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JEE Chemistry
Chemical and Ionic Equilibrium Previous Year Questions with Solutions

Chemical and Ionic Equilibrium Previous Year Questions with Solutions

1.0Chemical Equilibrium (Introduction)

Like a perfectly balanced seesaw – rate of forward reaction = rate of backward reaction.
A state in a reversible reaction where concentrations of reactants and products remain constant over time.

Dynamic Nature

  • Reactions continue in both directions.
  • No net change in concentration.

2.0Key Concepts

General Reaction & Rate Equality

aA+bB⇌cC+dD

Ratef=Rateb⇒kf[A]a[B]b=kb[C]c[D]d

Equilibrium Constant (Keq)

Keq​=[A]a[B]b[C]c[D]d​

  • Depends on temperature.
  • Expressed as Kc (concentration) or Kp (pressure for gases).

3.0Types of Equilibrium

  • Homogeneous: All reactants/products in same phase.

Ex: N2(g)+3H2(g)⇌2NH3(g)

N2​(g)+3H2​(g)⇌2NH3​(g)N2​(g)+3H2​(g)⇌2NH3​(g)

  • Heterogeneous: Reactants/products in different phases.
    Ex: H2O(g)+C(s)⇌H2(g)+CO(g)

H2​O(g)+C(s)⇌H2​(g)+CO(g)H2​O(g)+C(s)⇌H2​(g)+CO(g)

Factors Affecting Equilibrium (Le Chatelier’s Principle)

Factor

Effect on Equilibrium

Concentration

↑Reactants → shifts forward; ↑Products → shifts backward

Temperature

↑Temp → favors endothermic; ↓Temp → favors exothermic

Pressure (gases)

↑Pressure → shifts to side with fewer gas molecules

Volume (gases)

↓Volume → ↑Pressure → shift to side with fewer moles

Catalyst

No shift; speeds up attainment of equilibrium

4.0Ionic Equilibrium

Ionic equilibrium occurs when the rate of ion formation equals the rate of recombination, leading to constant ion concentration in solution.
Example: Acetic acid (CH₃COOH) in water establishes equilibrium:
CH₃COOH ⇌ CH₃COO⁻ + H⁺

5.0Key concepts

Electrolytes vs. Nonelectrolytes

Type

Behavior in Water

Examples

Electrolytes

Dissociate into ions; conduct electricity

NaCl, HCl

Nonelectrolytes

Do not dissociate; no conduction

Glucose, Urea

Degree of Dissociation (α)

Fraction of total molecules that dissociate into ions.
α=No. of molecules dissociated Total no. of molecules taken

α=Total no. of molecules takenNo. of molecules dissociated​

Strong Electrolytes

  • α ≈ 1
  • Irreversible ionization
  • High conductivity

Examples:

  • Acids: HCl, HNO₃, H₂SO₄
  • Bases: NaOH, KOH, Ba(OH)₂
  • Salts: NaCl, KNO₃
    Weak Electrolytes
  • α < 1
  • Reversible ionization
  • Low conductivity

Examples:

  • Acids: CH₃COOH, HF, H₂CO₃
  • Bases: NH₄OH, Aniline (C₆H₅NH₂)

Ostwald’s Dilution Law

For weak electrolytes:

α = KC

α=CK​​

Where:

  • α = degree of dissociation
  • K = dissociation constant
  • C = molar concentration

When α≪1

K≈Cα2

Ka and Kb Relationship

Ka×Kb=Kw=1.0×10−14

  • Ka: Dissociation constant of an acid
    Kb: Dissociation constant of a base
    Kw: Ionization constant of water

6.0Concepts of Acids and Bases

Arrhenius Concept

  • Acid: Produces H⁺ ions (e.g., HCl → H⁺ + Cl⁻)
  • Base: Produces OH⁻ ions (e.g., NaOH → Na⁺ + OH⁻)

Bronsted-Lowry Concept

  • Acid: Proton (H⁺) donor
  • Base: Proton acceptor

Example:
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

  • NH₃ = base, H₂O = acid (here)

Conjugate Pairs:

  • HCl / Cl⁻
  • NH₃ / NH₄⁺

Lewis Concept

  • Acid: Electron pair acceptor
  • Base: Electron pair donor

Examples:

  • BF₃ + NH₃ → H₃N→BF₃ (coordinate bond)
  • Ag⁺ + 2CN⁻ → [Ag(CN)₂]⁻

Acid & Base Strength

  • pH = −log[H⁺]
  • pOH = −log[OH⁻]
  • pH + pOH = 14

Buffer Solutions

  • Resist change in pH on addition of small amounts of acid or base.
  • Consist of a weak acid + conjugate base or weak base + conjugate acid.

Henderson-Hasselbalch Equation

  • For acid buffer: pH=pKa​+log([HA][A−]​)pH=pKa​+log([HA][A−]​)
  • For base buffer: pOH=pKb​+log([B][BH+]​)pOH=pKb​+log([B][BH+]​)

Salt Hydrolysis & pH

Type of Salt

Cation

Anion

Solution pH

From strong acid + strong base

Neutral (Na⁺, Cl⁻)

Neutral (NaCl)

≈ 7

From weak acid + strong base

Neutral (Na⁺)

Basic (CH₃COO⁻)

> 7

From strong acid + weak base

Acidic (NH₄⁺)

Neutral (Cl⁻)

< 7

From weak acid + weak base

Acidic or Basic (depends on Ka vs Kb)

Acidic or Basic

< or > 7

7.0Chemical and Ionic Equilibrium Important PYQs

Q.1 For the reaction ,  N2O2 (g)→2NO2 (g)    Kp= 0.492 atm at 300K. Kc for the reaction at the same temperature is ___ × 10–2 .

(Given : R = 0.082 L atm mol–1 K–1)]

Ans. (2)

Solution.  

Kp​=Kc​(RT)Δng​Δng​=1⇒Kc​=RTKp​​=0.082×3000.492​=2×10−2


Q.2   A(g)→B(g)+C22C​ (g) The correct relationship between KP​, α, and equilibrium pressure P is:      

(1)Kp​=(2+α)1/2a1/2P1/2​(2)Kp​=(2+α)1/2(1−α)a3/2P1/2​(3)Kp​=(2+α)1/2(1−α)a3/2P1/2​(4)Kp​=(2+α)3/2a3/2P1/2​

Ans. (2)      A(g)→B(g) + C2 (g)

t = teq         (1-ɑ)       ɑ       2a​ 

PB=   ɑ 1+   ɑ2.P

PB=   1-ɑ 1+   ɑ2.P

PC=   ɑ21+   ɑ2.P

KP= PB.PC.PA.

A(g)→B(g)+2C​(g)t=teq​A1−α​Bα​C2α​​PB​=(1+2α​)α​P,PA​=(1+2α​)1−α​P,PC​=(1+2α​)2α​​PKp​=PA​PB​⋅PC1/2​​=(1+2α​1−α​P)(1+2α​α​P)⋅(2(1+2α​)α​P)1/2​=(1−α)(2+α)1/2α3/2P1/2​PB​=(1+2α​)α​PPB​=(1+2α​)1−α​PPC​=(1+2α​)2α​​PKp​=PA​PB​⋅PC1/2​​


Q.3The equilibrium constant for the reaction SO3 →SO2(g) + 12 O2(g) is KC = 4.9 × 10–2. The value of KC for the reaction given below is 

2SO2 (g)+ O2(g) →   2SO2 (g) is

(1) 4.9                               

(2) 41.6

(3) 49                           

(4) 416

Ans.   (4)

Sol.

Kc′​={Kc​1​}2={4.9×10−21​}2

K'C = 416.49


Q.4  The effect of addition of helium gas to the following reaction in equilibrium state, is :

PCl5(g) →PCl3(g)  +Cl2(g)

(1) the equilibrium will shift in the forward direction and more of Cl2 and PCl3 gases will be produced.

(2) the equilibrium will go backward due to suppression of dissociation of PCl5.

(3) helium will deactivate PCl5 and reaction will stop.

(4) addition of helium will not affect the equilibrium.

Ans. (1 & 4)

Solution. PCl5(g) →PCl3(g)  +Cl2(g)

Case 1 : At constant P – volume will increase so reaction will shift in forward direction then answer will be A

Case 2 : At constant volume no change in active mass so reaction will not shift in any direction then answer will be D.


Q.5    For the given reaction  A → 2B(g)  +C(g), if the initial pressure is 450 mm Hg and the pressure at time t is 720 mm Hg at a constant temperature T and constant volume V. The fraction of A(g) decomposed under these conditions is x × 10­–1. The value of x is ________. (Nearest integer)

Ans.(3)  

Solution:                       

A →    2B(g)  +  C(g)

 t=0        450

time t    450-x     2x         x

PT =PA + PB +PC

720 = 450 – x + 2x + x

2x = 270

x = 135

Fraction of A decomposed =135450 \frac{135}{450}= 0.3 = 3×10–1

So, x = 3

     

Q.6  Rate law for a reaction between A and B is given by R = k [A]n[B]m  If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction is

(1) 2(n–m)                    

(2) (n–m)

(3) (m + n)                

(4) 2m+n1​

Ans.   (1)

Solution.

r1 = k[A]n [B]m

Now

 A is doubled & B is halved in concentration

r2 = k2n[A]n.2mBm​[B]m

r1​r2​​=2n−m    

 

Q.7 4.0 moles of argon and 5.0 moles of PCI5 are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The Kp for the reaction is [Given : R = 0.082L atm K–1 mol–1]

(1) 2.25              

(2) 6.24                           

(3) 12.13                           

(4) 15.24 

Ans. 1

Solution 

PCl5​=5moleAr=4molePtotal​=1009×0.82×610​=4.5,atmPPCl5​​=99×4.5​=2.5PAr​=94×4.5​=2

PCl5(g) →PCl3(g)  +Cl2(g)

 2.5 – P           P          P

Ptotal​=2.5−P+P+P+PAr​=6P=1.5Kp​=11.5×1.5​=2.25atm


Q.8 In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4 mol/L. The equilibrium constant at 30°C is ______ × 10–4.

2NOCl(g)→2NO(g) + Cl2(g)

Ans. by NTA (125)

Sol.      

2NOCl(g) → 2NO(g) + Cl2(g)

t = 0        2M                       ----

t=teq       (2–x)M          x M        12M

∵ x= 0.4M

[NOCl]eq = 1.6 M

[NO]eq = 0.4 M

[Cl2]eq = 0.2 M

Kc​=[chNOCl]2[chNO]2[chCl2​]​=[1.6]2[0.4]2[0.2]​ Kc​=2.5632​ ×10−3

Kc = 12.5 × 10-3

Kc = 125 × 10-4

Integer answer is 125


Q.9 Two solutions A and B, each of 100 L was made by dissolving 4g of NaOH and 9.8 g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is_______.

 Ans. (10.60)

Solution . 

4 gm of NaOH in 100 L sol. 10–3 M sol.

9.8 gm of H2SO4  in 100 L sol.10–3 M sol.

Mixture : 40L of 10–3 M NaOH and 10 L of 10–3 M H2SO4  sol.

Final Conc. of OH– 

=40+1010−3(40×1−10×1×2)​=6×10−4M

pOH = – log (6 ×10–4)

= 4 – log 6 = 4 – 0.60 = 3.40

pH = 14 – 3.40 = 10.60


Q.10  100 mL of 0.1 M HCl is taken in a beaker and to it 100 mL of 0.1 M NaOH is added in steps of 2 mL and the pH is continuously measured. Which of the following graphs correctly depicts the change in pH?

(1) 7pH vol. of NaOH                        

(2) pH 7vol. of NaOH

(3) pH 7vol. of NaOH                       

(4) pH7vol. of NaO

Ans. by NTA (3)

Solution. 

Steep rise in pH around the equivalence point  for titration of strong acid with strong base. 


Q.11 A solution is 0.1 M in Cl– and 0.001 M in chCrO42−.  Solid AgNO3 is gradually added to it. Assuming that the addition does not change in volume and Ksp(AgCl) = 1.7 × 10–10 M2 and Ksp(Ag2CrO4) = 1.9 × 10–12 M3. Select correct statement from the following :

(1)  AgCl precipitates first because its Ksp is high.

(2) Ag2CrO4 precipitates first as its Ksp is low.

(3) Ag2CrO4 precipitates first because the amount of Ag+ needed is low.

(4) AgCl will precipitate first as the amount of Ag+ needed to precipitate is low.

Ans. (4)

Solution.

(i)  [Ag+] required to ppt AgCl(s)

Ksp = IP = [Ag+] [Cl–] = 1.7 × 10–10

[Ag+] = 1.7 × 10–9

(ii) [Ag+] required to ppt Ag2CrO4(s)

Ksp = IP = [Ag+]2 [CrO4–2] = 1.9 × 10–12

[Ag+] = 4.3 × 10–5     

[Ag+] required to ppt AgCl is low so AgCl will ppt First.


Q.12 The pH of ammonium phosphate solution, if pKa of phosphoric acid and pkb of ammonium hydroxide are 5.23 and 4.75 respectively, is ______.

Ans  (7)

Solution . 

Since (NH4)3PO4 is salt of weak acid (H3PO4) & weak base (NH4OH).

pH=7+21​(pKa​−pKb​)=7+21​(5.23−4.75)=7.24≈7


Q.13 If the solubility product of PbS is 8 × 10–28, then the solubility of PbS in pure water at 298 K is x × 10–l6 mol L–1. The value of x is ________. (Nearest Integer)  [Given2 = 1.41]

Ans  (282)

Solution 

Ksp = S2

S=Ksp​​=8×10−28​=2​×10−14

= 2.82 × 10–14

= 282 ×10–16          

Ans. = 282

Q.14 Given below are two statements :

Statement (I) : A Buffer solution is the mixture of a salt and an acid or a base mixed in any particular quantities.

Statement (II) : Blood is naturally occurring buffer solution whose pH is maintained by H2CO3 concentrations.

Q.15 In the light of the above statements, choose the correct answer from the options given below.

(1) Statement I is false but Statement II is true

(2) Both Statement I and Statement II is true

(3) Both Statement I and Statement II is false

(4) Statement I is true but Statement II is false

Ans.  

Sol.

Buffer solution is a mixture of either weak acid / weak base and its respective conjugate.

Blood is a buffer solution of carbonic acid H2CO3 and bicarbonate HCO3- 

Statement 1 is false but Statement II is true.                                                                                                                                                                                                                                  

Table of Contents


  • 1.0Chemical Equilibrium (Introduction)
  • 1.1Dynamic Nature
  • 2.0Key Concepts
  • 2.1General Reaction & Rate Equality
  • 3.0Types of Equilibrium
  • 3.1Factors Affecting Equilibrium
  • 4.0Ionic Equilibrium
  • 5.0Key concepts
  • 5.1Electrolytes vs. Nonelectrolytes
  • 5.2Degree of Dissociation (α)
  • 6.0Concepts of Acids and Bases
  • 7.0Chemical and Ionic Equilibrium Important PYQs

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