Chemical Kinetics Test

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Chemical Kinetics Test - 7

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Weekly Quiz Competition

Question 1
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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 2
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Direction : 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 3
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Direction : 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 4
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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 5
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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 6
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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^-2 min^-1

B
0.0012 atm^-1 min^-1

C
0.0096 atm^-2 min^-1

D
0.0096 atm^-1 min^-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 7
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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^-2 M^-2 h^-1
Question 8
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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^-2 m^-2 h^-1
Question 9
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Direction : When worked out will result in one integer from 0 to 9 (both inclusive).

Graph between logT₅₀ (T₅₀ is half-life period) and (a = initial concentration) for n th order reaction is of the type.

What is the value of n?

A
2

B
3

C
4

D
5

Solution

For nth order reactions,
Question 10
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The acid catalysed hydrolysis of an organic compound A at 30 °C has a time for half change of 100 min when carried out in a buffer solution at pH 5 and 10 min when carried out at pH 4. Half-life in both cases are independent of initial concentration of A If rate law is , then what is the value of (a + b)?

A
1

B
2

C
3

D
5

Solution

Half-life

T50 is independent of [A], then

(a-1) = 0

a = 1

T₁ = 100 min, [H⁺ ]₁ = 10^-ph = 10^-5 M

T_{2 =} 100 min, [H⁺ ]_{2 =} 10^-ph = 10^-4 M

b - 1 = 1

b = 2

∴ a + b = 3
Question 11
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For a reaction, A →B, half-life time for a reaction at [A] = 0.1M is 200 s and at [A] = 0.4M it is 50 s. Thus, order of the reaction i s ......

A
1

B
4

C
5

D
2

Solution
Question 12
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For the reaction,

A + B → Product

Following four experiments were carried out:

Derive the value of y.

A
7

B
6

C
5

D
2

Solution