Equilibrium

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Equilibrium — Quick Facts

Chemical equilibrium occurs when the forward and reverse reaction rates become equal — the concentrations of reactants and products remain constant (though not necessarily equal).

Types of Equilibrium:

• Homogeneous: All species in same phase (gases or aqueous)
• Heterogeneous: Species in different phases (e.g., solid + gas)

Equilibrium Constant Expressions:

Type	Reaction	Kp	Kc	Relationship
Homogeneous (gases)	aA + bB ⇌ cC + dD	Kp = P_C^c P_D^d / P_A^a P_B^b	Kc = [C]^c[D]^d / [A]^a[B]^b	Kp = Kc(RT)^Δn
Heterogeneous	CaCO₃(s) ⇌ CaO(s) + CO₂(g)	Kp = P_CO₂	Kc = [CO₂]	Pure solids/liquids omitted

Δn = (c + d) − (a + b) = moles of gaseous products − moles of gaseous reactants

Le Chatelier’s Principle: When a system at equilibrium is disturbed, it shifts to counteract the change.

• Increase temperature → shift toward endothermic direction (ΔH +)
• Decrease temperature → shift toward exothermic direction (ΔH −)
• Increase pressure → shift toward fewer moles of gas
• Decrease pressure → shift toward more moles of gas
• Add reactant → shift right (toward products)
• Remove product → shift right
• Catalyst: No shift — equally speeds up forward and reverse reactions

⚡ Exam tip: For the reaction N₂O₄ ⇌ 2NO₂, Δn = 2 − 1 = +1. Increasing pressure shifts toward N₂O₄ (fewer gas moles). This was tested in NEET 2020.

⚡ Common mistake: Changing concentration/pressure does NOT change K. Only temperature changes K. However, adding a catalyst does not change K either.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Equilibrium — NEET/JEE Study Guide

Relation Between Kp and Kc: Kp = Kc(RT)^Δn where R = 0.0821 L atm mol⁻¹ K⁻¹ or 8.314 J mol⁻¹ K⁻¹

For the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Δn = 2 − 4 = −2 Therefore Kp = Kc(RT)^(−2) = Kc / (RT)²

Equilibrium Constant Significance:

• K >> 1: Products are heavily favoured
• K << 1: Reactants are heavily favoured
• K ≈ 1: Comparable amounts of both

Reaction Quotient (Q): Q = same expression as K but with initial concentrations (not equilibrium)

• Q < K: Reaction proceeds forward (products increase)
• Q > K: Reaction proceeds reverse (reactants increase)
• Q = K: System is at equilibrium

Ionic Equilibrium — Strong and Weak Electrolytes:

	Strong Electrolyte	Weak Electrolyte
Degree of dissociation (α)	≈ 1 (complete)	<< 1
Conductivity	High	Low
Examples	HCl, NaOH, NaCl	CH₃COOH, NH₄OH
K (for dissociation)	Very large	Small (Ka or Kb)

For weak electrolyte AxBy ⇌ xAʸ⁺ + yBˣ⁻: $$K = \frac{[A^{y+}]^x [B^{x-}]^y}{[A_xB_y]}$$ For weak acid HA ⇌ H⁺ + A⁻: Ka = [H⁺][A⁻]/[HA] = Cα²/(1−α) ≈ Cα² (when α << 1)

pH Calculations:

• Strong acid: pH = −log[H⁺]
• Strong base: pOH = −log[OH⁻]; pH = 14 − pOH
• Weak acid: pH = ½pKa − ½log C
• Weak base: pOH = ½pKb − ½log C

Buffer Solutions: Acidic buffer: pH = pKa + log([salt]/[acid]) Basic buffer: pOH = pKb + log([salt]/[base])

⚡ NEET 2021 Qn: Calculate pH of 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵): α = √(Ka/C) = √(1.8×10⁻⁵/0.1) = √1.8×10⁻⁴ = 1.34×10⁻² [H⁺] = Cα = 0.1 × 1.34×10⁻² = 1.34×10⁻³ pH = −log(1.34×10⁻³) = 3 − log(1.34) = 3 − 0.127 = 2.87

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Equilibrium — Comprehensive Notes

Solubility and Ksp: For sparingly soluble salt like Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq): Ksp = [Ag⁺]²[CrO₄²⁻] = s × (2s)² = 4s³, where s is solubility in mol L⁻¹

Common ion effect: Adding a common ion (AgNO₃ to Ag₂CrO₄) decreases solubility. The system obeys Le Chatelier’s principle — the product ion concentration increases, forcing the equilibrium left.

Hydrolysis of Salts: Salts of strong acid + weak base (e.g., NH₄Cl): solution is acidic (pH < 7) Salts of weak acid + strong base (e.g., Na₂CO₃): solution is basic (pH > 7) Salts of weak acid + weak base (e.g., CH₃COONH₄): pH depends on relative Ka and Kb

For salt of weak acid + strong base (e.g., CH₃COONa): CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ Kh = Kw/Ka (hydrolysis constant) [OH⁻] = √(Kh × C)

Henderson-Hasselbalch Equation: For acidic buffer: pH = pKa + log([A⁻]/[HA]) For basic buffer: pOH = pKb + log([salt]/[base])

Salt Solubility Rules: A salt dissolves if ionic product < Ksp. A salt precipitates if ionic product > Ksp.

For mixed salt solutions (e.g., adding K₂CrO₄ to AgNO₃): Precipitation of Ag₂CrO₄ occurs when [Ag⁺]²[CrO₄²⁻] > Ksp. Since [Ag⁺] from AgNO₃ is known, the [CrO₄²⁻] needed to start precipitation = Ksp/[Ag⁺]².

Degree of Dissociation (α) Derivation: For weak electrolyte: HA ⇌ H⁺ + A⁻ Initial: C…0…0 Change: −Cα…+Cα…+Cα Equilibrium: C(1−α)…Cα…Cα

Ka = (Cα × Cα) / C(1−α) = Cα²/(1−α) For α << 1: Ka ≈ Cα² → α = √(Ka/C)

Common Ion Effect — Numerical: Ksp of AgCl = 1.8 × 10⁻¹⁰ Solubility in pure water: s = √(1.8×10⁻¹⁰) = 1.34 × 10⁻⁵ M In 0.1 M NaCl: Ksp = [Ag⁺][Cl⁻] = s × (0.1 + s) ≈ s × 0.1 s = Ksp/0.1 = 1.8 × 10⁻⁹ M (significantly reduced!)

NEET Pattern Analysis: Equilibrium is one of the highest-weightage topics in NEET chemistry (3–4 questions per year). Key areas: Kp/Kc relationship, Le Chatelier applications, pH calculations, buffer solutions, and solubility/Ksp problems. The common ion effect on solubility is frequently combined with precipitation questions.

⚡ NEET 2023 Qn: For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kc = 0.04 at 300 K. If initial [PCl₅] = 1 M, find equilibrium concentrations. Solution: Let x dissociate. Kc = x²/(1−x) = 0.04 → x² + 0.04x − 0.04 = 0 x = [−0.04 + √(0.0016 + 0.16)]/2 = [−0.04 + 0.401]/2 = 0.18 M [PCl₅] = 0.82 M, [PCl₃] = [Cl₂] = 0.18 M