1. Introduction to Equilibrium

Chemical Equilibrium is the state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of reactants and products remain constant.

Types of Reactions

Irreversible: Proceed only in one direction (e.g., Precipitation). Cannot establish equilibrium.
Reversible: Proceed in both directions. Represented by double arrow (↔).

Nature: Equilibrium is Dynamic, not static. Reactions continue to occur at the molecular level, but net change is zero.

[Image of concentration vs time graph for equilibrium]

2. Law of Mass Action & Constants

Guldberg & Waage: The rate of a chemical reaction is directly proportional to the product of the active masses (molar concentrations) of reactants raised to the power of their stoichiometric coefficients.

aA + bB ↔ cC + dD

1. Concentration Constant (K_c):

K_c = [C]^c [D]^d / [A]^a [B]^b

(Active mass = Molar Conc. = n/V)

2. Pressure Constant (K_p):

K_p = (p_C)^c (p_D)^d / (p_A)^a (p_B)^b

(Valid only for gaseous species)

Relation Between K_p and K_c

K_p = K_c (RT)^Δn_g

Δn_g = (Moles of gaseous products) - (Moles of gaseous reactants).
R = 0.0821 L atm K^-1 mol^-1.

4. Characteristics of K

Operation on Reaction	New Constant (K')
Reaction Reversed	K' = 1/K
Multiplied by 'n'	K' = Kⁿ
Divided by 'n'	K' = K^1/n
Two reactions added	K' = K₁ × K₂

Note: The value of K depends ONLY on Temperature. It is independent of catalyst, pressure, or initial concentration.

5. Reaction Quotient (Q)

Q is calculated exactly like K_c but at any time (not necessarily equilibrium).

Q < K_c: Reactants → Products (Forward shift).
Q > K_c: Products → Reactants (Backward shift).
Q = K_c: Reaction is at Equilibrium.

6. Le Chatelier's Principle

"If a system at equilibrium is subjected to a change in conc, pressure, or temp, the equilibrium shifts in a direction that nullifies the effect of the change."

Summary of Effects

1. Concentration:

Add Reactant → Forward shift.
Add Product → Backward shift.

2. Pressure (For Gaseous Rxn):

Increase P → Shifts to side with lesser moles.
Decrease P → Shifts to side with more moles.
If Δn_g = 0, Pressure has NO effect.

3. Temperature (changes value of K):

Exothermic (ΔH < 0): Low Temp favors Forward. (K decreases as T increases).
Endothermic (ΔH > 0): High Temp favors Forward. (K increases as T increases).

4. Inert Gas Addition:

At Constant Volume: No effect.
At Constant Pressure: Shifts to side with more moles (Volume increases).

7. Van't Hoff Equation

Relates K with Temperature.

log(K₂/K₁) = (ΔH / 2.303 R) [ (T₂ - T₁) / (T₁ T₂) ]

8. Degree of Dissociation (α)

Fraction of 1 mole of reactant dissociated at equilibrium.

α = (Moles dissociated) / (Initial moles)

Relation with Vapor Density

α = (D - d) / ((n - 1) × d)

D = Initial VD (Theoretical) = Molar Mass / 2
d = Equilibrium VD (Observed)
n = Moles of product from 1 mole reactant

9. Ionic Equilibrium (Intro)

Strong Electrolyte: Ionizes completely (e.g., HCl, NaOH, NaCl). α ≈ 1.
Weak Electrolyte: Ionizes partially (e.g., CH₃COOH, NH₄OH). α << 1. Equilibrium exists between ions and unionized molecules.

10. Industrial Applications

1. Haber's Process (NH₃)

N₂ + 3H₂ ↔ 2NH₃ + Heat

Conditions for Max Yield: Low Temp, High Pressure.
Catalyst: Iron oxide (with Mo promoter).

2. Contact Process (H₂SO₄)

2SO₂ + O₂ ↔ 2SO₃ + Heat

Conditions: Low Temp, High Pressure.
Catalyst: V₂O₅.

Numericals & HOTS

Q1. Calculation of K_c

At equilibrium, a 10 L vessel contains 2 moles of A, 4 moles of B, and 6 moles of C. Calculate K_c for the reaction A + 2B ↔ 3C.

Solution:
Concentration = Moles / Volume (L)
[A] = 2/10 = 0.2 M
[B] = 4/10 = 0.4 M
[C] = 6/10 = 0.6 M

K_c = [C]³ / ([A][B]²)
K_c = (0.6)³ / (0.2 × (0.4)²)
K_c = 0.216 / (0.2 × 0.16)
K_c = 0.216 / 0.032
K_c = 6.75

Q2. K_p vs K_c

For the reaction N₂(g) + 3H₂(g) ↔ 2NH₃(g) at 500 K, the value of K_p is 1.44 × 10^-5. Calculate K_c. (R = 0.0821 L atm K^-1 mol^-1)

Solution:
Formula: K_p = K_c (RT)^Δn_g
Δn_g = 2 - (1 + 3) = -2

1.44 × 10^-5 = K_c (0.0821 × 500)^-2
1.44 × 10^-5 = K_c (41.05)^-2
K_c = 1.44 × 10^-5 × (41.05)²
K_c = 1.44 × 10^-5 × 1685.1
K_c ≈ 0.024

Q3. Degree of Dissociation (α)

HI is heated in a sealed tube at 440°C. At equilibrium, 22% of HI is dissociated. Calculate the equilibrium constant K_c for 2HI ↔ H₂ + I₂.

Solution:
Let initial moles of HI = 1, Volume = V.
α = 0.22 (22%)

Reaction: 2HI ↔ H₂ + I₂
Initial: 1 ..... 0 ..... 0
Eq: 1-α ... α/2 ... α/2

[HI] = (1 - 0.22)/V = 0.78/V
[H₂] = 0.11/V
[I₂] = 0.11/V

K_c = ([H₂][I₂]) / [HI]²
K_c = (0.11/V × 0.11/V) / (0.78/V)²
K_c = (0.11 × 0.11) / (0.78 × 0.78)
K_c = 0.0199 ≈ 0.02

Q4. Prediction of Direction (Q vs K)

For 2SO₂ + O₂ ↔ 2SO₃, K_c = 278. At a certain instant, [SO₂] = 0.4 M, [O₂] = 0.1 M, and [SO₃] = 2.0 M. Determine the direction of reaction.

Solution:
Q_c = [SO₃]² / ([SO₂]² [O₂])
Q_c = (2.0)² / ((0.4)² × 0.1)
Q_c = 4 / (0.16 × 0.1)
Q_c = 4 / 0.016
Q_c = 250

Here, Q_c (250) < K_c (278).
Since Q < K, the reaction will proceed in the Forward direction.

Q5. Vapor Density Method

For the reaction PCl₅ ↔ PCl₃ + Cl₂, the vapor density of the equilibrium mixture is 62. Calculate the degree of dissociation. (Molar mass of PCl₅ = 208.5)

Solution:
1. Theoretical V.D. (D) = M / 2 = 208.5 / 2 = 104.25
2. Observed V.D. (d) = 62
3. n = 2 (1 mole PCl₅ gives 1 PCl₃ + 1 Cl₂)

Formula: α = (D - d) / ((n - 1)d)
α = (104.25 - 62) / ((2 - 1) × 62)
α = 42.25 / 62
α = 0.68 or 68%

Q6. Thermodynamics (ΔG°)

Calculate the standard free energy change (ΔG°) for a reaction at 300 K if K = 100. (R = 8.314 J K^-1 mol^-1)

Solution:
Formula: ΔG° = -2.303 RT log K

ΔG° = -2.303 × 8.314 × 300 × log(100)
ΔG° = -2.303 × 2494.2 × 2
ΔG° = -11488 J/mol
ΔG° = -11.49 kJ/mol

Q7. Le Chatelier's (HOTS)

Given: 2NO₂(g) (Brown) ↔ N₂O₄(g) (Colorless); ΔH = -57 kJ.
What happens to the color of the mixture if:
(a) Pressure is increased?
(b) Temperature is increased?

Solution:
(a) Pressure Increase: Shifts to side with fewer moles (Product side, N₂O₄). Color fades (becomes lighter).

(b) Temp Increase: Reaction is Exothermic (ΔH < 0). High T favors backward reaction (Reactant NO₂). Color deepens (becomes darker brown).

Q8. K_p Calculation (Solid)

Solid Ammonium Carbamate dissociates as: NH₂COONH₄(s) ↔ 2NH₃(g) + CO₂(g). The total pressure at equilibrium is 0.3 atm. Calculate K_p.

Solution:
Let P be partial pressure of CO₂.
Then P(NH₃) = 2P (stoichiometry 2:1).
Total Pressure = 2P + P = 3P = 0.3 atm.
So, P = 0.1 atm.

p(NH₃) = 0.2 atm, p(CO₂) = 0.1 atm.
K_p = (pNH₃)² (pCO₂)
K_p = (0.2)² (0.1) = 0.04 × 0.1
K_p = 4 × 10^-3

Q9. Effect of T on K

For a reaction, K = 10 at 300 K and K = 100 at 400 K. Is the reaction Exothermic or Endothermic?

Solution:
Observation: As Temperature increases (300 → 400 K), the value of K increases (10 → 100).
K is directly proportional to T.

Conclusion: This behavior corresponds to an Endothermic Reaction (ΔH > 0). Heat acts as a reactant.

Q10. Combining Equations

Given:
1. N₂ + O₂ ↔ 2NO; K₁
2. 2NO + O₂ ↔ 2NO₂; K₂
Find K for: N₂ + 2O₂ ↔ 2NO₂.

Solution:
Add reaction (1) and (2):
(N₂ + O₂) + (2NO + O₂) ↔ 2NO + 2NO₂
Cancel 2NO on both sides:
N₂ + 2O₂ ↔ 2NO₂

When reactions are added, constants are multiplied.
K = K₁ × K₂

Important Formulae

1. Equilibrium Constant (K)

For reaction: aA + bB ↔ cC + dD

K_c = [C]^c[D]^d / [A]^a[B]^b

K_p = p_C^c p_D^d / p_A^a p_B^b

2. Relation between K_p & K_c

K_p = K_c (RT)^Δn_g

Δn_g = n_P(gas) - n_R(gas)
R = 0.0821 L atm K^-1 mol^-1

If Δn_g > 0 → K_p > K_c
If Δn_g < 0 → K_p < K_c
If Δn_g = 0 → K_p = K_c

3. Degree of Dissociation (α)

From Vapor Density (V.D.):

α = (D - d) / ((n - 1) d)

D = Initial V.D. (Theoretical)
d = Equilibrium V.D. (Observed)
n = Moles of product from 1 mole reactant

4. Thermodynamics of Equilibrium

ΔG° = -2.303 RT log K

At equilibrium, ΔG = 0.

5. Temperature Dependence

Van't Hoff Equation:

log(K₂/K₁) = (ΔH° / 2.303 R) [ (T₂ - T₁) / (T₁ T₂) ]

20 Golden Facts (NEET)

1. Only Temperature Matters: The value of the Equilibrium Constant (K) changes ONLY with temperature. It remains constant with changes in pressure, volume, concentration, or catalyst.
2. Catalyst Role: A catalyst does NOT change the value of K or the state of equilibrium. It only reduces the time taken to reach equilibrium by lowering activation energy.
3. Active Mass of Solids: The concentration (active mass) of pure solids and pure liquids is taken as UNITY (1) and ignored in K expression.
4. Unit of K: The unit of K_c is (mol/L)^Δn_g and K_p is (atm)^Δn_g. If Δn_g = 0, K has no unit.
5. Inert Gas (Const V): Adding an inert gas at constant volume has NO EFFECT on equilibrium because partial pressures of reactants/products do not change.
6. Inert Gas (Const P): Adding an inert gas at constant pressure increases volume. Equilibrium shifts to the side with more number of moles.
7. Reaction Quotient (Q): If Q < K, the reaction moves Forward (reactant consumption). If Q > K, it moves Backward (product consumption).
8. Exothermic Reaction: For ΔH < 0, increasing temperature decreases K. High yield is obtained at Low Temperature.
9. Endothermic Reaction: For ΔH > 0, increasing temperature increases K. High yield is obtained at High Temperature.
10. Pressure Effect: High pressure favors the reaction which proceeds with a decrease in the number of gaseous moles (Volume contraction).
11. Δn_g = 0 Case: Reactions like H₂ + I₂ ↔ 2HI have equal moles on both sides. Volume or Pressure changes have NO EFFECT on equilibrium position.
12. Reverse Reaction: If the equilibrium constant for a reaction is K, the constant for the reverse reaction is 1/K.
13. Multi-Step Rxn: If a reaction is the sum of two steps with constants K₁ and K₂, the overall constant is K = K₁ × K₂.
14. Boiling Point: At boiling point, Liquid ↔ Vapor are in equilibrium. Increasing pressure favors liquid state (less volume), so boiling point Increases.
15. Melting of Ice: Ice (more volume) ↔ Water (less volume). High pressure favors water (lower volume), so melting point of ice Decreases.
16. Vapor Density: As dissociation increases, the number of moles increases, so the average molar mass decreases. Thus, observed Vapor Density Decreases.
17. Strong Electrolytes: Substances like NaCl or HCl do not establish equilibrium between ions and molecules in solution because they are 100% ionized.
18. Meaning of K > 10³: If K is very large, the reaction proceeds almost to completion. Products predominate.
19. Meaning of K < 10^-3: If K is very small, the reaction hardly proceeds. Reactants predominate.
20. Stability: The larger the value of the stability constant (K for complex formation), the more stable is the complex ion.

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