Chemical Kinetics Test

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Chemical Kinetics Test - 11

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Question 1
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Direction (Q. Nos. 1-13) This section contains 13 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
For a given reaction,
A → Product,
rate = 1 x 10^-4 Ms^-1 at [A] = 0.01 M
rate = 1.41 x 10^-4Ms^-1 at [A] = 0.02 M
Hence, rate law is

A

B

C

D

Solution

Let rate law be
1 x 10^-4= K(0.01)ⁿ
1.41 x 10^-4= K(0.02)ⁿ
∴ 1.41 = (2)ⁿ
(2)^1/2 = (2)ⁿ(Since, √2 = 1.41)
∴ n= 1/2
Thus rate law is
Question 2
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The rate law for a reaction between the substances A and 8 is given by
rate = k[A]ⁿ [B]^m
If concentration of A is doubled and that of B is halved, the new rate as compared to the earlier rate would be

A

B
(m + n)

C
(n - m)

D
2^(n-m)

Solution
Question 3
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The rate equation for the reaction,
2A + B → C
is found to be, rate = k[A] [B]
Q. The correct statement in relation to this reaction is that the

A
unit of k must be s^-1

B
t_1/2 is constant

C
rate of formation of C is half of the rate of disappearance of A

D
value of k is independent of the initial concentration of A and 8

Solution

Rate = k (A)[B]
The given reaction is first order in A and first order is B.
Thus, total order = 2
(a) Unit of k = cone ^{1 - n}time ^-1 = conc^-1 time^-1 Thus, (a) is false.
(b) of second-order reaction, thus (b) is false.
Thus, (c) is correct.

Thus, value of k is dependent on the concentration of A and B. Thus, (d) is false.
Question 4
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For the reaction,
2 A+B → Product
The half-life period was independent of concentration of B. On doubling the concentration A, rate increases two times. Thus, unit of rate constant for this reaction is

A
L mol^-1s^-1

B
mol L^-1s^-1

C
unitless

D
S^-1

Solution

t₅₀ is independent of concentration of B then w.r.t. B, (n - 1) = 0. Thus, n = 1
Thus, order w.r.t. B = 1
Also, if concentration is made m-times, rate becomes, mⁿ-times.
Hence, w.r.t.A
m = 2, mⁿ= 2
∴ 2ⁿ = 2 = 2¹
Thus n = 1
Thus, order w.r.t. A = 1
Thus, total order = 2 = n
Unit of n th order reaction = conc^(1-n) time^-1
conc in mol L^-1 and time in seconds.
Then,
Question 5
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Which of the following statements appears to the first order reaction?
2N₂O₅(CCl₄) → 4NO₂(CCl₄) + O₂(g)

A
I, II and V

B
I, II and VI

C
II, V and VI

D
I, II and III

Solution

(a)
I, II, V - represent first order equation
III, VI - zeroth order equation
IV - second order equation
Question 6
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For a reaction, time of 75% reaction is thrice of time of 50% reaction. Thus, order of the reaction is

A
0

B
1

C
3

D
2

Solution

For n^th order reaction (n > 1)

This equation is true, if n = 2
Question 7
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For nth order reaction,
Graphs between log (rate) and log[A₀] are of the type ([A₀] is the initial concentration)
Lines P, Q, R and S are for the order

A

B

C

D

Solution

n = order of the reaction =slope of the line
Larger the value of the slope of the line, larger the order.
Thus, P = 3, Q = 2, R = 1, S = 0.
Question 8
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nA → Product
For the reaction, rate constant and rate of the reaction are equal, then on doubling the concentration of A, rate becomes,

A
doubled

B
four times

C
halved

D
constant

Solution

Thus, order = zero Rate is independent of concentration of A.
Thus, rate remains constant.
Question 9
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For the simple reaction,
A → B
When [A] was changed from 0.502 mol dm^-3 to 1.004 mol dm^-3 half-life dropped from 52 s to 26 s at 300 K. Thus, order of the reaction is

A
0

B
1

C
2

D
3

Solution

For nth order reaction,
(T₅₀ ) is t_1 ,at [A] = a₁and is t₂ at [A] = a₂
∴
(2) = (2)^n-1
∴ n - 1 = 1, n = 2
Question 10
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For the following reaction,
Variation of T50 with [A] is shown
Q.
After 10 min volume of N₂ (g) is 10 L and after complete reaction, volume of N₂ (g) is 50 L. Thus, T₅₀ is

A
10 min

B
20 min

C
31 min

D
41 min

Solution

T₅₀ is independent of concentration of [A], Thus, it follows first order kinetics.
Question 11
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Kinetics of the following reaction,
can be studied by

A
measurement of pH

B
titration with hypo after adding Kl

C
titration with AgNO₃ solution

D
All of the above

Solution

Cl-atom attached to N-atom, is an oxidising agent and can oxidise Kl to l₂.

l₂ can be titrated using hypo in iodometric titration.

Cl-atom attached to benzene nucleus is not an oxidising agent.
Question 12
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Graph between [B] and time t of the reaction of the type I for the reaction, A → B
Hence, graph between and time t will be of the type

A

B

C

D

Solution

A → B

[B] increases uniformly with time t. Order is zero. Thus, at every point, = constant
Question 13
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If in the fermentation of sugar in an enzymatic solution that is 0.12 M, the concentration of sugar is reduced to 0.06 M in 10 h and to 0.03 M in 20 h. Thus, order of the reaction is

A
3

B
2

C
1

D
0

Solution

We find that, t₇₅ = 2 x t₅₀
Thus, given reaction is of first-order.
Question 14
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Direction (Q. No. 14) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct.

A
a

B
b

C
c

D
d

Solution

(i) Rate remains constant
(dx/dt) = k(A)ⁿ = [A]⁰ = k
Thus order = 0
Thus,(i)→(s)
(ii) Half-life period is independent of concentration
∴ (n-1) = 0
∴ n = 1
Thus, (ii) →(r)

Thus, graph is for second-order reactions
Thus, (iii) → (q)
Question 15
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Direction (Q. Nos. 15-20) This section contains 3 paragraphs, each describing theory, experiments, data, etc. Six questions related to the paragraphs have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).
Passage I
Peroxy acetyl nitrite is an air pollutant, as it decomposes into radicals :
A sample of polluted air is analysed for its PAN content which is reported as molecules per litre of air at 298 K.
Q.
Determine the order of the PAN decomposition

A
0

B
1

C
2

D
3

Solution

0 min 5.0 x 10¹⁴ molecules L^-1
30 min 2.5 x 10¹⁴ molecules L^-1
60 min 1.25 x 10¹⁴ molecules L^-1
At 30 min, 50% reaction takes place.
At 60 min, 75% reaction takes place.
Thus, T₇₅ = 2 x T₅₀
it is valid for first order reaction. Thus, order = 1
Question 16
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Passage I
Peroxy acetyl nitrite is an air pollutant, as it decomposes into radicals :
A sample of polluted air is analysed for its PAN content which is reported as molecules per litre of air at 298 K.
Q.
Rate constant of the reaction is

A
0.009 min^-1

B
0.0067 L mol^-1 min^-1

C
0.0231 min^-1

D
0.0231 L mol^-1 min^-1

Solution

Time : [PAN]
0 min 5.0 x 10¹⁴ molecules L^-1
30 min 2.5 x 10¹⁴ molecules L^-1
60 min 2.5 x 10¹⁴ molecules L^-1
At 30 min, 50% reaction takes place.
At 60 min, 75% reaction takes place.
Thus, T₇₅ = 2 x t₅₀
It is valid for first order reaction. Thus, order = 1
For first order reactions, T₅₀= 0.693/k
Question 17
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Passage II
H The reaction between nitric oxide (NO) and hydrogen (H₂)has been investigated by measuring the initial rate of decrease of pressure in known mixture of gases. The following results were obtained at 1000 K.
Q.
Overall order of the reaction is

A
1

B
2

C
3

D
0

Solution

-dp/dt = k(P_NO)^a (P_H2)^b
0.0012 = k(0.25)^a(0.2)^b
0.0024 = k(0.50)^a(0.1)^b
0.0048 = k(0.50)^a(0.2)^b
From Eqs. (i) and (iii), we get
4 = (2)^a
∴ (2)² = (2)^a
∴ a = 2
Thus w.r.t. No, order = 2
From Eqs.)ii) and (iii), we get
2 = (2)^b
(2)¹ = (2)^b
∴ b = 1
Thus, w.r.t. H₂, order = 1
Total order = 3
From Eq. (i),
0.0012 = k(0.25)²(0.2)¹= k (0.25)²(0.2)
∴ k = 0.0012/(0.25)2(0.02) = 0.096 atm^-2 min^-1
Question 18
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Passage II
H The reaction between nitric oxide (NO) and hydrogen (H₂)has been investigated by measuring the initial rate of decrease of pressure in known mixture of gases. The following results were obtained at 1000 K.
Q.
Rate constant of the overall reaction is

A
0.0048 atm^-2min^-1

B
0.0012 atm^-1min^-1

C
0.0096 atm^-2min^-1

D
0.0096 atm^-1min^-1

Solution

0.0012 = k(0.25)²(0.2)¹ = k(0.25)²(0.2)
∴ k = 0.0012/(0.25)²(0.2) = 0.096 atm^-2 min^-1
Question 19
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Passage III
A reaction between substances A and B is represented as:
A+B → C
Observations on the rate of this reaction are obtained as :
Q.
Order of the reaction w.r.t. A and B respectively are

A
1,1

B
1,2

C
2,1

D
2,2

Solution

where, a == order w.r.t. A, 6 = order w.r.t. B
(i) 5.0 x 10^-3 = k[0.1]^a[1.0]^b
(ii) 2.0 x 1^-2 = k [0.1]^a[2.0]^b
(iii) 2.5 x 10^-3= k [0.05]^a[1.0]^b
From Eqs. (i) and (ii), we get
Order w.r.t.A = 1
From Eq. (ii), 5.0x10^-3 = K(0.1)(1.0)²
∴ k = 5.0 x 10^-2M^-2h^-1
Question 20
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Passage III
A reaction between substances A and B is represented as:
A+B → C
Observations on the rate of this reaction are obtained as :
Q.
Rate constant of the overall reaction is

A
5.0 * 10^-2 M^-1 h^-1

B
5.0 * 10^-2 M^-2 h^-1

C
5.0 * 10^-2 h^-1

D
5.0 * 10^-2 mh^-1

Solution

where, a == order w.r.t. A, 6 = order w.r.t. B
(i) 5.0 x 10^-3= K [0.1]^a [1.0]^b
(ii) 2.0 x 10^-2= k [ 0.1]^a[2.0]^b
(iii) 2.5 x 10^-3= k [0.05]^a[1.0]^b
From Eqs. (i) and (ii), we get
(2)²= (2)^b
∴ b = 2
Order w.r.t.B = 2
From Eqs. (i) and (iii), we get
(2) = (2)^a
a = 1
Order w.r.t.A
From Eq. (ii),
5.0 x 10^-3= k (0.1) (1.0)²
∴ k = 5.0 x 10^-2m^-2h^-1