Chemical Equilibrium

Concept Core

Dynamic equilibrium, constants, Le Chatelier, and acids/bases — the full picture.

Equilibrium Constants Kc and Kp

For aA + bB ⇌ cC + dD, at equilibrium:

Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ) (concentration terms) Kp = (Pᶜ_C × Pᵈ_D) / (Pᵃ_A × Pᵇ_B) (partial pressure terms)

Relation: Kp = Kc(RT)^Δn where Δn = (c+d) − (a+b) = change in moles of gas.

K >> 1: mostly products. K << 1: mostly reactants. K = 1: comparable amounts of both.

Reaction Quotient Q

Q has the same form as Kc but uses current (non-equilibrium) concentrations:

Q < K: reaction proceeds forward (products increasing)

Q > K: reaction proceeds backward (reactants increasing)

Q = K: system is at equilibrium

Le Chatelier's Principle

When a system at equilibrium is disturbed, it shifts to oppose the change and restore equilibrium.

Increase concentration of reactant → shifts forward

Increase pressure → shifts toward fewer moles of gas

Increase temperature → shifts in endothermic direction

Add catalyst → equilibrium position unchanged (faster, same K)

Acids, Bases, and pH

Arrhenius: acid gives H⁺, base gives OH⁻. Brønsted-Lowry: acid is proton donor, base is proton acceptor.

pH = −log[H⁺] pOH = −log[OH⁻] pH + pOH = 14 (at 25°C) Kw = [H⁺][OH⁻] = 10⁻¹⁴

Weak Acid Equilibrium & Ka

For weak acid HA ⇌ H⁺ + A⁻:

Ka = [H⁺][A⁻]/[HA] [H⁺] = √(Ka × C) (for weak acid, degree of ionisation α ≪ 1) pH = ½(pKa − log C)

Stronger acid = higher Ka = lower pKa.

Buffer Solutions & Solubility Product

Henderson-Hasselbalch:

pH = pKa + log([A⁻]/[HA])

Solubility product Ksp: For AgCl ⇌ Ag⁺ + Cl⁻, Ksp = [Ag⁺][Cl⁻]

If Q_ip (ion product) > Ksp: precipitation occurs. If Q_ip < Ksp: no precipitation.

Formula Vault

All equilibrium, acid-base, and solubility formulas.

Kc Expression

Kc = [C]ᶜ[D]ᵈ/([A]ᵃ[B]ᵇ)

Products over reactants; equilibrium conc.

Kp Expression

Kp = (P_C)ᶜ(P_D)ᵈ/...

Partial pressures at equilibrium

Kp and Kc Relation

Kp = Kc(RT)^Δn

Δn = moles gas (products−reactants)

pH and pOH

pH = −log[H⁺]; pOH = −log[OH⁻]

pH + pOH = 14 at 25°C

Water Ionisation

Kw = [H⁺][OH⁻] = 10⁻¹⁴

At 25°C; neutral pH = 7

Weak Acid [H⁺]

[H⁺] = √(Ka × C)

C = initial concentration; small α

Henderson-Hasselbalch

pH = pKa + log([A⁻]/[HA])

Buffer solution pH

Degree of Ionisation

α = √(Ka/C)

For weak acid; increases on dilution

Ksp

Ksp = [Mᵃ⁺]ᵃ[Xᵇ⁻]ᵇ

For MₐXᵦ ⇌ aM⁺ + bX⁻

Solubility from Ksp

For AB: s = √Ksp

For AB₂: s = (Ksp/4)^(1/3)

Worked Examples

5 problems — Kc, Le Chatelier, pH, weak acid, and buffer.

EasyWrite Kc for N₂ + 3H₂ ⇌ 2NH₃▾

Write the equilibrium constant expression Kc for: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Products over reactants, raised to stoichiometric powers:

Kc = [NH₃]² / ([N₂][H₂]³)

✓ Kc = [NH₃]² / ([N₂][H₂]³)

EasyFind pH of 0.01 M HCl (strong acid)▾

Calculate the pH of 0.01 M HCl solution.

HCl is a strong acid — completely ionises: [H⁺] = 0.01 = 10⁻² M

pH = −log[H⁺] = −log(10⁻²) = 2

✓ pH = 2

MediumFind [H⁺] and pH of 0.1 M acetic acid (Ka = 1.8×10⁻⁵)▾

Calculate [H⁺] and pH of 0.1 M CH₃COOH (Ka = 1.8 × 10⁻⁵).

Weak acid: [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 − 0.127 = 2.87

✓ [H⁺] = 1.34×10⁻³ M, pH ≈ 2.87

EAPCET LevelPredict shift in N₂+3H₂⇌2NH₃ when pressure is increased▾

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ. Predict equilibrium shift when: (a) pressure is increased, (b) temperature is increased.

(a) Pressure increase: Moles of gas: reactants = 1+3=4, products = 2. Fewer moles on product side. Le Chatelier → shifts forward (toward NH₃) to reduce pressure.

(b) Temperature increase: Reaction is exothermic (ΔH < 0). Le Chatelier → shifts backward (endothermic direction) to absorb heat. → Less NH₃ at higher T.

✓ (a) Forward (more NH₃); (b) Backward (less NH₃)

Trap QuestionDoes adding a catalyst change the equilibrium position?▾

A catalyst is added to an equilibrium system. Does the equilibrium shift? Does K change? ⚠️ Common EAPCET conceptual trap.

The trap: Students think catalyst shifts equilibrium toward products.

A catalyst increases BOTH forward and reverse rates equally by the same factor.

Since both rates increase equally, the equilibrium position (and K value) is unchanged.

The catalyst helps the system reach equilibrium faster — it doesn't change where equilibrium lies.

✓ No shift — catalyst does not change K or equilibrium position; only reaches it faster

Mistake DNA

5 equilibrium errors from EAPCET distractor analysis.

📊

Writing Kc with Concentrations Inverted

Kc = products/reactants. Students sometimes write reactants/products, especially when the equation is written in reverse.

❌ Wrong

For A + B ⇌ C: Kc = [A][B]/[C] ✗ (that's K for reverse!)

✓ Correct

Kc = [C]/([A][B]) ✓ Products always in numerator

Kc is always products over reactants as written. If the equation is reversed, the new K = 1/K_original.

🔢

Kp = Kc always (Forgetting (RT)^Δn)

Kp equals Kc only when Δn = 0 (no change in moles of gas). Otherwise Kp = Kc(RT)^Δn.

❌ Wrong

N₂+3H₂→2NH₃: Kp = Kc ✗ (Δn = 2−4 = −2 ≠ 0)

✓ Correct

Kp = Kc(RT)^Δn Δn = −2 Kp = Kc(RT)⁻² ✓

Calculate Δn first. Δn = 0 → Kp = Kc. Δn ≠ 0 → must include the (RT)^Δn correction.

🌡️

Le Chatelier Temperature Effect: Confusing Endothermic Direction

For exothermic reactions, increasing temperature shifts BACKWARD (favours reactants). Students often get this reversed.

❌ Wrong

Exothermic rxn, T↑: Shifts forward for more products ✗ (violates Le Chatelier)

✓ Correct

T↑ shifts toward endothermic direction ✓ For exothermic rxn: T↑ → shifts backward

Temperature is like adding/removing 'heat' as a product (exothermic) or reactant (endothermic). Increase T = add heat on product side → shifts backward.

🧪

Catalyst Shifts Equilibrium Position

A catalyst does not change where equilibrium lies — only how fast it is reached.

❌ Wrong

Catalyst added → more products formed ✗ (K or position unchanged)

✓ Correct

Catalyst → faster equilibrium but same position ✓ K unchanged

Catalyst lowers activation energy of BOTH forward and reverse reactions equally. The ratio of rate constants (= K) remains unchanged.

💧

pH of Neutral Water = 7 at ALL Temperatures

Kw changes with temperature. Neutral pH = 7 only at 25°C. At higher T, Kw increases, so neutral pH decreases below 7.

❌ Wrong

At 50°C: neutral pH = 7 ✗ (Kw > 10⁻¹⁴ at 50°C)

✓ Correct

Neutral: [H⁺]=[OH⁻] ✓ pH_neutral = pKw/2 ✓ At 50°C: pH_neutral≈6.6

pH 7 is neutral only at 25°C. At other temperatures, neutral pH = pKw/2. 'Neutral' means [H⁺] = [OH⁻], not pH = 7.

Chapter Intelligence

Equilibrium is the central chapter of Physical Chemistry — every calculation topic connects to it.

EAPCET Weightage (2019–2024)

pH and acid-base calculations

Le Chatelier predictions

Kc and Kp expressions

Weak acid [H⁺] calculations

Buffer solutions

Ksp and solubility

High-Yield PYQ Patterns

pH of weak acid given KaWrite Kc for given reactionLe Chatelier: P, T, concentrationKp from Kc calculationSolubility from KspBuffer pH using H-H equationEffect of catalyst on K

Exam Strategy

pH questions: first identify strong or weak acid/base. Strong → complete ionisation. Weak → use [H⁺] = √(Ka × C).
Le Chatelier: for pressure, count moles of gas on each side. For temperature, identify exothermic/endothermic direction. For concentration, apply forward/reverse logic.
Kp = Kc(RT)^Δn: always compute Δn first. If Δn = 0, Kp = Kc and the calculation is trivial.
Ksp questions: write the dissociation, set up ICE table (Initial, Change, Equilibrium), express Ksp in terms of s (solubility). For AB: s = √Ksp. For AB₂: 4s³ = Ksp.
This chapter connects to Thermodynamics (ΔG° = −RT ln K) and Electrochemistry (Nernst equation uses K). Strong connectivity across Physical Chemistry.

Concept Core

Formula Vault

Worked Examples

Mistake DNA

Chapter Intelligence

💡 Suggestions & Feedback

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