Chemical Equilibrium Test

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Chemical Equilibrium Test - 47

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Weekly Quiz Competition

Question 1
4 / -1

The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1 M) is 1/100th of that of a strong acid (HX, 1 M) at 25^o C. The K_a HA is equal to

A
1 × 10^-4

B
1 × 10^-5

C
1 × 10^-6

D
1 × 10^-3

Solution
Question 2
4 / -1

For the following three reactions I, II and III, equilibrium constants are given
I. CO (g) + H₂O ⇌ CO₂ (g) + H₂ (g), K₁
II. CH₄ (g) + H₂O (g) ⇌ CO (g) + 3H₂ (g), K₂
III. CH₄ (g) + 2H₂O (g) ⇌ CO₂ (g) + 4H₂ (g), K₃
Which of the following relations is correct ?

A
K₁ √K₂ = K₃

B
K₂ K₃ = K₁

C
K₃ = K₁ K₂

D
K₃ K₂² = K₁²
Question 3
4 / -1

A
1 × 10^-10

B
0.5

C
5 × 10^-8

D
1 × 10^-5

Solution
Question 4
4 / -1

2.5 mL of 2/5 M weak monoacidic base (K_b =1 × 10^-12 at 25^oC) is titrated with 2/15 M HCl in water at 25^oC. The concentration of H⁺ at equivalence point is (K_w = 1 × 10^-14 at 25^o C )

A
3.7 × 10^-13M

B
3.2 × 10^-7M

C
3.2 × 10^-2M

D
2.7 × 10^-2M
Question 5
4 / -1

Look at the following reactions carried out in closed vessels:
(i) PCl₅ → PCl₃ + Cl₂
(ii) N₂ + O₂ → 2NO
(iii) 2KClO₃ → 2KCl + 3O₂
(iV) Fe³⁺ + SCN^– → Fe(SCN)²⁺
Which one of them is irreversible ?

A
(i)

B
(ii)

C
(iii)

D
(iv)
Question 6
4 / -1

According to law of Mass Action, for the reaction 2A + B → Products

A
Rate = k [A] [B]

B
Rate = k [A]² [B]

C
Rate = k [A] [B]²

D
Rate = k [A]^1/2 [B]
Question 7
4 / -1

If Δn = (n_p – n_r) is the change in the number of moles of gaseous substances during a reaction, then equilibrium constants K_p and K_c are related as

A
K_p = K_c(RT)^–Δn

B
K_c = K_p(RT)^–Δn

C
K_p = (RT)^Δn/ K_c

D
K_c – K_p = (RT)^Δn

Solution
Question 8
4 / -1

The relation between K_p and K_c for the reaction

A
K_p = K_c

B
K_p = K_c (RT)

C
K_p = K_c / (RT)

D
K_p = K_c / (RT)²
Question 9
4 / -1

The ratio of K_p /K_c for the reaction:

A
1

B
RT

C
(RT)^1/2

D
(RT)^–1/2

Solution
Question 10
4 / -1

For a homogeneous chemical reaction, K_p = k_c when

A
Δn = 0

B
Δn = 1

C
Δn = 2

D
Δn = ∞

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Chemical Equilibrium Test - 47

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Question 1

1 × 10-4

1 × 10-5

1 × 10-6

1 × 10-3

Solution

Question 2

K1 √K2 = K3

K2 K3 = K1

K3 = K1 K2

K3 K22 = K12

Question 3

1 × 10-10

0.5

5 × 10-8

1 × 10-5

Solution

Question 4

3.7 × 10-13M

3.2 × 10-7M

3.2 × 10-2M

2.7 × 10-2M

Question 5

(i)

(ii)

(iii)

(iv)

Question 6

Rate = k [A] [B]

Rate = k [A]2 [B]

Rate = k [A] [B]2

Rate = k [A]1/2 [B]

Question 7

Kp = Kc(RT)–Δn

Kc = Kp(RT)–Δn

Kp = (RT)Δn/ Kc

Kc – Kp = (RT)Δn

Solution

Question 8

Kp = Kc

Kp = Kc (RT)

Kp = Kc / (RT)

Kp = Kc / (RT)2

Question 9

1

RT

(RT)1/2

(RT)–1/2

Solution

Question 10

Δn = 0

Δn = 1

Δn = 2

Δn = ∞

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1 × 10^-4

1 × 10^-5

1 × 10^-6

1 × 10^-3

K₁ √K₂ = K₃

K₂ K₃ = K₁

K₃ = K₁ K₂

K₃ K₂² = K₁²

1 × 10^-10

5 × 10^-8

1 × 10^-5

3.7 × 10^-13M

3.2 × 10^-7M

3.2 × 10^-2M

2.7 × 10^-2M

Rate = k [A]² [B]

Rate = k [A] [B]²

Rate = k [A]^1/2 [B]

K_p = K_c(RT)^–Δn

K_c = K_p(RT)^–Δn

K_p = (RT)^Δn/ K_c

K_c – K_p = (RT)^Δn

K_p = K_c

K_p = K_c (RT)

K_p = K_c / (RT)

K_p = K_c / (RT)²

(RT)^1/2

(RT)^–1/2