Chemical Equilibrium Notes – Complete Chapter

Types of Reactions

① Irreversible Reactions

Take place in one direction only
Products do NOT form reactants again
Take place in open vessel
e.g. C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

② Reversible Reactions

Take place in both directions simultaneously
Products make reactants again
Take place in closed vessel
e.g. CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Equilibrium

"State in a reversible reaction where rate of forward reaction becomes equal to backward reaction."

At Eqbm: Rate of forward rxn = Rate of backward rxn
Rate of change of reactants to products = Rate of change of products to reactant

⚠️ Important Notes about Equilibrium

After attaining equilibrium, all measurable properties of reaction like mol, pH, concentration etc. do NOT change
"Constant" does NOT mean "equal" at all
At equilibrium — concentrations of reactants and products are constant but not necessarily equal

Practice Questions

Q: Rxn 2NO + O₂ ⇌ 2NO₂ started with 4 mol NO & 6 mol O₂. At eqbm 2 mol NO₂ are present. Plot concn vs time graph.

2NO(g) + O₂(g) ⇌ 2NO₂(g); t=0: NO=4, O₂=6, NO₂=0; By problem: n_NO₂ = 2mol → 2x=2 → x=1; At eqbm: NO=4−2(1)=2mol; O₂=6−1=5mol; NO₂=2mol

Ans: At eqbm: [NO]=2mol, [O₂]=5mol, [NO₂]=2mol ✓

Q: Which of the following may be a reversible rxn? (a) 2KClO₃→2KCl+3O₂ (open vessel) (b) H₂(g)+I₂(g)⇌2HI(g) (c) HCl+NaOH→NaCl+H₂O (d) AgNO₃+NaCl→NaNO₃+AgCl

Ans: (b) H₂(g)+I₂(g)⇌2HI(g) ✓ — gases in closed vessel, all others irreversible

Q: S1: At eqbm ΔG=0. S2: Eqbm is state of minimum stability.

S1 is correct — at equilibrium ΔG=0; S2 is wrong — equilibrium is state of minimum Gibbs free energy (maximum stability)

Ans: (ii) S1✓, S2× ✓

Characteristics of Equilibrium

5 Key Characteristics

Stable in nature: Once rxn reaches state of eqbm, it always tends to stay there
Dynamic but quasistatic in nature: Reactions keep happening in both directions but macroscopic properties appear static
Concn, pH, colour, mol etc. become constant once eqbm is reached
Eqbm can be achieved from any direction: e.g. N₂O₄ (colourless) ⇌ 2NO₂ (brown) — equilibrium same from either side
Effect of Catalyst: In presence of catalyst, eqbm achieves sooner than expected; catalyst does NOT change equilibrium constant (Keq)

Types of Equilibrium

① Physical Equilibrium

Equilibrium in physical process

Melting: S ⇌ L
Boiling: L ⇌ V
Sublimation: S ⇌ V
Solubility equilibrium

② Chemical Equilibrium

Equilibrium achieved in chemical reactions

(a) Homogeneous equilibrium: All species in same phase

e.g. N₂(g) + O₂(g) ⇌ 2NO(g)
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

(b) Heterogeneous equilibrium: Species in different phases

e.g. CaCO₃(s) ⇌ CaO(s) + CO₂(g)
NH₄HS(s) ⇌ NH₃(g) + H₂S(g)

Law of Mass Action (LOMA) & Equilibrium Constant Kc

By Guldberg & Waage

Active mass = Concentration = mol/Volume
Active mass of pure liquid & pure solid = 1
Active mass = mol/Volume = wt/(mol wt × V) = ρ/(mol wt) = constant (for solids/liquids)

Statement of LOMA

"Rate of any reaction is directly proportional to product of active masses of its reactants raised to the power their stoichiometric coefficients"

For: aA + bB ⇌ cC + dD

Rate of forward rxn: r_f = k_F [A]ᵃ [B]ᵇ (where k_F = rate const of forward rxn)
Rate of backward rxn: r_b = k_b [C]ᶜ [D]ᵈ (where k_b = rate const of backward rxn)

At Equilibrium: r_f = r_b → k_F[A]ᵃ[B]ᵇ = k_b[C]ᶜ[D]ᵈ

\[K_c = \frac{k_F}{k_b} = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]

Kc = Equilibrium constant in terms of concentration | T → Temperature in K

Important Examples of Kc

N₂(g) + O₂(g) ⇌ 2NO(g): \(K_c = \dfrac{[NO]^2}{[N_2][O_2]}\)
CaCO₃(s) ⇌ CaO(s) + CO₂(g): \(K_c = \dfrac{[CO_2][CaO]^1}{[CaCO_3]^1} = [CO_2]\) (since active mass of solids = 1)
CH₃COO·C₅H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₅H₅OH(aq): \(K_c = \dfrac{[CH_3COOH][C_5H_5OH]}{[CH_3COOC_5H_5]}\) (since [H₂O]=1)

Practice Questions

Q: Active mass of 2g NaOH (solid) is?

Ans: (b) 1 ✓ — Active mass of pure solid = 1

Q: Active mass of 20g NaOH dissolved in 2L water?

Active mass = mol/V(L) = (20/40)/2 = 0.5/2 = 1/4

Ans: (c) 1/4 ✓

Q: Rate constants of forward & backward rxn are 2×10⁻³ & 5×10⁻⁴ respectively. Kc of rxn?

Kc = k_F/k_b = 2×10⁻³/5×10⁻⁴ = 4

Ans: (b) 4 ✓

Kp and Relation between Kp & Kc

Kp — Equilibrium Constant in terms of Partial Pressure

For gaseous rxn: aA(g) + bB(g) ⇌ cC(g) + dD(g)

At constant T: P ∝ C (from pV = nRT; p = n/V × RT = C×RT)

\[K_p = \frac{P_C^c \cdot P_D^d}{P_A^a \cdot P_B^b}\]

Kp = Equilibrium constant in terms of partial pressure

Derivation of Kp = Kc(RT)^Δng

Partial pressure: P_A = [A]RT; P_B = [B]RT etc.
\(K_p = \dfrac{([C]RT)^c([D]RT)^d}{([A]RT)^a([B]RT)^b} = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \times (RT)^{(c+d)-(a+b)}\)

\[K_p = K_c[RT]^{\Delta n_g}\]

Where T → Temperature in K | R → 0.082 = 1/12 Latm/molK | Δng = (c+d)−(a+b) = moles of gaseous products − moles of gaseous reactants

Kp vs Kc — Based on Δng

Δng > 0Kp/Kc > 1
Kp > Kc

Δng < 0Kp/Kc < 1
Kp < Kc

Δng = 0Kp/Kc = 1
Kp = Kc

Units of Kp and Kc

Unit of Kp = (atm)^Δng
Unit of Kc = (mol/L)^Δng
If Δng = 0 → both Kp and Kc are dimensionless

Practice Questions

Q: Match the following — (a) CaCO₃(s)⇌CaO(s)+CO₂(g) (b) N₂(g)+3H₂(g)⇌2NH₃(g) (c) H₂(g)+I₂(g)⇌2HI(g) (d) PCl₅(g)⇌PCl₃(g)+Cl₂(g); with (i)Kp>Kc (ii)Kp<Kc (iii)Kp=Kc

Δng: (a)=1→Kp>Kc; (b)=2−4=−2→Kp<Kc; (c)=2−2=0→Kp=Kc; (d)=2−1=1→Kp>Kc

Ans: (i) a&d, (ii) b, (iii) c ✓

Q: Correct relation between Kp & Kc for N₂(g)+3H₂(g)⇌2NH₃(g)?

Δng = 2−4 = −2; Kp = Kc[RT]⁻² → Kc = Kp[RT]²

Ans: (b) Kc = Kp(RT)² ✓

Q: For which rxn relation log Kp = log Kc + log RT is valid?

log Kp = log Kc + Δng log RT; given Δng = 1; so rxn with Δng=1 needed: PCl₅⇌PCl₃+Cl₂ has Δng=1

Ans: (a) PCl₅(g)⇌PCl₃(g)+Cl₂(g) ✓

Q: For which rxn unit of Kc is L/mol?

L/mol = [mol/L]⁻¹ → Δng = −1; need rxn with Δng=−1; 2SO₂(g)+O₂(g)⇌2SO₃(g) has Δng=2−3=−1

Ans: (d) 2SO₂(g)+O₂(g)⇌2SO₃(g) ✓

Q: Unit of Kp for N₂(g)+3H₂(g)⇌2NH₃(g)?

Δng = 2−4 = −2; Unit of Kp = atm^Δng = atm⁻²

Ans: (d) atm⁻² ✓

Factors Affecting Equilibrium Constant

(a) Changing Stoichiometric Coefficients

A ⇌ B → Kc = [B]/[A]
2A ⇌ 2B → Kc = [B]²/[A]² = (original Kc)²
In general: nA ⇌ nB → Kc' = [Kc]ⁿ

(b) Reversing the Direction

A ⇌ B → Kc = [B]/[A]
B ⇌ A → Kc' = [A]/[B] = 1/Kc

A ⇌ B → K₁ = [B]/[A]
B ⇌ C → K₂ = [C]/[B]
A ⇌ C → K₃ = [C]/[A] = [B]/[A] × [C]/[B] = K₃ = K₁ × K₂

(d) Subtracting Two Reactions

A ⇌ B → K₁ | C ⇌ B → K₂
eq(i) − eq(ii): A ⇌ C → K₃
Reverse K₂: B ⇌ C → K₂' = 1/K₂
Add: A + B ⇌ B + C → K₃ = K₁ × K₂' = K₁/K₂

Practice Questions

Q: If eqbm const of 2SO₂(g)+O₂(g)⇌2SO₃(g) is K, then eqbm const of SO₃(g)⇌SO₂(g)+½O₂(g) is?

Step1: 2SO₃⇌2SO₂+O₂ → K₁=1/K; Step2: ×½ → SO₃⇌SO₂+½O₂ → K₂=(1/K)^½ = 1/√K

Ans: (d) 1/√K ✓

Q: Given: N₂+O₂⇌2NO; K₁ and 2NO+O₂⇌2NO₂; K₂. Find eqbm const of 2NO₂(g)⇌N₂(g)+2O₂(g).

Add eq1+eq2: N₂+2O₂⇌2NO₂; K=K₁×K₂; Reverse: 2NO₂⇌N₂+2O₂; K=1/(K₁×K₂)

Ans: (d) 1/(K₁K₂) ✓

Q: Given: ①2SO₂⇌2SO₃+O₂; K₁ ②2SO₂+O₂⇌2SO₃; K₂ ③SO₃⇌SO₂+½O₂; K₃ ④SO₂+½O₂⇌SO₃; K₄. Find K₁=[K₂]ˣ=[K₃]ʸ=[K₄]ᶻ values of x,y,z.

By eq①&②: K₂=1/K₁; By eq①&③: K₃=√K₁; By eq①&④: K₄=1/√K₁; So K₁=[K₂]⁻¹=[K₃]²=[K₄]⁻²

Ans: x=−1, y=2, z=−2 ✓

Types of Problems on Equilibrium

Type 1 — When information at equilibrium is given

Directly substitute equilibrium concentrations/pressures in Kc/Kp expression.

Practice Questions — Type 1

Q: In rxn 2SO₂(g)+O₂(g)⇌2SO₃(g), eqbm conc of SO₂, O₂, SO₃ are 0.2, 0.25 & 0.5 mol/L. Find Kc.

Kc = [SO₃]²/[SO₂]²[O₂] = (0.5×0.5)/(0.2×0.2×0.25) = 0.25/0.01 = 25

Ans: (b) 25 ✓

Q: In rxn 3O₂(g)⇌2O₃(g), eqbm conc of O₂ is 0.04M. If Kc of rxn is 3×10⁻²⁹, find conc of O₃ at eqbm.

Kc=[O₃]²/[O₂]³; [O₃]²=3×10⁻²⁹×(4×10⁻²)³=3×10⁻²⁹×64×10⁻⁶=192×10⁻³⁵=19.2×10⁻³⁴; [O₃]=√(19.2×10⁻³⁴)=4.2×10⁻¹⁷M

Ans: O₃ = 4.2×10⁻¹⁷ M ✓

Q: In rxn 2A(g)+B(g)⇌4C(g); Kc=16 & 2 mol each of A,B,C not at eqbm. Volume of vessel is?

Kc=[C]⁴/[A]²[B]=(2/V)⁴/(2/V)²(2/V)=16; (2/V)⁴/(2/V)³=16; 2/V=16; V=2/16=1/8L=125mL

Ans: (c) 125 mL ✓

Type 2 — When initial data is given

Set up ICE table (Initial, Change, Equilibrium) and solve for x. Then calculate Kc or Kp.

⚡ Ritrick — ICE Method for Kp

For: aA(g) + bB(g) ⇌ cC(g) + dD(g)

\(K_p = \dfrac{n_C^c \cdot n_D^d}{n_A^a \cdot n_B^b} \left[\dfrac{P_{Total}}{n_{Total}}\right]^{\Delta n_g}\)

n = no. of moles

Ritrick: Inert Gas V No (No effect on equilibrium at constant V)

Practice Questions — Type 2

Q: Initial pressure of SO₂ & O₂ in rxn 2SO₂(g)+O₂(g)⇌2SO₃(g) are 4atm & 3atm. At eqbm, partial pressures of SO₂ & SO₃ are equal. Find Kp.

At t=0: SO₂=4, O₂=3, SO₃=0; At teq: SO₂=4−2p=2p → p=1; SO₂=2atm, O₂=3−1=2atm, SO₃=2atm; Kp=(2)²/(2)²(2)=4/8=1/2

Ans: Kp = 1/2 ✓

Q: PCl₅(g)⇌PCl₃(g)+Cl₂(g) started with 3 mol PCl₅. At eqbm 1/3rd of PCl₅ reacts. P_total = 8atm. Find Kp.

At t=0: 3mol; At teq: PCl₅=3−1=2mol, PCl₃=1mol, Cl₂=1mol; n_Total=4mol; Kp=(1×1/4)/(2/4)×[8/4]¹=(1/16)/(1/2)×2=1/4×2=1

Ans: Kp = 1 ✓

Q: 5 mol SO₃ in closed vessel: 2SO₃(g)⇌2SO₂(g)+O₂(g). At eqbm 2 mol SO₂ dissociated, total pressure = 1.8 atm. Find Kp.

At teq: SO₃=3mol, SO₂=2mol, O₂=1mol; n_Total=6mol; Kp=(2²×1)/3² × [1.8/6]¹=4/9×0.3=4×1.8/9×6=0.4/3=0.133

Ans: Kp = 0.133 ✓

Type 3 — Heterogeneous Equilibrium

Active mass of solid & liquid = 1; Active mass (partial pressure) of solid & liquid = 1.

For solids: use mole fraction or simply take activity = 1 in Kp expression.

Practice Questions — Type 3

Q: CaCO₃ dissociates as: CaCO₃(s)⇌CaO(s)+CO₂(g); Kp=0.5. P_CO₂ at eqbm?

Kp = P_CO₂ × [CaO]¹/[CaCO₃]¹ = P_CO₂ (since active mass of solids = 1); P_CO₂ = 0.5 atm

Ans: (d) 0.5 atm ✓

Q: NH₄HS(s)⇌NH₃(g)+H₂S(g); Kp=9. Total pressure at eqbm?

Kp = P_NH₃ × P_H₂S = p×p = p² = 9; p=3atm; P_Total = 3+3 = 6atm

Ans: (d) 6 atm ✓

Q: NH₂COONH₄(s)⇌2NH₃(g)+CO₂(g); P_total=1.5atm at eqbm. Find Kp.

Partial pressures: NH₃=2p, CO₂=p; P_total=3p=1.5 → p=0.5; Kp=[2p]²[p]=(1)²(0.5)=0.5

Ans: (b) 0.5 ✓

Degree of Dissociation (α)

Definition

No. of dissociated mol of reactant when initially 1 mol is taken is called degree of dissociation (α).

A ⇌ Product; At t=0: a mol; At t=teq: (a−x) mol

\[\alpha = \frac{x}{a} \times 100 \quad (\text{in \%}) \quad \text{or} \quad \alpha = \frac{x}{a} \text{ (as fraction)}\]

α < 1 always | α < 100% always | α: always in fraction, not percentage

⚡ Important Ritrick

When initial mol are NOT given → take them equal to stoichiometric coefficient of reactant

Kp for PCl₅ ⇌ PCl₃ + Cl₂ (α < 1)

At t=0: 1 mol PCl₅; At teq: (1−α) mol PCl₅, α mol PCl₃, α mol Cl₂
n_Total = 1−α+α+α = 1+α

\[K_p = \frac{\alpha^2 P}{1-\alpha^2}\]

where P = total pressure at eqbm

Practice Questions

Q: In rxn A⇌B, 2 mol of reactant taken. If at eqbm 0.8 mol of A reacted, degree of dissociation of A?

a=2, x=0.8; α = x/a × 100 = 0.8/2 × 100 = 40%

Ans: (c) 40% ✓

Q: A⇌2B in closed vessel with 2 mol A. At eqbm 0.8 mol B found. Degree of dissociation of A?

2x=0.8 → x=0.4; a=2; α = 0.4/2 × 100 = 20%

Ans: (a) 20% ✓

Q: SO₃ dissociates: 2SO₃⇌2SO₂+O₂ in closed vessel of 1L. Initially 5 mol SO₃, 40% dissociated at eqbm. Find Kc.

α=40/100=0.4; x=5×0.4=2mol; At teq: SO₃=3mol, SO₂=2mol, O₂=1mol; Kc=[SO₂]²[O₂]/[SO₃]²=(2)²(1)/(3)²=4/9

Ans: (c) 4/9 ✓

Q: 2NO(g)⇌N₂(g)+O₂(g) in closed vessel. At eqbm 50% NO dissociated, total pressure = 5atm. Find Kp.

Initial mol(take=2); x=2×50/100=1mol; At teq: NO=1, N₂=0.5, O₂=0.5; n_Total=2mol; Δng=0; Kp=[1/2×1/2]/(1/2)²×[5/2]⁰=(1/4)/(1/4)=1... wait Kp=(½)(½)/(1)²=1/4

Ans: Kp = 1/4 ✓

Q: Degree of dissociation of PCl₅ is α, total pressure P. Find Kp in terms of P and α.

From formula: Kp = α²P/(1−α²)

Ans: Kp = α²P/(1−α²) ✓

Q: For 2AB₂(g)⇌2AB(g)+B₂(g), degree of dissociation of AB₂ is α, total pressure P (α << 1). Find α.

After derivation with α<<1 approx: α³ = 2Kp/P → α = [2Kp/P]^(1/3)

Ans: (b) α ∝ [Kp/P]^(1/3) ✓

Applications of Equilibrium Constant

① To Predict Stability of Reactants & Products

R ⇌ P; Keq = [P]/[R]
Keq > 1: [P] > [R] → Product more stable
Keq < 1: [P] < [R] → Reactant more stable
Keq = 1: [P] = [R] → Both equally stable
Stability of reactant ∝ 1/Keq

② To Predict Direction of Reaction — Reaction Quotient Qc

At equilibrium: Kc = [P]/[R]
At any time: Qc = [P]/[R] (Reaction Quotient)
Note: Qc always approaches Kc
Case I: Qc < Kc → Rxn going in forward direction
Case II: Qc > Kc → Rxn going in backward direction
Case III: Qc = Kc → Rxn is at equilibrium

Practice Questions

Q: Which relation is incorrect? (a) Keq=10⁶ → Product more stable (b) Keq=10⁻⁶ → Product more stable (c) Keq=10⁴ → Product more stable (d) Keq=10¹ → Product more stable

Ans: (b) Keq=10⁻⁶ → Product more stable is INCORRECT ✓ (Keq<1 means reactant more stable)

Q: Eqbm const of rxn mixture is 10²². Mixture contains?

Ans: (a) Mostly products ✓ — Keq>>1 means products favoured

Q: Eqbm constants of dissociation of oxides: AO⇌B+½O₂; K=10⁻⁵ | BO⇌B+½O₂; K=10⁻⁶ | CO⇌C+½O₂; K=10⁻⁷. Correct order of stability of oxides?

Stability of oxide ∝ 1/Keq; Smaller Keq → more stable oxide; CO < BO < AO in terms of Keq → CO most stable

Ans: (d) CO > BO > AO (stability of oxide) ✓

Q: For rxn A⇌B; Kc=10². If at anytime rxn mixture has 0.02M B & 0.01M A. Rxn is moving?

Qc = [B]/[A] = 0.02/0.01 = 2; Kc=100; Qc < Kc → forward direction

Ans: (a) Forward direction ✓... wait Qc=2<100=Kc → forward ✓

Factors Affecting Equilibrium — Le Chatelier Principle

Le Chatelier Principle

"Whenever equilibrium is disturbed by any external factor, it shifts in that direction in which the effect of disturbance is cancelled out."

Effect of Concn → Does NOT affect Keq
Effect of Total Pressure → Does NOT affect Keq
Effect of Volume → Does NOT affect Keq
Effect of Inert Gas → Does NOT affect Keq
Effect of Temperature → AFFECTS Keq

① Effect of Concentration

Increasing conc of reactant → Forward shift
Increasing conc of product → Backward shift
Decreasing conc of product → Backward shift
Decreasing conc of product → Forward shift
Adding solid or liquid at eqbm → Eqbm NOT affected

② Effect of Increase in Pressure (P↑) & Volume (V)

P↑, V↓ → no. of molecules per unit volume↑ → Eqbm shifts towards less no. of gas molecules
V↑, P↓ → no. of molecules per unit volume↓ → Eqbm shifts towards more no. of gas moles

③ Addition of Inert Gas

(a) At Constant P (Flexible walls): Adding inert gas → n_Total↑, Volume↑ → Eqbm shifts towards more no. of gas moles
(b) At Constant V (Rigid walls): Adding inert gas → n_Total↑, P_Total↑; but mole fractions of reactants/products unchanged → No effect on equilibrium

⚡ Ritrick: Inert Gas → V (No effect at constant V)

④ Effect of Increase in Temperature

On increasing temp → Eqbm shifts towards endothermic direction
e.g. PCl₅(g)⇌PCl₃(g)+Cl₂(g) [ΔH>0] → T↑ → Forward shift
e.g. N₂(g)+3H₂(g)⇌2NH₃(g) [ΔH<0] → T↑ → Backward shift
For endothermic rxn: T↑ → K↑
For exothermic rxn: T↑ → K↓

Practice Questions

Q: Which of the following eqbm will shift nowhere on increasing V? (a)N₂+3H₂⇌2NH₃ (b)CaCO₃(s)⇌CaO(s)+CO₂(g) (c)2HI(g)⇌H₂(g)+I₂(g) (d)2SO₃⇌2SO₂+O₂

V↑ → shift towards more gas moles. In (c) 2HI⇌H₂+I₂: Δng=0 → no shift

Ans: (c) 2HI(g)⇌H₂(g)+I₂(g) ✓

Q: Best conditions for more production of NH₃: N₂+3H₂⇌2NH₃ [ΔH<0]?

ΔH<0 exothermic: Low T (favours forward); P↑: shift to less moles (products) → High P needed

Ans: (c) Low T, High P ✓

Q: N₂(g)+3H₂(g)⇌2NH₃(g). If some N₂(g) introduced at eqbm, the eqbm will shift?

Ans: (a) Forward direction ✓ — Adding reactant → forward shift

Q: CaCO₃(s)⇌CaO(s)+CO₂(g). If CaO introduced, eqbm shifts?

CaO is solid → active mass = 1; Adding solid doesn't change Kc; Eqbm not affected

Ans: (c) Nowhere ✓

Q: Rxn aA(g)⇌bB(g)+cC(g) shifts forward on increasing pressure. Which relation correct?

P↑ → shift towards less gas moles; For forward shift on P↑: reactant side must have MORE moles; a > b+c

Ans: (d) a > b+c ✓

Q: K₁ & K₂ are eqbm const of exothermic rxn at T₁ & T₂. If T₂ < T₁ then?

For exothermic rxn: T↑ → K↓; T₂<T₁ means T₂ is lower temp → K₂ is higher; K₁ < K₂... wait: T₂<T₁ → lower T → higher K for exothermic → K₂>K₁... no: if T₂<T₁, lower T gives higher K for exothermic rxn → K₂>K₁... so K₁<K₂... but if K₁ is at T₁ (higher T) → K₁ smaller. So K₁<K₂? Let re-check: higher T=T₁ → lower K for exothermic. Yes K₁<K₂ → K₁<K₂ but Q asks T₂<T₁ → T₂ lower → K₂ higher. Since K₁ at higher T₁: K₁<K₂... but ans written K₁>K₂. Re-read: T₂<T₁ means T₁ is larger. For exothermic rxn T↑→K↓. So K at T₁ (higher) is smaller = K₁<K₂. So K₁<K₂? Answer should be K₂>K₁ which is same as K₁<K₂

Ans: (a) K₁ > K₂ — wait: T₂<T₁; exothermic: lower T→higher K; so K at T₂(lower)>K at T₁(higher) → K₂>K₁ → correct answer is K₂>K₁, so if T₂<T₁ → K₁<K₂. But option says K₁>K₂. Since T₁>T₂ and exothermic → K₁<K₂. Ans: (b) actually let me just state: For exothermic rxn T↑→K↓; T₁>T₂ → K₁<K₂ ✓

Effect of Temperature on Keq — Vant Hoff Equation

Derivation

ΔG° = −RT ln K
ΔG° = ΔH° − TΔS°
−RT ln K = ΔH° − TΔS°

\[\ln K = -\frac{\Delta H°}{RT} + \frac{\Delta S°}{R}\]

\[\log K = \frac{-\Delta H°}{2.303R}\left[\frac{1}{T}\right] + \frac{\Delta S°}{2.303R}\]

Vant Hoff Equation — y = mx + c form (log K vs 1/T graph)

Graphs — log K vs 1/T

Endothermic rxn (ΔH>0): Slope = −ΔH°/2.303R < 0 → Negative slope (downward line)
Exothermic rxn (ΔH<0): Slope = −ΔH°/2.303R > 0 → Positive slope (upward line)
1/T ↑ → T↓ → for endothermic: log K↑ → K↑
For endothermic rxn: T↑ → K↑ | For exothermic rxn: T↑ → K↓

ΔG° = −RT ln K = −2.303 RT log K

Slope of log K vs 1/T graph = −ΔH°/2.303R

Practice Questions

Q: Which relation is correct? (a) ΔG°=−2.303 RT ln K (b) ΔG°=−RT ln Q (c) ΔG°=−2.303 RT log K (d) ΔG°=−2.303 RT log Q

Ans: (c) ΔG° = −2.303 RT log K ✓

Q: Correct value of slope of log K vs 1/T graph?

Ans: (d) −ΔH°/2.303R ✓

Relation between α and Vapour Density

For: A(g) ⇌ n·B(g)

D = Vapour density of reactant = Molar mass of reactant / 2
d = Vapour density of eqbm mixture = Molar mass of mixture / 2
n = no. of moles of gaseous products

\[\alpha = \frac{D - d}{(n-1)d}\]

\[\frac{D}{d} = 1 + (n-1)\alpha\]

α: always in fraction not percentage | tan θ of D/d vs α graph = n−1

⚡ Ritrick — α related to Pressure (for α << 1)

\(\alpha \propto \left[\dfrac{1}{P}\right]^{\dfrac{\Delta n_g}{\text{Sum of St. coeff of gaseous product}}}\)

e.g. for 2AB₂⇌2AB+B₂ with Δng=1: α ∝ [1/P]^(1/(2+1)) = [1/P]^(1/3)

Practice Questions

Q: Vapour density molar mass of Eqbm mixture of rxn N₂O₄(g)⇌2NO₂(g) is 70. Find degree of dissociation of N₂O₄.

M_N₂O₄ = 92; D = 92/2 = 46; M_mixture = 70; d = 70/2 = 35; n=2; α = (D−d)/((n−1)d) = (46−35)/((2−1)×35) = 11/35

Ans: α = 11/35 ✓

Q: For which rxn slope of D/d vs α curve may be 1/2 [tan θ = 1/2]?

tan θ = n−1; 1/2 = n−1; n = 3/2; Need rxn with gaseous product n=3/2; SO₃(g)⇌SO₂(g)+½O₂(g) has n=3/2

Ans: (b) SO₃(g)⇌SO₂(g)+½O₂(g) ✓

Le Chatelier on Physical Equilibrium

Case 1: Effect of P & T on Solubility of Gases in Liquid

Covered in Henry's Law — gas solubility ∝ partial pressure of gas above liquid

Case 2: Effect of P on Solid-Liquid Equilibrium

(i) H₂O(s) ⇌ H₂O(l): On P↑ → Forward shift (ice melts) since ρ_H₂O(s) < ρ_H₂O(l)
(ii) Wax(s) ⇌ Wax(l): On P↑ → Backward shift since ρ_wax(s) > ρ_wax(l)

⚡ Ritrick: On increasing P, solid-liquid eqbm shifts in direction of higher density