Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 7.1:

A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

a) What is the initial effect of the change on vapour pressure?

b) How do rates of evaporation and condensation change initially?

c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

Answer:

(a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.

(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.

(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

Question 7.2:

What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO₂]= 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M ?

Answer:

The equilibrium constant (Kc) for the give reaction is:

Hence, Kcfor the equilibrium is .

Question 7.3:

At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms

Calculate Kpfor the equilibrium.

Answer:

Partial pressure of I atoms,

Partial pressure of I₂ molecules,

Now, for the given reaction,

Question 7.4:

Write the expression for the equilibrium constant, Kc for each of the following

reactions:

(i) 

(ii) 

(iii) 

(iv) 

(v) 

Answer:

Question 7.5:

Find out the value of Kc for each of the following equilibria from the value of Kp:

Answer:

The relation between Kp and Kc is given as:

Kp = Kc (RT)^Δn

(a) Here,

Δn = 3 – 2 = 1

R = 0.0831 barLmol^–1K^–1

T = 500 K

Kp = 1.8 × 10^–2

Now,

Kp = Kc (RT) ^Δn

(b) Here,

Δn = 2 – 1 = 1

R = 0.0831 barLmol^–1K^–1

T = 1073 K

Kp= 167

Now,

Kp = Kc (RT) ^Δn

Question 7.6:

For the following equilibrium, 

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?

Answer:

It is given that   for the forward reaction is

Then,  for the reverse reaction will be,