Chemistry NCERT Exemplar Solutions Class 11th Chapter Seven: Overview, Questions, Preparation

1. How can you predict the following stages of a reaction by comparing the value of K_c and Q_c ?

(i) Net reaction proceeds in the forward direction

(ii) Net reaction proceeds in the backward direction.

(iii) No net reaction occurs.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: The values of K_c and Q_c are itself sufficient to explain the direction of reaction and less than or greater than one another decides the direction in which reaction will proceed as follows-

(i) As Qc < Kc, the reaction proceeds in the forward direction. 

(ii) If Qc > Kc, the reaction will proceed in the direction of reactants (reverse reaction). 

(iii) If Qc = Kc , no net reaction occurs.

2. On the basis of Le Chatelier principle explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ∆H = – 92.38 kJ mol^–1

What will be the effect of addition of argon to the above reaction mixture at constant volume?

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: According to Le Chatelier’s principle, when we raise the temperature, it shifts the equilibrium to left and decreases the equilibrium concentration of ammonia since it is an exothermic reaction. In other words, low temperature and high pressure is favourable for high yield of ammonia. There will be no change in equilibria on addition of argon (Ar).

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: The equation can be written as-

A^p+_x  +  B^q- _y ⇌ xA^p+ (aq) +  yB^q-

S moles of A and B dissolves to give x S moles of A^p+ and y S moles of Bq-.

K_sp = [A^p+]^x  [B^q-]^y = [x5]^x [y5]^y

= x^x y ^y 5 ^x+y

4. Write a relation between ∆G and Q and define the meaning of each term and answer the following :

(a) Why a reaction proceeds forward when Q < K and no net reaction occurs when Q = K.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: (a) As we know, 

ΔG=  ∆Gᶱ +RTlnQ

∆Gᶱ = Change in free energy as the reaction proceeds,

ΔG = Standard free energy change,

Q = Reaction quotient,

R = Gas constant,

T = Absolute temperature.

Since,  ∆Gᶱ =−RTlnK

∴ΔG=−RTlnK + RTlnQ = RTlnKQ​

If Q

If Q=K, ΔG=0, reaction is in equilibrium and no net reaction is there.

(b) Explain the effect of increase in pressure in terms of reaction quotient Q. for the reaction :

 CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: When we increase the pressure equilibrium will shift in forward direction it means Q

Types of Chemical Reactions :

There are 4 main types of chemical reactions. These are-

1. Combination Reactions- Combination Reactions are chemical reactions with 2 or more substances combining to form one single compound.
2. Decomposition Reactions- Decomposition Reactions are chemical reactions with one compound decomposing to produce 2 or more different substances. The digestion process of food in our body is a Decomposition reaction.
3. Displacement Reactions- Displacement reactions are reactions that involve the displacement of 1 element or group by another. 
4. Double Displacement- Displacement reactions are reactions with 2 compounds that react to form 2 new compounds. In the displacement reaction, no change in oxidation state occurs.

1. How can you predict the following stages of a reaction by comparing the value of K_c and Q_c ?

(i) Net reaction proceeds in the forward direction

(ii) Net reaction proceeds in the backward direction.

(iii) No net reaction occurs.

Ans: The values of K_c and Q_c are itself sufficient to explain the direction of reaction and less than or greater than one another decides the direction in which reaction will proceed as follows-

(i) As Qc < Kc, the reaction proceeds in the forward direction. 

(ii) If Qc > Kc, the reaction will proceed in the direction of reactants (reverse reaction). 

(iii) If Qc = Kc , no net reaction occurs.

2. On the basis of Le Chatelier principle explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ∆H = – 92.38 kJ mol^–1

What will be the effect of addition of argon to the above reaction mixture at constant volume?

Ans: The equation can be written as-

A^p+_x  +  B^q- _y ⇌ xA^p+ (aq) +  yB^q-

S moles of A and B dissolves to give x S moles of A^p+ and y S moles of Bq-.

K_sp = [A^p+]^x  [B^q-]^y = [x5]^x [y5]^y

= x^x y ^y 5 ^x+y

4. Write a relation between ∆G and Q and define the meaning of each term and answer the following :

(a) Why a reaction proceeds forward when Q < K and no net reaction occurs when Q = K.

Ans: (a) As we know, 

ΔG=  ∆Gᶱ +RTlnQ

∆Gᶱ = Change in free energy as the reaction proceeds,

ΔG = Standard free energy change,

Q = Reaction quotient,

R = Gas constant,

T = Absolute temperature.

Since,  ∆Gᶱ =−RTlnK

∴ΔG=−RTlnK + RTlnQ = RTlnKQ​

If Q

If Q=K, ΔG=0, reaction is in equilibrium and no net reaction is there.

(b) Explain the effect of increase in pressure in terms of reaction quotient Q. for the reaction :

 CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)

Ans: When we increase the pressure equilibrium will shift in forward direction it means Q

Types of Chemical Reactions :

There are 4 main types of chemical reactions. These are-

1. Combination Reactions- Combination Reactions are chemical reactions with 2 or more substances combining to form one single compound.
2. Decomposition Reactions- Decomposition Reactions are chemical reactions with one compound decomposing to produce 2 or more different substances. The digestion process of food in our body is a Decomposition reaction.
3. Displacement Reactions- Displacement reactions are reactions that involve the displacement of 1 element or group by another. 
4. Double Displacement- Displacement reactions are reactions with 2 compounds that react to form 2 new compounds. In the displacement reaction, no change in oxidation state occurs.

1. The ionisation of hydrochloric in water is given below:

HCl(aq) + H₂O (l ) ⇌ H₃O+ (aq) + Cl^– (aq)

Label two conjugate acid-base pairs in this ionization.

Ans: Conjugate acid and conjugate base can be shown as-

HCl(aq) + H₂O (l) ⇌ H₃O⁺(aq) + Cl^-(aq) 

 acid        base        conjugate    conjugate

                             acid          base

Ans: BF₃ does not have a proton but still it acts as a Lewis acid as it is an electron pair acceptor and reacts with NH₃by accepting its lone pair of electrons. The reaction can be represented by Coordinate bond is formed between BF₃ and NH₃. It can be represented as-

 [ BF₃ ← :NH₃]

Nitrogen acts as lone pair donator in this combination

4. Ionisation constant of a weak base MOH, is given by the expression

k_b= $\frac{[\mathbf{M}+][\mathbf{O}\mathbf{H}-\mathbf{}]}{\left[\mathbf{M}\mathbf{O}\mathbf{H}\right]\mathbf{}}$

Values of ionisation constant of some weak bases at a particular temperature are given below:

Arrange the bases in decreasing order of the extent of their ionisation at equilibrium. Which of the above base is the strongest?

Ans: Higher the value of Kb stronger will be the base. 

K_b-  5.4 × 10^-4 >1.77 × 10^-5 > 1.77 × 10^-9> 1.3 × 10^-14

Decreasing order of basic strength. 

Dimethylamine > Ammonia > Pyridine > Urea

Hence among the given bases, the strongest base is Dimethylamine. 

5. Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases?

OH^– , RO^– , CH₃COO^– , Cl^–

Ans: Conjugate acid of a weak base is always stronger and conjugate base of weak acid is always strong.

The weak acids are:

⇒H₂O, ROH, CH₃COOH, HCl

⇒HCl > CH₃COOH > ROH > H₂O

⇒ Conjugate base order ⇒ OH⁻>RO⁻>CH₃COO⁻>Cl⁻

6. Arrange the following in increasing order of pH.

KNO₃ (aq), CH₃COONa (aq), NH₄Cl (aq), C₆H₅COONH₄ (aq)

Ans: Correct order of pH is -

NH₄Cl  C₆H₅COONH₄  KNO₃  CH₃COONa

Salts of strong acid and strong base do not hydrolyse and form neutral solution thus, pH will be nearly 7 of KNO₃​. 

In sodium acetate, acetic acid remains unionised this results in increase in OH- concentration and pH will be more than 7. 

NH₄Cl formed from weak base, NH₄OH and strong acid, HCl, in water dissociates completely, aq. ammonium ions undergo hydrolysis with water to form NH₄OHand H₊ ions resulting in less pH value.

7. The value of K_c for the reaction 2HI (g) ⇌ H₂ (g) + I₂ (g) is 1 × 10^–4

At a given time, the composition of reaction mixture is

[HI] = 2 × 10^–5 mol, [H₂] = 1 × 10^–5 mol and [I₂] = 1 × 10^–5 mol

In which direction will the reaction proceed?

Ans: At a given time the reaction quotient Q for the reaction will be given by the expression:

Q= $\frac{\left[\mathbf{H}2\right]\left[\mathbf{I}2\right]}{\left[\mathbf{H}\mathbf{I}\right]\mathrm{}}$  = $\frac{[1\times 10⁻⁵\times 1\times 10⁻⁵]}{{\left(2\times \mathrm{}10^{-5}\right)}^2}$  = $\frac{1}{2}\mathrm{}$ = 0.25

 = 2.5×10⁻¹

As the value of the reaction quotient is greater than the value of K_c i.e. 1×10⁻⁴ the reaction will proceed in the reverse direction.

Ans: Here concentration of water cannot be neglected since the solution is very dilute. pH will be less than 7.0. Hence, the total concentration is given as -

 [ H₃O⁺ ] =10^-8+10^-7M.

Ans: Given with the pH = 5 as, pH= - log [ H⁺ ] 

 [ H⁺ ] =10^-5M

On 100 times dilution-

 [ H⁺ ] =10^-7M.

Ans: The chemical equation is given below-  

                                                           BaSO4(s) → Ba2+(aq) + SO42-(aq)

At t = 0                                                    1                 0                 0 

At equilibrium  in water                       1-S             S                 S 

At equilibrium in the presence           1-S             S                (S+0.01) 

of sulphuric acid                                                       

K_sp for BaSO₄ in water = [ Ba₂⁺ ] [ SO₂^4- ] = (S) (S) = S₂

But S = 8 x 10^-4 mol dm-3

K_sp = (8 × 10^-4)2 = 64 × 10^-8 ... (1) 

The expression for K_sp in the presence of sulphuric acid will be as follows: 

K_sp = (S) (5 + 0.01) ... (2) 

Since value of K_sp will not change in the presence of sulphuric acid, therefore from (1) and (2) 

(S) (S + 0.01) = 64 × 10^-8

S² + 0.01 S = 64 × 10^-8

S² + 0.01 S - 64 × 10^-8 = 0

S = $\frac{-0.01\pm \sqrt[]{{\left(0.01\right)}^2}+\mathrm{}{\left(4\times 64\mathrm{ }\times \mathrm{ }10-8\right)}^2\mathrm{ }}{2}$

When we solve the above equation, we will get the value of

 S is= 6 × 10^-4 mol^-1 dm^-3

Ans: pH of HOCL = 2.85

But ,- pH = log [H⁺]

 -2.85=log[H⁺]

= -3.15 = log[H⁺]

 [H⁺]=anti log3.15

 [H⁺]= 1.413× 10^-3 

or weak mono basic acid

[H⁺]= $\sqrt[]{K_a\times }$  C

= $K_a$  = $\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{\mathrm{C}}$  = $\frac{{\left(\mathrm{}1.43\times {10}^{-3}\right)}^2\mathrm{ }\mathrm{ }}{0.08}$

=24.957 $\times \mathrm{}$ 10^-6

=2.4957 $\times \mathrm{}$ 10^-5

Ans: we can find the pH of the solution as-

pH of solution A=6

Hence, concentration of ^[H⁺] ion in solution A=10-6mol L^-1

pH of solution B=4

Therefore, concentration of [H]⁺ ion in solution B=10^-4mol L^-1

On mixing one litre of each solution, total volume = 1L+1L=2L.

Amount of H⁺ ions in 1L of solution

 A= concentration× Volume(V)= 10^-6mol× IL 

Amount of H⁺ ions in 1L of solution B=10^-4mol×1L

∴ Total amount of H⁺ ions in the solution formed by mixing solutions A and B is (10-5mol+10-4mol)

This amount is present in 2L solution,

∴ Total H⁺

=210^-4 (1+0.01)​=21.01×10-4​mol L-1

=21.01×10^-4​mol L^-1 =0.5×10⁻⁴mol L−1 =5×10^-5mol L^-1.

pH=−log[H]⁺=−log(5×10^-5)=−log5+(−5log10)

= −log5+5=5−log5=5−0.6990=4.3010=4.3

Thus, the pH of the solution will be 4.3.

Ans: Let us assume S be the solubility of Al(OH)³.

K_sp=[Al³⁺][OH^-]3=(S)(3S)³=27S⁴

S⁴=27K_sp=27 × 10²⁷ × 10^-11=1 × 10^-12

S=1×10⁻³ mol L^-1.

(i) Solubility of Al(OH)³ ​: Molar mass of Al(OH)³  is 78g. Hence, solubility of Al(OH)³ ​in gL^-1=1 × 10³ × 78gL^-1=78 × 10-3gL¹=7.8 × 10^-2gL^-1
(ii) pH of the solution: S=1× 10⁻³mol L^-1

OH=3S= 3×1× 10⁻³ = 3× 10⁻³

pOH=3−log³

pH=14−pOH= 11+log³= 11.4771.

Ans:

K_sp=3.2 × 10^-8

PbCl₂⇌Pb₂⁺ +2Cl^-

K_sp=(S)(2S)²

K_sp=4S³

s³=43.2 × 10^-8

s³=8 × 10^-9

s=2 × 10^-3

Molar   mass   of   PbCl₂=278g/mol   

By using Molarity formula,     

2 × 10^-3= 278 ×  x0.1

x= $\frac{0.1}{278\mathrm{ }\times \mathrm{ }2\mathrm{ }\times \mathrm{ }{10}^{-3}}$

= $\frac{100}{278\mathrm{ }\times \mathrm{ }2\mathrm{ }}$

x=0.18L   

15. A reaction between ammonia and boron trifluoride is given below: :

 NH₃ + BF₃ → H₃N : BF₃ Identify the acid and base in this reaction. Which theory explains it? What is the hybridisation of B and N in the reactants?

16. Following data is given for the reaction: CaCO₃ (s) → CaO (s) + CO₂ (g)

 ∆_fHᶱ [CaO(s)] = – 635.1 kJ mol^–1

 ∆_fHᶱ[CO₂(g)] = – 393.5 kJ mol^–1

∆_fHᶱ[CaCO₃(s)] = – 1206.9 kJ mol^–1

Predict the effect of temperature on the equilibrium constant of the above reaction.

Ans: Δ_rH^ᶱ= ∆_fHᶱ [CaO(s)]+ ∆_fHᶱ[CO2(g)]− ∆_fHᶱ[ [CaCO₃(s)]

∴ ∆_fHᶱ =178.3kJ mol⁻¹.

The reaction is endothermic. Hence, according to Le Chatelier’s principle, reaction will proceed in forward direction on increasing temperature.

1. We know that the relationship between Kc and Kp is

Kp = Kc (RT)^∆n What would be the value of ∆n for the reaction

NH4Cl (s) ⇌ NH3 (g) + HCl (g)

(i) 1

(ii) 5

(iii) 5

(iv) 2

Ans:  option(iv)

The relationship between Kp and Kc is

Kp=Kc(RT)Δn

Where ∆n = (number of moles of gaseous products) – (number of moles of gaseous reactants)

For the reaction,

NH4C1(s)⇆NH3(g)+HCl(g)

Δn=2–0=2

2. For the reaction H₂(g) + I₂(g) ⇌ 2HI (g), the standard free energy is ∆Gᶱ > 0. The equilibrium constant (K ) would be __________.

(i) K = 0

(ii) K > 1

(iii) K = 1

(iv) K < 1

Ans: option(iv)

 The given chemical reaction is:

H₂(g) + I₂(g) ⇌ 2HI (g)

By, thermodynamical equation of equilibrium,

∆Gᶱ= -RTlnk>0

Now, R>0 & T>0

⇒ −lnK > 0

⇒    K<1

3. Which of the following is not a general characteristic of equilibria involving physical processes?

(i) Equilibrium is possible only in a closed system at a given temperature.

(ii) All measurable properties of the system remain constant.

(iii) All the physical processes stop at equilibrium.

(iv) The opposing processes occur at the same rate and there is dynamic but stable condition.

Ans: option (iii)

A physical equilibrium is having dynamic nature, both forward and reverse process occur at equal rates. Some kind of motion is always present. The individual molecules continuously move from one phase to another.

4. PCl₅, PCl₃ and Cl₂ are at equilibrium at 500K in a closed container and their concentrations are 0.8 × 10^–3 mol L^–1, 1.2 × 10^–3 mol L^–1 and 1.2 × 10^–3 mol L^–1 respectively. The value of Kc for the reaction PCl5 (g) ⇌
PCl3 (g) + Cl2 (g) will be

(i) 1.8 × 10³ mol L^–1

(ii) 1.8 × 10^–3

(iii) 1.8 × 10^–3 L mol^–1

(iv) 0.55 × 10⁴

Ans: The given reaction is :-

PCl5(g)⇌PCl3(g)+Cl2(g)

KC= $\frac{\left[\mathrm{P}\mathrm{C}\mathrm{l}3\right]\left[\mathrm{C}\mathrm{l}2\right]}{\left[\mathrm{P}\mathrm{C}\mathrm{l}5\right]}$

=1.2×10⁻³ ×1.2×10⁻³ 0.8×10−3 $\frac{[1.2\times 10-3\times 1.2\times 10-3\mathrm{}]}{[0.8\times 10-3\mathrm{}]}$

=1.8×10⁻³molL⁻¹

5. Which of the following statements is incorrect?

(i) In equilibrium mixture of ice and water kept in perfectly insulated flask mass of ice and water does not change with time.

(ii) The intensity of red colour increases when oxalic acid is added to a solution containing iron

(III) nitrate and potassium thiocyanate.

(iii) On addition of catalyst the equilibrium constant value is not affected.

(iv) Equilibrium constant for a reaction with negative ∆H value decreases as the temperature increases.

Ans:  option (ii)

The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate.
Iron (III) nitrate and potassium thiocyanate react to give deep red coloration.
The reaction is:
Fe³ + aq + SCN⁻↔ [FeSCN]² + aq
yellow deep red
Now, to this solution, if oxalic acid (H₂C₂O₄) is added, then it reacts with Fe³⁺ ions to form the stable complex ion [Fe(C₂O₄)3]³⁻, thus decreasing the concentration of Fe³⁺ ions. In accordance with the Le-Chatelier's principle, the decrease in the concentration of Fe³⁺ ions, leads to the decrease in the concentration of [FeSCN]²⁺ ions which is indicated by the fact that the colour of the solution becomes lighter.

6. When hydrochloric acid is added to cobalt nitrate solution at room temperature, the following reaction takes place and the reaction mixture becomes blue. On cooling the mixture it becomes pink. On the basis of this information mark the correct answer.

[Co (H₂O)₆] ³⁺ (aq) + 4Cl^– (aq) ⇌ [CoCl₄]^2– (aq) + 6H₂O (l)

(pink)                                  (blue)

(i) ∆H > 0 for the reaction

(ii) ∆H < 0 for the reaction

(iii) ∆H = 0 for the reaction

(iv) The sign of ∆H cannot be predicted on the basis of this information.

Ans: option (i)

On cooling the mixture reaction moves towards backward direction it means it is an endothermic reaction i.e.,

7. The pH of neutral water at 25°C is 7.0. As the temperature increases, ionisation of water increases, however, the concentration of H⁺ ions and OH^– ions are equal. What will be the pH of pure water at 60°C?

(i) Equal to 7.0

(ii) Greater than 7.0

(iii) Less than 7.0

(iv) Equal to zero

Ans: As Kw increases Ans: As Kw increases [H+][OH^-] $>$ 10^-14

[H⁺]=[OH^-]

[H⁺]=[OH^-] [H⁺]_{2 = 110}^-14

[H⁺] $>$ 10^-7

8. The ionisation constant of an acid, Ka , is the measure of strength of an acid. The Ka values of acetic acid, hypochlorous acid and formic acid are 1.74 × 10^–5, 3.0 × 10^–8 and 1.8 × 10^–4 respectively. Which of the following orders of pH of 0.1 mol dm^–3 solutions of these acids is correct?

(i) acetic acid > hypochlorous acid > formic acid

(ii) hypochlorous acid > acetic acid > formic acid

(iii) formic acid > hypochlorous acid > acetic acid

(iv) formic acid > acetic acid > hypochlorous acid

Ans: option(iv)

Higher the value of Ka higher will be the acidic strength Ka value for formic acid is highest and that will be of hypochlorous acid is lowest.

9. K_a1 , K_a2 and K_a3 are the respective ionisation constants for the following reactions.

H₂S $\mathit{}\to$   H⁺ + HS

HS^–  $\mathit{}\to$ H⁺ + S²

 H₂S $\mathit{}\to$ 2H⁺ + S²

The correct relationship between is K_a1 , K_a2 and K_a3

1. K_a3= K_a1 × K_a2
2. K_a3= K_a1 + K_a2
3. K_a3= K_a1 – K_a2
4. K_a3= K_a1 / K_a2

Ans: option(i)  

H₂S⇌H⁺ +H^S− K_a1

HS−⇌H⁺+S²− K_a2

H₂S⇌2H⁺+S²− K_a3

K_a1= $\frac{\left[{\mathrm{H}\mathrm{S}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{H}}_2\mathrm{S}\right]\mathrm{}}$

K_a2= $\frac{\left[{\mathrm{H}\mathrm{S}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{H}}_2\mathrm{S}\right]\mathrm{}}$

K_a3=K_a1.K_a2

10. Acidity of BF₃ can be explained on the basis of which of the following concepts?

(i) Arrhenius concept

(ii) Bronsted Lowry concept

(iii) Lewis concept

(iv) Bronsted Lowry as well as Lewis concept

Ans: option (iii)

 BF₃ is an electron deficient species that is why it is a Lewis acid. (all Lewis acid can accept a pair of electrons.

11. Which of the following will produce a buffer solution when mixed in equal volumes?

(i) 0.1 mol dm^–3 NH₄OH and 0.1 mol dm^–3 HCl

(ii) 0.05 mol dm^–3 NH₄OH and 0.1 mol dm^–3 HCl

(iii) 0.1 mol dm^–3 NH₄OH and 0.05 mol dm^–3 HCl

(iv) 0.1 mol dm^–3 CH₄COONa and 0.1 mol dm^–3 NaOH

Ans: option(iii)

When 0.1 mol dm^-3 NH₄OH and 0.05 mol dm^-3 HCl react total amount of HCl reacts with NH₄OH to form NH₄Cl and some NH₄OH will be left unreacted. Thus, the resultant solution contains NH₄Cl and NH₄OH which will produce a buffer solution.

12. In which of the following solvents is silver chloride most soluble?

(i) 0.1 mol dm^–3 AgNO3 solution

(ii) 0.1 mol dm^–3 HCl solution

(iii) H₂O

(iv) Aqueous ammonia

Ans: option(iv)

Aqueous ammonia will absorb chloride ions and thus the equilibria will shift in forward direction and solubility of silver chloride will increase.

13. What will be the value of pH of 0.01 mol dm^–3 CH₃COOH (Ka = 1.74 × 10^–5 )?

(i) 3.4

(ii) 3.6

(iii) 3.9

(iv) 3.0

Ans: option (i)

pH for weak acid is given by the following equation:

pH= $\frac{1}{2}$ [pK_a−logC]

= $\frac{1}{2}$ [−log(1.74×10⁻⁵)−log0.01]

=3.4

14. Ka for CH₃COOH is 1.8 × 10^–5 and K b for NH₄OH is 1.8 × 10^–5 . The pH of ammonium acetate will be

(i) 7.005

(ii) 4.75

(iii) 7.0

(iv) Between 6 and 7

Ans: option(iii)

The equation for weak acid and weak base is given as-

pH=7+ $\frac{1}{2}$  ​(pKa​−pKb​)

K_a=1.8×10^-5

pK_a=−log (1.8×10-5) =5−log1.8 =4.744

pK_b​=−log (1.8×10-5) =4.744

pH=7+ $\frac{1}{2}$  ​(pKa​−pK_b​) =7

15. Which of the following options will be correct for the stage of half completion of the reaction A ⇌ B.

(i) ∆Gᶱ = 0

(ii) ∆Gᶱ> 0

(iii) ∆Gᶱ< 0

(iv) ∆Gᶱ= –RT ln2

Ans: option(i)

When the reaction is half completed, [A] = [B].
Thus, [B][A]=1
△G⁰=−RTInK=−RTIn(1)=0

6. On increasing the pressure, in which direction will the gas phase reaction proceed to re-establish equilibrium, is predicted by applying the Le Chatelier’s principle. Consider the reaction.

N₂ (g) + 3H₂ (g) ⇌2NH₃ (g)

Which of the following is correct, if the total pressure at which the equilibrium is established, is increased without changing the temperature?

(i) K will remain same

(ii) K will decrease

(iii) K will increase

(iv) K will increase initially and decrease when pressure is very high

Ans: option(i)

The given reaction is :-

N₂ (g) + 3H₂ (g) ⇌2NH₃ (g)

A/C to vart Hoff's eqn,

$\frac{\mathrm{d}\mathrm{l}\mathrm{n}\mathrm{K}}{\mathrm{d}\mathrm{T}\mathrm{}}\mathrm{}$ = $\frac{△\mathrm{M}}{\mathrm{R}\mathrm{T}2}$

K is dependent on temperature. As temperature remains constant at equilibrium, so K will remain constant in the given conditions.

17. What will be the correct order of vapour pressure of water, acetone and ether at 30°C. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point?

(i) Water < ether < acetone

(ii) Water < acetone < ether

(iii) Ether < acetone < water

(iv) Acetone < ether < water

Ans: option(ii)

 Greater the boiling point, lower will be the vapour pressure.

18. At 500 K, equilibrium constant, K_c , for the following reaction is 5.

$\frac{1}{2}{\mathbf{H}}_2$ (g) + $\frac{1}{2}{\mathbf{I}}_2$  (g) ⇌ HI (g)

What would be the equilibrium constant K_c for the reaction

2HI(g) ⇌ H₂ (g) + I₂ (g)

(i) 04

(ii) 4

(iii) 25

(iv) 5

Ans: option(i) 

If the equation is multiplied by 2, the equilibrium constant for the new equation is the square of K it means we must do the square of 5 that is 25. Then and on reversing the reaction the value of the equilibrium constant is inversed means the value of K comes out to be 1/25= 0.04. 

19. In which of the following reactions, the equilibrium remains unaffected on addition of small amount of argon at constant volume?

(i) H₂ (g) + I₂ (g) ⇌ 2HI (g)

(ii) PCl₅ (g) ⇌  PCl₃ (g) + Cl₂ (g)

(iii) N₂ (g) + 3H₂ (g) ⇌
2NH₃ (g)

(iv) The equilibrium will remain unaffected in all the three cases.

Ans: option(iv) 

Since in all the given reaction, the volume is to be kept constant and an inert gas argon is added which will do not take part in reaction. So, the equilibrium will remain unaffected in all the given cases.

In the following questions two or more options may be correct.

20. For the reaction N₂O₄ (g) ⇌2NO₂ (g), the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct?

(i) The reaction is endothermic

(ii) The reaction is exothermic

(iii) If NO₂ (g) and N₂O₄ (g) are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more N₂O₄ (g) will be formed.

(iv) The entropy of the system increases.

Ans: , option (i), (iii) and (iv)

As, K increases with increase in temperature. K increases, it shows that the reaction must be endothermic. No. of molecules increases thus, there is increase in entropy.

21. At a particular temperature and atmospheric pressure, the solid and liquid phases of a pure substance can exist in equilibrium. Which of the following term defines this temperature?

(i) Normal melting point

(ii) Equilibrium temperature

(iii) Boiling point

(iv) Freezing point

Ans: options (i) and (iv)

Ice and water are in equilibrium only at particular temperature and pressure. For any pure substance at atmospheric pressure, the temperature at which the solid and liquid phases are at equilibrium is called the normal melting point or normal freezing point of the substance. For example, it is 4°C in case of water. The system here is in dynamic equilibrium.

1. Match the following equilibria with the corresponding condition.

Ans: (i) → (b); (ii) → (d); (iii) → (c); (iv) → (a).

2. For the reaction : N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Equilibrium constant K_c = $\frac{{\left[{\mathbf{N}\mathbf{H}}_3\mathbf{}\right]}^2}{\left[{\mathbf{N}}_2\mathbf{}\right]{\left[{\mathbf{H}}_3\mathbf{}\right]}^3\mathbf{ }}$

Some reactions are written below in Column I and their equilibrium constants in terms of K_c are written in Column II. Match the following reactions with the corresponding equilibrium constant.

Ans: (i) → (d); (ii) → (c); (iii) → (b)

3. Match standard free energy of the reaction with the corresponding equilibrium constant.

Ans: (i) → (d); (ii) → (a); (iii) → (b).

4. Match the following species with the corresponding conjugate acid.

Ans: (i) → (b); (ii) → (e); (iii) → (c); (iv) → (d)

5. Match the following graphical variation with their description-

                A                                                                                                B 

	(a) Variation in product concentration with time
	(b) Reaction at equilibrium
	(c) Variation in reactant concentration with time

Ans: (i) → (c); (ii) → (a); (iii) → (b).

6. Match Column (I) with Column (II). Column I Column II-

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

1. Assertion (A) : Increasing order of acidity of hydrogen halides is HF < HCl < HBr < HI

Reason (R) : While comparing acids formed by the elements belonging to the same group of periodic table, H–A bond strength is a more important factor in determining acidity of an acid than the polar nature of the bond.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

Ans: option(i)

 As, H-A bond strength is a more important factor in determining acidity than its polar nature. As the size of A increases down the group, H-A bond strength decreases.

2. Assertion (A) : A solution containing a mixture of acetic acid and sodium acetate maintains a constant value of pH on addition of small amounts of acid or alkali.

Reason (R) : A solution containing a mixture of acetic acid and sodium acetate acts as a buffer solution around pH 4.75.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

Ans: option(ii)

The solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali are called buffer solutions.

3. Assertion (A): The ionisation of hydrogen sulphide in water is low in the presence of hydrochloric acid.

 Reason (R) : Hydrogen sulphide is a weak acid.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not correct explanation of A.

(iii) A is true but R is false

(iv) Both A and R are false

Ans: option(ii)

In presence of H⁺ ion, equilibrium shifts in backward direction.

4. Assertion (A): For any chemical reaction at a particular temperature, the equilibrium constant is fixed and is a characteristic property.

Reason (R) : Equilibrium constant is independent of temperature.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false

Ans: option(iii)

The equilibrium constant for an exothermic reaction (negative H) decreases as the temperature increases. The equilibrium constant for an endothermic reaction (positive H) increases as the temperature increases.

5. Assertion (A) : Aqueous solution of ammonium carbonate is basic.

Reason (R) : Acidic/basic nature of a salt solution of a salt of weak acid and weak base depends on K_a and K_b value of the acid and the base forming it.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

6. Assertion (A): An aqueous solution of ammonium acetate can act as a buffer.

 Reason (R) : Acetic acid is a weak acid and NH₄OH is a weak base.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not correct explanation of A.

(iii) A is false but R is true.

(iv) Both A and R are false.

Ans:  option (iii)

Salt of weak acid and weak base can form buffer solution.

7. Assertion (A): In the dissociation of PCl₅ at constant pressure and temperature addition of helium at equilibrium increases the dissociation of PCl₅ .

Reason (R) : Helium removes Cl₂ from the field of action.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

Ans: option(iv) 

If the volume is kept constant and an inert gas such as argon is added which does not take part in the reaction, the equilibrium remains undisturbed.