CHAPTER 5_SEM1_LALITHAPRAKASH

CHAPTER 5:
REACTION KINETICS

LALITHA / KTESI

ANALYSIS OF STPM QUESTION

YEAR 2013 2014 2015 2016 2017 2018 2019

BC B C B C B C B C B C B C

SECTION √√ √√

LALITHA / KTESI

5.1 RATE OF REACTION

CHEMICAL REACTION : a process in which one or more substances;

the reactants are converted to one or more different substances; the
products.

All chemical reactions are classified to 2 categories:
a) Fast
b)Slow

LALITHA / KTESI

❑CHEMICAL KINETICS is the study and discussion

of chemical reactions with respect to reaction rates,
effect of various variables, re-arrangement of atoms,
formation of intermediates etc..

❑ RATE OF REACTION IS THE CHANGE IN THE

CONCENTRATION OF A REACTANT OR PRODUCT

WITH TIME.

Rate ∝ 1

LALITHA / K T( )ESI

Rate of Reaction = changes of concentration of a reactant
‘or’ product per unit time

LALITHA / KTESI

x A(aq) yB(aq)

RATE OF REACTION (moldm-3s-1)

RATE OF DISAPEARANCE RATE OF FORMATION

ℎ ℎ

ℎ ℎ

1 − [ ] 1 + [ ]


LALITHA / KTESI

2A + B C + D + 3E

2 mol of A reacts with 1 mol of B to form 1 mol of C, 1 mol D and 3 mol of E

Rate of disappearance of A is 2 times disappearance of B //
Rate of formation of E is 3 times the rate of disappearance of B

Rate of Reaction : − [ ] ≡ − [ ] ≡ + [ ]



LALITHA / KTESIRate of formation of E is 3 times higher than rate of formation of C and D

Example:

Dinitrogen pentoxide decomposes according to the equation below:

2 2 5 4 2 + 2( )
At a particular instance the rate of disappearance of dinitrogen pentoxide is

1.65 x 10-3 moldm-3s-1determine the rate formation of NO2 and O2 at that
instance.

Rate of reaction: − [ 2 5] ≡ + [ ] ≡ + [ ]


ANSWER: d[O2] / dt = 8.25 x 10-4 moldm-3s-1
ANSWER: d[NO2]/dt = 3.30 x 10-3 moldm-3s-1

LALITHA / KTESI

− [ 2 5] = 1.65 x 10-3



− [ 2 5]


− [ 2 5]



4 1.65 × 10−3
2
= 3.30 × 10−3 −3

LALITHA / KTESI

AVERAGE RATE INSTANTANEOUS INITIAL RATE
RATE
Changes in concentration Instantaneous rate of
of a substance in a fixed Rate at specific time reaction at the time is zero
time interval

Calculated from Calculated from only Calculated from
• Data gradient of graphs • Gradient of graph at t=0
• Graph • Rate equation
• formula

LALITHA / KTESI

For a reaction to take place:-

a) Reacting molecules must collide

b) Molecules must collide in correct
orientation

c) Reacting molecules must posses
kinetic energy same or greater than
activation energy

LALITHA / KTESI

ACTIVATION ENERGY

IS THE MINIMUM ENERGY
REQUIRED TO INITIATE A

CHEMICAL REACTION

LALITHA / KTESI

• When collision of particles obey both rule (b) and (c), collisions are
called as EFFECTIVE COLLISION

• When frequency of collision increases Rate of Reaction also
INCREASES

LALITHA / KTESI

FACTORSAFFECTING
RATE OF REACTION

LALITHA / KTESI

❖ MATHEMATIC AL RELATIONSHIP OF THE RATE OF
REACTION TO THE RATE CONSTANT AND THE
CONCENTRATIONS OF EACH REACTANTS RAISED TO
POWER OF ITS ORDER OF REACTION

❖ EQUATION THAT SHOWS HOW RATE IS AFFECTED
BY THE CONCENTRATION OF REACTANTS

Rate = k [ ] [ ]

LALITHA / KTESI

Rate = k[ ] [ ]

km

Rate constant Order of reaction with respect to A

a coefficient of proportionality Exponent of the reactant's
relating the rate of a chemical concentration in the rate equation
reaction at a given temperature to which shows how the rate of
the concentration of reactants reaction when concentration
(each raised to the power of its changes
order of reaction)
Rate α [ ]

➢Same value for the same reaction Determined only by experiment

➢Only influenced by temp

➢Unit depends on overall order of

reaction

LALITHA / KTESI➢ If k is high, rate of reaction is high

2 2 + 2 − +2 + 2 2 + 2

The above reaction is first order respect to hydrogen peroxide

and iodide ions.

a)Write the rate equation for the reaction above

Rate = k[H2O2][I-]

b)What is the unit of rate constant, k?

k = ( − − = dm3mol-1s-1
− )( − )

LALITHA / KTESI

2 2 + 2 − +2 + 2 2 + 2

The rate equation for the reaction above

Rate = k[H2O2][I-]

What is the

Reaction order respect to H2 2 = first order respect to H2O2
a) Reaction order respect to − = first order respect to I-

c) Reaction order respect to + = zero order respect to H+

d) Overall order of reaction = 1+1 =2
LALITHA / KTESI

Zero order reaction First order reaction Second order
reaction

Rate = k[ ]0 Rate = k[A] Rate = k[ ]2

It’s a zero order It’s a first order respect It’s a second order
respect to A to A respect to A

Rate of reaction is Rate of reaction Rate of reaction
independents of the
concentration of A doubles when increases 4 times if

concentration doubles concentration doubles

LALITHA / KTESI

CALCULATION OF RATE CONSTANT , INITIAL
RATE , ORDER OF REACTION & RATE LAW
4 METHODS TO DETERMINE THESE VALUES:
1. INITIAL RATE METHOD

2. GRAPHICAL METHOD

3. HALF LIFE

4. RATE CONSTANT UNIT

LALITHA / KTESI

1. INITIAL RATE METHOD

0.24

(a) Determine
(i) The order of reaction with respect to A and B,
(ii) What is the overall order?

(b) Write the rate equation for the reaction.
(c) Using experiment (I); calculate the rate constant and state its unit.
(d) Calculate the rate when the concentrations of A and B are 0.34 moldm-3

and 0.52 moldm-3 respectively.

LALITHA / KTESI

Let the rate equation be:

Rate = k[A]x[B]y

From expt 1: 0.514 = k(0.12)x(0.26)y …………….(1)
From expt II: 2.06 = k(0.12)x(0.52)y …………….(2)
From expt III: 1.03 = k(0.24)x(0.26)y …………….(3)

Experiment II ÷ I . = k(0.12)x(0.26)y

0.514 = k(0.12)x(0.52)y
= ; =

Experiment III ÷ I . = k(0.24)x(0.26)y

0.514 = k(0.12)x(0.26)y
= ; =

LALITHA / KTESI

(ii) Overall order = 1+2 =3

(b) Rate = k[A][B]2

(c) 0.514 = k(0.12)(0.26)2
k = 63 dm6 mol-2 min-1

(d) Rate = 63 x (0.34)(0.52)2
= 5.8 mol dm-3 min-1

LALITHA / KTESI

2. GRAPHICAL METHOD // 3. HALF LIFE // 4.UNIT K

RATE AGAINTS CONCENTRATION CONCENTRATION AGAINTS

TIME

Rate

k Zero order concentration Zero order
k=mol dm-3 s-1

time

Rate First order First order concentration t1= t2
k =s-1
m= k

Rate Rate concentration t2= 2t1

Second order m= k Second

k = dm3mol-1s-1 [A]2 order

LALITHA / KTESI[A] time

gradient = -k

LALITHA / KTESI

2. GRAPHICAL METHOD

The data below refers to the following reaction
A products

Plot a graph of 1 against time

[ ]

a) Use your graph to determine the rate constant.
b) Calculate the half life of the reaction when the concentration of A is

(i) 1.00 moldm-3
(ii) 0.500 moldm-3

Time / s 0 40 80 120 160
24.9
− − − 63.0 11.0 15.6 20.2
[ ]

LALITHA / KTESI

Gradient = k
k = 1.17 x 10-2 dm3mol-1s-1

1
1/2 = [ ]

1
1/2 = 1.17 x 10−2 (1.00) = 85.5

1
1/2 = [ ]

1
1/2 = 1.17 x 10−2 (0.500)

= 171

3. RATE CONSTANT METHOD

The rate constant for the decomposition of H2O2 is 6.33 x 10-3 min-1
a) What is the half life of the decomposition of H2O2 ?
b) How long will it take for H2O2 concentration to fall to 12.5% of its

initial value?

From the unit of k it confirms

the reaction is first order:

a) 1/2= ln 2 1 2 3

0.693 ) 100% ՜ 50% ՜ 25% ՜ 12.5%
1/2 = 1.70 10−3 3(110) = 330 min

= 110

LALITHA / KTESI

LALITHA / KTESImol dm-3 s-1
s-1 dm3mol -1s-1

1. Reaction rates increases as Temperature increases.

2. Reaction rate becomes 2x // DOUBLES when temperature increases 10℃.

3. When T increases , Average Kinetic Energy of particles increases.
4. Frequency of collision between reacting particles increases.
5. Frequency of effective collision increases.

6. Fraction of molecules with higher energies or greater than activation energy, Ea

LALITHAalso increases. / KTESI

MAXWELL BOLTZMANN
DISTRIBUTION CURVE

T2 > T1

LALITHA / KTESI

AVERAGE KINETIC ENERGY

K = 3 T

2
SPEED OF MOLECULE

v = 3



speed depends on molecular weight

• DIFFERENT GASES AT SAME TEMPERATURE has same
AVERAGE KINETIC ENERGY BUT AVERAGE SPEED not same

LALITHA / KTESI

ARRHENIUS EQUATION

Shows how:
• To analyze effect of T on the

rate constant, k and rate of
reaction
• Analyze the effect of Ea on the
reaction rate
• Allow to determine the Ea if k is
known with T

Rate constant, k increases with:

LALITHA / KTESIT increases
Ea decreases

LALITHA / KTESI

LALITHA / K 1TE S 2 I

The rate constant of a first order of reaction is 3.64 x 10-2 s-1 at 298 K.
What is the rate constant at 350 K if the activation energy 50.2 kJmol-1?

ln 1 = ( 1 − 1 )
2 2 1

ln 3.64 10−2 = 50.2 103 ( 1 − 1 )

2 (8.31) 350 298

k = 7.40 x 10-1 s-1

LALITHA / KTESI

LALITHA / KTESI

CHARACTERISTICS OF CATALYST

❑ Highly specific
❑ Small amount only used
❑ No effect on yield of product
❑ Can be poisoned by As or - C=N
❑ Increases efficiency by adding promoter
❑ Exist from transition metals

LALITHA / KTESI

CATALYST MAY CATALYST
ALTER DOESN’T ALTER

• The rate of reaction • The yield
• The mechanism of reaction • The overall

• The order of reaction stoichiometry

• The activation energy of • The enthalpy of the
reaction reaction

• Rate LawLALITHA / KTESI

Ea decreases ; k increases

LALITHA / KTESI

• catalyst INCREASES THE RATE OF REACTION by providing an ALTERNATIVE
PATHWAY which requires LOWER ACTIVATION ENERGY

• LOWERING THE ACTIVATION ENERGY, INCREASES THE NUMBER OF

LALITHA / KTESIPARTICLES WITH MINIMUM ENERGY FOR A EFFECTIVE COLLISON

LALITHA / KTESIEnergy profile diagram for an exothermic reaction
Energy profile diagram for an endothermic reaction

2 TYPES OF CATALYST

HOMOGENOUS CATALYST HETREOGENOUS CATALYST

• CATALYST EXISTS AT SAME • CATALYST AND REACTANTS ARE
PHASE AS REACTANTS IN DIFFERENT PHASE

• INTERMEDIATE PRODUCT • ADSORPTION THEORY
THEORY

LALITHA / KTESI

EXAMPLE OF REACTION

• Reactant : SO2 (g) & O2 (g)
NO2 (g)
• Homogenous Catalyst:

• Mechanism

• STEP 1 : 2NO2(g) + 2SO2(g) 2NO(g) +2SO3(g)
• STEP 2 : 2NO(g) + O2(g)
2NO2(g)
2SO2(g) + O2(g) 2SO3(g)

LALITHA / KTESI

• Intermediate product which shown in the valley (between 2 peaks)
• Many transition metal ions are homogenous catalyst because the

intermediate compounds involved a different oxidation state of the

LALITHA / KTESItransition metal ion.

EXAMPLE OF REACTION

• Reactant : SO2 (g) & O2 (g)

• Heterogenous Catalyst: V2O5 (s)

• SO2 & O2 adsorb onto the surface of V2O5 catalyst facilitates the
oxidation of SO2 to SO3

• Reaction takes place where the covalent bonds in the SO2 & O2
molecules are weaken and the Ea is lower. It holds the reactant

molecules in the right orientation for new bonds to be formed.

• When completely old bonds broken & new bonds formed , SO3
release from V2O5 surface which is called desorption.

LALITHA / KTESI

3 STEPS;

1. ADSORPTION

chemisorption

2. REACTION

Lower Ea Temporary bonds

[Reactant] increases

Correct orientation

3. DESORPTION

Transition element are common heterogenous
catalyst bcoz they have empty d-orbitals used
to form temporary bonds with reacting
molecules during adsorption.

LALITHA / KTESI

POSITIVE CATALYST FOR ‘REVERSIBLE REACTION’

❑CATALYST LOWERS THE ACTIVATION ENERGY OF THE FORWARD AND REVERSE
REACTION BY THE SAME AMOUNT.

❑ INCREASES THE RATE CONSTANT FORWARD AND RATE CONSTANT REVERSE
( −1 and 1)
❑CATALYST INCREASES THE RATE OF REACTION FORWARD AND REVERSE BY THE

SAME AMOUNT
❑CATALYST DOES NOT CHANGE THE COMPOSITION OF PRODUCT AND

LALITHA / KTESIREACTANT THAT EXIST AT EQUILIBRIUM

IMPORTANT DEFINITIONS:

TERMS DEFINITION

Rate of reaction Is the change in the concentration of a reactant or a product with
time.

Rate law An expression relation the rate of reaction and rate constant and
concentration of reactants.

Rate constant Rate constant, k is a constant in the rate equation which relate the
rate of reaction to the reactant’s concentration.

Order of reaction Sum of the powers of the concentration of the reactants in the rate

equation of the chemical reaction.

Half life Time required for the concentration decrease to half of its initial
concentration.

Activation energy Minimum energy required to initiate a chemical reaction.

Catalyst Substance that increase the rate of reaction by providing an

alternative pathway with lower activation energy
Catalyst n reactants are in the same phase.
LALITHA / KTESIHomogenous