Page No 228: - Chapter 7 Equilibrium class 11 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 7.31:

Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that  4.0 bar, what will be the partial pressure of H₂ at equilibrium? Kp= 10.1 at 400°C

Answer:

Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is:

It is 

Now,

Hence, at equilibrium, the partial pressure of H₂ will be 3.04 bar.

Question 7.32:

Predict which of the following reaction will have appreciable concentration of reactants and products:

a) 

b) 

c) 

Answer:

If the value of Kc lies between 10^–3 and 10³, a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products.

Question 7.33:

The value of Kc for the reaction

3O₂ (g)   2O₃ (g)

is 2.0 ×10^–50 at 25°C. If the equilibrium concentration of O₂ in air at 25°C is 1.6 ×10^–2, what is the concentration of O₃?

Answer:

The given reaction is:

Then, we have,

Hence, the concentration of

Question 7.34:

The reaction, CO(g) + 3H₂(g) CH₄(g) + H₂O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.

Answer:

Let the concentration of methane at equilibrium be x.

It is given that Kc = 3.90.

Therefore,

Hence, the concentration of CH₄ at equilibrium is 5.85 × 10^–2 M.

Question 7.35:

What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:

Answer:

A conjugate acid-base pair is a pair that differs only by one proton.

The conjugate acid-base for the given species is mentioned in the table below.

Which of the followings are Lewis acids? H₂O, BF₃, H⁺, and Question 7.36:

Answer:

Lewis acids are those acids which can accept a pair of electrons. For example, BF₃, H⁺, and  are Lewis acids.

Question 7.37:

What will be the conjugate bases for the Brönsted acids: HF, H₂SO₄ and HCO₃?

Answer:

The table below lists the conjugate bases for the given Bronsted acids.

Write the conjugate acids for the following Brönsted bases: NH₂^–, NH₃ and HCOO^–.Question 7.38:

Answer:

The table below lists the conjugate acids for the given Bronsted bases.

The species: H₂O, , and NH₃ can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.Question 7.39:

Answer:

The table below lists the conjugate acids and conjugate bases for the given species.

Question 7.40:

Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH^– (b) F^– (c) H⁺ (d) BCl₃.

Answer:

(a) OH^– is a Lewis base since it can donate its lone pair of electrons.

(b) F^– is a Lewis base since it can donate a pair of electrons.

(c) H⁺ is a Lewis acid since it can accept a pair of electrons.

(d) BCl₃ is a Lewis acid since it can accept a pair of electrons.

Question 7.41:

The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^–3 M. what is its pH?

Answer:

Given,

pH value of soft drink

Question 7.42:

The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

Answer:

Given,

pH = 3.76

It is known that,

Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74 × 10^–4 M.

Question 7.43:

The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10^–4, 1.8 × 10^–4 and 4.8 × 10^–9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Answer:

It is known that,

Given,

Ka of HF = 6.8 × 10^–4

Hence, Kb of its conjugate base F^–

Given,

Ka of HCOOH = 1.8 × 10^–4

Hence, Kb of its conjugate base HCOO^–

Given,

Ka of HCN = 4.8 × 10^–9

Hence, Kb of its conjugate base CN^–

Question 7.44:

The ionization constant of phenol is 1.0 × 10^–10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

Answer:

Ionization of phenol:

Now, let ∝ be the degree of ionization of phenol in the presence of 0.01 M C₆H₅ONa.

Also,

Question 7.45:

The first ionization constant of H_2­­S is 9.1 × 10^–8. Calculate the concentration of HS^– ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H₂S is 1.2 × 10^–13, calculate the concentration of S^2– under both conditions.

Answer:

(i) To calculate the concentration of HS^– ion:

Case I (in the absence of HCl):

Let the concentration of HS^– be x M.

Case II (in the presence of HCl):

In the presence of 0.1 M of HCl, let   be y M.

(ii) To calculate the concentration of :

Case I (in the absence of 0.1 M HCl):

 (From first ionization, case I)

Let 

Also,   (From first ionization, case I)

Case II (in the presence of 0.1 M HCl):

Again, let the concentration of HS^– be X’ M.

 (From first ionization, case II)

 (From HCl, case II)