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CHEMICAL KINETICS- Previous HSE Questions And Answers
1. An archeological substance contained wood had only 66.66% of the 14C found in a tree. Calculate the
age of the sample if the half life of 14C is 5730 years. (3) [ March 2008]
Ans: We know that radioactive decay follows first order kinetics.
For a first order reaction, k= 2.303 log [R]0
t [R]
Here t½ = 5730 years, [R]0 = 100 and [R] = 66.66
k = 0.693/t½ = 0.693/5730 = 1.21 x 10-4
Age of the sample, t = 2.303 log [R]0 = (2.303/1.21 x 10-4)log(100/66.66) = 3352.38 years
k [R]
2. Unit of rate constant (k) of a reaction depends on the order of the reaction. If concentration is
expressed in mol L-1 and time in seconds (s), find the unit of k for zero, first and second order reaction.
(3) [March 2009]
Ans:
Reaction Unit of rate
constant
Zero order reaction mol L-1s-1
First order reaction s-1
Second order reaction mol-1L s-1
3. The order of a reaction can be zero and even a fraction but Molecularity cannot be zero or a non-
integer.
i) What do you mean by the order of a reaction? (1)
ii) What is Molecularity of a reaction? (1)
iii) The conversion of molecules A to B follows second order kinetics. If concentration of A is
increased to three times, how will it affect the rate of formation of B? (2) [March 2010]
Ans: i) Order of a reaction is the sum of the powers of the concentration terms of the reactants in the
rate law.
ii) Molecularity of a reaction is the total number of reacting species collides simultaneously in a
chemical reaction.
iii) Let the initial concentration of A be x. Then the rate law for this reaction is r = k[x]2
When the concentration of A is increased to three times, the final concentration becomes 3x.
Now the rate law is r1 = k[3x]2 = 9.k[x]2
So r1 = 9 x r
i.e. the rate formation of B is increased by 9 times.
4. The value of rate constant k of a reaction depends on temperature. From the values of k at two
different temperatures, the Arrhenius parameters Ea and A can be calculated.
The rate constants of a reaction at 1000K and 1060K are 0.01M-1S-1 and 0.10M-1S-1 respectively. Find
the values of Ea and A. (3) [March 2010]
Ans: We know that,
log k2/k1 = Ea [T2 - T1]
2.303 R T1.T2
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Here T1 = 1000 K, k1 = 0.01M-1S-1, T2 = 1060 K, k2 = 0.1 M-1 s-1 and R = 8.314 J K-1 mol-1
log (0.1) = Ea [1060 – 1000]
(0.01) 2.303 x 8.314 1000 x 1060
Ea = log (10) x 2.303 x 8.314 x 1000 x 1060 = 338266 J mol-1 = 338.266 kJ mol-1
60
Also, from Arrhenius equation, k = A.e-Ea/RT
A = k/ e-Ea/RT
= 0.01/e –(338266/8.314 x 1000)
OR,
From logarithmic form of Arrhenius equation, log k = log A – Ea/2.303RT,
log A = log k + Ea/2.303RT
= log (0.01) + 338266/(2.303x8.314x1000) = 15.67
A = Anti-log (15.67)
= 4.67 x 1015
5. The hydrolysis of an ester in acidic medium is a first order reaction.
a) What do you mean by a first order reaction? (½ )
b) What is the relation between Rate constant and half life period of a first order reaction? (½ )
c) Half life period of a first order reaction is 20 seconds. How much time will it take to complete 90%
of the reaction? (3) [March 2011]
Ans: a) Order of the reaction = 1. OR, it is a reaction in which rate of the reaction is directly
proportional to the concentration of the reactant. i.e. r = k[R].
b) Rate constant, k = 0.693/t½
c) Here t½ = 20 s.
So k = 0.693/20 = 0.03465 s-1
For a first order reaction, k= 2.303 log [R]0
t [R]
Let [R]0 = 100. Then [R] = 100 – 90 = 10
So t = 2.303 log [R]0
k [R]
= (2.303/0.03465)x log(100/10)
= 66.46 s
6. The value of rate constant k of a reaction depends on temperature. From the values of k at two
different temperatures, the Arrhenius parameters Ea and A can be calculated.
a) The rate constants of a reaction at 600K and 900K are 0.02s-1 and 0.06s-1 respectively. Find the
values of Ea and A. (3)
b) Write the unit of rate constant of a 2nd order reaction if concentration is in mol L-1 and time in s.
(1) [SAY 2011]
Ans: a) We know that, log k2/k1 = Ea [T2 - T1]
2.303 R T1.T2
Here T1 = 600 K, k1 = 0.02 s-1, T2 = 900 K, k2 = 0.06 s-1 and R = 8.314 J K-1 mol-1
log (0.06) = Ea [900 – 600]
(0.02) 2.303 x 8.314 900 x 600
Ea = 0.4771 x 2.303 x 8.314 x 900 x 600 = 16443 J mol-1 = 16.443 kJ mol-1
300
From logarithmic form of Arrhenius equation, log k = log A – Ea/2.303RT,
CHEMICAL KINETICS – Prepared by ANIL KUMAR K L, APHSS ADICHANALLOOR, KOLLAM Page 2
CHEMICAL KINETICS- Previous HSE Questions And Answers
1. An archeological substance contained wood had only 66.66% of the 14C found in a tree. Calculate the
age of the sample if the half life of 14C is 5730 years. (3) [ March 2008]
Ans: We know that radioactive decay follows first order kinetics.
For a first order reaction, k= 2.303 log [R]0
t [R]
Here t½ = 5730 years, [R]0 = 100 and [R] = 66.66
k = 0.693/t½ = 0.693/5730 = 1.21 x 10-4
Age of the sample, t = 2.303 log [R]0 = (2.303/1.21 x 10-4)log(100/66.66) = 3352.38 years
k [R]
2. Unit of rate constant (k) of a reaction depends on the order of the reaction. If concentration is
expressed in mol L-1 and time in seconds (s), find the unit of k for zero, first and second order reaction.
(3) [March 2009]
Ans:
Reaction Unit of rate
constant
Zero order reaction mol L-1s-1
First order reaction s-1
Second order reaction mol-1L s-1
3. The order of a reaction can be zero and even a fraction but Molecularity cannot be zero or a non-
integer.
i) What do you mean by the order of a reaction? (1)
ii) What is Molecularity of a reaction? (1)
iii) The conversion of molecules A to B follows second order kinetics. If concentration of A is
increased to three times, how will it affect the rate of formation of B? (2) [March 2010]
Ans: i) Order of a reaction is the sum of the powers of the concentration terms of the reactants in the
rate law.
ii) Molecularity of a reaction is the total number of reacting species collides simultaneously in a
chemical reaction.
iii) Let the initial concentration of A be x. Then the rate law for this reaction is r = k[x]2
When the concentration of A is increased to three times, the final concentration becomes 3x.
Now the rate law is r1 = k[3x]2 = 9.k[x]2
So r1 = 9 x r
i.e. the rate formation of B is increased by 9 times.
4. The value of rate constant k of a reaction depends on temperature. From the values of k at two
different temperatures, the Arrhenius parameters Ea and A can be calculated.
The rate constants of a reaction at 1000K and 1060K are 0.01M-1S-1 and 0.10M-1S-1 respectively. Find
the values of Ea and A. (3) [March 2010]
Ans: We know that,
log k2/k1 = Ea [T2 - T1]
2.303 R T1.T2
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Here T1 = 1000 K, k1 = 0.01M-1S-1, T2 = 1060 K, k2 = 0.1 M-1 s-1 and R = 8.314 J K-1 mol-1
log (0.1) = Ea [1060 – 1000]
(0.01) 2.303 x 8.314 1000 x 1060
Ea = log (10) x 2.303 x 8.314 x 1000 x 1060 = 338266 J mol-1 = 338.266 kJ mol-1
60
Also, from Arrhenius equation, k = A.e-Ea/RT
A = k/ e-Ea/RT
= 0.01/e –(338266/8.314 x 1000)
OR,
From logarithmic form of Arrhenius equation, log k = log A – Ea/2.303RT,
log A = log k + Ea/2.303RT
= log (0.01) + 338266/(2.303x8.314x1000) = 15.67
A = Anti-log (15.67)
= 4.67 x 1015
5. The hydrolysis of an ester in acidic medium is a first order reaction.
a) What do you mean by a first order reaction? (½ )
b) What is the relation between Rate constant and half life period of a first order reaction? (½ )
c) Half life period of a first order reaction is 20 seconds. How much time will it take to complete 90%
of the reaction? (3) [March 2011]
Ans: a) Order of the reaction = 1. OR, it is a reaction in which rate of the reaction is directly
proportional to the concentration of the reactant. i.e. r = k[R].
b) Rate constant, k = 0.693/t½
c) Here t½ = 20 s.
So k = 0.693/20 = 0.03465 s-1
For a first order reaction, k= 2.303 log [R]0
t [R]
Let [R]0 = 100. Then [R] = 100 – 90 = 10
So t = 2.303 log [R]0
k [R]
= (2.303/0.03465)x log(100/10)
= 66.46 s
6. The value of rate constant k of a reaction depends on temperature. From the values of k at two
different temperatures, the Arrhenius parameters Ea and A can be calculated.
a) The rate constants of a reaction at 600K and 900K are 0.02s-1 and 0.06s-1 respectively. Find the
values of Ea and A. (3)
b) Write the unit of rate constant of a 2nd order reaction if concentration is in mol L-1 and time in s.
(1) [SAY 2011]
Ans: a) We know that, log k2/k1 = Ea [T2 - T1]
2.303 R T1.T2
Here T1 = 600 K, k1 = 0.02 s-1, T2 = 900 K, k2 = 0.06 s-1 and R = 8.314 J K-1 mol-1
log (0.06) = Ea [900 – 600]
(0.02) 2.303 x 8.314 900 x 600
Ea = 0.4771 x 2.303 x 8.314 x 900 x 600 = 16443 J mol-1 = 16.443 kJ mol-1
300
From logarithmic form of Arrhenius equation, log k = log A – Ea/2.303RT,
CHEMICAL KINETICS – Prepared by ANIL KUMAR K L, APHSS ADICHANALLOOR, KOLLAM Page 2
