Equilibrium: Chemical & Ionic

Equilibrium is one of the highest-yield Physical Chemistry topics for NEET, spanning both chemical and ionic equilibrium. At dynamic equilibrium, the rate of forward reaction equals the rate of backward reaction, and concentrations remain constant (not necessarily equal).

For the reaction aA + bB ⇌ cC + dD, the equilibrium constant Kc = [C]^c[D]^d / [A]^a[B]^b (using equilibrium concentrations only). For gaseous reactions, Kp uses partial pressures.

Derivation of Kp = Kc(RT)^Δn: Since partial pressure P = CRT (from ideal gas law PV = nRT → P = (n/V)RT = CRT), substituting into the Kp expression gives Kp = Kc × (RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants). Dimensional analysis: Kp has units of (atm)^Δn; RT has units of L·atm/mol when R = 0.0821 L·atm/(mol·K) ✓

The reaction quotient Q has the same expression as K but uses non-equilibrium concentrations. If Q < K → reaction proceeds forward; Q > K → reaction proceeds backward; Q = K → at equilibrium.

ΔG° and K relationship: ΔG° = −RT ln K = −2.303 RT log K. When K > 1, ΔG° < 0 (products favored). When K < 1, ΔG° > 0 (reactants favored).

Solved Example 1: For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = 0.5 at 400 K. Calculate Kp. Δn = 2 − (1+3) = −2 Kp = Kc(RT)^Δn = 0.5 × (0.0821 × 400)^(−2) = 0.5 × (32.84)^(−2) = 0.5 / 1078.5 = 4.63 × 10⁻⁴

Le Chatelier's principle predicts the shift in equilibrium when conditions change:

• Concentration: Adding reactant → shifts forward; adding product → shifts backward
• Pressure: Increase pressure → shifts toward fewer moles of gas; no effect if Δn = 0
• Temperature: Increase T → shifts toward endothermic direction; K changes with temperature
• Catalyst: No effect on equilibrium position or K; only increases rate of attainment
• Inert gas: At constant volume → no effect (concentrations unchanged); at constant pressure → shifts toward more moles of gas (volume increases, concentrations decrease)

Ionic equilibrium: Arrhenius (H⁺/OH⁻ in water), Bronsted-Lowry (proton donor/acceptor), Lewis (electron pair acceptor/donor). Water ionization: Kw = [H⁺][OH⁻] = 10⁻¹⁴ at 25°C. pH = −log[H⁺]; pH + pOH = 14 (at 25°C).

For weak acids: Ka = Cα²; when α << 1, [H⁺] = √(Ka × C). Conjugate pair: Ka × Kb = Kw.

Solved Example 2: pH of 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵). [H⁺] = √(Ka × C) = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M pH = −log(1.34 × 10⁻³) = 3 − log 1.34 = 3 − 0.13 = 2.87

Henderson-Hasselbalch equation: pH = pKa + log([salt]/[acid]) for acidic buffers; pOH = pKb + log([salt]/[base]) for basic buffers. Buffers resist pH change: acidic buffer (weak acid + conjugate salt, e.g., CH₃COOH + CH₃COONa), basic buffer (weak base + conjugate salt, e.g., NH₄OH + NH₄Cl).

Common ion effect: Adding a common ion (e.g., Cl⁻ to HCl + CH₃COOH mixture) suppresses the ionization of the weak electrolyte.

Solubility product: For sparingly soluble salt $A_{mB_n}$: Ksp = [A^n⁺]^m[B^m⁻]^n. If ionic product > Ksp → precipitation occurs.

Solved Example 3: Solubility of AgCl (Ksp = 1.8 × 10⁻¹⁰) in 0.1 M NaCl. AgCl ⇌ Ag⁺ + Cl⁻; Ksp = [Ag⁺][Cl⁻]; [Cl⁻] = 0.1 + s ≈ 0.1 M (s << 0.1) s = Ksp/0.1 = 1.8 × 10⁻¹⁰/0.1 = 1.8 × 10⁻⁹ M (solubility reduced dramatically by common ion effect)

The key testable concept is Le Chatelier's principle applications and pH calculations using Ka, Henderson-Hasselbalch equation, and Ksp with the common ion effect.