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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.
2.303 [𝑅]0
For a first order reaction, k = log
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
2.303 [𝑅]0
Age of the sample, t = log
𝑘 [𝑅]
2.303 100
t= log
1.21 𝑥 10−4 66.66
= 3352.38 years
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-1 s-1 and 0.10M-1 s-1 respectively. Find
the values of Ea and A. (3) [March 2010]

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Ans: We know that,

k2 Ea T2 − T1
log =
k1 2.303 R T1 T2

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
0.1 Ea 1060 − 1000
log =
0.01 2.303 x 8.314 1000 x 1060
log 10 x 2.303 x 8.314 x 1000 x 1060
Ea = = 338266 J mol-1 = 338.266 kJ mol-1
60
−𝐸𝑎
Also, from Arrhenius equation, k = A.𝑒 𝑅𝑇
𝑘 0.01
A = −𝐸𝑎 = −338266 = 4.67 x 1015
𝑒 𝑅𝑇 e 2.303 x 8.314 x 1000
OR,
𝐸𝑎
From logarithmic form of Arrhenius equation, log k = log A –
2.303 𝑅𝑇
𝐸𝑎
log A = log k +
2.303 𝑅𝑇
338266
= log (0.01) + = 15.67
2.303 𝑥 8.314 𝑥 1000
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].
0.693
b) 𝑡1 =
2 𝑘
c) Here t½ = 20 s.
So k = 0.693/20 = 0.03465 s-1
2.303 [𝑅]0
For a first order reaction, k = log
𝑡 [𝑅]
Let [R]0 = 100. Then [R] = 100 – 90 = 10
2.303 [𝑅]0
So, t = log
𝑘 [𝑅]
2.303 100
= log
0.03465 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 2 order reaction if concentration is in mol L-1 and time in
nd

second. (1) [SAY 2011]

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k E T2 − T1
Ans: a) We know that, log k2 = 2.303
a
R 1 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
0.06 E 900−600
log 0.02 = 2.303 xa8.314 600 x 900

0.4771 x 2.303 x 8.314 x 600 x 900

Ea = = 16443 J mol-1 = 16.443 kJ mol-1
300
𝐸
From logarithmic form of Arrhenius equation, log k = log A – 2.303𝑎 𝑅𝑇
𝐸
So, log A = log k + 2.303𝑎 𝑅𝑇
16443
= log (0.02) + 2.303 𝑥 8.314 𝑥 600 = - 0.267
A = Anti-log (- 0.267) = 0.54
b) Mol-1L s-1
7. Rate of a reaction is the change in concentration of any one of the reactants or any one of the
products in unit time.
i) Express the rate of the following reaction in terms of reactants and products: 2HI → H2 + I2 (1½)
ii) If the rate expression for the above reaction is rate = k[HI]2, what is the order of the reaction ? (½ )
iii) Define order of a reaction. (1)
iv) Whether the Molecularity and order of the above reaction are the same? Give reason (1)
[March 2012]
Ans:
d[HI] d[H2 ] d[I2 ]
i) 𝑟𝑖𝑛𝑠𝑡 = – ½ = =
dt dt dt

∆[HI] ∆[H2 ] ∆[I2 ]

OR, 𝑟𝑎𝑣 = – ½ = =
∆t ∆t ∆t
ii) Order = 2
iii) It is the sum of the powers of the concentration terms of the reactants in the rate law.
iv) Yes. Here the power of the concentration term in the rate law = 2, so order =2 . The no. of
reactant species = 2. So the molecularity = 2.
8. For a first order reaction half-life period is independent of initial concentration of its reacting species.
i) What is mean by half-life period of a reaction? (1)
ii) By deriving the equation for t½ of first order reaction, prove that t½ is independent initial
concentration of reacting species. (3) [SAY 2012]
Ans: i) It is the time taken to reduce the concentration of reactants to half of its initial concentration.
2.303 [𝑅]0
ii) For a first order reaction, k = log
𝑡 [𝑅]
[𝑅]0
When t = 𝑡1 , [R] =
2 2
Substitute these values in the above equation:
2.303 [𝑅]0
k= log [𝑅]0
𝑡1
2 2
2.303 x 𝑙𝑜𝑔2 2.303 x 0.3010
Or, 𝑡1 = =
2 𝑘 𝑘
0.693
Or, 𝑡1 =
2 𝑘

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Thus for a first order reaction, half-life period is independent of initial concentration of the reacting
species.
9. a) Zero order reaction means that the rate of a reaction is independent of the concentration of the
reactants.
i) Write an example for a zero order reaction. (1)
ii) Write the integral rate expression for the zero order reaction, R → P. (1)
b) The temperature dependence of rate of a chemical reaction can be accurately explained by
Arrhenius equation. With the help of Arrhenius equation, calculate the rate constant for the first
order reaction C2H5I → C2H4 + HI at 700K. Energy of activation (Ea) for the reaction is 209 kJ/mol
and rate constant at 600 K is 1.6x10-5 s-1 (R = 8.314 J/K/mol). (2) [March 2013]
Ans: a) i) Decomposition of ammonia at the surface of pt metal at high pressure.
[𝑅]0 −[𝑅]
ii) k =
𝑡
k2 Ea T2 − T1
b) We know that, log =
k1 2.303 R T1 T2

Here T1 = 600 K, k1 = 1.6 x 10-5 s-1, T2 = 700 K, 𝐸𝑎 = 209 kJ/mol = 209 x 103 J/mol = 209000 J mol-1 and
R = 8.314 J K-1 mol-1
k2 209000 700 − 600
log =
1.6 x 10−5 2.303 x 8.314 600 x 700
= 2.6
i.e. log k2 – log(1.6 x 10-5) = 2.6
log k2 = 2.6 + log(1.6 x 10-5)
= -2.195
So, k2 = Anti log(-2.195) = 6.38 x 10-3 s-1
10. The conversion of a molecule A to B follows second order kinetics.
a) Write the rate equation for the second order reaction. (1)
b) If the concentration of A is increased to four times, how will it affect the formation of B. (2)
c) Indicate the order and Molecularity of the reaction given below:
𝐻+
C12H22O11 → C6H12O6 + C6H12O6 (1) [SAY 2013]
Ans: a) r = k[A]2
b) When the concentration is increased by 4 times, the new concentration of A = 4A
So, r = k[4A]2 = 16k[A]2
So the rate of formation of B is increased by 16 times.
c) Order = 1 and molecularity = 2 [It is a pseudo first order reaction].

11. a) Consider a general reaction aA + bB → cC + dD. The rate expression for the reaction is r = k[A]x[b]y
i) Establish the significance of (a+b) and (x+y) in terms of order and molecularity. (1)
ii) Write any two differences between order and molecularity. (2)
b) “Reactions with zero order are possible, but zero molecularity is not”. Justify the statement. (1)
[March 2014]
Ans: a) i) (a+b) indicates molecularity and (x+y) indicates order.
ii)

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Order Molecularity
1. It is the sum of the powers of the concentration It is the total number of reactant species
terms in the rate law expression collide simultaneously in a chemical reaction
2. It is an experimental quantity It is a theoretical quantity
3. It can be zero or fractional It cannot be zero or fractional
b) Zero order reaction means the rate of the reaction is independent of the concentration of the
reactants. So it is possible. But zero molecularity means there is no reactants. This is not possible.
12. a) Unit of rate constant (k) of a reaction depends on the order of the reactions. Values of ‘k’ of two
reactions are given below. Find the order of each reaction.
i) k = 3 x 10-2 mol L-1 s-1
ii) k = 5 x 10-3 mol-1 L s-1 (1)
b) i) Write integrated rate equation for a first order reaction. (1)
ii) Write the relation between half-life (t½) and rate constant (k) of a first order reaction. `(1)
iii) Rate constant of a reaction is 5 x 10-2 s-1. Find the half-life (t½) of the reaction. (1) [SAY 2014]
Ans: a) i) Zero order ii) Second order
2.303 [𝑅]0
b) i) For first order reaction, the integrated rate equation is: k = log
𝑡 [𝑅]
0.693
ii) k=
𝑡1
2
iii) Here k = 5 x 10-2 s-1.
So 𝑡1 = 0.693/(5x10-2) = 13.86 s
2
13. The terms order and molecularity are common in chemical kinetics.
a) What do you mean by order and molecularity? (2)
b) i) Write two factors influencing rate of a reaction. (1)
ii) Write Arrhenius equation. (1) [March 2015]
Ans: a) Refer the answer of the Question no. 3
b) i) Concentration of the reactants and temperature.
−𝐸𝑎
ii) Arrhenius equation is k = A.𝑒 𝑅𝑇
Ea
Or, log k = log A –
2.303RT
14. Integrated rate expression for rate constant of a first order reaction R → P is given by
2.303 [𝑅]0
k= log . Derive an expression for half-life period of first order reaction. (2)
𝑡 [𝑅]
i) A first order reaction has a rate constant 1.15 x 10-3 s-1. How long will 5 g of the reactant take to
reduce 3g? (2) [SAY 2015]
Ans: i) Refer the answer of the Question no. 8 (ii)
2.303 [𝑅]0
ii) For a first order reaction, k = log
𝑡 [𝑅]
Here k = 1.15 x 10-3 s-1, [R]0 = 5g and [R] = 3g
2.303 [𝑅]0
So, t = log
𝑘 [𝑅]
2.303 5
= log
1.15 𝑥 10−3 3
= 440.5 s
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15. (i) The molecularity of the reaction 2NO + O2 → 2NO, is: a) 5 b) 2 c) 3 d) 0 (1)
(ii) What do you mean by rate of a reaction ? (1)
(iii) What will be the effect of temperature on rate of a reaction ? (1)
-14 -1
(iv) A first order reaction is found to have a rate constant, k = 5.5 x 10 s . Find out the half-life of the
reaction. (1) [March 2016]
Ans: i) 3
ii) It is the change in concentration of any one of the reactant or product in unit time.
iii) When temperature increases, rate of the reaction also increases.
iv) Here k = 5.5 x 10-14 s-1.
For a first order reaction, t½ = 0.693/k = 0.693/(5.5 x 10-14) = 1.26 x 1013 s
16. Rate of a reaction is the change in concentration of any one of the reactants or products in unit time.
a) Express the rate of the following reaction in terms of reactants and products
2NO(g) + O2(g) → 2NO2(g) (1)
b) (i) N2O5(g) → 2NO2(g) + ½ O2(g) is a first order reaction. Find the unit of k. (1)
(ii) Calculate the time required for the completion of 90% of a first order reaction. (k = 0.2303 s-1)
(2) [SAY 2016]
d[NO] d[O2 ] d[NO2 ]
Ans: a) 𝑟𝑖𝑛𝑠𝑡 = – ½ =– =½
dt dt dt
∆[NO] ∆[O2 ] ∆[NO2 ]
OR, 𝑟𝑎𝑣 = – ½ =– =½
∆t ∆t ∆t
b) (i) For a first order reaction, unit of k = s -1
2.303 [𝑅]0
(ii) For a first order reaction, k = log
𝑡 [𝑅]
For 90% completion, [R]0 = 100 and [R] = 100 – 90 = 10
2.303 [𝑅]0
So, t = log
𝑘 [𝑅]
2.303 100
𝑡90% = log
0.2303 10
= 10 s
17. a) Plot a graph showing variation in the concentration of reactants against time for a zero order
reaction. (1)
b) What do you mean by zero order reaction? (1)
c) The initial concentration of the first order reaction, N 2O5(g) → 2 NO2(g) + ½ O2(g), was 1.24 x 10-2
mol L-1 at 300 K. The concentration of N2O5 after 1 hour was 0.20 x 10-2 mol L-1. Calculate the rate
constant of the reaction at 300 K. (2) [March 2017]
Ans: (a)

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(b) If the order of a reaction is zero, it is called zero order reaction. Or, these are reactions in which
the rate of reaction is independent of concentration of the reactants.
2.303 [𝑅]0
(c) For a first order reaction, k = log
𝑡 [𝑅]
Here [R]0 = 1.24 x 10-2 mol/L, [R] = 0.2 x 10-2 mol/L, t = 1 hr = 60 min.
2.303 1.24 𝑥 10−2
k= log = = 0.0304 min-1
60 0.2 𝑥 10−2
18. The effect of temperature on rate of reaction is given by Arrhenius equation.
i) Write Arrhenius equation. (1)
ii) Define activation energy (Ea) (1)
iii) Rate constant k2 of a reaction at 310K is two times of its rate constant k1 at 300 K. Calculate
activation energy of the reaction. (log 2 = 0.3010 and log 1 = 0) (2) [SAY 2017]
−𝐸𝑎
Ans: i) Arrhenius equation is k = A.𝑒 𝑅𝑇
ii) It is the minimum amount of kinetic energy required for effective collision during a reaction.
k E T2 − T1
iii) We know that, log k2 = 2.303
a
R 1 T1 T2

Here T1 = 300 K, k1 = x, T2 = 310 K, k2 = 2x and R = 8.314 J K-1 mol-1

2x Ea 310 − 300
log =
x 2.303 x 8.314 300 𝑥 310
2.303 x 8.314 x 300 x 310 x log 2
So, 𝐸𝑎 = = 53598 J/mol
10
19. Identify the order of reaction if the unit of rate constant is mol L-l s-l. (1)
Ans: Zero order
20. For hydrolysis of methyl acetate in aqueous solution, the following results were observed.

t/s 0 30 60
CH3COOH
0.60 0.30 0.15
C/mol L-1
Show that it follows pseudo first order reaction as the concentration of water remains constant.
(3) [March 2018]

Ans: Here the concentration of water remains constant. So for being pseudo first order, the reaction
should be first order with respect to the concentration of ester (i.e. methyl acetate). The rate
constant for pseudo first order reaction is:
2.303 [𝑅]0
k= log where k = k’[H2O]
𝑡 [𝑅]
Here [R]0 = 0.6 mol/L.
When t = 30 s, [R] = 0.3 mol/L
2.303 0.6
So, 𝑘1 = log = 0.0231 s-1
30 0.3
When t = 60 s, [R] = 0.15 mol/L
2.303 0.6
So, 𝑘2 = log = 0.0231 s-1
60 0.15
Since k1 = k2, it is a pseudo first order reaction.

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21. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the
energy of activation of the reaction assuming that it does not change with temperature. (3)
k2 Ea T2 − T1
Ans: We know that, log =
k1 2.303 R T1 T2

Here T1 = 293 K, k1 = x, T2 = 313 K, k2 = 4x and R = 8.314 J K-1 mol-1

4x Ea 313 − 293
log =
x 2.303 x 8.314 293 x 313
2.303 x 8.314 x 293 x 313 x log4
So, Ea = = 52854 J/mol
20
22. What is the order of a reaction, if its half-life is independent of initial concentration? (1) [SAY 2018]
Ans: First order.
23. For the reaction, 2NO(g) + O2(g) → 2NO2(g), the rate law is given as, Rate = k[NO]2 [O2]. The order of
the reaction with respect to O2 is …………. (1)
Ans: With respect to O2, the order of the reaction = 1
24. Examine the graph given below. Identify the integrated rate equation and the order of the reaction
corresponding to it.

(2)

Ans: It’s a zero order reaction.

[𝑅]0 −[𝑅]
Its integrated rate equation is k =
𝑡
25. The rate constant of a reaction at 293K is 1.7 x 105 s-1. When the temperature is increased by 20 K, the
rate constant is increased to 2.57 x 106 s-1. Calculate Ea and A of the reaction. (3) [March 2019]

k2 Ea T2 − T1
Ans: We know that, log =
k1 2.303 R T1 T2

Here T1 = 293K, k1 = 1.7 x 105s-1, T2 = 293 + 20 =313 K, k2 = 2.57 x 106 s-1 and R = 8.314 J K-1 mol-1
2.57 x 106 Ea 313 − 293
log =
1.7 x 105 2.303 x 8.314 293 x 313
1.179 x 2.303 x 8.314 x 313 x 293
Ea = = 103514 J mol-1 = 103.514 kJ mol-1
20

−𝐸𝑎
Also, from Arrhenius equation, k = A.𝑒 𝑅𝑇
𝑘 1.7 𝑥 105
A = −𝐸𝑎 = − 103514 = 4.8 x 1023
𝑒 8.314 𝑥 293
𝑒 𝑅𝑇
OR,
𝐸𝑎
From logarithmic form of Arrhenius equation, log k = log A –
2.303 𝑅𝑇
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𝐸𝑎
log A = log k +
2.303 𝑅𝑇
103514
= log (0.01) + = 23.68
2.303 𝑥 8.314 𝑥 293
A = Anti-log (23.68)
= 4.8 x 1023
26. Differentiate molecularity and order of a reaction. (2)
Ans: Refer the answer of the Question no. 11 (a)
27. Write the Arrhenius equation and identify the terms in it. (2) [SAY 2019]
Ans: Arrhenius equation is k = A.e -Ea/RT

Where k – Rate constant of the reaction, A – Arrhenius factor, Ea – Energy of activation, R –

Universal gas constant and T – absolute temperature.
28. For a reaction A + B → C + D, the rate equation is r = K [A]3/2[B]1/2. Give the overall order and
molecularity of the reaction. (2)
Ans: Overall Order = 2
Molecularity = 2
29. The temperature dependence of the rate of a chemical reaction can be explained by Arrhenius
equation.
a) Give Arrhenius equation. (1)
b) The rate of a chemical reaction doubles for an increase of 10K in absolute temperature from 300K.
Calculate the activation energy (Ea)? [R = 8.314 J/K/mol and log 2 = 0.3010] (3) [March 2020]

Ans: a) The Arrhenius equation is k = A.e-Ea/RT

𝐸𝑎
Or, log k = log A –
2.303 𝑅𝑇

k2 Ea T2 − T1
b) We know that, log =
k1 2.303 R T1 T2
Here T1 = 300K, T2 = 300 + 10 =310 K, and R = 8.314 J K-1 mol-1
Suppose k1 = x, then k2 = 2x
Then,
2x Ea 310 − 300
log =
x 2.303 x 8.314 300 x 310
0.3010 x 2.303 x 8.314 x 300 x 310
Ea = = 53598.6 J mol-1 = 53.599 kJ mol-1
10
30. (a) Define molecularity. (1)
(b) Give an example of pseudo first order reaction. (1)
Ans: (a) Molecularity of a reaction is the total number of reacting species collides simultaneously in a
chemical reaction.
(b) Hydrolysis of ester or inversion of cane sugar
31. (a) Write a relation which connects rate constant with temperature. (1)
(b) The rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to
400 K. Calculate the activation energy for the reaction. (R = 8.314 J/K/mol ) (2) [SAY 2020]
−𝐸𝑎
Ans: (a) k = A.𝑒 𝑅𝑇 (The equation is known as Arrhenius equation)
k E T2 − T1
(b) We know that, log k2 = 2.303
a
R 1 T1 T2

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Here T1 = 350 K, T2 = 400 K, and R = 8.314 J K-1 mol-1

Suppose k1 = x, then k2 = 6x
On substituting in the equation, we get:
6x Ea 400 − 350
log =
x 2.303 x 8.314 350 x 400
log 6 x 2.303 x 8.314 x 350 x 400
Ea = = 41715 J/mol = 41.715 kJ mol-1
50
32. What is a zero order reaction? Give the unit of rate constant for zero order reaction. (2)
Ans: Zero order reaction means the rate of the reaction is independent of the concentration of the
reactants. Or, if the order of a reaction is zero, it is called zero order reaction. The unit of rate
constant for a zero order reaction is mol L-1 s-1.
33. The integrated rate equation for a first order reaction is:
k = 2.303 log [R]0
t [R]
(i) What is half-life period? (1)
(ii) Derive an expression for the half-life period of a first order reaction. (2)
0.693
Ans: (i) t½ =
𝑘
(ii) Refer the answer of question no. 8 (ii).
34. (i) Write Arrhenius equation. (1)
(ii) The rate of a reaction doubles when the temperature is increased from 298K to 308K. Calculate the
activation energy. (2)
(iii) Give two differences between order and molecularity. (1) [March 2021]
−𝐸𝑎
Ans: (i) The Arrhenius equation is k = A.𝑒 𝑅𝑇
𝐸𝑎
Or, logk = logA –
2.303 𝑅𝑇

k2 Ea T2 − T1
(ii) We know that, log =
k1 2.303 R T1 T2
Here T1 = 298 K, T2 = 308 K, and R = 8.314 J K-1 mol-1
Suppose k1 = x, then k2 = 2x
Then,
2x Ea 308 − 298
log =
x 2.303 x 8.314 298 x 308
0.3010 x 2.303 x 8.314 x 298 x 308
Ea = = 52897.7 J mol-1 = 52.898 kJ mol-1
10
(iii) Refer the answer of the Question no. 11 (a) ii)
35. Differentiate order and molecularity of a reaction. (2)
Ans: Refer the answer of the Question no. 11 (a) ii)
36. Derive an equation for the half-life of a first order reaction. (3)
Refer the answer of the Question no. 33 (ii)
H+
37. Consider the pseudo order reaction, CH3COOC2H5 + H2O → CH3COOH + C2H5OH
(i) Identify the order and molecularity of the above reaction. (2)
(ii) Give the rate law expression of the above reaction. (1)

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(iii) What happens to the rate of the above reaction when the concentration of CH3COOC2H5 is doubled? (1)
[SAY 2021]
Ans: (i) Order = 1 and Molecularity = 2
(ii) r = k[CH3COOC2H5]
(iii) The rate of reaction also gets doubled.
38. The unit of rate constant of a first order chemical reaction is:
(a) mol–1Ls–1 (b) s–1 (c) molL–1s–1 (d) mol–2L2s–1 (1)
Ans: (b) s-1
39. (i) Write Arrhenius equation. (1)
(ii) How will you obtain the value of activation energy (Ea) from a graphical plot using Arrhenius equation? (1)
Ans: (i) k = A.e-Ea/RT
(ii) We can obtain the value of Ea by plotting a graph between log k against 1/T. From the slop of
this graph we can calculate activation energy as:
− 𝐸𝑎
Slope =
2.303 𝑅
So, Activation energy, Ea = – slope x 2.303 x R
40. (i) Mention any two factors which influence the rate of a chemical reaction. (1)
(ii) Derive an expression for half-life of a first order reaction from its integrated rate equation. (2) [March 2022]
Ans: (i) Nature of reactants, Concentration of reactants, Temperature, Pressure, Effect of catalyst
and influence of radiation [Any 2 factors are required].
(ii) Refer the answer of the Question no. 33 (ii)
41. The unit of rate constant of zero order reaction is _______. (1)
Ans: mol L-1 S-1
42. (i) Define half-life period of a reaction. (1)
(ii) How half-life period of a first order reaction is related to the rate constant of the reaction? (1)
Ans: (i) It is the time taken to reduce the concentration of reactants to half of its initial
concentration.
0.693
(ii) 𝑡1 =
2 𝑘
Or, for a first order reaction, half-life period is inversely proportional to the rate constant.
43. (i) Give Arrhenius equation. Explain the terms in it. (2)
(ii) What is the significance of Arrhenius equation? (1) [SAY 2022]
−𝐸𝑎
Ans: (i) The Arrhenius equation is k = A.𝑒 𝑅𝑇
𝐸𝑎
Or, logk = logA –
2.303 𝑅𝑇
Where k is the rate constant, A is the Arrhenius factor or pre-exponential factor, Ea is the activation
energy, R is the Universal gas constant and T is the absolute temperature.
(ii) This reaction gives the temperature dependence of rate of a reaction.
44. The unit of rate constant of a zero order chemical reaction is …………… (1)
Ans: mol L-1 s-1 OR, M s-1
45. (i) What are pseudo first order reactions? (1)
(ii) Write one example for pseudo first order reaction. (1)
Ans: (i) These are reactions which appears to follow higher order, but actually follows first order.
(ii) Hydrolysis of ester OR, Inversion of cane sugar
46. (i) A first order reaction is found to have a rate constant, k = 6.8 x 1014 s. Find the half-life of the reaction. (1)
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(ii) Write the integrated rate equation for a first order reaction and explain the terms in it. (2)
Ans: (i) Here k = 6.8 x 10-14 s-1
0.693 0.693
𝑡1 = = = 1.019 x 1013 s
2 𝑘 6.8 𝑥 10−14
(ii) Integrated rate equation for a first order reaction is:
2.303 [𝑅]0
k= log
𝑡 [𝑅]
Where k – rate constant, t- time taken, [R]0 – Initial concentration of the reactant and [R] is the
concentration of the reactant at time ‘t’.
47. Explain the effect of temperature and catalyst on the rate of chemical reaction. (3) [March 2023]
Ans: When temperature increases, the fraction of molecules with energy equal to or greater than
activation energy increases. So, the rate of reaction also increases. A catalyst increases the rate of
a reaction by providing a new path for the reaction with low activation energy.
48. Give an example for a Pseudo first order reaction. (1)
Ans: Hydrolysis of ester OR, Inversion of cane sugar
49. Calculate the half-life period of a first order reaction whose rate constant is 200 s–1 . (2)
Ans: Here k = 200 s-1
For a first order reaction,
0.693 0.693
𝑡1 = = = 0.0035 s
2 𝑘 200
50. Write three differences between order and molecularity. (3) [SAY 2023]
Ans: Refer the Answer of the Question number 11 (a) (ii).
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