Chemical Equilibrium Test

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Chemical Equilibrium Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

The equilibrium, is attained at 25^oC in a closed container and an inert gas He is introduced. Choose the correct statement.

A
Concentration of N₂, 3H₂ and 2NH₃ are changed.

B
More N₂ is formed

C
Concentration of NH₃ is reduced

D
Nothing happens

Solution

Since formation of NH₃ is carried out in a closed container. When Inert gas (He) is introduced to the system in equilibriumatconstant volume then no any change occurs in the partial pressures of N₂, 3H₂ and 2NH₃.
Question 2
1 / -0

A vessel at 1000 K contains CO₂with a pressure of 0.5 atm. Some of the CO₂ is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K_P is:

A
3atm

B
0.3atm

C
0.18atm

D
1.8 atm

Solution

CO₂(g) + C(s) ⇌ 2CO(g)

at t = 0 0.5 atm – 0

at eqb. (0.5 x)atm – 2x atm

As graphite is a pure solid, its concentration is invariable (or constant) during the course of reaction, it will not be a part of K_p 's expression. At equilibrium,

Total pressure, P_T = 0.8 atm

P_T = 0.5 − x + 2x = 0.8atm

0.5 + x = 0.8

x = 0.3 atm
Question 3
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40% of a mixture of 0.2 mol of N₂ and 0.6 mol of H₂ react to give NH₃ according to the equation, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

A
4:5

B
5:4

C
7:10

D
8:5

Solution
Question 4
1 / -0

One mole of ethanol is treated with one mole of ethanoic acid at 25°C. One-fourth of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be

A
1/9

B
4/9

C
9

D
9/4

Solution

C₂H₅OH + CH₃COOH ⇌ CH₃COOC₂H₅ + H₂O
Question 5
1 / -0

One mole of N₂O₄(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when N₂O₄ (g) 23 decomposes to NO₂(g). If the resultant pressure is 2.4 atm, the percentage dissociation by mass of N₂O₄ (g) is

A
10%

B
20%

C
30%

D
40%

Solution

N₂O₄(g) ⇌ 2NO₂(g)

Total pressure = 2 – 2α + 4α

= 2 + 2α

2 + 2α = 2.4 (Given)

α = 0.2

∴ Percentage dissociation = 20%
Question 6
1 / -0

K for the synthesis of HI(g) is 50. The degree of dissociation of HI is

A
0.10

B
0.14

C
0.18

D
0.22

Solution
Question 7
1 / -0

For the reaction PCl₃ (g) + Cl₂(g) ⇌ PCl₅ (g), the value of Kp at 250°C is 0.61 atm^–1. The value of K_c at this temperature will be

A
15 (mol/l)^–1

B
26 (mol/l)^–1

C
35 (mol/l)^–1

D
52 (mol/l)^–1

Solution
Question 8
1 / -0

In the reaction A₂(g) + 4B₂(g) ⇌ 2AB₄(g) , ∆H > 0. The decomposition of AB₄ (g) will be favoured at

A
low temperature and high pressure

B
high temperature and low pressure

C
low temperature and low pressure

D
high temperature and high pressure

Solution

2AB₄ (g) ⇌ A₂ (g) + 4B₂ (g) ∆H = –ve

It is an exothermic reaction and hence favoured at low temperature. ∆n for the reaction is +3. Therefore low pressure will favour the forward reaction
Question 9
1 / -0

In a system, A (s) ⇌ 2B (g) + 3C (g)

If the conc. of C at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration of B to change to

A
two times the original value

B
one half of its original value

C
2√2 times the original value

D
1/2√2 times the original value

Solution

A(s) ⇌ 2B(g) + 3C(g)

Let, x and y be the concentrations of B and C at equilibrium respectively.

∴ K_c = x²y³ …. (1)

Now, the concentration of C is changed from y to y′ such that y′ = 2y.

If x′ is the new concentration of B

∴ K_c = (x′)² (y′)³ = (x′)²(2y)³ …. (2)

From Eqs. (1) and (2)

(x′) ² (8y³ ) = x²y³

∴Equilibrium concentration of B changes to 1/2√2 times the original value.
Question 10
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For the following three reactions (i), (ii) and (iii), equilibrium constants are given

(i) CO (g) + H₂O (g) ⇌ 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

What is the relation between equilibrium constants of the three reactions?

A
K₁√K₂ = K₃

B
K₂K₃ = K₁

C
K₃ = K₁K₂

D

Solution

Reaction (iii) can be obtained by adding reactions (i) and (ii) therefore K₃ = K₁. K₂.