UNIT 8: Physical and Chemical Equilibrium - Online Test

Q1. For the reaction AB (g) A(g) + B(g), at equilibrium, AB is 20% dissociated at a total pressure of P, The equilibrium constant KP is related to the total pressure by the expression
Answer : Option A
Explaination / Solution:


Total no. of moles at equilibrium = 80 + 20 + 20 = 120



Q2. In which of the following equilibrium, KP and KC are not equal?
Answer : Option D
Explaination / Solution:

For reaction given in option (a), (b) & (c) Δng = 0

For option (d) Δng = 2 – 1 = 1

KP = KC (RT)


Q3.

If x is the fraction of PCl5 dissociated at equilibrium in the reaction

PCl5 PCl3 + Cl2

then starting with 0.5 mole of PCl5, the total number of moles of reactants and products at equilibrium is
Answer : Option B
Explaination / Solution:


Total no. of moles at equilibrium = 0.5–x + x + x

= 0.5 + x

Q4.

The values of K P1 and KP2 for the reactions

X Y + Z

A 2B are in the ratio 9 : 1 if degree of dissociation and initial concentration of X and A be equal then total pressure at equilibrium P1, and P2 are in the ratio

Answer : Option A
Explaination / Solution:



Q5.

In the reaction,

Fe (OH)3 (s) Fe3+(aq) + 3OH(aq),

if the concentration of OH ions is decreased by ¼ times, then the equilibrium concentration of Fe3+ will

Answer : Option D
Explaination / Solution:


To maintain KC as constant, concentration of Fe3+ will increase by 64 times.

Q6.

Consider the reaction where KP = 0.5 at a particular temperature

PCl5(g) PCl3 (g) + Cl2 (g)

if the three gases are mixed in a container so that the partial pressure of each gas is initially 1 atm, then which one of the following is true

Answer : Option C
Explaination / Solution:

KP = 0.5


Q > KP Reverse reaction is favoured ; i.e. more PCl5 will be produced. option (c)


Q7. Equimolar concentrations of H2 and I2 are heated to equilibrium in a 1 litre flask. What percentage of initial concentration of H2 has reacted at equilibrium if rate constant for both forward and reverse reactions are equal
Answer : Option A
Explaination / Solution:

V = 1L

H2 + I2 2HI

[H2]initial = [I2]initial = a

[H2]eq = [I2]eq = (a – x)

and [HI]eq = 2x


4x2 = (a – x)2

4x2 = a2 + x2 – 2ax

3x2 + 2ax – a2 = 0

x = –a & x = a/3

degree of dessociation = a/3× 100

= 33.33 %


Q8. In a chemical equilibrium, the rate constant for the forward reaction is 2.5 × 102 and the equilibrium constant is 50. The rate constant for the reverse reaction is,
Answer : Option B
Explaination / Solution:

Kf = 2.5 × 102

KC = 50

Kr = ?

Kc = Kf/Kr

50 = 2.5 x 102/ Kr

Kr = 5 


Q9. Which of the following is not a general characteristic of equilibrium involving physical process
Answer : Option C
Explaination / Solution:

correct statement : Physical processes occurs at the same rate at equilibrium

option (c) is incorrect statement


Q10. For the formation of Two moles of SO3(g) from SO2 and O2, the equilibrium constant is K1. The equilibrium constant for the dissociation of one mole of SO3 into SO2 and O2 is
Answer : Option C
Explaination / Solution: