REACTION RATES

CHAPTER 10
REACTION KINETICS

10.1 Reaction rate
10.2 Collision theory and

transition state theory
10.3 Factors affecting the

reaction rate.

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CHAPTER 10
REACTION KINETICS

10.1 Reaction rate

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Learning Outcomes

At the end of this lesson, students should be able to:

(a) Define reaction rate.
(b) Explain the graph of concentration against time

in relation to reaction rate.
(c) Write differential rate equation.

aA + bB → cC
Differential rate equation :

Rate = - 1 d[A] = - 1 d[B] = + 1 d[C]

a dt b dt c dt

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Learning Outcomes

At the end of this lesson, students should be able to:

(d) Determine the reaction rate based on a
differential equation.

(e) Define: (i) rate law

(ii) order of reaction

(iii) half-life, t½

(f) Write rate law with respect to the order of
reaction

(g) Write the integrated rate equation for zero, first

and second order reaction. 4

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Learning Outcomes

At the end of this lesson, students should be able to:

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

(i) initial rate method

(ii) the units of rate constants, k

(iii) half-life based on the graph of concentration
against time.

(iv) linear graph method based on integrated
rate equation and rate law.

(i) Perform calculations using the integrated rate

equations. 5

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Reaction Kinetics

Chemical kinetics is the study of the rates of chemical
reactions, the factors that affect these rates, and the
reaction mechanisms by which reactions occur.

Important

industrial process 6

Time
Optimum yield
Optimum conditions
control over reaction,
obtain products economically,
using optimum conditions

Rate of Reaction

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

Example : A  B

rate = - d[A] d[A] = change in concentration of A
dt dt = period of time
d[B] = change in concentration of B
rate = d[B]

dt

Because [A] decreases with time, d[A] is negative.

Unit of rate (mol L-1 s-1) or Ms-1 7

Rate of Reaction

Reaction rate is inversely proportional to time.

rate  1
time

The shorter the time taken for the reaction to progress,
the higher the rate of reaction.

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Rate of Reaction

AB

time

rate = - d[A] [B] ↑
dt

rate = d[B] [A] ↓
dt

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Rate of Reaction

• The average rate is the rate over a period of time.
• The rate of reaction at a given time is called an

instantaneous rate of reaction.
• The instantaneous rate at the beginning of a reaction

is called the initial rate of reaction.
• Instantaneous rate is determined from a graph of

concentration vs time by drawing a line tangent to
the curve at that particular time.

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purple Rate of Reaction

blue Reaction :

red H2O2(aq)  H2O(l) + ½ O2(g)

Reaction rates are obtained
from the slopes of the straight
lines;
An average rate from the
purple line.
The instantaneous rate at
t =300 s from the red line.
The initial rate from the blue
line.

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Rate of Reaction

AB [A]
[A] (M)

Instantaneous rate = Rate at a specific time

Average rate = - d[A] =- [A]final – [A]initial
dt12 tfinal - tinitial
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The differential Rate Equation

A differential rate equation :

the relationship between the rate of disappearance of
reactants and the rate of appearance (formation) of
products.

Consider this reaction :

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 13

The differential Rate Equation

Example : 1

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 means that the rate of disappearance of

N2 is 1/3 of the rate of disappearance of H2 and ½ of the
rate of formation of NH3.
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The differential Rate Equation

Example : 2

Consider the reaction between Zn and AgNO3 to form
Zn(NO3)2 and Ag.

a) Write the differential rate equation for the above
reaction.

Zn(s) + 2AgNO3(aq)  Zn(NO3)2 (aq) + 2Ag(s)

Rate = - 1 d[ AgNO3 ] = d [ Zn( NO3 )2 ]
2 dt dt

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The differential Rate Equation

Example : 2

b) When [Zn2+] is increasing at 0.25 Ms-1, what is the rate
of decrease of [Ag+]?

Rate = -1 d[ AgNO3 ] = d [ Zn( NO3 )2 ]
2 dt dt

- d[ AgNO3 ] = 2 d [ Zn( NO3 )2 ]
dt dt

= 2 x 0.25 Ms-1 16
= 0.50 Ms-1

The differential Rate Equation

Example : 3

Consider the reaction, 2HI  H2 + I2

Determine the rate of disappearance of HI when the rate

of formation of I2 is 1.8 x 10-6 M s-1.

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

d[I2 ] = 1.8  10-6  1 d[HI]  d[I2 ]
dt 2 dt dt

- d[HI] = 2 d[I2 ] 17
dt dt
= 2  1.8  10-6
= 3.6  10-6 M s-1

The differential Rate Equation

Try this … 1
Consider the reaction between Mg and HCl to form
MgCl2 and H2 .

Mg(s) + 2HCl (aq)  MgCl2 (aq) + H2(g)

a) Write the differential rate equation for the above
reaction.

b) When [H2] is increasing at 0.32 moldm-3s-1, what is the
rate of decrease of hydrochloric acid?

(0.64 moldm-3s-1)

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The differential Rate Equation

Try this … 1 : solution

Mg(s) + 2HCl (aq)  MgCl2 (aq) + H2(g)

a) Write the differential rate equation for the above
reaction.

Rate =- 1 d[HCl] = d[MgCl2] = + d[H2]
2 dt dt dt

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The differential Rate Equation

Try this … 1 : solution
b) When [H2] is increasing at 0.32 moldm-3s-1, what is the

rate of decrease of hydrochloric acid?

Given that, d[H2] = 0.32 moldm-3s-1
dt

Rate =- 1 d[HCl] = d[MgCl2] = + d[H2]
2 dt dt dt

- d[HCl] = d[H2] x 2
dt dt

= 0.64 moldm-3s-1

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The differential Rate Equation

Try this … 2
Hydrogen gas produced nonpolluting product water
vapour when react with O2. The reaction is as follows …

2H2(g) + O2(g)  2H2O(g)

a) Express the rate in terms of changes in [H2], [O2] and
[H2O] with time.

b) When [O2] is decreasing at 0.23 mol L-1 s-1, at what
rate is [H2O] increasing?

(0.46 mol L-1 s-1) 21

The differential Rate Equation

Try this … 2 : solution

2H2(g) + O2(g)  2H2O(g)

a) Express the rate in terms of changes in [H2], [O2] and
[H2O] with time.

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

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The differential Rate Equation

Try this … 2 : solution

2H2(g) + O2(g)  2H2O(g)

b) When [O2] is decreasing at 0.23 mol L-1 s-1, at what
rate is [H2O] increasing?

- d[O 2] = - 0.23 M s-1
dt
1 d[H2O]
- d[O2] = + 2 dt

dt

d[H2O] = 0.46 mol L-1 s-1 23
dt

The differential Rate Equation

Try this … 3

Consider the reaction:

4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g)
Nitrogen gas was formed at a rate of 0.72 mol L1 s1.

i. Write the rate differential equation for the above
reaction.

ii. Calculate the rate of:

(a) water formation

(b) oxygen consumption (a) 2.16 mol L1 s1
(c) ammonia disappearance (b) 1.08 mol L1 s1
(c) 1.44 mol L1 s1

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The differential Rate Equation

Try this … 3 : solution
(i) Write the rate differential equation for the reaction.

4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g)

rate = - 1 d[NH3] = - 1 d[O2] =+ 1 d[N2] =+ 1 d[H2O]
4 dt 3 dt 2 dt 6 dt

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The differential Rate Equation

Try this … 3 : solution

(ii) (a) Given that, d[N2] = 0.72 mol L1 s1
dt

Rate = + 1 d[N2] = + 1 d[H2O]
2 dt 6 dt

d[H2O] = 6 x ½ x 0.72 mol L1 s1
dt

= 2.16 mol L1 s1

Rate of formation of water = 2.16 mol L1 s1
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The differential Rate Equation

Try this … 3 : solution

(ii) (b) Given that, d[N2] = 0.72 mol L1 s1
dt

Rate = - 1 d[O2] =+ 1 d[N2]
3 dt 2 dt

- d[O2] = 3 x ½ x 0.72 mol L1 s1 )
dt

= 1.08 mol L1 s1

Rate of oxygen consumption = 1.08 mol L1 s1

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Next Lecture …

THE
RATE LAW