Topic: UNIT 8: Physical and Chemical Equilibrium (Test 1)



Topic: UNIT 8: Physical and Chemical Equilibrium
Q.1
If Kb and Kf for a reversible reactions are 0.8 ×10–5 and 1.6 × 10–4 respectively, the value of the equilibrium constant is,
A. 20
B. 0.2 × 10–1
C. 0.05
D. none of these
Answer : Option A
Explaination / Solution:

Kb = 0.8 × 10–5

Kf = 1.6 × 10–4

Keq = Kf / Kb = 1.6X10-4 / 0.8X10-5 = 20


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Q.2

At a given temperature and pressure, the equilibrium constant values for the equilibria


The relation between K1 and K2 is

A. K1 = 1/ √K2
B. K2 = (K1)-1/2
C. K12 = 2K2
D. K1/2 = K2
Answer : Option B
Explaination / Solution:



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Q.3
The equilibrium constant for a reaction at room temperature is K1 and that at 700 K is K2. If K1 > K2, then
A. The forward reaction is exothermic
B. The forward reaction is endothermic
C. The reaction does not attain equilibrium
D. The reverse reaction is exothermic
Answer : Option A
Explaination / Solution:

T1 = 25 + 273 = 298 K

T2 = 700 K


ΔHº is –ve ie., forward reaction is exothermic



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Q.4

The formation of ammonia from N2(g) and H2(g) is a reversible reaction

N2(g) + 3H2(g) 2NH3(g) + Heat

What is the effect of increase of temperature on this equilibrium reaction
A. equilibrium is unaltered
B. formation of ammonia is favoured
C. equilibrium is shifted to the left
D. reaction rate does not change
Answer : Option C
Explaination / Solution:

Increase in temperature, favours the endothermic reaction,

Given that formation of NH3 is exothermic ie., the reverse reaction is endothermic.

increase in temperature, shift the equilibrium to left option (c)


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Q.5
Solubility of carbon dioxide gas in cold water can be increased by
A. increase in pressure
B. decrease in pressure
C. increase in volume
D. none of these
Answer : Option A
Explaination / Solution:


increase in pressure, favours the forward reaction.

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Q.6
Which one of the following is incorrect statement ?
A. for a system at equilibrium, Q is always less than the equilibrium constant
B. equilibrium can be attained from either side of the reaction
C. presence of catalyst affects both the forward reaction and reverse reaction to the same extent
D. Equilibrium constant varied with temperature
Answer : Option A
Explaination / Solution:

option (a) : wrong statement

Correct statement is, for a system at equilibrium, Q = Keq


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Q.7

K1 and K2 are the equilibrium constants for the reactions respectively.


What is the equilibrium constant for the reaction NO2(g) ½N2(g) + O2(g)

A. 1/√k1√k2
B. (K1=K2)1/2
C. 1/2K1K2 
D. (1 / K1K2)1/2
Answer : Option A
Explaination / Solution:



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Q.8

In the equilibrium,

2A(g) 2B(g) + C2(g)

the equilibrium concentrations of A, B and C2 at 400 K are 1 × 10–4 M, 2.0 × 10–3 M, 1.5 × 10–4 M respectively. The value of Kc for the equilibrium at 400 K is

A. 0.06
B. 0.09
C. 0.62
D. 3 × 10–2
Answer : Option A
Explaination / Solution:

 [A] = 1 × 10–4 M ;

[B] = 2 × 10–3 M

[C] = 1.5 × 10–4 M

2A (g) 2B(g) + C2(g)



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Q.9
An equilibrium constant of 3.2 × 10–6 for a reaction means, the equilibrium is
A. largely towards forward direction
B. largely towards reverse direction
C. never established
D. none of these
Answer : Option B
Explaination / Solution:

Kc = [Products] / [Reactants]

3.2 x 10-6 = [Products] / [Reactants]

KC < 10-3 ; indicates

that [Reactant] >> [Product]

option (b) is correct,largely towards reverse direction.


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Q.10
KC/ KP for the reaction, N2(g) + 3H2(g) 2NH3(g) is
A. 1 / RT
B. √(RT)
C. RT
D. (RT)2
Answer : Option D
Explaination / Solution:

for the reaction,

N2(g) + 3H2(g) 2NH3(g) ;

Δng = 2 – 4 = – 2

KP = KC (RT)–2

Kc/Kp = (RT)2


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