first

and second order reactions.

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Learning Outcomes:

(g) Determine the order of reaction involving a single
reactant using:

i. initial rate method
ii. half-life based on the graph of concentration

against time
iii. linear graph method based on the integrated

rate equation and rate law.
(h) Perform calculation using the integrated rate

equationS.

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Reaction Kinetics

Reaction Kinetics is the study of the rates
of chemical reactions and the factors that
affect these rates.

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8.1 Reaction Rate

Reaction rate is the change in the concentration
of a reactant or a product with time.

• Consider a reaction : A  B
• When reaction occur, the concentration of

A is decreased while the concentration of
B increased with time.

6

Concentration Consider the reaction :
AB

B (product)

A (reactant)

time

Graph of Concentration versus Time

7

Reaction rate = change in concentration of a reactant or product

time taken

Consider the reaction : A  B

Reaction rate = d[A] The minus sign indicates concentration
dt of reactant, A decreases with time.

Reaction rate = + d[B] The positive sign indicates concentration
of product,B increases with time.
dt

The unit for the reaction rate is usually mol dm–3 s–1

However, other unit for time can also be used. 8
Example: mol dm–3 min–1

Reaction rate can be expressed as :

i) average rate
ii) instantaneous rate
iii) initial rate

i) The average rate of reaction is how fast a
reaction progresses over an interval or in a
period of time

Average rate = change in concentration of a reactant or product
of reaction time taken for the change to happen

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Example :

Consider the following reaction:

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

By using the following data, calculate the average
rate of reaction during the first 50s.

10

Answer :
Average rate = change in concentration of a reactant or product

time taken for the change to happen

Average rate = 0.0101 – 0.0120
50.0

= 3.80 x 105 Ms1

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ii) Instantaneous rate of reaction is the rate of
reaction at any particular time.

• The instantaneous rate is determined from a
graph of concentration versus time by the
gradient of the tangent to the curve at a
particular time.

iii) Initial rate of reaction is instantaneous rate
at the beginning of a reaction (t=0), i.e
immediately after the reactants are mixed.

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Graph of [reactant] versus Time

[reactant]

y1 t2 time
t1
13
t1

Rate at t1 = y1
t1

Graph of [product] versus Time

[product]

y2

t2

t2 time

Rate at t2 = y2
t2
14

Example : 2Br- (aq) + 2H+ (aq) + CO2 (g)
Br2 (aq) + HCOOH (aq)

The instantaneous rate at 100 s = 2.96 x 105 Ms 1

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The Differential Rate Equation
 shows the relationship between the rate of

disappearance of reactants and formation of
products.

aA + bB  cC + dD

Rate =  1 d[A] =  1 d[B] = + 1 d[C] = + 1 d[D]
a dt b dt c dt d dt

a,b,c and d are the stoichiometric coefficients

16

Example :

The formation of NH3,
N2(g) + 3H2(g)  2NH3(g)

The differential rate equation is;

Rate =  d[N2 ]   1 d[H2 ]  1 d[NH3]
dt 3 dt 2 dt

The equation shows

i. the rate of disappearance of N2 is ⅓ the rate of

disappearance of H2 and ½ the rate of formation of

NH3. 3

ii. the rate of disappearance of H2 is 2 the rate of

formation of NH3 and 3 times the rate of disappearance

of N2. 17

Example :
Consider the reaction,

2HI(g)  H2(g) + I2(g)
Determine the rate of disappearance of HI when
the rate of I2 formation is 1.8 x 10-6 M s-1.

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Answer:

Rate = 1 d[HI]  d[H2 ]  d[I2 ]
2 dt dt dt

given: d[I2 ] 1.810-6 Ms-1
dt

Rate =  1 d[HI]  d[I2 ]
2 dt dt

 d[HI] = 2  1.8  10-6
dt

= 3.6  10-6 M s-1

19

Example :
Consider the following reaction :

2H2(g) + O2(g)  2H2O(g)
When [O2] is decreasing at 0.23 Ms1,
what is the rate of change of the [H2O]?

20

Answer :

Rate =  1 d[H2] =  d[O2] = + 1 d[H2O]
2 dt dt 2 dt

 d[O2] = 0.23 Ms1
dt

+ 1 d[H2O] =  d[O2] = ( 0.23 Ms1)
2 dt dt

+ d[H2O] = 2(0.23 Ms1)
dt 0.46 Ms1

=

 [H2O] changes twice as fast as [O2] 21

REACTION RATE EQUATION, RATE CONSTANT &
ORDER OF REACTION

Rate Equation or Rate Law

• An equation that relates the rate of a reaction to the
rate constant, k and the concentrations of the
reactants raised to some powers.

aA + bB  cC + dD

The rate equation or the rate law :

Rate = k [A]m [B]n

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Where :

[A] = concentration of A
[B] = concentration of B
k = rate constant
m = order of reactions with respect to A
n = order of reaction with respect to B
m + n = overall order of reaction

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Order of reaction

The power to which the concentration of a reactant
is raised in the rate law, i.e

Rate = k [A]m [B]n
• They are usually integers: 0,1,2….
• Order of reaction is usually defined in terms

of reactant concentrations.

Note:

The reaction orders cannot be predicted from the
stoichiometry of the balanced equation; it must be
determined experimentally.

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Example:
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)

Rate = k[NO]2[H2]

The reaction is:
 second order with respect to NO,
 first order with respect to H2
 Overall reaction = third (2 + 1) order reaction

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Rate constant, k

• A constant of proportionality between the reaction
rate and the concentrations of the reactants that
appear in the rate law.

• The magnitude of k changes with temperature

• Units of the rate constants depend on the overall
reaction order of the rate law.

(a) Zero-order Reactions : Ms1

(b) First-order Reactions : s1

(c) Second-order Reaction : M1s1

Note:

(The unit of rate constant, k can be used to determine

the order of reaction.) 26

Example :
Consider the following reaction :
H2 + I2  2HI

Rate = k[H2][I2]
What is the unit of the rate constant, k ?

Answer: Rate = k[H2][I2]

k = Ms1
(M)(M)

= M1s1

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Integrated Rate Equation

•The relationship between the concentration of reactant
and time is represented by the integrated rate equation.

(a) Zero-order Reaction

 In a zero-order reaction, the rate is constant and
independent to the concentration of the reactant.

e.g. If the reaction , A  products 28
is a zero-order reaction, therefore;
Rate = k[A]o

Rate = k or -d[A] = k
dt

The integrated rate equation for zero order
reaction is:

[A]o – [A] = kt
OR

[A] = [A]o – kt

Where:
[A]o = the initial concentration
[A] = concentration at particular time
t = time of reaction
k = rate constant

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Graph of zero order reaction:

(a) From the rate law of zero order :Rate = k[A]o
Rate

[A]

(b) From the integrated rate law:

(i) [A] = [A]o – kt (ii) [A]o – [A] = kt
[A]o – [A]
y = c + mx

[A]

gradient = k gradient = k

t
t 30

Half life ( t½ )

• is the time taken for the concentration of a reactant
to reach half of its initial concentration.
Half life for zero-order reaction

[A]o – [A] = kt

when t = t½ , The initial concentration decreases

to half : [A] = [A]o
2

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Substitute [A] = [A]o into the integrated law:

2

[A]o – [A] = kt
[A]o – [A]o = kt½

2

[A]o = kt½

2

t1  Ao
2
2k

The half life of zero-order reaction depends on

the initial concentration of the reactant. 32

Example :
Consider the reaction:

B(aq)  C(aq)

The rate of the reaction does not depend on the
concentration of the reactant. If the rate constant
is 2.95 x 10-2 mol dm-3 s-1 and the initial
concentration of the reactant is 4.0 mol dm-3.
Determine the half-life for the reaction.

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Answer :

t1Τ2= [A]
2k

= 4.0
2(2.95×10−2)

= 67.85s

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(b) First-order Reactions

• In a first-order reaction, the rate depends
on the concentration of a single reactant
raised to the power of one.

If the reaction : A  products
is a first-order reaction, therefore;

Rate = k[A]1 or -d[A] = k [A]
dt

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The integrated rate law for first-order reaction is:

ln[A]o – ln[A] = kt

OR

ln [A]o  kt
[A]

OR

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

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Graph of first order reaction:
[A]

(a)
[A]o

gradient = rate

t
(b) From rate law of first order : Rate = k[A]1

Rate

gradient = k

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[A]

(c) From the integrated rate law:

i) ln [A]o  kt ln Ao
A
[A]

gradient = k

(ii) ln[A] = ln[A]o - kt ln[A] t

ln[A]o gradient = -k

t 38

Half life for first-order reaction

When t = t½ , [A] = [A]o
2

By substituting the value into the integrated law:

ln [A]o  kt
[A]

ln [A]o = kt½
½[A]o

ln 2 = kt½

t½ = ln 2 = 0.693
kk

The half life of a first-order reaction is independent of the

initial concentration of the reactant. 39

Example :

The dissociation of SO2Cl2 is a first order reaction.
SO2Cl2(g)  SO2(g) + Cl2(g)

i Write the differential rate equation for the above
reaction.

ii Calculate the rate constant, k at 500 K if 5% of
SO2Cl2 decomposes in 6.75 minutes.

iii Calculate the half life for the decomposition.

(PSPM 2001/2002)

(Ans:7.6 x 103 min 1 , 91.2 min )

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ANSWER
i) Rate =

41

Example :

Radioactive decay is a first order reaction. The
half life of a radioactive element, astatine is 8.3
hours. Starting with 2.0 g of astatine, calculate
the mass of astatine remains after 33.2 hours.

[Ans : 0.125 g]

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For first order reaction:

t1Τ2= ln 2
k

ln 2
k = t1ൗ2

= 0.693 = 0.0835hr−1
8.3

ln [A] = kt
[A]

ln2[A.0] = 0.0835 × 33.2

[A] = 0.125

Mass A = 0.125g 43

(c) Second-order Reaction

• In a second-order reaction, the rate depends
on the concentration of two reactants or
to the square of the concentration of a single
reactant
If the reaction : A  products
is a second-order reaction, therefore;

Rate = k[A]2

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Graph of second order reaction:
Rate = k[A]2

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The integrated rate law for second-order reaction is:

1  kt  1

A AO

or

1  1  kt

A AO

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Graph of second order:

Rate = k[A]2

y =m x + c

Rate Rate

[A] [A]2

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1  kt  1

A AO

y =m x + c

1

A

1

Ao

time,t

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1  1  kt

A AO

1  1

A AO

time,t

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Half life for second-order reaction

When t = t½ , [A] = [A]o
2

substitute [A] into the integrated law:

1 = kt + 1
[A ] [A]o

11
= kt½ +
½[A]
[A]o

t1  1
2
k Ao

The half life of the second-order reaction is

inversely proportional to the initial concentration. 50