Chemical equilibrium for JEE Main

Equilibrium questions in JEE Main are mostly bookkeeping. Write the reaction, set up a table of initial, change and equilibrium amounts, substitute into K and solve. The concept questions (what happens when you add an inert gas, raise the temperature or add a catalyst) take seconds once the rules are clear. The traps are in the details: which species appear in K, whether you are working with concentrations or pressures, and whether the volume is fixed.

The official 2026 syllabus covers dynamic equilibrium; physical equilibria (solid–liquid, liquid–gas, gas–gas and solid–gas) and Henry's law; the law of chemical equilibrium, Kp and Kc and their significance; the significance of ΔG and ΔG°; the effect of concentration, pressure, temperature and catalysts; and Le Chatelier's principle. Ionic equilibrium is in the same unit and has its own guide.

Dynamic equilibrium

At equilibrium the forward and reverse reactions continue at equal rates, so concentrations stay constant. Equilibrium can be reached from either side, and it is possible only in a closed system.

Physical equilibria follow the same idea:

• ice and water coexist at 0 °C (273 K) and 1 atm
• a liquid in a closed vessel reaches a constant vapour pressure at a given temperature
• Henry's law: the partial pressure of a gas over a solution is proportional to its mole fraction in the solution, p = K_H x. Gas solubility rises with pressure and falls with temperature, which is why a soda fizzes when opened

The equilibrium constant

For aA + bB ⇌ cC + dD:

Kc = [C]ᶜ[D]ᵈ/([A]ᵃ[B]ᵇ), and Kp is the same expression in partial pressures.

• Pure solids and pure liquids are left out of K. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = p(CO₂).
• K depends only on temperature, not on starting amounts, pressure or a catalyst.
• A large K (above about 10³) means products dominate; a small one (below about 10⁻³) means reactants dominate.

Kp = Kc(RT) raised to the power Δn_g, where Δn_g is moles of gaseous products minus gaseous reactants. So Kp = Kc when Δn_g = 0, Kp = Kc·RT when Δn_g = 1, and Kp = Kc/(RT)² when Δn_g = −2. Use R = 0.0831 L bar K⁻¹ mol⁻¹ with pressures in bar.

Reaction quotient

Q has the same form as K but uses the current amounts.

• Q < K: the reaction moves forward
• Q > K: the reaction moves backward
• Q = K: the system is at equilibrium

Equilibrium and Gibbs energy

ΔG = ΔG° + RT ln Q. At equilibrium ΔG = 0, so ΔG° = −RT ln K = −2.303RT log K. A negative ΔG° means K > 1. Combining this with ΔG° = ΔH° − TΔS° shows why K changes with temperature: for an exothermic reaction, K falls as T rises.

Degree of dissociation

The degree of dissociation α is the fraction of a reactant that has dissociated. For PCl₅ ⇌ PCl₃ + Cl₂ starting with 1 mol:

With total pressure P, each partial pressure is (moles/total moles) × P, which gives Kp = α²P/(1 − α²). For N₂O₄ ⇌ 2NO₂, the same method gives Kp = 4α²P/(1 − α²).

Because dissociation increases the number of gas molecules, the vapour density falls. If D is the theoretical and d the observed vapour density, and one molecule gives n molecules, α = (D − d)/[(n − 1)d].

Le Chatelier's principle

If a system at equilibrium is disturbed, it shifts to reduce the effect of the disturbance.

For the Haber process, N₂ + 3H₂ ⇌ 2NH₃ (exothermic), high pressure and low temperature favour ammonia. In practice a moderate temperature and a catalyst are used, because the reaction is too slow when cold.

Worked problems

Problem 1: manipulating K. For N₂ + 3H₂ ⇌ 2NH₃, K = 4 at some temperature. Find K for 2NH₃ ⇌ N₂ + 3H₂ and for ½N₂ + 3/2 H₂ ⇌ NH₃.

Reversed: 1/4. Halved: √4 = 2.

Problem 2: an ICE table. 1 mol each of H₂ and I₂ are heated in a 1 L flask. At equilibrium, Kc = 64 for H₂ + I₂ ⇌ 2HI. Find the equilibrium concentrations.

Let x mol/L react: H₂ = I₂ = 1 − x, HI = 2x.
(2x)²/(1 − x)² = 64, so 2x/(1 − x) = 8, giving 2x = 8 − 8x and x = 0.8.
[HI] = 1.6 M, [H₂] = [I₂] = 0.2 M.

Problem 3: Kp from Kc. For N₂O₄(g) ⇌ 2NO₂(g), Kc = 4.0 × 10⁻² at 400 K. Find Kp (R = 0.0831 L bar K⁻¹ mol⁻¹).

Δn_g = 2 − 1 = 1, so Kp = Kc × RT = 0.04 × 0.0831 × 400 = 0.04 × 33.24 ≈ 1.33 bar.

Problem 4 (numerical answer): Kp from α. PCl₅ is 50% dissociated at 3 bar total pressure. Find Kp in bar.

Kp = α²P/(1 − α²) = (0.25 × 3)/(1 − 0.25) = 0.75/0.75 = 1.

Traps to watch

• Including solids or liquids in K. Leave them out.
• Using moles instead of concentrations. Divide by volume unless the volume is 1 L or Δn_g = 0, when the volume cancels.
• Thinking pressure changes K. Pressure shifts the position; only temperature changes K.
• Inert gas. At constant volume, nothing happens.
• Catalyst. It changes neither K nor the equilibrium amounts.

Practice set

1. For which reaction is Kp = Kc: H₂ + I₂ ⇌ 2HI or N₂ + 3H₂ ⇌ 2NH₃?
2. What is the effect of a catalyst on K?
3. For 2SO₂ + O₂ ⇌ 2SO₃ (exothermic), what happens to the equilibrium when temperature is raised, and when pressure is raised?
4. For A ⇌ B, Kc = 10. If [A] = 0.2 M and [B] = 1.0 M, which way does the reaction move?
5. K for A ⇌ B is 10 and K for B ⇌ C is 5. Find K for A ⇌ C.
6. N₂O₄ (theoretical vapour density 46) has an observed vapour density of 40 at some temperature. Find the degree of dissociation.
7. K = 10 for a reaction at 300 K. Find ΔG°.

Answers

1. H₂ + I₂ ⇌ 2HI, since Δn_g = 0.
2. None. It speeds up both directions equally.
3. Raising T shifts it backward (and K falls); raising P shifts it forward (3 gas moles to 2).
4. Q = 1.0/0.2 = 5, which is less than K, so it moves forward.
5. 10 × 5 = 50.
6. α = (46 − 40)/[(2 − 1) × 40] = 0.15.
7. ΔG° = −2.303 × 8.314 × 300 × log 10 ≈ −5.74 kJ/mol.

What to do next

• Solve ten ICE-table problems, including ones where the volume is not 1 L.
• Derive Kp in terms of α and P for PCl₅ and N₂O₄ yourself once.
• Revise chemical thermodynamics for the ΔG° link.
• Move on to ionic equilibrium, which applies the same ideas to acids, bases and salts.