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NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Q6.1: 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?
Ans:  (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.

 

Q6.2: 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?
NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium
Ans:  The equilibrium constant (Kc) for the give reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, Kcfor the equilibrium is NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium .

 

Q6.3: At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium . Calculate Kp for the equilibrium. 
Ans: Partial pressure of I atoms,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Partial pressure of I2 molecules,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now, for the given reaction,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.4: Write the expression for the equilibrium constant, Kc for each of the following reactions:
NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium
Ans:  

(i)

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(ii)

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(iii)

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(iv)

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(v)

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.5: Find out the value of Kc for each of the following equilibria from the value of Kp:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium
Ans:  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

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(b) Here, Δn = 2 – 1 = 1

R = 0.0831 barLmol–1K–1

T = 1073 K

Kp= 167

Now,

Kp = Kc (RT) Δn

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.6: For the following equilibrium, NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium
NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

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

Ans:  

It is given that KC for the forward reaction is 6.3 X1014.

Then, KC for the reverse reaction will be,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.7: Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Ans: For a pure substance (both solids and liquids),

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.

 

Q6.8: Reaction between N2 and O2 takes place as follows:
NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 × 10–37, determine the composition of equilibrium mixture.

 Ans:  Let the concentration of N2O at equilibrium be x.

The given reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Therefore, at equilibrium, in the 10 L vessel:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

The value of equilibrium constant i.e., NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium = 2.0 × 10–37 is very small. Therefore, the amount of N2 and O2 reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N2 and O2.

Then,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now,

  NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.9: Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:
NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.
Ans:  

The given reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now, 2 mol of NOBr are formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr are formed from 0.0518 mol of NO.

Again, 2 mol of NOBr are formed from 1 mol of Br.

Therefore, 0.0518 mol of NOBr are formed from NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium mol of Br, or

  0.0259 mol of NO.

The amount of NO and Br present initially is as follows:

[NO] = 0.087 mol [Br2] = 0.0437 mol

Therefore, the amount of NO present at equilibrium is:

[NO] = 0.087 – 0.0518

= 0.0352 mol

And, the amount of Br present at equilibrium is:

[Br2] = 0.0437 – 0.0259

= 0.0178 mol

 

Q6.10: At 450 K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium.

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

What is Kc at this temperature? 
Ans: For the given reaction,

Δn = 2 – 3 = – 1

T = 450 K

R = 0.0831 bar L bar K–1 mol–1

 = 2.0 × 1010 bar –1

We know that,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.11: A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 Ans:  

The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Therefore,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, the value of Kp for the given equilibrium is 4.0.

 

Q6.12: A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction 

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Ans:  

The given reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now, reaction quotient Qc is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Since NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium, the reaction mixture is not at equilibrium.

Again, NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium . Hence, the reaction will proceed in the reverse direction.

 

Q6.13: The equilibrium constant expression for a gas reaction is,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Write the balanced chemical equation corresponding to this expression.
Ans:  The balanced chemical equation corresponding to the given expression can be written as:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.14: One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Calculate the equilibrium constant for the reaction.
Ans: The given reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Therefore, the equilibrium constant for the reaction,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.15: At 700 K, equilibrium constant for the reaction

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

is 54.8. If 0.5 molL–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?
Ans:  

It is given that equilibrium constant NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium for the reaction

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium is 54.8.

Therefore, at equilibrium, the equilibrium constant NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium for the reaction

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium will be NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium .

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Let the concentrations of hydrogen and iodine at equilibrium be x molL–1

.NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, at equilibrium, NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.16: What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium
Ans:  

The given reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

  NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, at equilibrium,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.17: Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium
Ans:  Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium.

Now, according to the reaction,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

We can write,

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, at equilibrium,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.18: Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)

(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

Ans:  (i) Reaction quotient, NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess.

The given reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Therefore, equilibrium constant for the given reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(iii) Let the volume of the reaction mixture be V.

  NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Therefore, the reaction quotient is,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Since NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium, equilibrium has not been reached.

 

Q6.19: A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be 0.5 × 10–1 mol L–1. If value of Kcis 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium?

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium
Ans: Let the concentrations of both PCl3 and Cl2 at equilibrium be x molL–1. The given reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now we can write the expression for equilibrium as:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Therefore, at equilibrium,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.20: One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO2.

FeO (s) CO (g)   Fe (s) CO2 (g); Kp= 0.265 at 1050 K.

What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO = 1.4 atm and  NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium= 0.80 atm?
Ans: For the given reaction,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Since NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium , the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of CO will increase while the pressure of CO2 will decrease.

Now, let the increase in pressure of CO = decrease in pressure of CO2 be p.

Then, we can write,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Therefore, equilibrium partial of NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

And, equilibrium partial pressure of NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.21: Equilibrium constant, Kc for the reaction

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibriumat 500 K is 0.061.

At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?
Ans: The given reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now, we know that,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Since NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium, the reaction is not at equilibrium.

Since NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium , the reaction will proceed in the forward direction to reach equilibrium.

 

Q6.22: Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

for which K= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10–3 molL–1, what is its molar concentration in the mixture at equilibrium?
Ans:  Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now, we can write,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Therefore, at equilibrium,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.23: At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by mass

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Calculate Kc for this reaction at the above temperature.
Ans:  Let the total mass of the gaseous mixture be 100 g.

Mass of CO = 90.55 g

And, mass of CO2 = (100 – 90.55) = 9.45 g

Now, number of moles of CO, NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Number of moles of CO­2, NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Partial pressure of CO,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Partial pressure of CO2,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

For the given reaction,

Δn = 2 – 1 = 1

We know that,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.24: Calculate a) ΔG°and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

where ΔfG° (NO2) = 52.0 kJ/mol

ΔfG° (NO) = 87.0 kJ/mol

ΔfG° (O2) = 0 kJ/mol
Ans:  

(a) For the given reaction,

ΔG° = ΔG°( Products) – ΔG°( Reactants)

ΔG° = 52.0 – {87.0 0}

= – 35.0 kJ mol–1

(b) We know that,

ΔG° = RT log Kc

ΔG° = 2.303 RT log Kc

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, the equilibrium constant for the given reaction Kc is 1.36 × 106

 

Q6.25: Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

(a) NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(b) NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(c) NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium
Ans:  

 (a) The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.

(b) The number of moles of reaction products will decrease.

(c) The number of moles of reaction products remains the same.

 

Q6.26: Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.

(i)NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium                          

(ii) NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(iii)     NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium                      

(iv) NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(v)  NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium                  

(vi) NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium
Ans:  The reactions given in (i), (iii), (iv), (v), and (vi) will get affected by increasing the pressure.

The reaction given in (iv) will proceed in the forward direction because the number of moles of gaseous reactants is more than that of gaseous products.

The reactions given in (i), (iii), (v), and (vi) will shift in the backward direction because the number of moles of gaseous reactants is less than that of gaseous products.

 

Q6.27: The equilibrium constant for the following reaction is 1.6 ×105 at 1024 K.

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.
Ans: Given,

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium for the reaction i.e., NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Therefore, for the reaction NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium the equilibrium constant will be,

  NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now, let p be the pressure of both H2 and Br2 at equilibrium.

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now, we can write,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Therefore, at equilibrium,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.28: Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(a) Write as expression for Kp for the above reaction.

(b) How will the values of Kp and composition of equilibrium mixture be affected by

(i) Increasing the pressure              

(ii) Increasing the temperature                   

(iii) Using a catalyst?
Ans: (a) For the given reaction,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(b) (i) According to Le Chatelier’s principle, the equilibrium will shift in the backward direction.

(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.

(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

 

Q6.29: Describe the effect of:

(a) Addition of H2                                                                                                         

(b) Addition of CH3OH

(c) Removal of CO                                                  

(d) Removal of CH3OH

on the equilibrium of the reaction:          

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Ans:   

(a) According to Le Chatelier’s principle, on addition of H2, the equilibrium of the given reaction will shift in the forward direction.

(b) On addition of CH3OH, the equilibrium will shift in the backward direction.

(c) On removing CO, the equilibrium will shift in the backward direction.

(d) On removing CH3OH, the equilibrium will shift in the forward direction.

 

Q6.30: At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 ×10-3. If decomposition is depicted as,

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 ΔrH° = 124.0 kJmol–1

(a) Write an expression for Kc for the reaction.

(b) What is the value of Kc for the reverse reaction at the same temperature?

(c) What would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased? (iii) The temperature is increased?
Ans:  

(a)NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(b) Value of Kc for the reverse reaction at the same temperature is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(c) (i) Kc would remain the same because in this case, the temperature remains the same.

(ii) Kc is constant at constant temperature. Thus, in this case, Kc would not change.

(iii) In an endothermic reaction, the value of Kc increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of Kc will increase if the temperature is increased.

 

Q6.31: 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,

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp= 10.1 at 400°C
Ans: Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

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

 

Q6.32: Predict which of the following reaction will have appreciable concentration of reactants and products:

a) NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

b) NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

c) NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium
Ans: 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.

 

Q6.33: The value of Kc for the reaction 3O2 (g) NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium  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?
Ans: The given reaction is:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Then, we have,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, the concentration of NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.34: The reaction, CO(g) +3H2(g)NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium 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.
Ans: Let the concentration of methane at equilibrium be x.

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

It is given that Kc = 3.90.

Therefore,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

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

 

Q6.35: What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium
Ans:  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.
NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.36: Which of the followings are Lewis acids? H2O, BF3 H+  and NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium
Ans:  Lewis acids are those acids which can accept a pair of electrons. For example, BF3, H+ and NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium are Lewis acids.

 

Q6.37: What will be the conjugate bases for the Brönsted acids: HF, H2SO4 and HCO3?
Ans:  The table below lists the conjugate bases for the given Bronsted acids.
NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium


Q6.38: Write the conjugate acids for the following Brönsted bases: NH2, NH3 and HCOO.
Ans:  The table below lists the conjugate acids for the given Bronsted bases.
NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.39: The species: H2O, NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium and NH3 can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.
Ans:  The table below lists the conjugate acids and conjugate bases for the given species.

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.40: 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.
Ans:   (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.

 

Q6.41: The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. what is its pH?
Ans:  Given,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

pH value of soft drink

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.42: The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Ans: Given,

pH = 3.76

It is known that,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

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

 

Q6.43: The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4, 1.8 × 10 and 4.8 × 10–9 respectively. Calculate the ionization constants of the corresponding conjugate base.
Ans:  It is known that,

  NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Given,

Ka of HF = 6.8 × 10–4

Hence, Kb of its conjugate base F

  NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Given,

Ka of HCOOH = 1.8 × 10–4

Hence, Kb of its conjugate base HCOO

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium 

Given,

Ka of HCN = 4.8 × 10–9

Hence, Kb of its conjugate base CN

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.44: The ionization constant of phenol is 1.0×10–10. What is the concentration of phenolate ion in 0.05M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?
Ans:  Ionization of phenol:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

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

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Also,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.45: The first ionization constant of H2­­S 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 S2– under both conditions.
Ans:   (i) To calculate the concentration of HS ion:

Case I (in the absence of HCl):

Let the concentration of HS be x M.

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Case II (in the presence of HCl):

In the presence of 0.1 M of HCl, let  NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium be y M.

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(ii) To calculate the concentration of NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium:

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

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium(From first ionization, case I)

Let NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Also, NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium (From first ionization, case I)

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

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

Again, let the concentration of HS be X' M.

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium(From first ionization, case II)

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium(From HCl, case II)

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.46: The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
Ans:  Method 1

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Since Ka >> Kw, :

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Method 2

Degree of dissociation,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

c = 0.05 M

Ka = 1.74 × 10–5

   NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Thus, concentration of CH3COO– = c.α 

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.

 

Q6.47: It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.
Ans:  Let the organic acid be HA.

 NCERT Solutions Class 11 Chemistry Chapter 6 - EquilibriumNCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Concentration of HA = 0.01 M

pH = 4.15

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now, NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Then,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.48: Assuming complete dissociation, calculate the pH of the following solutions:

 (a) 0.003 M HCl      (b) 0.005 M NaOH      (c) 0.002 M HBr      (d) 0.002 M KOH
Ans:   (a) 0.003MHCl:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Since HCl is completely ionized,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now,

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, the pH of the solution is 2.52.

(b) 0.005MNaOH:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, the pH of the solution is 11.70.

(c) 0.002 HBr:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, the pH of the solution is 2.69.

(d) 0.002 M KOH:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, the pH of the solution is 11.31.

 

Q6.49: Calculate the pH of the following solutions:

a) 2 g of TlOH dissolved in water to give 2 litre of solution. 

b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution.

c) 0.3 g of NaOH dissolved in water to give 200 mL of solution. 

d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

Ans:   (a) For 2g of TlOH dissolved in water to give 2 L of solution:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium  

(b) For 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(d) For 1mL of 13.6 M HCl diluted with water to give 1 L of solution:

13.6 × 1 mL = M2 × 1000 mL

(Before dilution) (After dilution)

13.6 × 10–3 = M2 × 1L

M2 = 1.36 × 10–2

[H ] = 1.36 × 10–2

pH = – log (1.36 × 10–2)

= (– 0.1335 2) = 1.866 ∼ 1.87

 

Q6.50: The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.
Ans:  Degree of ionization, α = 0.132

Concentration, c = 0.1 M

Thus, the concentration of H3O  = c.α

= 0.1 × 0.132

= 0.0132

 NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now,

  NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium


Q6.51: The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.
Ans:

c = 0.005

pH = 9.95

pOH = 4.05

pH = – log (4.105)

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.52: What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Ans:

Kb = 4.27 × 10–10

c = 0.001M

pH =?

α =?

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10–5

 

Q6.53: Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl?
Ans:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

When HCl is added to the solution, the concentration of H+ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

Case I: When 0.01 M HCl is taken.

Let x be the amount of acetic acid dissociated after the addition of HCl.

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 – x and 0.01 + x can be taken as 0.05 and 0.01 respectively.

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Case II: When 0.1 M HCl is taken.

Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.54: The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?
Ans:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

 

Q6.55: Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

(a) Human muscle-fluid, 6.83

(b) Human stomach fluid, 1.2

(c) Human blood, 7.38

(d) Human saliva, 6.4.
Ans:

(a) Human muscle fluid 6.83:

pH = 6.83

pH = – log [H+]

∴6.83 = – log [H+]

[H+] =1.48 × 10–7 M

(b) Human stomach fluid, 1.2:

pH =1.2

1.2 = – log [H+]

∴[H+] = 0.063

(c) Human blood, 7.38:

pH = 7.38 = – log [H+]

∴ [H+] = 4.17 × 10–8 M

(d) Human saliva, 6.4:

pH = 6.4

6.4 = – log [H+]

[H+] = 3.98 × 10–7

 

Q6.56: The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
Ans: The hydrogen ion concentration in the given substances can be calculated by using the given relation:

pH = –log [H+]

(i) pH of milk = 6.8

Since, pH = –log [H+]

6.8 = –log [H+]

log [H+] = –6.8

[H+] = anitlog(–6.8)

= NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(ii) pH of black coffee = 5.0

Since, pH = –log [H+]

5.0 = –log [H+]

log [H+] = –5.0

[H+] = anitlog(–5.0)

= NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(iii) pH of tomato juice = 4.2

Since, pH = –log [H+]

4.2 = –log [H+]

log [H+] = –4.2

[H+] = anitlog(–4.2)

= NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(iv) pH of lemon juice = 2.2

Since, pH = –log [H+]

2.2 = –log [H+]

log [H+] = –2.2

[H+] = anitlog(–2.2)

= NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(v) pH of egg white = 7.8

Since, pH = –log [H+]

7.8 = –log [H+]

log [H+] = –7.8

[H+] = anitlog(–7.8)

= NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.57: If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
Ans:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.58: The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Ans:
 Solubility of Sr(OH)2 = 19.23 g/L

Then, concentration of Sr(OH)2

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.59: The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
Ans:
 Let the degree of ionization of propanoic acid be α.

Then, representing propionic acid as HA, we have:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

In the presence of 0.1M of HCl, let α´ be the degree of ionization.

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.60: The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Ans:

c = 0.1 M

pH = 2.34

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.61: The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Ans:

NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2).

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.62: A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Ans:
 pH = 3.44

We know that,

pH = – log [H+]

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Q6.63: Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF
Ans:

  NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium


Q6.64: The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?
Ans:

It is given that Ka for ClCH2COOH is 1.35 × 10–3.

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

ClCH2COONa is the salt of a weak acid i.e., ClCH2COOH and a strong base i.e., NaOH.

NCERT Solutions Class 11 Chemistry Chapter 6 - EquilibriumNCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.65: Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?
Ans:

Ionic product,

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, the pH of neutral water is 6.78.

 

Q6.66: Calculate the pH of the resultant mixtures:

a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl

b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2

c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH

Ans:

  NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.67: Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.
Ans:

(1) Silver chromate:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Molarity of NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium= 2s = 2 × 0.65 × 10–4 = 1.30 × 10–4 M

Molarity of NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium= s = 0.65 × 10–4 M

(2) Barium chromate:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(3) Ferric hydroxide:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(4) Lead chloride:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

(5) Mercurous iodide:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.68: The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Ans:

Let s be the solubility of Ag2CrO4.

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Therefore, the ratio of the molarities of their saturated solution is  

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

 

Q6.69: Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate Ksp = 7.4 × 10–8).
Ans: When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., 0.001 M.

Then,

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Now, the solubility equilibrium for copper iodate can be written as:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Since the ionic product (1 × 10–9) is less than Ksp (7.4 × 10–8), precipitation will not occur.

 

Q6.70: The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
Ans:

Since pH = 3.19,

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Let the solubility of C6H5COOAg be x mol/L.

Then,

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 × 10–6 mol/L. 

Now, let the solubility of C6H5COOAg be x’ mol/L.

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Hence, C6H5COOAg is approximately 3.317 times more soluble in a low pH solution.

 

Q6.71: What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10–18).
Ans: Let the maximum concentration of each solution be x mol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., x/2

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

If the concentrations of both solutions are equal to or less than 5.02 × 10–9 M, then there will be no precipitation of iron sulphide.


Q6.72: What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6).
Ans:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Let the solubility of CaSO4 be s.

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Molecular mass of CaSO4 = 136 g/mol

Solubility of  CaSOin gram/L = 3.02 × 10–3 × 136

= 0.41 g/L

This means that we need 1L of water to dissolve 0.41g of CaSO4

Therefore, to dissolve 1g of CaSO4 we require NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibriumof water.

 

Q6.73: The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place?

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

Ans: For precipitation to take place, it is required that the calculated ionic product exceeds the Ksp value.

Before mixing:

NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

This ionic product exceeds the Ksp of Zns and CdS. Therefore, precipitation will occur in CdCl2 and ZnCl2 solutions.

The document NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium is a part of the NEET Course Chemistry Class 11.
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FAQs on NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

1. What is the condition for equilibrium in a system?
Ans. In a system, equilibrium is achieved when the net force acting on an object is zero, and the net torque acting on the object is also zero.
2. How does a change in temperature affect equilibrium in a chemical reaction?
Ans. A change in temperature can shift the equilibrium position of a chemical reaction. An increase in temperature typically favors endothermic reactions, while a decrease in temperature favors exothermic reactions.
3. Can a system be in both static and dynamic equilibrium at the same time?
Ans. Yes, a system can be in both static and dynamic equilibrium simultaneously. Static equilibrium refers to the balance of forces, while dynamic equilibrium involves the balance of rates of reaction.
4. How does a change in pressure affect equilibrium in a system?
Ans. A change in pressure can affect the equilibrium position of a system in which gases are involved. According to Le Chatelier's Principle, an increase in pressure will shift the equilibrium towards the side with fewer moles of gas, while a decrease in pressure will shift the equilibrium towards the side with more moles of gas.
5. What is the significance of the equilibrium constant in a chemical reaction?
Ans. The equilibrium constant (K) indicates the position of equilibrium in a chemical reaction. It helps in determining the extent to which reactants are converted into products at equilibrium and provides valuable information about the reaction conditions.
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