Test Level 2: Chemical Equilibrium


30 Questions MCQ Test Chemistry for JEE | Test Level 2: Chemical Equilibrium


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This mock test of Test Level 2: Chemical Equilibrium for Class 11 helps you for every Class 11 entrance exam. This contains 30 Multiple Choice Questions for Class 11 Test Level 2: Chemical Equilibrium (mcq) to study with solutions a complete question bank. The solved questions answers in this Test Level 2: Chemical Equilibrium quiz give you a good mix of easy questions and tough questions. Class 11 students definitely take this Test Level 2: Chemical Equilibrium exercise for a better result in the exam. You can find other Test Level 2: Chemical Equilibrium extra questions, long questions & short questions for Class 11 on EduRev as well by searching above.
QUESTION: 1

For the reaction equilibrium :

N2O4 (g)  2NO2(g) ; the concentration of N2O4 and NO2 at equilibrium are 4.8 × 10-2 and 1.2 × 10-2 mol/L respectively. The value of Kc for the reaction is :

Solution:


conc. of N2O4 = 4.8 × 10–2 mol/lit
NO2 = 1.2 × 10–2 mol/lit

QUESTION: 2

What is the equilibrium constant for the reaction P4(s) + 5O2(g)  P4O10(s) :

Solution:


QUESTION: 3

The equilibrium constant for the reaction :

N2(g) + O2(g)  2NO(g) at temperature

T is 4 × 10-4. The value of Kc for the reaction.

NO(g)  N2(g) + O2(g) at the same temperature is :

Solution:


QUESTION: 4

The equilibrium constant for the given reaction :

SO3(g)  SO2(g) + O2(g); Kc = 4.9 × 10-2

The value of Kc for the reaction :

2SO2(g) + O2(g)  2SO3(g), will be

Solution:


QUESTION: 5

For the following three reactions 1, 2 and 3, equilibrium constants are given :

(1) CO(g) + H2O(g)CO2(g)+H2(g) ; K1 

(2) CH4(g)+H2O(g)CO(g)+3H2(g) ; K2

(3) CH4(g)+2H2O(g)CO2(g)+4H2(g) ; K3

Which of the following relations is correct ?

Solution:

CO(g) + H2O(g) ⇋  CO2(g)+H2(g) ; K1  --------(i)
CH4(g)+H2O(g)  ⇋ CO(g)+3H2(g) ; K2  --------(ii)
CH4(g)+2H2O(g)  ⇋ CO2(g)+4H2(g) ; K3 ------(iii)
We can see that by adding eqn i and ii, we get eqn iii. So for K, on addition of reaction, the k value gets multiplied.
Therefore, k3 = k1k2

QUESTION: 6

Consider following reactions in equilbrium with equilibrium concentration 0.01M of every species

(I) PCl5(g)  PCl3(g) + Cl2(g)

(II) 2HI(g)  H2(g) + I2(g)

(III) N2(g) + 3H2 (g)  2NH3(g)

Extent of the reactions taking place is :

Solution:




QUESTION: 7

A definite amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. NH4HS decomposes to give NH3 and H2S and at equilibrium total pressure in flask is 0.84 atm. The equilibrium constant for the reaction is :

Solution:



QUESTION: 8

For the reaction 3A(g) + B(g)  2C(g) at a given temperature, Kc = 9.0. What must be the volume of the flask, if a mixture of 2.0 mol each of A, B and C exist in equilibrium ?

Solution:



QUESTION: 9

Sulfide ion in alkaline solution reacts with solid sulfur to form polysulfide ions having formulas S22-, S32-, S42- and so on. The equilibrium constant for the formation of S22- is 12 (K1) & for the formation of S32- is 132 (K2), both from S and S2-. What is the equilibrium constant for the formation of S32- from S22- and S

Solution:


QUESTION: 10

1 mole N2 and 3 mol H2 are placed in a closed container at a pressure of 4 atm. The pressure falls to 3 atm at the same temperature when the following equilibrium is attained.

N2(g) + 3H2(g)  2NH3(g). The equilibrium constant Kp for dissociation of NH3 is :

Solution:





 

QUESTION: 11

One mole of N2O4(g) at 300 K is left in a closed container under one atm. It is heated to 600 K when 20% by mass of N2O4 (g) decomposes to NO2(g). The resultant pressure is :

Solution:


QUESTION: 12

For the following gases equilibrium. N2O4(g)  2NO2(g) Kp is found to be equal to Kc. This is attained when temperature is

Solution:


QUESTION: 13

For the reaction :

CO(g) + O2(g)  CO2(g), Kp/Kc is :

Solution:



QUESTION: 14

For the reaction :

2NO2(g)  2NO(g) + O2(g) Kc = 1.8 × 10-6 at 184°C and R = 0.083 JK-1mol-1. When Kp and Kc are compared at 184°C, it is found that :

Solution:



QUESTION: 15

PCl5 dissociation a closed container as :

PCl5(g)  PCl3(g) + Cl2(g)

If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is α, the partial pressure of PCl3 will be :

Solution:

PCl5 dissociation a closed container

QUESTION: 16

For the reaction : 2HI (g)  H2(g) + I2(g), the degree of dissociated (α) of HI(g) is related to equilibrium constant Kp by the expression :

Solution:





QUESTION: 17

The equilibrium constant for the reaction

A(g) + 2B(g)  C(g) is 0.25 dm6 mol-2. In a volume of 5 dm3, what amount of A must be mixed with 4 mol of B to yield 1 mol of C at equilibrium

Solution:



QUESTION: 18

For the reaction

A(g) + 2B(g)  C(g) + D(g); Kc = 1012 If the initial moles of A, B, C and D are 0.5, 1, 0.5 and 3.5 moles respectively in a one litre vessel. What is the equilibrium concentration of B ?

Solution:





QUESTION: 19

The equilibrium constant Kc for the reaction,

A(g) + 2B(g)  3C(g) is 2 × 10-3

What would be the equilibrium partial pressure of gas C if initial pressure of gas A & B are 1 & 2 atm respectively.

Solution:





QUESTION: 20

A 20.0 litre vessel initially contains 0.50 mole each of H2 and I2 gases. These substances react and finally reach an equilibrium condition. Calculate the equilibrium concentration of HI if Keq = 49 for the reaction H2 + I2  2HI.

Solution:




QUESTION: 21

A vessel of 250 litre was filled with 0.01 mole of Sb2S3 and 0.01 mole of H2 to attain the equilibrium at 440°c as

Sb2S3(s) + 3H2(g)  2Sb(s) + 3H2S(g).

After equilibrium the H2S formed was analysed by dissolving it in water and treating with excess of Pb2+ to give. 1.195 g of PbS (Molecular weight = 239) precipitate.

What is value of Kc of the reaction at 440°C ?

Solution:



QUESTION: 22

The equilibrium constant for the reaction

CO(g) + H2O(g)  CO2(g) + H2(g) is 3 at 500 K. In a 2 litre vessel 60 gm of water gas [equimolar mixture of CO(g) and H2(g)] and 90 gm of steam is initially taken.

What is the equilibrium concentration of H2(g) at equilibrium (mole/L) ?

Solution:







QUESTION: 23

At 87°C, the following equilibrium is established

H2(g) + S(s)  H2S(g) Kp = 7 × 10-2

If 0.50 mole of hydrogen and 1.0 mole of sulfur are heated to 87°C in 1.0 L vessel, what will be the partial pressure of H2S at equilibrium ?

Solution:





QUESTION: 24

At certain temperature (T) for the gas phase reaction

2H2O(g)+2Cl2(g) 4HCl(g)+O2(g)       ; Kp =12×108 atm

If Cl2, HCl & O2 are mixed in such a manner that the partial pressure of each is 2 atm and the mixture is brought into contact with excess of liquid water. What would be approximate partial pressure of Cl2 when equilibrium is attained at temperature (T) ?

[Given : Vapour pressure of water is 380 mm Hg at temperature (T)]

Solution:




QUESTION: 25

At 675 K, H2(g) and CO2(g) react to form CO(g) and H2O(g), Kp for the reaction is 0.16. If a mixture of 0.25 mole of H2(g) and 0.25 mol of CO2 is heated at 675 K, mole% of CO(g) in equilibrium mixture is :

Solution:





QUESTION: 26

In which of the following reactions, increase in the pressure at constant temperature does not affect the moles at equilibrium.

Solution:

QUESTION: 27

 Change in volume of the system does not alter the number of moles in which of the following equilibrium

Solution:

QUESTION: 28

The conditions favourable for the reaction :

2SO2(g)+O2(g)  2SO3(g) ; ΔH° = -198 kJ  are :

Solution:


Exothermic reaction favourable at low temperature Δng < 0
⇒ favourable at high pressure

QUESTION: 29

The exothermic formation of ClF3 is represented by the equation :

Cl2(g) + 3F2(g)  2ClF3(g)  ;  ΔH = -329 kJ

Which of the following will increase the quantity of ClF3 in an equilibrium mixture of Cl2, F2 and ClF3 :

Solution:


⇒ Exothermic low temperature
∴ Addition of reactant Δng < 0
⇒ Increase in pressure lower the volume.

QUESTION: 30

Densities of diamond and graphite are 3.5 and 2.3 gm/mL. respectively and for

C(diamond)  C (graphite) ; ΔH= -1.9 kJ/mole favourable conditions for formation of diamond are

Solution:


It is exothermic reaction so it favours low temperature density of graphite is less than diamondlow pessure.

⇒ but for formation of diamond Reverse condition high temperature & high pressure.