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)
1
Products over reactants, raised to stoichiometric powers:
2
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.
1
HCl is a strong acid — completely ionises: [H⁺] = 0.01 = 10⁻² M
2
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).
1
Weak acid: [H⁺] = √(Ka × C) = √(1.8×10⁻⁵ × 0.1) = √(1.8×10⁻⁶)
3
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.
1
(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.
2
(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.
1
The trap: Students think catalyst shifts equilibrium toward products.
2
A catalyst increases BOTH forward and reverse rates equally by the same factor.
3
Since both rates increase equally, the equilibrium position (and K value) is unchanged.
4
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~9 Le Chatelier predictions~7 Weak acid [H⁺] calculations~5 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.