Kp, Kc & Equilibrium Calculations: JEE Main

Equilibrium calculations — finding Kc, Kp, degree of dissociation, and equilibrium concentrations — appear in two to three JEE Main questions per session and are among the most formula-dependent problems in Physical Chemistry. Students who confuse Kp and Kc, or who cannot set up the ICE table quickly, consistently drop marks here. This guide takes you through every calculation type the exam uses.

Kc and Kp: Definitions and Relationship

For a general reaction: aA + bB ⇌ cC + dD. Kc = [C]^c [D]^d / ([A]^a [B]^b) (concentrations in mol/L, at equilibrium). Kp = (P_C)^c (P_D)^d / ((P_A)^a (P_B)^b) (partial pressures in atm or bar). Relationship: Kp = Kc × (RT)^Δn where Δn = (c + d) − (a + b) = change in moles of gas. R = 0.0821 L·atm/(mol·K). JEE uses this in two ways: (1) give Kc and T, find Kp; (2) give Kp and T, find Kc. If Δn = 0 (same moles of gas on both sides): Kp = Kc. If Δn > 0: Kp > Kc. If Δn < 0: Kp < Kc.

Kx (mole fraction equilibrium constant): Kp = Kx × P^Δn where P is total pressure. Kx depends on pressure (unlike Kc and Kp which depend only on temperature). JEE tests the effect of increasing total pressure on Kx: if Δn < 0, increasing P shifts equilibrium toward products (more gas moles combine), increasing Kx; if Δn > 0, increasing P decreases Kx. Take a free equilibrium mock. See our equilibrium guide.

ICE Table Method for Equilibrium Calculations

ICE = Initial, Change, Equilibrium. Template: set up concentration (or partial pressure) terms for each species. Let x = change in one species. Express all equilibrium concentrations in terms of x. Substitute into K expression, solve for x. Example: H₂ + I₂ ⇌ 2HI, Kc = 50, initial [H₂] = [I₂] = 1 M, [HI]₀ = 0. ICE: H₂: 1-x, I₂: 1-x, HI: 2x. Kc = (2x)²/((1-x)²) = 50 → 2x/(1-x) = √50 = 7.07 → x = 7.07/9.07 ≈ 0.78. Equilibrium: [H₂] = [I₂] = 0.22 M, [HI] = 1.56 M. JEE uses this template directly. Simplification: if K is very large (reaction almost complete), x ≈ initial concentration; if K is very small, approximate (1-x) ≈ 1 and solve the simplified equation.

Degree of Dissociation and Vapour Density

Degree of dissociation α = fraction of initial moles that have dissociated. For N₂O₄ ⇌ 2NO₂ (1 mole → 2 moles, Δn = 1): at pressure P, if initial moles = 1, equilibrium moles = (1 − α) + 2α = 1 + α. Mole fractions: X(N₂O₄) = (1−α)/(1+α), X(NO₂) = 2α/(1+α). Partial pressures: P(N₂O₄) = P(1−α)/(1+α), P(NO₂) = 2Pα/(1+α). Kp = [P(NO₂)]²/P(N₂O₄) = 4α²P/(1−α²). Solving for α given Kp and P: α = √(Kp/(4P + Kp)). Vapour density method: D (before dissociation) = M_r/2; d (after dissociation) = M_observed/2. α = (D−d)/((n−1)d) where n = moles of product per mole of reactant. For N₂O₄: n = 2, so α = (D−d)/d.

Reaction Quotient Q and Predicting Direction

Q is calculated the same way as Kc or Kp but using current (non-equilibrium) concentrations. If Q < Kc: reaction proceeds forward (toward products) to reach equilibrium. If Q > Kc: reaction proceeds backward (toward reactants). If Q = Kc: at equilibrium. JEE question: given initial concentrations of all species and Kc, find Q and state which direction the reaction proceeds. This is a direct one-step comparison — no calculation required beyond computing Q. For the ionic equilibrium analogue (Ksp and Q for precipitation), see our solubility product guide and for Le Chatelier's applications see our Le Chatelier guide.