Chemical Kinetics

ACTIVE SITE Chemical Kinetics
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3 Chemical Kinetics
The branch of chemistry which deals with study of rates of chemical reactions, the factors affecting the rates
of the reactions and the mechanism by which the reactions proceed is known as chemical kinetics.
Rate of Reaction
The rate of a reaction is defined as “the change in concentration of any one of the reactants or products per
unit time”.
Let the change in the concentration with respect to the given reactant or product is dx in a small interval of
dx
time dt. The rate of reaction (r) is given by, rate of reaction r = 
dt
Consider a reaction of the type: A → B
dx d[A] decrease in concentration of reactant
rate of reaction r =  =– =
dt dt time interval
d[B] increase in concentration of product
=+ =
dt time interval
Average Rate of Reaction

According to the law of mass action, the rate of a reaction depends upon the molar concentrations of the
reactants. But the concentrations of the reactants keep on decreasing with the passage of time while those of
the products keep on increasing. Hence, the rate of the reaction does not remain constant throughout. Thus
the rate of the reaction is the ‘average rate of reaction’ during the time interval chosen.
change in concentration during the time interval chosen
 Average rate of reaction =
time interval
Calculation of the Average Rate of Reaction

A→B
To calculate the average rate of reaction between any two 0.03 B products
conc.(mol dm–3)
instants of time t1 and t2, the corresponding concentrations 0.02 x1

x1 and x2 are as shown in the figure. 0.014 x2
0.01
x − x1 A reactants
Then the average rate of reaction r = 2 t1 t 2
(t 2 − t1 ) O
0.0 5 10 15 20
dx ( 0.014 − 0.020 ) mol dm −3 −0.006 time (minutes)
r= − = =
dt (10 − 5)  60 s 300
= –2.0  10–5 mol dm–3 s–1
Similar method is used to calculate the rate of the reaction with respect to product by using curve B.
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Instantaneous Rate of a Reaction
The instantaneous rate of a reaction i.e. the rate of a reaction at any instant of time is the rate of change of
concentration of any one of the reactants or products at that particular instant of time.
To express the instantaneous rate of reaction a very small interval of time (dt) is chosen. At that particular
instant of time the rate of the reaction is supposed to be almost constant. The small change in concentration
(dx) is determined during this small interval of time (dt). The rate of reaction at that instant of time is given
dx
by .
dt
Calculation of Instantaneous Rate of Reaction
The change in the concentration of reactants
A→B
(curve A) or products (curve B) with progress of reaction M
0.03
is shown in the figure.
conc.(mol dm–3)
0.029
B products
To know the rate of the reaction at any time t, a tangent is
drawn to the curve at the point corresponding to that time 0.02
(as shown in diagram at time 7.5 minutes) and it is

0.01
extended on either side so as to cut the axes, A reactants
rate of reaction O
dx change in concentration 0.0 5 10 15 20 N 25
r=– = t (7.5)
dt time time (minutes)
x OM
= = slope of the tangent =
t ON
At time, t = 7.5 minutes, x = OM = 0.029 mol dm–3
t = ON = 23 minutes = 23  60 = 1380 seconds
dx 0.029 mol dm −3
The rate of reaction at 450th second (at 7.5 min) = = = 2.1  10–5 mol dm–3 s–1
dt 1380 s
Thus the relation between average and instantaneous rates is as follows:
 x  dx
Instantaneous rate = average rate as t approaches to zero =   =
 t  t →0 dt
Expression for the Rates of Reaction

Consider the reaction, 2N2O5 ⎯⎯
→ 2N2O4 + O2
In this reaction 2 moles of N2O5 decompose to give 2 moles of N2O4 and 1 mole of O2. Hence, if we express
the rate of the reaction in terms of O2, it will be half of that expressed in terms of
N2O5 or N2O4. To overcome this difficulty, the rates of reactions are expressed not only by dividing the small
change in concentration by a small interval of time but by further dividing it by the stoichiometric
co-efficient. Hence for the above reaction
1 d[N 2 O5 ] 1 d[N 2 O 4 ] d[O 2 ]
rate = – =+ =+
2 dt 2 dt dt
This will give identical value of the rate of reaction irrespective of the reactant or product in terms of which
it has been expressed.
The instantaneous rate of reaction in terms of the given reactants or products is expressed by using general
equation of the type
aA + bB + … → mM + nN + … where a, b, … and m, n, … are the stoichiometric coefficients of reactants
and products.
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dx 1 d[A] 1 d[B] 1 d[M] 1 d[N]
rate of reaction r =  =– =– =–… =+ =+ =+…
dt a dt b dt m dt n dt
Unit of Rate of Reaction

dx mol dm −3
r= = = mol dm–3 s–1 or mol L–1 s–1
dt s
For gaseous reaction, units are atm s–1
Factors Affecting Rate of Reaction
• Nature of reactants: Reactions involving lesser bond rearrangements proceed much faster than those
which involve larger bond rearrangements. Gaseous, liquid or aqueous reactants react faster than solids.
• Concentration of reactants: At constant temperature and in absence of catalyst, the rate of reaction
increases with increase in concentration of reactants.
• Temperature: In most of the cases the rate of reaction in a homogeneous system is approximately
doubled or even tripled by an increase in temperature by 10 K.
• Catalyst: The catalyst increases the rate of reaction by taking the reaction in an alternate path for which
the energy of activation is low.
• Presence of light: Photochemical reactions are influenced by the incident radiation of specific
wavelength.
Law of Mass Action

According to the law of mass action, the rate of a chemical reaction is directly proportional to the product of
molar concentrations (active masses) of the reactants with each concentration term being raised to the power
of its coefficients.
Consider the reaction aA + bB → products.
Rate of reaction  [A]a [B]b or Rate = k [A]a [B]b where k is constant of proportionality.
It is written from balanced chemical equation.
Rate Law Expression

The rate law states that “the rate of a reaction is proportional to the product of molar concentrations of
reactants, each concentration term being raised to power equal to stoichiometric coefficient in the
experimentally determined rate equation at constant temperature”.
Consider reaction of type : aA + bB → products
The rate of the reaction may not depend upon all the ‘a’ concentration terms of A and all the ‘b’
concentration terms of B. By the experiment if it is found that the rate of reaction depends upon 
concentration terms of A and  concentration terms of B, then by applying rate law, we get
dx
r=  [A] [B]; at constant temperature.
dt
 r = k[A] [B]
where k is a constant called rate constant or velocity constant or specific reaction rate.
If [A] = [B] = 1.0 mol dm–3, then
r = k[1] [1] = k
From this expression, the rate constant is defined as the rate of the reaction when the concentration of
each reactant is taken as unity at the given temperature.
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Characteristics of Rate Constant (k)
The characteristics of rate constant are
• Each reaction has a definite value of the rate constant at a particular temperature.
• Rate constant is a measure of the rate of the reaction. Greater the value of rate constant, faster is the
reaction.
• The value of rate constant for the reaction changes with temperature.
• The value of rate constant for the reaction does not depend on the concentration of reactants.
• The units of rate constants depend upon the order of the reaction.
Differences between rate constant (k) and rate of reaction (r)

Rate constant (k) Rate of reaction (r)
• It is equal to rate of the reaction when the It is the change in concentration of reactants or
concentration of each reactant is unity. products per unit time.
• Units of rate constant is (mol dm–3)1–n s–1 Units of rate of reaction is mol dm–3 s–1
i.e. it depends on order of reaction.
• It depends only upon temperature. It depends upon both temperature and
concentration.
Difference between rate law and law of mass action

Rate law states that the rate of reaction depends upon the concentration term on which the rate of reaction
actually depends, as observed experimentally. The law of mass action is simply based on the stoichiometry of
 the equation.
For the reaction of type aA + bB → products
According to rate law, rate = k[A] [B]. Here,  and  need not be equal to a and b respectively.
According to law of mass action, rate = k[A] a [B]b
Order of Reaction (n)

The order of reaction is defined as “the sum of all the powers to which the concentration terms are
raised in rate law equation”. For general reaction: aA + bB → cC + dD,
The rate law equation is r = k[A]x [B]y
order with respect to A = x and order with respect to B = y
 Overall order of reaction, n = (x + y)
The characteristics of order of reaction are as follows:
• The order of reaction is an experimentally determined quantity.
• It cannot be written from the balanced chemical equation.
• It is evaluated only from the rate law equation.
• It depends upon the number molecules undergoing change in concentration.
• Order may be zero, whole number, fractional or even negative.
• The reactions with order  3 are rare.
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Rate Determining Step (rds)
Some reactions take place in more than one step. Each step has its own rate, the slowest step is called rate
determining step (abbreviated as rds) or rate controlling step.
Example: 2 NO2 + F2 → 2 NO2F
Mechanism: NO2 + F2 ⎯⎯⎯ slow
→ NO2F + F … (1)
NO2 + F ⎯⎯→ NO2 F
fast
… (2)
2NO2 + F2 → 2NO2F
The first step (1) is slow and hence is the rate determining step. (rds)
The rate law equation is written from the slowest step.
r = k[NO2] [F2]; order of reaction = (1 + 1) = 2.
Molecularity of a Reaction
The molecularity is defined as “the sum of the molecules of different or same reactant species as
represented by balanced chemical equation”. In case of multi step reaction, each reaction has its own
molecularity.
Example: 4HBr + O2 → 2H2O + 2Br2
slow
Probable mechanism: HBr + O2 HOOBr molecularity two (bimolecular)
HOOBr + HBr ⎯⎯→ fast
2HOBr molecularity two
(HOBr + HBr ⎯⎯→ H2O + Br2 )  2
fast
molecularity two
4HBr + O2 → 2H2O + 2Br2 molecularity five
The rate law equation is r = k[HBr] [O2].
 Overall order of reaction, n = (1 + 1) = 2
Difference between order and molecularity of reaction

Order of reaction Molecularity of reaction
• It is experimentally determined quantity. It is calculated from balanced chemical
equation.
• It can have zero, whole number, fractional It can only be a whole number.
and sometimes even negative value.
• It tells us about the slowest step in the It does not tell anything about the mechanism.
mechanism.
• It is the sum of the powers of concentration It is the number of reacting species undergoing
terms in the rate law equation. simultaneous collision in the reaction.
• It is changed when one reactant is taken in It does not undergo any change.
large excess.
• Order can change with conditions such as Molecularity cannot change with conditions
pressure, temperature, etc. such as pressure, temperature, etc.
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Kinetic Equations of Different Orders
Zero Order Reactions
The reactions in which reaction rate is independent of the concentration of reactants (i.e., the concentration
of reactants does not change with time) are called zero order reactions.
For a reaction
dx
A → products; rate law of equation, r = k[A]0 or – = k[A]0 = k i.e., rate = rate constant
dt
On integration, x = kt + c (c = integration constant)
when t = 0, x = 0, hence c = 0  x = kt
x
k= for zero order reactions
t
The same conclusion can also be given as
1
Rate constant k =  [A]0 − [A] 
t
Units of k
mol dm −3
k= = mol dm–3 s–1
s
For zero order reactions, unit of rate constant (k) is same as that of rate of the reaction.
Half life period
a
When t = t1/2; x =
2
a
t1/2 = or t1/2  a
2k
The time taken for the completion of half of the reaction is directly proportional to the initial concentration
of the reactants.
The characteristics of zero order reactions

• Rate of reaction is independent of concentration of reactants.
• The change in concentration of the reactants do not affect the rate of the reaction.
• The rate of reaction does not vary with time.
• The rate of reaction is always equal to rate constant.
Examples for zero order reaction

h
H2(g) + Cl2(g) ⎯⎯⎯ → 2HCl(g) ; r = k[H2]0 [Cl2]0; order = (0 + 0) = 0
Mo / h
2 NH3(g) → N2(g) + 3H2(g); r = k[NH3]0; order = 0
heat
Au
2 HI(g) → H2(g) + I2(g); r = k[HI]0; order = 0
heat
First Order Reactions

The reaction rate is determined by change in concentration of one reactant.
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d[A]
A → products; rate law equation; r = – = k [A]
dt
Let, initial concentration of reactant A = ‘a’ mol dm–3
Number of moles reacted in time t = ‘x’ mol dm–3
The concentration of reactant at time t = (a – x) mol dm–3
2.303 a
Expression for rate constant k is given by k = log . This is called the kinetic equation of a first
t (a − x)
order reaction.
Units of k
1 mol dm −3 1
k= log −3
= = (time–1) = s–1
time mol dm time
Half life period (t1/2)

The time taken for the completion of half of the reaction is known as half life period.
a
When t = t1/2; x =
2
2.303 a 2.303 a 2.303 a
k= log = log = log
t (a − x) t1/2  a t1/2 a/2
a − 
 2
2.303 2.303  0.3010 0.693
k= log 2 = =
t1/2 t1/2 t1/2
0.693
k= for first order reactions.
t1/2
Therefore, half life period for first order reactions is independent of initial concentration of the reactants. For
the first order reaction, not only the half life period, but also any definite fraction of the reaction to be
completed is independent of initial concentration. The amount of reactant left (a – x) at nth half life is
calculated using following equation.
n
1
(a – x) = a   where n = number of half life periods.
2
total time lapsed
n=
half life period
Relationship between half life period and initial concentration

The time required to complete a definite fraction of a reaction depends on order of reaction. The relation
between half life period (t1/2) and order (n) of the reaction is given by:
1  2n −1 − 1  1
t1/2 =  n −1  ; Units of k = (mol dm–3)1–n s–1 or t1/2  n −1 ,
k(n − 1)  a  a
where a = initial concentration of reactants, n = order of reaction.
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If (t1/2)1 and (t1/2)2 are the half life period of a reaction for the initial concentrations a1 and a2 respectively,
( t1/2 )1  a 2  ( t1/ 2 )1
n −1
a
then =  or log = (n – 1)log 2
( t1/2 )2  a1  ( t1/ 2 )2 a1
This equation is used to calculate the order of the reaction by determining half life periods for two initial
concentrations.
Characteristics of first order reactions
• The unit of rate constant (k) of first order reaction is time–1 i.e. s–1 or min–1 or hr–1 or yr−1.
• The half life period of first order reaction is independent of initial concentration of reactants.
• The rate of reaction is proportional to the concentration of reactant at time t.
• The concentration of reactant decreases with progress of the reaction.
Radioactive decay is a first order reaction
First order growth kinetics
 It is used in population growth and bacteria multiplication.
k=
2.303
log
(a + x)
t a
where ‘a’ is initial population and (a + x) is population after time ‘t’.
Examples for first order reactions

2N2O5 → 4NO2 + O2 r = k[N2O5]; order = 1
2H2O2 → 2H2O + O2; r = k[H2O2]; order = 1

NH4NO2 ⎯⎯ → 2H2O + N2; r = k[NH4NO2]; order = 1

SO2Cl2 ⎯⎯ → SO2 + Cl2; r = k[SO2Cl2]; order = 1
2O3 ⎯⎯
h
→ 3O2; r = k[O3]2 [O2]–1 order = (2 – 1) = 1
+
H (aq)
CH3COOC2H5 + H2O ⎯⎯⎯→ CH3.COOH + C2H5OH
+
H (aq)
C12H22O11 + H2O ⎯⎯⎯→ C6H12O6 + C6H12O5 (inversion of sugar)
sucrose glucose fructose
O

2 + C2 H6
O O
ditertiarybutylperoxide
h
N N + N2
diazoisopropane
COOH
O2 N NO2 O2 N NO2
+ CO2
NO2 NO2
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Pseudo unimolecular reactions

The chemical reactions in which molecularity of the reaction is more than one whereas the order is one are
called pseudo unimolecular reactions. Higher order reactions are converted into first order by taking
reactants in excess except one.
H+ (aq)
Example: CH3COOCH3 + H 2 O ⎯⎯⎯⎯→ CH3.COOH + CH3OH
excess
Rate law equation: r = k[CH3.COOCH3] [H2O]0; order = (1 + 0) = 1 molecularity = (1 + 1) = 2

Examples for second order reactions
2 HI → H2 + I2
K2S2O8 + 2KI → 2K2SO4 + I2
CH3.COOC2H5 + NaOH → CH3.COONa + C2H5OH
Examples for third order reactions
2 NO + O2 → 2NO2
2 NO + Cl2 → 2NOCl
2 FeCl3 + SnCl2 → 2FeCl2 + SnCl4
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Additional Information
Fractional order reactions
Examples of fractional order reactions are given below:
h  1 3
H2(g) + Br2(g) ⎯⎯⎯ → 2HBr; r = k[H2] [Br2]1/2 order = 1 +  =
 2 2
h  3 5
CO(g) + Cl2(g) → COCl2; r = k[CO] [Cl2]3/2; order = 1 +  =
activated carbon  2 2
heat → CO + Cl ; 3
COCl2(g) ⎯⎯⎯ (g) 2(g) r = k[COCl2]3/2; order =
2
400 C 3
CH3.CHO(g) ⎯⎯⎯⎯→ CH4(g) + CO(g); r = k[CH3CHO]3/2; order =
2
Negative order reaction

Some times rate of the reactions decrease with increase in concentration of reactants. Consider the
decomposition of ozone which is multistep reaction.
fast
O3 O2 + [O] … (1)
slow
[O] + O3 ⎯⎯⎯→ 2O2 … (2)
2O3 3O2
The equilibrium constant K of the reaction (1) is
[O2 ] [O] [O 3 ]
K=  [O] = K
[O3 ] [O 2 ]
The slowest step (2) of the reaction determines the rate of overall reaction. Hence, rate equation is
d[O 3 ]
r=– = k[O3] [O]
dt
By substituting the value of [O] we get

d[O 3 ] K[O3 ] kK[O3 ]2
r=– = k[O3]. =
dt [O2 ] [O 2 ]
k [O3 ]2
r= = k  [O3]2 [O2]–1 [Q kK = k  ]
[O 2 ]
Hence order with respect to O3 = 2 , order with respect to O2 = –1, overall order = (2 – 1) = 1
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Graphical Representations of various order reactions
(a) The plot of half life period (t1/2) against initial concentration (a) for different order (n) of the reactants is
 1 
as follows:  t1/2  n −1 
 a 
t1/2 t1/2 t1/2 t1/2
a a 1/a 1/a2
Zero order First order Second order Third order
t1/2  a t1/2  a0 = 1 t1/2  (1/a) t1/2  (1/a2)
dx
(b) The graphs between rate of reaction (r) i.e.  against the concentration of reactants at the given time
dt
(a – x) for the different order (n) of reactions are as follows: The rate of reaction is related to order of
dx
reaction as r =  = k(a – x)n where n = order of reaction
dt
dx dx dx dx
dt dt dt dt
(a–x) (a–x) (a–x)2 (a–x)3

Zero order First order Second order Third order
r = (dx/dt)  (a – x)0 r = (dx/dt)  (a – x) r = (dx/dt)  (a – x)2 r = (dx/dt)  (a – x)3
(c) The integrated form of rate law equations are used to evaluate the rate constant (k) for different
orders (n) of reactions by using following graphs:
Zero order reaction First order reaction

dx 2.303 a
t r= = ka t t= log
dt k (a − x )
1
 t  x or t  2.303
(a − x ) slope =
k
x log (a-x)
t = (x/k)  slope = (1/k)
Second order reaction Third order reaction

x 1 x (2a − x )
kt = t kt =
t a (a − x ) 2 a 2 (a − x ) 2
1 1
t=
1
−
1 t= −
k (a − x ) ka 1/(a – x)2
2 k (a − x ) 2
2ka 2
1/(a – x)
slope = (1/k) slope = (1/2k)
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Quick quiz
1. The rate constant of a reaction depends on
(A) temperature (B) initial concentration of the reactants
(C) time taken for the reaction (D) extent of reaction
2. Order and molecularity is same for the reaction
(A) CH3COOCH3 + H2O ⎯⎯ → CH3COOH + CH3OH
(B) CH3COOC2H5 + NaOH ⎯⎯ → CH3COONa + C2H5OH
(C) C12H22O11 + H2O ⎯⎯ → C6H12O6 + C6H12O6
(D) 2H2O2 ⎯⎯ → 2H2O + O2
3. In the inversion of cane sugar

(A) the order of the reaction is one but molecularity is two
(B) the order of the reaction is two but molecularity is one
(C) both order and molecularity of the reaction are two
(D) both order and molecularity of the reaction are one
4. For a single step reaction, A + 2B → Products, the molecularity is
(A) zero (B) three (C) two (D) one
5. Which of the following reactions would match with the relation of rate of a reaction
= 10−6 mol dm−3 s−1 ?
(A) 2NH3 → N2 + 3H2 (B) N2 + 3H2 → 2NH3
1 1 1 1
(C) N 2 + H 2 → NH 3 (D) NH 3 → N 2 + H 2
3 2 2 3
6. Which of the following does not affect the order of a reaction?

(A) Temperature (B) Nature of reactant
(C) pH of solution (D) concentration of reactants
7. Which is wrong w.r.t molecularity?
(A) It is a theoretical value
(B) It can be zero
(C) It is the number of species colliding at a same moment to result into the products.
(D) Molecularity of single step reaction is the order of reaction also
8. The molecularity of a reaction is
(A) same as its order (B) different from order
(C) may be same or different from order (D) always zero
9. The rate of a reaction can be increased in general by all the factors, except
(A) by increasing the temperature
(B) using a suitable catalyst
(C) by increasing the concentration of reactants
(D) by an increase in activation energy of the reaction
10. Which of the following does not influence the rate of a reaction?
(A) nature of the reactants (B) concentration of the reactants
(C) temperature of the reaction (D) molecularity
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Illustrations
 −d[N 2 ] 
1. Consider the reaction, N2(g) + 3H2(g) → 2NH3(g). The rate of this reaction   is
 dt 
− d[H 2 ]
3  10−3 mol L−1 min−1. What is the value of in mol L−1 min−1?
dt
(A) 3  10−3 (B) 1  10−3 (C) 1.5  10−3 (D) 9  10−3
Ans (D)
−d[N 2 ] 1 d[H 2 ] 1 d[NH 3 ]
=− =+
dt 3 dt 2 dt
−d[H2 ]  −d[N2 ]  −3 −3 −1 −1
 = 3   = 3 3 10 = 9  10 mol L min
dt  dt 
2. The overall order of a reaction which has the rate expression, Rate = k[CHCl3] [Cl2]1/2 is
(A) 1.0 (B) 1.5 (C) 0.5 (D) 2.0
Ans (B)
3. Consider a reaction aA + bB → Products. When the concentration of both the reactants A and B is
doubled, the rate increases 8 times. However, when concentration of A is doubled, keeping the
concentration of B fixed, the rate is doubled. The overall order of the reaction is
(A) 0 (B) 1 (C) 2 (D) 3
Ans (D)
rate = k[A]a [B]b … (1)
a
8r = k[2A] [2B] b
… (2)
a
2r = k[2A] [B] b
… (3)
Divide equations (2)  (3) 4 = 2 b
 b=2
From equations (1) and (2)
8 = 2a + b = 2 a + 2
 overall order of the reaction = a + b = 1 + 2 = 3
4. The half life period of a first order reaction is 6.93 minutes. The time required for the completion of 99%
of the chemical reaction will be
(A) 230.3 min (B) 23.03 min (C) 46.03 min (D) 4606.6 min
Ans (C)
0.693 0.693 0.693
t1 = or k = = = 0.1
2 k t1 6.93
2
2.303 a
k= log
t a−x
2.303 100 2.303
0.1 = log  t 99% =  2 = 46.06 min
t 99% 1 0.1
Think
Further
In the above problem, find
(a) time required for 75 % completion, and
(b) time required for 99.9 % completion of the reaction.
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5. Consider a reaction, A → B + C. If the initial concentration of A was reduced from 2M to 1 M in 1 hour
and from 1 M to 0.25 M in 2 hours. The order of the reaction is
(A) 1 (B) 0 (C) 2 (D) 3
Ans (A)
Half life of the given reaction is independent of initial concentration, hence it is a first order reaction.
+
6. For a reaction C12 H22O11 + H2O ⎯⎯
H
→ C6 H12O6 + C6 H12O6 rate of disappearance of A is related to the
rate of appearance of B by the expression
−d[A] d[B] −d[A] 1 [B] 1 1 3
(A) =4 (B) = (C) b = (D) 1 + =
dt dt dt 2 dt 2 2 2
Ans (C)
The reaction may be written as A → 4 B
a
2k
7. The specific reaction rate of a first order reaction depends on

(A) the concentration of reactants (B) concentration of products
(C) time of measurement (D) temperature at which the reaction is carried out.
Ans (D)
The specific reaction rate (i.e., the rate constant) depends on temperature, and not on other factors such
as given in (A), (B) and (C).
Think
Further
The specific reaction rate of a first order reaction does not depend on concentration. It is a constant
through out the progress of the reaction. However, the value of k is dependent on temperature. What
other factor influences the value of rate constant k of a reaction?
8. Inversion of cane sugar in dilute acid is

(A) bimolecular reaction (B) pseudo – first order reaction
(C) first order reaction (D) third order reaction
Ans (B)
+
C12 H22O11 + H2O ⎯⎯
H
→ C6 H12O6 + C6 H12O6
Water is in excess, its concentration remains constant.
 Rate = k [C12 H22O11]
9. The rate equation for the reaction, 2A + B → C is found to be, rate = K[A][B]. The correct statement in
relation to this reaction is
(A) K is independent of [A] and [B]
(B) t1/2 is constant
(C) rate of formation of C is twice the rate of disappearance of A
(D) unit of K is s−1
Ans (A)
K is characteristic constant for a given reaction.
It is a II order reaction, hence t1/2 is not constant. Rate of formation of C = Half the rate of disappearance
of A.
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10. The rate law for a reaction between the substances A and B is given by rate = K[A] n [B]m. On doubling
the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of
the reaction will be as
1
(A) m + n (B) (m + n)
2
(C) (n − m) (D) 2(n − m)
Ans (D)
r = K[A]n [B]m
m
nB r
r1 = K[2A]    1 = 2(n − m )
2 r
Temperature dependence of rate of a reaction
The rate of reaction depends on temperature. This is expressed in terms of temperature coefficient which is a
ratio of two rate constants of the reaction differing by a temperature of 10 K. Generally, the temperatures
selected are 298 K and 308 K. It is mathematically expressed as,
rate constant at 308 K k
Temperature coefficient = = 308
rate constant at 298 K k 298
k T + 10
Temperature coefficient =  2or 3
kT
The value of temperature coefficient is more at lower temperature.
The value of temperature coefficient is more for the reactions having higher activation energy. With increase
in temperature the time taken for the reactants to cross the energy barrier and form products decreases. The
fraction of molecules possessing extra energy equal to or greater than Ea increases and as a result, the number
of effective collisions increases. This results in the increase of collision frequency and hence the reaction rate
increases.
Arrhenius rate equation: Energy Diagram
Arrhenius Rate Equation
Number of molecules (n)
T1K
Arrhenius rate equation is k = Ae− Ea /RT (T2 – T1) = 10 K
E
ln k = ln A – a
RT T2 = (T1 + 10)K
Ea
log k = log A –
2.303 RT
Energy Diagram Energy (E) → ET
A diagram showing relative energies of reactants and products and activated complex is called energy
diagram or energy profile of the reaction. It is shown for endothermic and exothermic reactions of the type:
A → B.
ET activated complex ET activated complex
(Ea)b
(Ea)f
PE (Ea)f PE (Ea)b
EP B ER
ER H
A Products
Reactants EP
Reaction co-ordinate Reaction co-ordinate
Endothermic reaction Exothermic reaction

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In endothermic reactions activation energy of forward reaction [(Ea)f] is greater than that of backward
reaction [(Ea)b]. Whereas in exothermic reactions (Ea)b is greater than (Ea)f. From the energy profile of the
reactions it is clear that (Ea)f = (ET – ER) and (Ea)b = (ET – TP), H = (Ea)b – (Ea)f = (HP – HR)
Graphical Representation of the Effect of Temperature on the Rate of Reactions

The fraction of the molecules having energy equal to or greater than the threshold energy (E T) is shown by
shaded portion.
From this diagram it is clear that the fraction of the molecules become almost double for 10 K rise in
temperature (shaded portion is almost double). As a consequence the rate of reaction almost doubles for 10 K
rise in temperature (number of effective collision doubles).
Collision theory of chemical Reactions

Though Arrhenius equation is applicable under a wide range of circumstances, collision theory, which was
developed by Max Trautz and William Lewis in 1916 − 18, provides a greater insight into the energetic and
mechanistic aspects of reactions. It is based on kinetic theory of gases. According to this theory, the reactant
molecules are assumed to be hard spheres and reaction is postulated to occur when molecules collide with
each other. The number of collisions per second per unit volume of the reaction mixture is known as
collision frequency (Z). Another factor which affects the rate of chemical reactions is activation energy (as
we have already studied).
For a bimolecular elementary reaction
A + B → products
rate of reaction can be expressed as

Rate = ZAB e− Ea /RT
Where ZAB represents the collision frequency of reactants, A and B and e− Ea /RT represents the fraction of
molecules with energies equal to or greater than Ea. Comparing with Arrhenius equation, we can say that A
is related to collision frequency.
Equation predicts the value of rate constants fairly accurately for the reactions that involve atomic species or
simple molecules but for complex molecules significant deviations are observed. The reason could be that all
collisions do not lead to the formation of products. The collisions in which molecules collide with sufficient
kinetic energy (called threshold energy) and proper orientation, so as to facilitate breaking of bonds between
reacting species and formation of new bonds to form products are called as effective collisions
For example, formation of methanol from bromoethane depends upon the orientation of reactant molecules
as shown in figure. The proper orientation of reactant molecules lead to bond formation whereas improper
orientation makes them simply bounce back and no products are formed.
To account for effective collisions, another factor P, called the probability or steric factor is introduced. It
takes into account the fact that in a collision, molecules must be property oriented i.e.,
Rate = PZABe− Ea /RT
Thus, in collision theory activation energy and proper orientation of the molecules together determine the
criteria for an effective collision and hence the rate of a chemical reaction.
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Collision theory also has certain drawbacks as it considers atoms/molecules to be hard spheres and ignores
their structural aspect. Your will study details about this theory and more on other theories in your higher
classes.
..
CH3Br + OH CH3OH + Br
Improper H + −
H C Br OH No
Orientation H products
H + − ..
H C Br + OH
H
H Proper −
H + −
HO C Br HO C H + Br
Orientation H H
H
Intermediate
Collision Theory
According to collision theory a chemical reaction occurs as a result of effective collisions between reacting
molecules. For this, two factors are important.
• Proper orientation of reacting molecules.
• Possession of certain minimum amount of energy by reacting molecules.
Threshold Energy
The minimum amount of energy possessed by the reacting molecules to have effective collisions, resulting in
the formation of product is called threshold energy.
Activation Energy
The minimum extra energy over and above the average potential energy of the reactants which must be
supplied to the reactants to enable them to cross over the energy barrier between reactants and products is
called activation energy.
• The reactions with smaller activation energy are fast.

• The activation energy depends on nature of reactants.
• Activation energy (Ea) = threshold energy (ET) – average potential energy of reactants.
• The activation energy does not change with temperature.
• The activation energy is always positive.
• The state of the reactants corresponding to threshold energy is called transition state or activated
complex.
Calculation of Activation Energy

1. Fractional Change Method
If k1 and k2 are the rate constants of the reaction at temperatures T1 and T2 respectively, then
k
log 2  2.303 RT1T2
k E a  T2 − T1  k1
log 2 =   or Ea =
k1 2.303 R  T1T2  (T2 − T1 )
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2. Graphical Method
Ea 1
log k = log A – 
2.303R T
This equation corresponds to the equation of a straight line. A plot of log k against (1/T) gives a straight
line with negative slope.
It is important to note that as the value of Ea increases, the value of k decreases and the rate of reaction
decreases. With increase in temperature, T the value of k increases and the rate of reaction also
increases.
slope = –Ea/2.303 R
log k
or Ea = –2.303 R  slope
1/T
Effect of Catalyst on the Reaction Rate

A catalysed reaction provides a new mechanism (new pathway) by which the potential energy barrier
between the reactants and products is lowered. The graphical presentation is shown in the diagram.
ET activated complex
Ea
uncatalysed reaction
ETc Activated complex
PE (Ea)c
ER catalysed reaction
EP H
Reaction co-ordinate
.
The enthalpy of reaction (H) remains the same for the catalysed as well as uncatalysed reaction.
The energy of activation [(Ea)c] is lowered for catalysed forward and backward reactions.
A positive catalyst increases the rate of a reaction by lowering the activation energy and by providing a new
path for the reaction. The catalyst may form suitable intermediates during the reaction.
Greater the lowering of Ea of catalysed reaction, higher will be the efficiency of catalyst.
Quick quiz
11. In the presence of a catalyst, the reaction enthalpy

(A) increases (B) decreases
(C) remains the same (D) may increase or decrease
12. A substance reacts with initial concentration of ‘a’ mol dm −3 according to zero order kinetics. The time
it takes for the completion of the reaction is [k = rate constant]
k a a 2k
(A) (B) (C) (D)
a 2k k a
13. The half-life period for nth order reaction is inversely proportional to
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(A) an + 1 (B) an − 1 (C) an − 2 (D) an
14. For a reaction, 3A + 2B → Products,
The rate equation is rate = k[A] [B]2. If A is taken in excess, the order of the reaction is
(A) 3 (B) 2 (C) 1 (D) 5
15. The half life period of a first order reaction is 1 min 40 seconds. Its rate constant is
(A) 6.93  10−3 min−1 (B) 6.93  10−3 sec−1
(C) 6.93  10−3 sec (D) 6.93  103 sec
16. The activation energy of a reaction is zero. The rate constant of the reaction
(A) increases with increase of temperature
(B) decreases with decrease of temperature
(C) decreases with increase of temperature
(D) is nearly independent of temperature
17. A plot of log10(a −x) against time ‘t’ is a straight line with positive slope. This indicates that the reaction
is of
(A) zero order (B) first order (C) second order (D) third order
18. The rates of chemical reactions increase vary rapidly with increase in temperature because:
(A) The fraction of total number of molecules with kinetic energy greater than the activation energy
increases very rapidly as temperature increases.
(B) The average kinetic energy increases as the temperature increases
(C) The activation energy decreases as the temperature increases
(D) More collisions take place with particles placed so that reaction can occur
19. According to the collision theory,

(A) all collisions are sufficiently violent
(B) all collisions are responsible for product formation
(C) all collisions are effective
(D) only a fraction of collisions are effective which have enough energy to form products
a
20. In a first order reaction the was found to be 8 after 10 minute. The rate constant is
(a − x)
2.303  3log 2 2.303  2log 3
(A) (B) (C) 10  2.303  2 log 3 (D) 10  2.303  3 log 2
10 10
Illustrations
11. For an elementary process of a chemical reaction,

(A) the order and the moelcularity are identical (B) the order is greater than the molecularity
(C) the order is lesser than the moleculartiy (D) the order is always fractional
Ans (A)
The order and molecularity are the same for an elementary step of a reaction.
Think
Further
For an elementary reaction (single-step reaction), can we apply the law mass action to write the rate law
expression?
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12. For a reaction Ea = 0 and k = 3.2  104 s−1 at 300 K. The value of k at 310 K would be
(A) 6.4  104 s−1 (B) 3.2  104 s−1 (C) 3.2  108 s−1 (D) 3.2  105 s−1
Ans (B)
Ea
−
k = Ae RT
When Ea = 0, k = A = constant,  k310 = k300 = 3.2 104 s−1
13. If doubling the initial concentration of a reactant doubles t1/2 of the reaction, the order of the reaction is
(A) 3 (B) 2 (C) 1 (D) 0
Ans (D)
Think
Further
On the other hand, what happens to t1/2 of a first order reaction, when the initial concentration is
doubled?
14. For the reaction N2(g) + 3H2(g) → 2NH3(g) + 22 kcal, Ea for the reaction is 70 K cal. The activation
energy for 2NH3(g) → N2(g) + 3H2(g) is
(A) 92 K cal (B) 70 K cal (C) 48 K cal (D) 22 K cal
Ans (A)
H = Ef − Eb
Where H = Enthalpy of the reaction
Ef = Activation energy of a forward reaction
Eb = Activation energy of a backward reaction
H = −22 K cal, Ef = 70 K cal, Eb = ?
H = Ef − Eb
−22 = 70 − Eb
 Eb = 92 k cal
15. For the decomposition of HI at 1000 K, 2HI → H2 + I2, the following data were obtained.
[HI] Rate of decomposition of HI (mol L−1 s−1)
0.1 2.75  10−8
0.2 11  10−8
0.3 24.75  10−8
The order of a reaction is
(A) 1 (B) 2 (C) 0 (D) 1.5
Ans (B)
Rate = k[HI]n
11  10−8 = k[0.2]n … (1)
2.75  10 = k[0.1]
−8 n
… (2)
Dividing equation (1) by equation (2)
4 = 2n  n = 2.
16. The rate of a gaseous reaction is given by the expression k[A]2 [B]3 . The volume of the reaction vessel is
reduced to one half of the initial volume. The reaction rate relative to the original rate will be
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32
(A) Q (1 − n) = −2 (B) t 50% = = 16 min (C) 32
2
(D) 24
Ans (C)
Rate = k[A]2 [B]3 … (1)
When volume is halved, the concentration doubles.
 Rate = k[2A]2 [2B]3
= 32 k[A]2 [B]3
= 32  original rate from (1)
17. For a reaction A + B → C + D, if the concentration of A is doubled without altering the concentration of
B, the rate gets doubled. If the concentration of B is increased by nine times without altering the
concentration of A, the rate gets tripled. The order of the reaction is
3 4
(A) 2 (B) 1 (C) (D)
2 3
Ans (B)
Rate = k[A]a [B]b … (1)
2  rate = k[2A] [B]
a b
… (2)
3  rate = k[A] [9B]
a b
… (3)
From equations (1) and (2)
a=1
1 1 3
From equations (1) and (3), b =  order = a + b = 1 + =
2 2 2
18. If the temperature of a reaction A → Products ; Ea = 12.49 K cal mol−1 is increased from 295 to 305 K,
the rate of reaction increases by
(A) 2 times (B) 3 times (C) 4 times (D) 0.5 times
Ans (A)
k  E a  T2 − T1 
log10  2  =  
 k1  2.303 R  T1T2 
 x  12.49  10  10 
3
log10   =  
 1  2.303 2  295  305 
x=2
19. The activation energy for a reaction is 9.0 kcal/mol. the increase in the rate constant when its
temperature is increased from 298 K to 308 K is
(A) 10 % (B) 100 % (C) 50 % (D) 63 %
Ans (D)
K E [T − T ]
2.303 log 2 = a 2 1 ;
K1 R T1T2
K2 9  10  K2 K2
 2.303 log =  298  308  ;  log = 0.2125  = 1.63 ; i.e., 63% increase
K1 2  10−3   K1 K1
20. In a first order reaction, the concentration of the reactant is decreased from 1.0 M to 0.25 M in
20 minute. The rate constant of the reaction would be
(A) 10 min−1 (B) 6.931 min−1 (C) 0.6931 min−1 (D) 0.06931 min−1
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Ans (D)
2.303 a
t= log OR 1.0 → 0.5 → 0.25
K (a − x) M M M
2t1/2 = 20 mins  t1/ 2 = 10 min

0.693 0.693 2.303 1
K= = = 0.0693 min −1  K= log = 0.06931min −1
t1/2 10 20 0.25
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Exercise - 1
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1. Rate constant of a reaction is 175 L2 mol−2 s−1. What is the order of reaction?
(A) 1 (B) 2 (C) 3 (D) 0
2. A certain zero order reaction has k = 0.025 mol s−1 for the disappearance of A. What will be the
concentration of A after 15 seconds if the initial concentration is 0.5 M?
(A) 0.5 M (B) 0.32 M (C) 0.12 M (D) 0.06 M
3. If a first order reaction takes 32 minutes for 75% completion, then time required for 50% completion is
(A) 32 min (B) 16 min (C) 8 min (D) 4 min
4. At 500 K, the half-life period of a gaseous reaction at an initial pressure of 80 kPa is 350 sec. When the
pressure is 40 kPa the half life period is 175 sec, the order of the reaction is
(A) zero (B) one (C) two (D) three
5. The ratio of the time required for 75% of the reaction of first order to complete to that required for half
of the reaction is
(A) 4 : 3 (B) 3 : 2 (C) 2 : 1 (D) 1 : 2
6. The specific reaction rate of a first order reaction depends on
(A) concentration of the reactants (B) concentration of the products
(C) time (D) temperature
7. According to Arrhenius equation, rate constant of a chemical reaction is equal to
(A) Ae− Ea /RT (B) AeEa /RT (C) AeRT/Ea (D) Ae− RT/Ea
1
8. If we plot a graph between log10k and by Arrhenius equation, the slope is
T
Ea E −E a Ea
(A) − (B) + a (C) (D) +
R R 2.303 R 2.303R
9. The activation energy for a chemical reaction depends upon

(A) temperature (B) nature of reactants
(C) concentration of reactants (D) collision frequency
10. In a reaction, the threshold energy is equal to
(A) activation energy + normal energy of reactants
(B) activation energy − normal energy of reactants
(C) normal energy of reactants − activation energy
(D) average kinetic energy of molecules of reactants
11. For an endothermic reaction, where H represents the enthalpy of the reaction in kJ mol−1, the minimum
value for the energy of activation will be
(A) less than H (B) zero
(C) more than H (D) equal to H
12. Threshold energy ETh can never be
(A) > ER (B) > EP (C) < EP (D) >ER and >EP
13. t50% of first order reaction is 10 min. Starting with 10 mol L−1, rate after 20 min is
(A) 0.0693 mol L−1 min−1 (B) 0.0693  2.5 mol L−1 min−1
(C) 0.0693  5 mol L−1 min−1 (D) 0.0693  10 mol L−1 min−1
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14. At 25C, the values of rate constant, activation energy and Arrhenius constant of a reaction are
3  10−4 s−1, 129 kJ mol–1 and 2  1015 s−1 respectively. The value of rate constant at T →  is
(A) zero (B) 2  1015 (C) 3  10−4 (D) 6  1011
15. For a certain reaction, a plot of concentration of the reactant against time gives a straight line passing
through the origin. The order of the reaction is
(A) zero (B) one (C) two (D) four
16. In a first order reaction, the concentration of the reactant decreases from 8 mol dm −3 to 0.5 mol dm−3 in
2  104 s. Rate constant of the reaction is
(A) 2  104 s−1 (B) 6.2  10−6 s−1 (C) 1.38  10−4 s−1 (D) 2.8  10−4 s−1
17. The half-life period of a first order reaction is 12 hours, starting from 100 g of the reactant amount of the
reactant left over at the end of 60 hours is nearly
(A) 12.5 g (B) 6.3 g (C) 3 g (D) 1.5 g
18. Among the following reactions, the fastest reaction is

(A) burning of coal
(B) rusting of iron
(C) conversion of water to water vapour
(D) precipitation of AgCl by mixing AgNO3 and NaCl solutions
19. The rate at which a substance reacts depends on its
(A) atomic mass (B) equivalent mass (C) molecular mass (D) active mass
20. The correct statement about molecularity of a reaction is
(A) molecularity of a reaction may be zero
(B) molecularity of a reaction may be a fraction
(C) molecularity of a reaction is always different from the order for the same reaction
(D) molecularity of a reaction is always a whole number
21. The hydrolysis of ethyl acetate in an alkaline medium is

(A) zero order (B) first order (C) second order (D) third order
22. The reaction; 2N2O5 ⎯⎯
→ 4NO2 + O2 follows first order kinetics. Hence, it is
(A) unimolecular (B) pseudo unimolecular
(C) bimolecular (D) none of (A), (B), (C)
23. Which of the following statement is wrong?
(A) the order of the reaction is the sum of the powers of the concentration terms in the rate law
representing the reaction
(B) the order of a reaction is an experimental quantity
(C) the order of a reaction depends on the stoichiometry of the reaction
(D) the order of the reaction may be zero or even fraction
24. The rate of a gaseous reaction is given by the expression k[A][B]. If the volume of the reaction vessel is
reduced to one fourth its original volume, the rate will be
1 1
(A) (B) (C) 16 (D) 6
10 8
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25. In the catalytic conversion of N2 to NH3 by Haber process, the rate of the reaction was expressed as
change in the concentration of ammonia per time as 40  10−3 mol L−1 s−1. If there are no side reactions,
the rate of the reaction as expressed in terms of hydrogen is
(A) 60  10−3 mol L−1 s−1 (B) 20  10−3 mol L−1 s−1
(C) 1200 mol L−1 s−1 (D) 10.3  10−3 mol L−1 s−1
26. For the reaction, Cl2(g) + 2NO(g) ⎯⎯ → 2NOCl(g) ; doubling the concentration of both the reactants
increase the rate by a factor of eight. If only the concentration of Cl2 is doubled, the rate increases by a
factor of two. The order of the reaction with respect to NO is
(A) 0.5 (B) 3 (C) 1 (D) 2
27. Given, A + B ⎯⎯ → products
Trial [A] [B] Initial rate
1 0.012 0.035 0.1
2 0.024 0.070 0.8
3 0.024 0.035 0.1
4 0.012 0.070 0.8
The rate law for the reaction is
(A) rate = k[B]3 (B) rate = k[B]4 (C) rate = k[A][B]3 (D) rate = k[A]2[B]2
28. Rate of chemical reaction at constant temperature is
(A) inversely proportional to the product of active masses of the reactants
(B) equal to the sum of the active masses of the reactants
(C) proportional to the product of active masses of the reactants
(D) independent of the concentration of the reactants
29. Half life period of a first order reaction is
(A) inversely proportional to the concentration of the reactants
(B) directly proportional to the concentration of the reactants
(C) independent of the initial concentration of the reactants
(D) none of (A), (B), (C)
30. For a certain reaction it takes ten minutes for the initial concentration of 1.0 mol/litre to become
0.5 mol/litre and another 10 minutes to become 0.25 mol/litre. From the above data, the order of
reaction is
(A) 1 (B) 2 (C) zero (D) 3
31. In a second order reaction, the time needed for the initial concentration of the reactant to reduce to half
that value is
(A) independent of the initial concentration
(B) proportional to the initial concentration
(C) inversely proportional to the initial concentration
(D) proportional to the square of the initial concentration
dc
32. For a reaction, = k.C 1n1 C n22 C 3n3 . Then the order of the reaction is
dt
n + n2 + n3 n + n2 + n3 n1 + n 2 + n 3
(A) 1 (B) n1 + n2 + n3 (C) 1 (D)
3 k C1 + C2 + C3
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33. In a 1st order reaction, the reacting substance has a half life period of ten minutes. The fraction of the
substance left after one hour is
1 1 1 1
(A) (B) (C) (D)
6 64 12 32
1
34. The half-life for the reaction N2O5 ‡ˆ ˆˆ †ˆ 2NO2 + O2 is 24 hrs at 30 C. Starting with 10 g of N2O5,
2
amount of N2O5 remaining after a period of 96 hours is
(A) 1.25 g (B) 0.63 g (C) 1.77 g (D) 0.5 g
35. 8 g of the radioactive isotope, caesium-137 were collected on February 1 last year and kept in a sealed
tube. On July 1 of the same year, it was found that only 0.25 g of it remained. So the half-life period of
the isotope is
(A) 25 days (B) 30 days (C) 37.5 days (D) 50 days
36. A first order reaction is half completed in 45 minutes. The time required for 99.9 % of the reaction to be
complete is
(A) 10 hours (B) 20 hours (C) 5 hours (D) 7 ½ hours
37. For a first order reaction t0.75 is 138.6s. Its specific rate constant is
(A) 10−2 s−1 (B) 10−4s−1 (C) 10−5 s−1 (D) 10−6 s−1
38. The velocity of a chemical reaction doubles for every 10C rise of temperature. If the temperature is
raised by 50C, the velocity of the reaction increases nearly by
(A) 10 times (B) 20 times (C) 30 times (D) 50 times
39. From the plot given below for the conversion of A to B (kJ) mole, the energy of activation is
20
16
Energy
(kJ) 12
8
A
4 B
Reaction coordinate
(A) +16 kJ (B) +8 kJ (C) +12 kJ (D) +4 kJ

40. If rate law for a reaction is r = k [A]p [B]−p/q, then order of the reaction is
p p p p
(A) q − (B) p + (C) (q − 1) (D) (q + 1)
q q q q
41. In 2NO + O2 → 2NO2, rate of formation of NO2 is 92 g s−1. The rate of consumption of NO in g s−1 is
(A) 92 (B) – 92 (C) 60 (D) 80
42. Units of rate constants for first and zero order reactions in terms of molarity M units are respectively
(A) s−1, Ms−1 (B) s−1, M (C) Ms−1, s−1 (D) M, s−1
43. A first order reaction is 20 % complete in 20 minutes. Time taken in minutes for 60 % completion of this
reaction will be
(A) 42 (B) 62 (C) 72 (D) 82
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44. If the rate law for a reaction is rate = k[A] [B]2, the units of k would be
(A) mol L–1 s–1 (B) mol s–1 (C) L2 mol–2 s–1 (D) mol2 L–2 s–1
45. The rate law for the reaction NH+4 (aq) + NO2− (aq) → N2 (g) + 2H2O(l) is given by rate =
k  NH +4   NO −2  . The rate constant is 3  0 10−4 L mol−1 s−1 at 25 C. Calculate the rate of the reaction
at this temperature if  NH +4  = 0.30 M and  NO 2−  = 0.10 M.
(A) 9  10−5 mol L−1 s−1 (B) 910−4 mol L−1 s−1
(C) 9 10−6 mol L−1 s−1 (D) 9 10−7 mol L−1 s−1
Each of the following questions consists of a statement−I and a Statement−II. Examine

both of them and select one of the options using the following codes:
(A) Both the Statement-I and Statement-II are true and the statement-II is the correct explanation of
the statement-I
(B) Both the statements are true, but the Statement-II is not the correct explanation of the
Statement -I
(C) The Statement-I is true, but the Statement -II is false
(D) Both the statements are false or Statement -I is false and the Statement -II is true on its own.
46. Statement-I: For an endothermic reaction, the activation energy is less than the change in enthalpy.
Statement-II: Activation energy of a reaction, at a given temperature can never be greater than its
change in enthalpy.
47. Statement-I: The unit of frequency factor for a second order reaction is mol L−1 s−1.
Statement-II: Unit of Arrhenius factor is same as that of the rate of the reaction.
48. Statement-I: On increasing the temperature by 10K the rate of the reaction doubles.
Statement-II: The collisions become more energetic at higher temperatures..
49. Statement-I : A catalyst may increase or decrease the rate of a reaction

Statement-II: The catalyst increases the average kinetic energy of the colliding molecules.
50. Statement-I: Fractional order reactions are not elementary reactions.
Statement-II: For an elementary reaction, order must be same as molecularity.
Exercise - 2
1. If the rate of a reaction is r = k[X]1/3 [Y] the unit of rate constant is

(A) mol1/3 L−2/3 time−1 (B) mol2/3 L1/3 time–1
(C) mol1/6 L−1/6 time−1 (D) mol−1/3 L1/3 time−1
2. The specific reaction rate of a reaction is 2.0  10−2 mol−1 L sec−1. The order of reaction is
(A) 1 (B) 2 (C) 3 (D) 4
3. Rate of reaction aA + bB → Products, becomes 8 times if the concentration of A is doubled but remains
unchanged if the concentration of B is doubled. The order of reaction is
(A) 1 (B) 2 (C) 3 (D) 0
4. Half-life time of a first order reaction is 69.3 sec. The rate constant will be
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(A) 10−4 s−1 (B) 10−2 s−1 (C) 10 s−1 (D) 102 s−1
5. Half-life time of a reaction is 15 sec. The time for its 99.9% completion will be
(A) Infinity (B) 150 s (C) 45 s (D) 60 s
6. For the following reaction, 2P + Q + 3R → S + T

The elementary steps of the reaction is
(i) P + Q ‡ˆ ˆˆ †ˆ A (fast)
(ii) A + R → B (slow)
(iii) B + Q → S + T (fast)
The rate law of the reaction is
(A) r = k[P] [Q] (B) r = k[P]2 [Q] [R]3 (C) r = k[P]1/2 [Q] [R]1/3 (D) r = k[P][Q][R]
7. On increasing the pressure 3 folds, the increase in the rate of gaseous reaction represented below is
2H2S + O2 → products
8. For a reaction A + B → products, the order w.r.t A is 1 and w.r.t B is 2. If concentrations of both are
doubled the rate of reaction increases by
9. After 4 half-lives, the amount of a substance left over was 5g , the initial amount of the substance taken
is
(A) 20 g (B) 40 g (C) 80 g (D) 100 g
10. If the rate of a reaction at 50 C is 2.6  10−3 mol L−1 s−1. Given that the temperature coefficient is 3. The
rate of reaction at 80 C would be
(A) 7.02  10−2 (B) 7.02 10−3 (C) 7.8  10−3 (D) 7.8 10−2
11. For a reaction, the activation energy is zero. At 300 K its value is 2.5  105 s−1, the rate constant at 400 K
is
1 1
(A) 2.5  105 s−1 (B) 5.0  105 s−1 (C) s −1 (D) s −1
2.5 10 5
5 105
12. The rate constant k1 and k2 for two different reactions are 1016 e−2000/T and 1015 e−1000/T respectively. The
temperature at which k1 = k2 is
2000 1000
(A) K (B) 2000 K (C) K (D) 1000 K
2.303 2.303
13. 1 M, 0.5 M, 2 M H2O2 solutions are separately allowed to decompose under same conditions of
temperature and pressure. The half life period is
(A) moderate in 1 M (B) least in 0.5 M
(C) moderate in 2 M (D) same for all of (A), (B), (C)
14. Bimolecular reactions which follow first order kinetics are called
(A) pseudo unimolecular reactions (B) second order reactions
(C) unimolecular reactions (D) first order reactions
15. The reactions of a higher molecularity are rare because
(A) many body collisions have a low probability
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(B) many body collisions are not favoured energetically
(C) activation energy of many body collisions is very large
(D) activation energy of many body collisions is very small
16. An endothermic reaction A ⎯⎯ → B, has an activation energy as x kJ mol–1. If enthalpy of the reaction is
y kJ, the activation energy of reverse reaction is
(A) –x (B) x – y (C) x + y (D) y – x
17. During a particular reaction, 10 % of the reactant decomposes in one hour, 20 % in two hours, 30 % in
three hours and so on. The unit of the rate constant is
(A) hour–1 (B) L mol–1 hour–1 (C) mol L–1 hour–1 (D) mol hour–1
18. In the fermentation of sugar, the concentration is reduced from 0.12 M to 0.06 M in 10 h and to 0.03 M
in 20 h. The order of the reaction is
(A) 1 (B) 2 (C) 3 (D) zero
19. For a given reaction, the activation energy Ea = 0 and the rate constant k = 3.2  106 s–1 at 300 K. What is
the value of the rate constant at 310 K?
(A) 3.2  106 s–1 (B) 0 (C) 6.4  106 s–1 (D) 6.4  1012 s–1
20. A particular reaction rate increases by a factor of 2, when the temperature is raised from 27C to 37C.
The Ea of the reaction is
(A) 12.9 k cal (B) 0.114 k cal (C) 0.14 k cal (D) 1.29 k cal
21. If the initial concentration is tripled, the time for half reaction is also tripled. The order of the reaction is
(A) zero (B) one (C) two (D) three
22. The rate constant for the forward reaction, A ⎯⎯ → products is given by
1.25  10 4
kJ
log k(sec–1) = 14.34 – , H = – 478 kJ mol–1 at 50 C
T
Energy of activation for the backward reaction at the same temperature is
(A) 239 kJ mol–1 (B) 478 kJ mol–1 (C) 717 kJ mol–1 (D) 1195 kJ mol–1
23. In the reaction, 3Fe(s) + 4H2O(g) ⎯⎯
→ Fe3O4(s) + 4H2(g) , the rate of formation of hydrogen is equal to
(A) the rate of disappearance of iron (B) the rate of disappearance of water vapour
(C) the rate of formation of Fe3O4 (D) all of (A), (B), (C)
24. For the reaction, 2NO + O2 ⎯⎯
→ 2NO2 at a given temperature rate is r. When the volume of reaction
vessel is reduced to half, the rate of reaction becomes (given rate = k [NO]2 [O2]1]
8 r
(A) (B) (C) 8 + r (D) 8r
r 8
25. In the reaction, aA + bB + cC ⎯⎯
→ products
(i) if concentration of A is doubled, keeping conc. of B and C constant, the rate of reaction becomes
double
(ii) if concentration of B is halved keeping conc. of A and C constant, the rate of reaction remains
unaffected.
(iii) if concentration of C is made 1.5 times, the rate of reaction becomes 2.25 times.
The order of the reaction is
(A) 1 (B) 2.5 (C) 3 (D) 3.5
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26. The activation energy of exothermic reaction, A ⎯⎯ → B is 80 kJ. The heat of reaction is –200 kJ. The
activation energy for the reaction, B ⎯⎯
→ A will be
(A) 80 kJ (B) 120 kJ (C) 280 kJ (D) 200 kJ
27. The reaction is fastest in

(A) k = 102 (B) k = 10–2 (C) k = 10 (D) k = 1
28. In a reaction, P → Q, the rate of the reaction doubles when the molar concentration of the reactant P is
increased by four times. Order of the reaction is
(A) 2 (B) 4 (C) ½ (D) ¼
29. According to Collision theory, the pre-exponential factor is
(A) independent of temperature
(B) directly proportional to temperature
(C) directly proportional to the square of temperature
(D) directly proportional to square root of temperature
30. The rate constant of a reaction increases by 5% when its temperature is raised from 27 to 28 C. The
activation energy of the reaction is
(A) 36.6 kJ/mol (B) 16.6 kJ/mol
(C) 46.6 kJ/mol (D) 26.6 kJ/mol
31. The rates (dc/dt) of a certain reaction at 10-second intervals are as follows:
Time (s) Rate (mol L−1s−1)
0 4.80  10−2
10 4.78  10−2
20 4.80  10−2
30 4.82  10−2
The order of the reaction is
(A) 1 (B) 2 (C) 0 (D) 3
32. The half-life period for a first order reaction is 10 minutes. The time required to change the concentration
of the reactant from 0.08 M to 0.01 M is
(A) 20 min (B) 40 min (C) 30 min (D) 50 min
33. For a reaction, the activation energies of forward and reverse reactions are equal in value. Then, we can
conclude that
(A) the reaction is stoichiometrically balanced
(B) H = 0
(C) the order of the reaction is zero
(D) there is no catalyst.
34. If the temperature of a reaction is increased from 25C to 75C,
(A) the reaction rate decreases, but k remains the same
(B) the reaction rate and k both decrease
(C) the reaction rate increases, but k remains the same
(D) the reaction rate and k both increase.
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1
35. The slope of the plot ln k versus gives
T
Ea A Ea
(A) + (B) (C) H (D) −
R R R
36. The decay constant of a radioactive substance having a half-life period of 2.95 days is
(A) 2.9  10−5 s−1 (B) 2.9  106 s−1
(C) 2.7  10−6 s−1 (D) 3.0  105 s−1
37. The role of a catalyst in a chemical reaction is to change the
(A) activation energy (B) equilibrium concentration
(C) heat of the reaction (D) yield of the final products.
38. The energy of activation of a forward reaction is 200 kJ. The energy of activation of its backward
reaction is
(A) equal to 200 kJ (B) greater than 200 kJ
(C) less than 200 kJ (D) either greater or less than 200 kJ
39. The rate of the simple reaction 2NO + O2 → 2NO2, when the volume of the reaction vessel is doubled,
1
(A) will increase by 8 times of its initial rate (B) will reduce to th of its initial rate
8
1
(C) will enhance by 4 times of its initial rate (D) will reduce to th of its initial rate
4
40. All the following statements are correct except one. The incorrect one is
(A) kinetic order is an experimental property derived from the experimental rate law
(B) the order of a reaction cannot be changed by varying the conditions such as pressure, temperature,
etc
(C) in an elementary reaction, the order and the molecularity are generally the same.
(D) in complex reactions, the order of the slowest elementary reaction gives the order of the complex
reaction
41. The rate of formation of O3(g) is 2.0  10−7 mol L−1 s−1 for the reaction
3O2(g) → 2O3(g)
What is the rate of disappearance of O2(g) in mol L−1s−1?
(A) 1.3  10−7 (B) 2.0 10−7 (C) 3.0  10−7 (D) 4.5  10−7
42. The reaction between NO(g) and O2(g) is second order in NO(g) and first order in O2(g). By what factor
will the reaction rate change if the concentrations of both reactants are doubled?
(A) 2 (B) 4 (C) 6 (D) 8
43. The dependence of the rate constant of a reaction on temperature is given by the relation k = e− Ea /RT .
Under what conditions is k the smallest?
(A) High T and large Ea (B) High T and small Ea
(C) Low T and large Ea (D) Low T and small Ea
44. The energy profile diagram show above relates to the reaction
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CO(g) + NO 2 (g) CO 2 (g) + NO(g)

Which of the following statements follows from this?
(A) The activation energy of the forward reaction is (x + y)
(B) H for the reverse reaction is x
(C) The forward reaction is exothermic
(D) The activation energy for the reverse reaction is (x −y)
45. For the reaction 2A + 2B → product, the rate law is rate = k[A][B]2. Which mechanism is consistent with
this information?
(A) B + B C (B) A + B → C (slow)
C + A → Product (slow) C + B → product
(C) A + A C (D) A + B C
B+B D B + C → C (slow)
C + D → product (slow) D + A → product
46. The activation energy for a simple chemical reaction A → B is Ea in forward direction. The activation
energy for reverse reaction: [AIPMT 2003]
(A) is negative of Ea (B) is always less than Ea
(C) can be less than or more than Ea (D) is always double of Ea
47. An elementary reaction is given as 2P + Q → products. If concentration of Q is kept constant and

concentration of P is doubled then rate of reaction is: [AFMC 2001]
(A) doubled (B) halved (C) quadrupled (D) remains same
48. In a first order reaction A → B, if K is the rate constant and initial concentration of the reactant is 0.5 M,
then half-life is: [AIPMT 2007]
ln 2 ln 2 log10 2 0.693
(A) (B) (C) (D)
K K 0.5 K 0.5 K
49. If 60% of a first order reaction was completed in 60 minute, 50% of the same reaction would be
completed in approximately: [AIPMT 2007]
(A) 45 minute (B) 60 minute (C) 40 minute (D) 50 minute
50. The rate constants k1 and k2 for two different reactions are 1016 e−2000/T and 1015 e−1000/T , respectively. The
temperature at which k1 = k2 is: [AIPMT 2008]
2000 1000
(A) K (B) 2000 K (C) K (D) 1000 K
2.303 2.303
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Chemical kinetics
Quick Quiz
Q.No. Ans. Q.No. Ans. Q.No. Ans. Q.No. Ans. Q.No. Ans.
1. (A) 2. (B) 3. (A) 4. (B) 5. (B)
6. (A) 7. (B) 8. (C) 9. (D) 10. (D)
11. (C) 12. (C) 13. (B) 14. (B) 15. (B)
16. (D) 17. (B) 18. (A) 19. (D) 20. (A)
Exercise - 1
1. (C) 2. (C) 3. (B) 4. (A) 5. (C)
6. (D) 7. (A) 8. (C) 9. (B) 10. (A)
11. (C) 12. (C) 13. (B) 14. (B) 15. (A)
16. (C) 17. (C) 18. (D) 19. (D) 20. (D)
21. (C) 22. (C) 23. (C) 24. (C) 25. (A)
26. (D) 27. (A) 28. (C) 29. (C) 30. (A)
31. (C) 32. (B) 33. (B) 34. (B) 35. (B)
36. (D) 37. (A) 38. (C) 39. (C) 40. (C)
41. (C) 42. (A) 43. (D) 44. (C) 45. (C)
46. (D) 47. (D) 48. (C) 49. (C) 50. (A)
Exercise - 2
1. (D) 2. (B) 3. (C) 4. (B) 5. (B)
6. (D) 7. (A) 8. (C) 9. (C) 10. (C)
11. (A) 12. (C) 13. (D) 14. (A) 15. (A)
16. (B) 17. (C) 18. (A) 19. (A) 20. (A)
21. (A) 22. (C) 23. (B) 24. (D) 25. (C)
26. (C) 27. (A) 28. (C) 29. (D) 30. (A)
31. (C) 32. (C) 33. (B) 34. (D) 35. (D)
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36. (C) 37. (A) 38. (D) 39. (B) 40. (B)
41. (D) 42. (D) 43. (C) 44. (C) 45. (D)
46. (C) 47. (C) 48. (A) 49. (C) 50. (C)
SAT – 4
1. (D) 2. (C) 3. (D) 4. (A) 5. (C)
6. (D) 7. (A) 8. (A) 9. (D) 10. (C)
11. (C) 12. (A) 13. (D) 14. (C) 15. (C)
16. (D) 17. (B) 18. (C) 19. (D) 20. (B)
21. (A) 22. (A) 23. (B) 24. (B) 25. (D)
26. (A) 27. (B) 28. (C) 29. (C) 30. (B)
31. (A) 32. (C) 33. (B) 34. (C) 35. (C)
36. (C) 37. (C) 38. (C) 39. (D) 40. (B)
41. (A) 42. (A) 43. (B) 44. (A) 45. (C)
Chemical Kinetics
Quick quiz
1. (A)
2. (B)
3. (A)
4. (B)
5. (B)
6. (A)
7. (B)
8. (C)
9. (D)
10. (D)
Solution to Think Further

4. (a) We know that t75 % = 2t1/2
= 2  6.93
= 13.86 min
(b) We know that t99.9 % = 10  t1/2
= 10  6.93
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= 69.3
min
7. The presence of a positive catalyst increases the value of the specific reaction rate.
Quick Quiz
11. (C)
H is a thermodynamic quantity, which is unaffected by the presence of a catalyst
12. (C)
13. (B)
14. (B)
15. (B)
0.693 0.693
k= = = 6.9310−3 sec−1
t1/2 100
16. (D)
17. (B)
18. (A)
19. (D)
20. (A)
2.303 a
t= log
K (a − x)
2.303 (2.303 3 log 2)
Thus, K = log8 =
10 10
Exercise - 1
1. (C)
Unit of k = (mol L−1)1 − n s−1
Q (1 − n) = −2
n=3
2. (C)
x = kt = 0.025  15 = 0.375 M
Remaining concentration = 0.5 − 0.375 = 0.125 M.
3. (B)
t75% = 2t50%
t75% = 32 min
32
 t 50% = = 16 min
2
4. (A)
Pa
1
t 1  where a is concentration
2
a n −1
n −1
(t )  P 
 1/2 =  2 
(t1/2 ) 2  P1 
n −1
350  40 
= 
175  80 
n −1
1
2= 
2
Taking log and then on simplification, n − 1 = −1
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n=0
5. (C)
t75% = 2t50%
6. (D)
7. (A)
k = Ae−Ea /RT
8. (C)
9. (B)
10. (A)
11. (C)
12. (C)
13. (B)
1
Remaining concentration of reactants after 20 min =  10 = 2.5 mol L−1
4
Rate = k[reactant]
0.693 0.693
= [reactant] =  2.5 = 0.0693  2.5 mol L−1 min −1
t1/2 10
14. (B)
When T → in Arrhenius equation,
1 
Ea
−
k = Ae RT
 = 0
 
k =
A.e0
k =
A (as
zero = 1)
15. (A)
dx
For a zero order reaction, = k, dx = kdt, x = kt, x  t.  plot of x against t will be a straight line passing through the origin.
dt
16. (C)
Since it is a first order reaction, half-life period is a constant. In this reaction, the change of concentration can be represented as
8 → 4 → 2 → 1 → 0.5.
Half-life period from 8 mol dm−3 to 0.5 mol dm−3. Then
2 104
t1 = = 5  103 s
2
4
0.693
k= = 1.38 10−4 s −1
5 103
17. (C)
Total time = 60 hours , t1/2 = 12 hours.
60
Number of t 1 = =5
2
12
In 5 steps, the change is concentration is 100 → 50 → 25 → 12.5 → 6.25 → 3.125
18. (D)
It is an ionic reaction.
19. (D)
20. (D)
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21. (C)
22. (C)
23. (C)
24. (C)
rate1 = k[A][B]. When volume is reduced to 1/4 its original value, concentration increases by 4 times.
 rate2 = k 4[A]4[B] = 16 k[A][B] rate2 = 16  (rate1)
25. (A)
N2 + 3H2 2NH3
3 3
Rate with respect to H2 = [rate with respect to ammonia] =  40  10−3 = 60  10−3 mol L−1s−1
2 2
26. (D)
rate1 = k[Cl2]x[NO]y
rate2 = k [2Cl2]x[2NO]y = 8 rate1
k. 2x [Cl2]x 2y [NO]y = 2x. 2y rate
2x. 2y = 8
…
(i)
rate3 = k[2Cl2]x[NO]y = 2 rate1
= 2x k [Cl2]x[NO]y = 2 rate1
x
2 =2
…
(ii)
x=1
From equation (i) 2.2y = 8 order w.r.t. NO, i.e., y = 2.
27. (A)
From experiment (1) and (3) when [A] is doubled, rate does not increase. Hence, order with respect to A is zero.
From experiment (1) and (4) when [B] is doubled rate increases by 8 times
k[2B]x = k8[B]x ; 2x[B]x = 8[B]x ; 2x = 8, x = 3.
Hence, order with respect to B = 3;  Rate = k[B]3.
28. (C)
29. (C)
30. (A)
Half life period is independent of initial concentration for a first order reaction.
31. (C)
1
t1/2  , n=2
a n −1
1
 t1/2 
a
32. (B)
Order is the sum of the powers of concentration terms in the rate equation.
33. (B)
One hour has 6 half life periods.
1 1
The amount left after 6 half life periods = =
26 64
34. (B)
96 hours has 4 half life periods
10
The amount left after 4 half life period = = 0.625
24
35. (B)
Feb 1st to July 1st has 150 days
t1/ 2 t1/ 2 t1/ 2 t1/ 2 t1/ 2
8 ⎯⎯⎯ → 4 ⎯⎯⎯ → 2 ⎯⎯⎯ → 1 ⎯⎯⎯ → 0.5 ⎯⎯⎯ → 0.25
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150
5 half life periods = 150 days; half life period = = 30 days
5
36. (D)
0.693
k= , t =
45
2.303 100
log OR
k 100 − 99.9
t99.9%
= 10t50%
2.303  45  3 2.303  450
= = = 450 min ;
0.693 2.31
450
t= = 7 ½ hours
60
37. (A)
t0.75 = 138.6 s
138.6 0.693 0.693
t0.5 = = 69.3 s ; k = = = 1  10−2.
2 t 0.5 69.3
38. (C)
The rate of reaction doubles, for every 10 C. In this case, it doubles 5 times, when temperature increases by 50 C.
The increase in rate = 25 = 32  30.
39. (C)
40. (C)
41. (C)
2NO+ O2 → 2NO2
230 246
42. (A)
43. (D)
44. (C)
45. (C)
Solutions to reasoning and assertion questions

46. (D)
47. (D)
48. (C)
49. (C)
50. (A)
Exercise – 2
1. (D)
2. (B)
3. (C)
4. (B)
5. (B)
6. (D)
7. (A)
8. (C)
9. (C)
10. (C)
11. (A)
163
EDUTECH
12. (C)
13. (D)
H2O2 decomposition follows 1st order kinetics and t1/2 is independent of the initial concentration of H2O2.
14. (A)
15. (A)
16. (B)
17. (C)
The rate is independent of concentration. Hence it is a zero order reaction.
18. (A)
0.12 M changes to 0.06 M in 10 h ; t1/2 = 10h
0.12 M changes to 0.03 M in 20 h ; t 75% = 20h
since t 75% = 2  t1/2 , the order of the reaction is 1.
19. (A)
When Ea = 0; k = a constant and independent of temperature.
20. (A)
k2 Ea  T2 − T1 
log10 =  
k1 2.303 R  T1T2 
k2
= 2; T1 = 27 + 273 = 300 K ; T2 = 37 + 273 = 310 K ;
k1
R = 1.89 cal or approximately 0.002 kcal.
Ea 310 − 300
log 2 = 
2.303  0.002 310  300
2.303  0.002  310  300  log 2
Ea = = 12.894  12.9 k cal
10
21. (A)
22. (C)
Forward reaction is exothermic and backward reaction is endothermic.
Ea (endothermic) = Ea (exothermic) + H
Ea 1.25  104
=
2.303 RT T
Ea = 1.25 104  2.303  8.314 10−3 kJ = 239 kJ
Ea (endothermic) = 239 + 478 = 717 kJ.
23. (B)
24. (D)
Rate1 = k[NO]2 [O2 ]
2
 NO   O 2 
Rate2 = k     = 8 Rate1
 0.5   0.5 
25. (C)
(i) order w. r. t. A = 1
(ii) order w. r. t. B = 0
(iii) order w. r. t. C = 2
overall order = 1 + 0 + 2 = 3
26. (C)
27. (A)
28. (C)
164
EDUTECH
29. (D)
30. (A)
31. (C)
In a zero order reaction, rate is independent of the concentration of the reactant. Thus at 10 s intervals, the rate remains the
same.
32. (C)
0.693 0.693
k= = = 0.0693 min −1
t1/2 10
2.303 a
k= log
t a−x
2.303 0.08 2.303 2.303  0.903
0.0693 = log t = log8 = = 29.97  30min
t 0.01 0.0693 0.0693
33. (B)
R P
Ea (forward) = Ea (reverse)
 H = H(Products) − H(Reactants) = 0
34. (D)
The increase in temperature increases both the rate and rate constant of a reaction.
35. (D)
k = A.e− Ea /RT
Ea
Lnk = l n A −
RT
Thus, a plot of ln k versus
1 E
gives a slopeof − a
T R
36. (C)
0.693
=
t1/2
0.693
= = 2.7 10−6 s −1
2.95  24  60  60
37. (A)
38. (D)
For an exothermic reaction, Ea of forward reaction is greater than 200 kJ whereas for an endothermic reaction Ea of backward
reaction is greater than 200 kJ
39. (B)
When the volume of the reaction vessel is doubled, the concentration of each of the reactants become half of their initial values.
Thus, Initial rate = k[NO]2 [O2] ... (1)
2
1  1 
New rate = k  NO   O 2  ... (2)
2  2 
2
1 1 1
 New rate will be     or th of its initial rate.
2 2 8
40. (B)
165
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Statements (A), (C) and (D) are correct. Statement (B) is incorrect. The order of a reaction can be changed by varying the
conditions such as pressure, temperature etc.
41. (D)
42. (D)
43. (C)
44. (C)
45. (D)
46. (C)
It is a fact
47. (C)
r = K[P]2 [Q]
r1 = K[2P]2[Q]
r1
 =4
r
48. (A)
2.303 log10 2
For I order reaction, K =
t1
2
0.693 ln 2
t1 = =
2
K K
49. (A)
2.303 100
60 = log
K 40
2.303 100
t= log
K 50

60 0.40
=
t 0.3010
 t = 45 minute
50. (C)
K1 = 1016 e−2000/T ; K 2 = 1015 e−1000/T
if
K1 = K 2 then 106 e−2000/T = 1015 e−1000/T

or
2000 1000 1000

log10 − =− or T = K
T T 2.303
Self Assessment Test – 4

1. (D)
2A2 + B2 → 2A2B , for every two moles of A2, only one mole of B2 will disappear. Thus, the reactant A will disappear at twice
the rate that B will decrease.

2. (C)
0.693 0.693 0.693
t1 = or k = = = 0.1
2 k t1 6.93
2
166
EDUTECH
2.303 a
k= log
t a−x
2.303 100
0.1 = log
t 99% 1
2.303
 t 99% =  2 = 46.06 min
0.1
3. (D)
The specific reaction rate (i.e., the rate constant) depends on temperature, and not on other factors such as given in (A), (B) and
(C).
4. (A)
In the presence of a catalyst, the reaction follows a different path involving less activation energy. Hence, the reaction will
become faster.
5. (C)
H is a thermodynamic quantity, which is unaffected by the presence of a catalyst.
6. (D)
Without the mechanism, it is not possible to determine the slowest step and the order of the reaction.
7. (A)
1/[A]
8. (A)
The order and molecularity are the same for an elementary step of a reaction.
9. (D)
2.303 a
k= log
t a−x
2.3 100
k= log
20 100 − 20
2.3 100 2.3
= log   0.0969 = 0.01114 min−1
20 80 20
2.3 100
k= log
t 60% 100 − 60
2.3 100
0.01114 = log
t 60% 40
2.3
t 60% =  0.3979 = 82 min
0.01114
10. (C)
Rate = k [A] [B]2
dx
= k[A][B]2
dt
mol L−1
= k[mol L−1 ][mol L−1 ]2
s
mol L−1 1
k=  k= = mol−2 L2 s −1
s(mol L−1 )3 s(molL−1 ) 2
11. (C)
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T99.9% = 10 half lives.
12. (A)
Ea
slope = −
2.303R
Ea
−5.64 = −
2.303R
 Ea = 5.64  2.303  8.314
= 107.9898 ; 108 J mol–1
13. (D)
t1/2  a for a zero order reaction.
14. (C)
a a
t1 = t100% =  t100% = 20 min
2 2k k
15. (C)
2N2O5 → 4NO2 + O2 , is the first order reaction, as indicated by the unit for rate constant.
Rate = k[N2O5]
Rate 1.02  10 −4
 [N 2O5 ] = = = 3 mol L−1
k 3.4  10−5
16. (D)
From 1 and 4, by keeping [B] constant and increasing [A] by 4 times, the rate is also increased by 4 times.
 Order w.r.t. A is = 1
From 2 and 3, by keeping [A] constant and doubling the concentration of B, the rate increased by 4 times.
 Order w.r.t B = 2
Thus, the overall order of the reaction = 1 + 2 = 3.
17. (B)
Since rate is proportional to [N2O5], the reaction is of first order.
2.303 a
Rate constant k = log
t a−x
2.303 100
= log = 6.38 10−4 s−1
3600 100 − 90
18. (C)
Rate of appearance of D is double the rate of disappearance of B.
19. (D)
20. (B)
After every 30 minutes amount is reduced to half.
1
 t1/2 is 30 min. In 90 min the amount is reduced to
8
This is true for first order reaction.
21. (A)
2.303 a 2.303 0.5
t= log = log = 0.384 min.
k a−x 6 0.05
22. (A)
For the change 2A + 3B → products
1 d[A] 1 dt / [B]
− =−
2 dt 3 dt
1 1
r1 = r2 or 3r1 = 2r2 .
2 3
23. (B)
168
EDUTECH
[O][O2 ]
From equation (A) K =
[O3 ]
K[O3 ]
i.e. [O] =
[O2 ]
Rate law = k [O3] [O]
[O ][O ]
rate = Kk 3 3
[O2 ]
rate = k' [O3 ]2[O 2 ]−1
24. (B)
As the conversion of X to AB is fast, it means the process has very low activation energy.
25. (D)
2
1 1 1
r = [SO2 ]2[O2 ] . When volume doubles, concentration reduces to half.     =  
 2  2 8
26. (A)
27. (B)
28. (C)
29. (C)
The activation energy for the reverse reaction can be less than or more than E a depending upon whether the reaction is
endothermic or exothermic.
30. (B)
12 days means 4 half lives. From 48g after 4 half lives remaining is 3g.
31. (A)
32. (C)
33. (B)
2.303  a  2.303
k= log  = log8 = 0.03466 min−1
t a−x 60
Rate
= k[A]1
= 0.03466  0.1 = 3.466  10−3 mol L−1 min−1

34. (C)
35. (C)
36. (C)
37. (C)
38. (C)
39. (D)
Rate = k[A], i.e., 2.0  10–5 = k(0.01) or k = 2.0  10–3
0.693
t1 = = 347 s
2
2.0 10−3
40. (B)
41. (A)
42. (A)
0.6 3 0.675 3
In case I, fraction of A reacted = = . In case II, fraction of A reacted = =
0.8 4 0.9 4
For a first order reaction, time taken for the same fraction of reaction is independent of initial concentration.
43. (B)
Rate = k[conc.]n
169
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2.4 = k(2.2)n … (1)
0.6 = k(1.1)n … (2)
n
2.4  2.2 
 = n
 or 4 = (2) or n = 2
0.6  1.1 
44. (A)
ln 2
t1/2 =
k
ln 2
Half-life of a first order reaction does not depend upon initial concentration. It is equal to .
k
45. (C)
2.303 a 2.303 1 2.303 10
k= log = log = log
60 min a − 0.6a 60 0.4 60 4
2.303
= (1 − 0.60) = 0.0153 min −1
60
0.693
t1/2 = = 45.3 min
0.0153
Test paper
1. (D)
2. (B)
2r+ + 2r – = 400
or r+ +
r – = 200
–
r =
200 – r+
–
r =
200 – 75
–
r =
175 pm
3. (A)
1
In bcc lattice distance of closest approach is of body diagonal
2
3
i.e., a = 1.73 Å or
2
1.73
a=  2 = 2 Å = 200 pm
3
4. (C)
8 6
Z atoms = + = 4 ; X atoms = 8
8 2
Ration X : Z is 2 : 1
 formula is X2Z
5. (B)
6. (C)
3a
As CsCl is body centred, d =
2
7. (D)
8. (D)
Lower is the number of particles in solution, lower will be the depression in freezing point and higher will be the freezing point.
9. (B)
2-
Fe2(SO4)3 is very dilute solution ionises as Fe2(SO4)3 2Fe3+ + 3SO4
170
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 one formula unit of ferric sulphate gives 5 ions in solution. Hence, i = 5.
10. (C)
For association i < 1
11. (B)
Minimum depression in freezing point, highest is the freezing point. Lowest depression in freezing point will be in 0.1 M
glucose.
12. (D)
K  w 2 1000 2.16  0.15 1000
M2 = b = = 100
Tb  w1 0.216 15
13. (A)
Tf = Kf.m = 1.86  0.018 = 0.151
 Freezing point of the solution is 273 – 0.151 = 272.849 K
14. (A)
500
w2 = 3.42 g; V = = 0.5 L , T = 300 K; R = 0.0821 L atm K−1 mol−1, M2 = 342
1000
w RT 3.42  0.0821 300
 = 2 = = 0.492 atm
M2V 342  0.5
15. (D)
Tb K b
=
Tf K f
K b  Tb 0.512  0.186
Tb = = = 0.0512 C
Kf 1.86
1 1
Alternatively, freezing point is th of Kf and there by elevation is th of Kb.
10 10
16. (B)
17. (A)
By definition of Kf, 1 mole of any solute produces a depression of Kf in 1 kg of solvent. X produces the largest depression for a
given mass of any solute
18. (D)
19. (A)
20. (C)
1 mol of monovalent ions carry a charge of 96500 C.
2 mol of divalent ions carry a charge of 2  96500 C or 2F.
21. (A)
This is the definition of electrochemical equivalent of a substance.
22. (A)
By IUPAC convention, anode is defined as the site of oxidation.
23. (A)
This is a consequence of Faraday’s first law of electrolysis.
24. (B)
SO24− is quite stable and is not oxidized. Instead, OH− is oxidized to O2 at the anode.
25. (A)
Discharge of chlorine requires less energy than discharge of oxygen. Also anions are oxidized to non-metals at the anode.
26. (A)
Conductance of solutions of electrolytes increases with increase in temperature and consequently resistance decreases.
27. (B)
Degree of dissociation of a weak electrolyte increases with dilution. It is a fraction. The highest value of a fraction is 1.
28. (D)
171
EDUTECH
29. (D)
1g eq. mass of hydrogen occupies 11,200 cm3 at STP.
To liberate 11,200 cm3 of H2 at STP, 96,500 C of electricity is required
 To liberate 1.12 cc of H2 at STP, 9.65 C of electricity is required
Q = I  t  9.65 = I  i I = 9.65 amperes
30. (C)
31. (D)
Specific conductance is conductance per unit volume. The higher is the concentration, the higher will be the number of ions per
unit volume and consequently, the higher will be the specific conductance.
32. (B)
a. N HNO3 is 0.1 M HNO3
0.1 mole of HNO3 is present in 1 litre
 1 mole of HNO3 will be present in 10 litres = 10  103 cm3
Molar conductivity = specific conductance  volume in cm3 containing 1 mole
= (6.3  10−2 ohm−1 cm−1)  10  103 cm3 mol−1
= 6.3  102 ohm−1 cm2 mol−1
= 630 ohm−1 cm2 mol−1
33. (C)
34. (D)
35. (D)
36. (B)
37. (C)
38. (C)
39. (B)
40. (D)
1 1
t1/2  n −1 when n = 2, t1/2 
a a
41. (C)
42. (D)
rate1 = kp 2H 2S p O 2
rate2 = k(3p H2S ) 2 (3pO2 ) = 27 rate1
43. (C)
1
t1/2  n −1
a
When n = 0, t1/2  a
When a = 1, t1/2  1
1 1
When a = , t1/2 
4 4
44. (C)
45. (A)
2.303 100 2.303 100
t= log = log t =
k (100 − 90) k 10
2.303
k
46. (C)
172
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a
For fcc structure r =
2 2
a 408
 diameter = 2r = = = 288.5 pm
2 1.414
47. (D)
In a cubic close packing, the number of octahedral voids is equal to number of atoms and number of tetrahedral voids is equal to
the twice the number of atoms.
Number of atoms in a ccp array = 1
A2+ B+ O2−
1 1 1
1 2 
4
1 1 1
2
1 2 2
AB2O2
48. (B)
n  atomic mass
For fcc lattice  =
average no  a 3
4  atomic mass
2.72 =
6.02  1023  (404  10−10 )3
 atomic mass = 27 g mol−1
49. (A)
Diamond has zinc blende (ZnS) structure. Carbon has ccp(fcc) structure having half of the tetrahedral voids occupied. Total
1 1
number of carbon atoms per unit cell = 8  (corners) +6  + 4(tetrahedral voids) = 8
8 2
50. (C)
100  K f  w
T = i
m.W
T = 0 − (–3.82) = 3.82C
1000 1.86  5  i
3.82 =  i = 2.63
142  45
51. (B)
PM = PA + PB
=
PA.xA + PB.xB
=
PA.xA + PB(1 − xA) = pB + xA(pA − pB)
52. (B)
Ethylene glycol is used as antifreeze compound in automobile radiators under the name “coolant prestone’
53. (B)
and Fe ( CN )6 are oxidants. Higher is ERP stronger is oxidant
3−
Fe3+
Fe3+ + e → Fe2+ ; ERP = 0.77 V
[Fe(CN)6]3− + e → [Fe(CN)6]4−; ERP =
0.35 V
54. (C)
m it
Eq of Al formed = =
E 96500
4 104  6  60  60  27  27 
 mAl = E = 
96500  3  3 
 mAl = 8.05  104 g
55. (C)
173
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2
3
( Al3+ )2 + 4e− → 23 Al
or ( O2− ) → 4e− + O2
2
3 3
−G = nFE (E
is potential required)
960 1000
E= = 2.5 V
4  96500
56. (D)
E cell = E OP + E RP
Zn/Zn +2 Ag + /Ag
= EOPZn + ERPAg
= 0.76 + 0.34
Ecell = 1.10 V
57. (A)
K = Ae− Ea /RT
58. (D)
r = K[CH3COCH3]a[Br2]b[H+]c
 5.7 
10−5 = K[0.30]a [0.05]b [0.05]c
…(i)
5.7  10−5 = K[0.30]a [0.10]b [0.05]c
…(ii)
1.2  10−4 = K[0.30]a [0.10]b [0.10]c
…(iii)
3.1  10−4 = K[0.40]a [0.05]b [0.25]c
…(iv)
By (i) and (ii) b = 0
By (ii) and (iii) c = 1
By (i) and (iv) a = 1
3.1  10 −4  0.4   0.2 
a 1
=   
5.7  10 −5  0.3   0.05 
5.44 = (1.33)a  4
a=1
59. (B)
1 d[NH3 ] 1 d[H 2 ]
=−
2 dt 3 dt
d[H 2 ] 3 d[NH3 ] 3
− =  =  2 10−4 = 3 10−4
dt 2 dt 2
60. (D)
r(t1 + 10)
= 2 for each 10 rise in temperature
rt
r100
 = (2)9 = 512
r10
174

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Chemical Kinetics

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ACTIVE SITE Chemical Kinetics

Average Rate of Reaction

Calculation of the Average Rate of Reaction

instants of time t1 and t2, the corresponding concentrations 0.02 x1

(as shown in diagram at time 7.5 minutes) and it is

Expression for the Rates of Reaction

Unit of Rate of Reaction

Law of Mass Action

Rate Law Expression

Differences between rate constant (k) and rate of reaction (r)

Difference between rate law and law of mass action

Order of Reaction (n)

Difference between order and molecularity of reaction

The characteristics of zero order reactions

Examples for zero order reaction

First Order Reactions

Half life period (t1/2)

Relationship between half life period and initial concentration

 It is used in population growth and bacteria multiplication.

Examples for first order reactions

2H2O2 → 2H2O + O2; r = k[H2O2]; order = 1

Pseudo unimolecular reactions

Rate law equation: r = k[CH3.COOCH3] [H2O]0; order = (1 + 0) = 1 molecularity = (1 + 1) = 2

Negative order reaction

By substituting the value of [O] we get

t1/2 t1/2 t1/2 t1/2

(a–x) (a–x) (a–x)2 (a–x)3

Zero order reaction First order reaction

Second order reaction Third order reaction

3. In the inversion of cane sugar

6. Which of the following does not affect the order of a reaction?

7. The specific reaction rate of a first order reaction depends on

8. Inversion of cane sugar in dilute acid is

Energy Diagram Energy (E) → ET

Reaction co-ordinate Reaction co-ordinate

Endothermic reaction Exothermic reaction

Graphical Representation of the Effect of Temperature on the Rate of Reactions

Collision theory of chemical Reactions

rate of reaction can be expressed as

• The reactions with smaller activation energy are fast.

Calculation of Activation Energy

Effect of Catalyst on the Reaction Rate

11. In the presence of a catalyst, the reaction enthalpy

19. According to the collision theory,

11. For an elementary process of a chemical reaction,

2t1/2 = 20 mins  t1/ 2 = 10 min

9. The activation energy for a chemical reaction depends upon

18. Among the following reactions, the fastest reaction is

21. The hydrolysis of ethyl acetate in an alkaline medium is

(A) +16 kJ (B) +8 kJ (C) +12 kJ (D) +4 kJ

Each of the following questions consists of a statement−I and a Statement−II. Examine

49. Statement-I : A catalyst may increase or decrease the rate of a reaction

1. If the rate of a reaction is r = k[X]1/3 [Y] the unit of rate constant is

6. For the following reaction, 2P + Q + 3R → S + T

27. The reaction is fastest in

CO(g) + NO 2 (g) CO 2 (g) + NO(g)

47. An elementary reaction is given as 2P + Q → products. If concentration of Q is kept constant and

1. (A) 2. (B) 3. (A) 4. (B) 5. (B)

6. (A) 7. (B) 8. (C) 9. (D) 10. (D)