Chemical Equilibrium is a small but reliable scorer, and the first thing to fix is what "weightage" actually refers to. Almost every article online quotes a combined "Chemical + Ionic Equilibrium" figure (~6–7%, 2 questions) — but those are two separate chapters. This page is Chemical Equilibrium proper: the law of mass action, Kc/Kp, the reaction quotient and Le Chatelier's principle. On its own it is about 1 question per JEE Main shift, almost always formula-direct. The pH, buffer and solubility-product material that inflates the "equilibrium" count lives in Ionic Equilibrium, a different (and harder) chapter.

How to Study Chemical Equilibrium — In Order

1. Law of mass action and the equilibrium constant. Write Kc for homogeneous and heterogeneous equilibria first — and learn the single rule that catches most students: pure solids and pure liquids are left out of the expression. Everything else builds on writing K correctly.
2. Kp, and Kp = Kc(RT)^Δn. Δn = (moles of gaseous product) − (moles of gaseous reactant). Get the sign of Δn right and this conversion is a guaranteed mark; get it wrong and the whole numerical collapses.
3. Reaction quotient Q vs K — predicting direction. If Q < K the reaction goes forward, if Q > K it goes backward, if Q = K it is at equilibrium. This "which way does it shift" logic is a favourite single-correct question and takes ten seconds once the habit is built.
4. Le Chatelier's principle. Apply it to changes in concentration, pressure/volume and temperature. The system shifts to oppose the change — toward fewer gas moles under higher pressure, toward the endothermic side on heating. A catalyst changes neither K nor the position of equilibrium.
5. Degree of dissociation (α). Express Kp/Kc in terms of α and initial pressure/concentration for a dissociation like PCl₅ ⇌ PCl₃ + Cl₂. This ties the chapter together and is the most common "show your working" numerical type.

High-Yield Sub-Topics (most-asked first)

1. Kp–Kc relation and the sign of Δn. Kp = Kc(RT)^Δn. For N₂ + 3H₂ ⇌ 2NH₃, Δn = 2 − 4 = −2, so Kp = Kc(RT)⁻². Reactions with equal gas moles on both sides (e.g. H₂ + I₂ ⇌ 2HI) have Δn = 0 and Kp = Kc — a frequently-tested special case.
2. Degree of dissociation and Kp(α). For a one-into-two gas dissociation A ⇌ B + C at total pressure P, Kp = α²P/(1 − α²). For small α this approximates Kp ≈ α²P, so α ∝ 1/√P — increasing pressure suppresses dissociation, which is Le Chatelier in algebraic form.
3. Le Chatelier predictions (qualitative). Increase pressure → shift toward fewer gas moles; add an inert gas at constant volume → no shift (partial pressures unchanged); add inert gas at constant pressure → shift toward more moles; raise temperature → shift in the endothermic direction and change K. These "predict the shift" questions are pure marks.
4. ΔG° = −RT ln K (the thermodynamics bridge). Connects equilibrium to thermodynamics: a large positive K means a large negative ΔG° (spontaneous), K = 1 means ΔG° = 0. JEE Advanced loves to ask for K given ΔG° (or vice versa) inside a thermodynamics problem.

Mistakes Students Repeatedly Make

• Putting pure solids or liquids into the equilibrium expression. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = p(CO₂) only — the two solids do not appear.
• Thinking a catalyst shifts the equilibrium. A catalyst speeds up forward and backward reactions equally; it changes neither K nor the equilibrium position — only how fast equilibrium is reached.
• Getting the sign of Δn wrong in Kp = Kc(RT)^Δn. Count gaseous moles only, products minus reactants — liquids and solids are not counted.
• Assuming adding an inert gas always shifts the equilibrium. At constant volume it does nothing (partial pressures unchanged); only at constant pressure (which increases the volume) does it shift the equilibrium toward more gaseous moles.