Chemical Kinetics Test

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Chemical Kinetics Test - 14

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Question 1
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Direction (Q. Nos. 1-21) This section contains 21 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q. For zeroth order reaction,
A → B
[A]₀ = 0.01 M, [A]_t = 0.008Matter 10 min.
Thus, half-life is

A
5 min

B
12 min

C
16 min

D
25 min

Solution

For zeroth order reaction
When reaction is 50% completed.
Question 2
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A Certain Zeroth Order reaction has k = 0.025 Ms^-1for the disappearance of A. What will be the concentration of A after 15s, if the intial concentration is 0.50 M?

A
0.50 M

B
0.375 M

C
0.125 M

D
0.060 M

Solution

x(product) formed after 15 s
= 0.025 Ms^-1 x 15s
= 0.375 M
Then, reactant left
= 0.500 - 0.375
= 0.125 M
Question 3
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For Zeroth Order Reaction, variation of x with time is shown as
Q. At initial concenration of the reactant as 17.32 min dm^-1, half - life period is

A
5.0 min

B
10.0 min

C
8.66 min

D
15.0 min

Solution
Question 4
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For the first order reaction, rate constant k is given by
kt = logC₀- log C_t
where, C₀ is the intial concenration and C_tis the concentration at time t. Graph between log C_tand time t is

A

B

C

D

Solution

This represents a straight line
Question 5
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In a first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M is 15 min. The time taken for the concentration to change from 0.1 M to 0.025 M is

A
30 min

B
15 min

C
7.5 min

D
60 min

Solution

For first order reaction,
time = 15 min to change its concentration from 0.8 m to 0.4 m Thus, 50% reaction.
Thus, t₅₀ = 15 min
Question 6
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For the First order reaction, concentration of the reactant after two averagelives is reduced to

A
25 %

B
75%

C

D

Solution

For first order reaction,
Question 7
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The half-life period of a first order chemical reaction is 6.93 min. The time required for the completion of 99% of the chemical reaction will be (log2 = 0.3010)
[AIEEE 2009]

A
230.3 min

B
23.03 min

C
46.06 min

D
460.6 min

Solution

∴
when, a = 100,
(a-x) = 100-99 = 1, t = t₉₉
Question 8
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For the first order reaction,
A → Product
Q. The concentration of A changes from 0.1 M to 0.025 M in 40 min. The rate of the reaction when the concentration of A is 0.01 M is
[AIEEE 2012]

A
1.73 x 10^-6 Mmin^-1

B
3.47 x 10^-4 M min^-1

C
3.47 x 10^-5 M min^-1

D
1.73 x 10^-6 M min^-1

Solution

For the first order kinetic,
Question 9
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For the reaction,
A → Products,
Concentration of A at different time intervals are given :
At 20 min, rate will be

A
12 M min^-1

B
10 M min^-1

C
8 M min^-1

D
0.4 M min^-1

Solution
Question 10
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For the first order reaction with, C₀ as the initial concentration and C at time t, (C₀ - C) =

A
C₀ [1-e^-kt]

B
C₀e^-^kt

C
C₀ [1+e^kt]

D
C₀e^kt

Solution
Question 11
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The radioisotope N-13 which has a half-life of 10 min is used to image organs in the body. If the injected sample has an activity of 40 μCi (microcurie), what is the activity after 30 min?

A
10μCi

B
20μCi

C
5 μCi

D
2.5 μCi

Solution

Disintegration follows first-order kinetics
t = total time = 30 min
t ₅₀ = half-life = 10 min
Question 12
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The following data were obtained during the first order decomposition of
2A(g) → B(g) + C(s)
at constant volume and at a particular temperature :
Thus, rate constant is

A
0.0693 min^-1

B
69.3 min^-1

C
6.93 min^-1

D
6.93 x 10^-4min^-1

Solution
Question 13
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A G .M. Counter is used to study the process of first order of radioactive disintegration. In the absence of radioactive substance A, it counts 3 dps (disintegration per second). When A is placed in G.M Counter it records 23 dps at the start and 13 dps after 10 min. It records x dps after next 10 min. Half-life period of A is y min. Thus, x and y are

A
8 dps, 10 min

B
5 dps, 10 min

C
5 dps, 20 min

D
5 dps, 5 min

Solution

When there is no radioactive substance in G.M. Counter, it records = 3 dps. Thus, it is the zero-error in the counter and in each it is added. Thus, initial counts = 23 - 3 = 20 dps.
Attimef (after 10min) = 13- 3 = 10dps
Thus, half-life is 10 min = y
Hence, after next 10 min, dps = 5
Hence, reading in G.M. Counter = 5 + 3 = 8 dps = x
Question 14
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Inversion of sucrose (C₁₂H₂₂O₁₁) is a first order reaction and is studied by measuring angle of rotation at different interval of time
r₀ = angle of rotation at the start, r_t = angle of rotation at time t
r_∞ = angle of rotation at the complete reaction
There is 50% inversion, when

A
r₀ = 2r_t - r_∞

B
r₀ = r_t - r_∞

C
r₀ = r_t - 2r_∞

D
r₀ = r_t + r_∞

Solution

r₀ = rotation at the start, r_t = rotation at time t
r_∞ = rotation at the end then (r_∞ - r₀) = a and (r_∞ - r_t) = (a - x)
When 50% inversion has taken place

∴ 2 (a - x) = a
2(r_∞ - r_t) = r_∞ - r₀
2r_∞ - 2r_t = r_∞ - r₀
∴ r₀ = (2r_t - r_∞)
Question 15
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Following radioactive disintegration follows first order kinetics :
Starting with 1 mol of A in a 1L flask at 300 K, pressure set up after two half-lifes is approximately

A
25 atm

B
12 atm

C
37 atm

D
40 atm

Solution

When B is formed from A, two a-particies (He) are formed.
A taken initially = 1 mol

This helium remains in closed vessel.
Thus, pressure set up is due to helium.
Question 16
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For the first order reaction T_av (average life), T₅₀ (half-life) and T₇₅ (time 75% reaction) are in the ratio of

A
1 : 2 : 3

B
1.44 : 2 : 3

C
1.44 : 1 : 2

D
1 : 2 : 1.44

Solution
Question 17
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For the first order reaction,

Following equations are given :

Select the correct equations.

A
I and II

B
II and III

C
I and III

D
I, II and III

Solution

For the first order reaction,

∴ (a-x) = a (1/2)^y(Equation II)
Question 18
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Rate constant of a first order reaction (A → B) is 0.0693 min^-1. If initial concentration is 1.0 M, rate after 30 min is

A
0.0693 M min^-1

B
8.66 x 10^-3 M min^-1

C
0.0233 M min^-1

D
0.0466 M min^-1

Solution

C₀ = 1 M

∴
Question 19
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Following reaction is pseudo-unimolecular w.r.t. C₆H₅N₂CI (A)
50 mL of 1M benzene diazonium chloride (A) is taken. After 1 h, 1.226 L of N₂ gas at 1 atm and 300 K is obtained. Thus, half-life of the reaction is (log 250 = 2.40)

A
7.53 min

B
0.092 min

C
10.86 min

D
None of these

Solution

Half-life(t₅₀)=
Question 20
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Minimum half-life of an isotope needed so that not more than 0.1 % of the nuclei undergo decay during a 3.0 h laboratory experiment is

A
72 days

B
80 days

C
80.7 days

D
86.5 days

Solution
Question 21
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Naturally occurring potassium consists of 0.01% at 40 K which has a half-life of 1.28 x 10⁹ yr. Activity of 1.00 g sample of KCl is

A
8 x 10¹⁷ yr^-1

B
4 x 10⁸ yr^-1

C
4.37 x 10⁸ yr^-1

D
2.286 x 10^-9yr^-1

Solution

Number of 40 K atoms in 1g KCI