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Equilibrium
Chemistry · Class 11NCERT pp. 168–2072 sets × 25 questions
Key facts and formulas for Equilibrium, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| Equilibrium constants | K_p = K_c (RT)^Δn_g; halve the equation: K → √K; add equations: K = K₁ × K₂; reverse: 1/K |
| Reaction quotient | Q < K: forward; Q > K: backward; Q = K: equilibrium. Pure solids and liquids are left out |
| Free energy and K | ΔG° = −2.303 RT log K; ΔG = ΔG° + 2.303 RT log Q; 2.303 RT = 5.71 kJ mol⁻¹ at 298 K |
| Temperature effect | log(K₂/K₁) = (ΔH°/2.303R)(1/T₁ − 1/T₂); endothermic: K rises with T; exothermic: K falls |
| pH basics | pH = −log [H⁺]; pH + pOH = pK_w = 14 at 298 K; K_a × K_b = K_w |
| Weak acid or base | [H⁺] = √(K_a c); [OH⁻] = √(K_b c); α = √(K_a / c) |
| Buffers | pH = pK_a + log([salt]/[acid]); pOH = pK_b + log([salt]/[base]); best at ratio 1, range pK_a ± 1 |
| Salts and amphiprotic ions | Weak acid + strong base: pH = 7 + ½(pK_a + log c); amphiprotic ion: pH = ½(pK_a1 + pK_a2) |
| Solubility product | AB: S = √K_sp; A₂B or AB₂: K_sp = 4S³; AB₃: K_sp = 27S⁴; precipitate if Q_sp > K_sp |
| Le Chatelier | Inert gas at constant V: no shift; at constant P: acts like lower pressure; catalyst: no change in K or position |
Three sample questions
Q1. At chemical equilibrium in a closed vessel at constant temperature, which statement is correct?
- The reaction has stopped
- The forward and reverse rates are equal and the concentrations stay constant
- The concentrations of reactants and products are equal
- The forward rate is greater than the reverse rate
Hint and solution
Hint: Equilibrium is dynamic. Think about what both reactions are doing and what stays fixed.
- Equilibrium = dynamic state — At equilibrium both the forward and reverse reactions continue.
- rate(forward) = rate(reverse) — Since the two rates are equal, no net change occurs in the closed vessel.
- Macroscopic properties constant — Concentrations, colour and pressure stay constant, but not necessarily equal.
- Answer: rates equal, concentrations constant — So the correct statement is the one with equal rates and constant concentrations. Option B.
Answer: B. Equilibrium is dynamic. Both reactions continue at equal rates, so macroscopic properties stay fixed.
Q2. For N₂ + 3H₂ ⇌ 2NH₃, the unit of K_c (concentrations in mol L⁻¹) is
- mol² L⁻²
- L mol⁻¹
- it has no unit
- L² mol⁻²
Hint and solution
Hint: Find the change in moles of gas, then raise the concentration unit to that power.
- K_c = [NH₃]² / ([N₂][H₂]³) — Write the expression: products in the numerator, reactants in the denominator, powers equal to coefficients.
- Δn = 2 − (1 + 3) = −2 — Subtract the moles of gaseous reactants from those of gaseous products.
- unit = (mol L⁻¹)^Δn = (mol L⁻¹)⁻² — Each concentration carries mol L⁻¹, so the unit is that raised to Δn.
- L² mol⁻² — Inverting the power gives litre squared per mole squared. Option D.
Answer: D. Δn = 2 − 4 = −2, so K_c has units (mol L⁻¹)^Δn = L² mol⁻².
Q3. For N₂O₄ ⇌ 2NO₂, K_c = 0.05 at 400 K. K_p (R = 0.0821 L atm K⁻¹ mol⁻¹) is about
- 1.64
- 0.0015
- 20.3
- 0.05
Hint and solution
Hint: Use K_p = K_c (RT)^Δn and first find Δn for gases.
- Δn = 2 − 1 = 1 — Count gas moles: 2 mol NO₂ in products, 1 mol N₂O₄ in reactants.
- RT = 0.0821 × 400 = 32.84 — Multiply the gas constant by temperature in kelvin.
- K_p = 0.05 × 32.84 — Apply K_p = K_c (RT)^1 with the value of RT.
- K_p = 1.64 — 0.05 times 32.84 is 1.642. Option A.
Answer: A. K_p = K_c (RT)^Δn with Δn = 1: 0.05 × 0.0821 × 400 = 1.64.
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.