Dynamic equilibrium in physical and chemical processes
Equilibrium means a balance. It is reached only in a closed system (nothing enters or leaves).
Physical processes: in a closed bottle, water evaporates and vapour condenses. After some time both happen at the same speed, so the vapour pressure becomes constant. Ice and water at 0 °C, and a saturated sugar solution with extra sugar at the bottom, are other examples.
Chemical processes: in a reversible reaction, reactants make products and products make reactants back. At first the forward rate is high. As products build up, the backward rate rises. When the two rates become equal, amounts stop changing.
It is called dynamic because the reaction never stops. Molecules keep reacting both ways; we just do not see any net change. You can reach the same equilibrium starting from either side.
Try it: in the 3D at step 1, watch the grey and brown counts. They level off, yet pairs and singles keep swapping.
Law of equilibrium and the equilibrium constant Kc
For a general reaction aA + bB ⇌ cC + dD, at a fixed temperature:
Kc = [C]c[D]d / [A]a[B]b
Square brackets mean molar concentration (mol/L) at equilibrium. This is the law of chemical equilibrium (law of mass action). Products go on top, reactants below, and each is raised to its coefficient.
Rules to remember:
- Reverse the reaction → new K = 1/K.
- Multiply the equation by n → new K = Kⁿ. Divide by 2 → new K = √K.
- Add two reactions → multiply their K values.
- K changes only with temperature, not with starting amounts, pressure or a catalyst.
Try it: at step 2 move the start sliders. The final Q always settles on the same Kc.
Kc and Kp: relation Kp = Kc(RT)^Δn
For gases, partial pressure p is proportional to concentration: p = (n/V)RT = [X]RT. Putting this in the expression gives:
Kp = Kc (RT)Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants).
- Δn = 0 → Kp = Kc (example: H₂ + I₂ ⇌ 2HI).
- Δn > 0 → Kp > Kc (when RT > 1).
- Δn < 0 → Kp < Kc (example: N₂ + 3H₂ ⇌ 2NH₃, Δn = −2).
Use R = 0.0831 L bar K⁻¹ mol⁻¹ with pressure in bar and T in kelvin. Only gases are counted in Δn.
Homogeneous and heterogeneous equilibria
Homogeneous equilibrium: everything is in one phase. Examples: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); CH₃COOH(aq) + C₂H₅OH(aq) ⇌ CH₃COOC₂H₅(aq) + H₂O(l) in solution.
Heterogeneous equilibrium: more than one phase. Examples: CaCO₃(s) ⇌ CaO(s) + CO₂(g); H₂O(l) ⇌ H₂O(g); Ni(s) + 4CO(g) ⇌ Ni(CO)₄(g).
The concentration of a pure solid or pure liquid is fixed (its density does not change), so it is taken as 1 and left out of K. So for CaCO₃ ⇌ CaO + CO₂, Kp = p(CO₂). But the solid must be present for the equilibrium to exist.
Try it: at step 4 change the amount of CaCO₃. The CO₂ reading does not move.
Uses of the equilibrium constant
1. Extent of reaction. K > 10³: mostly products (reaction nearly complete). K < 10⁻³: mostly reactants (hardly goes). K between 10⁻³ and 10³: good amounts of both.
2. Direction of reaction. Write Q with the present (not equilibrium) amounts. Q < K → goes forward. Q > K → goes backward. Q = K → at equilibrium.
3. Equilibrium amounts. Use an ICE table: Initial, Change, Equilibrium. Put the equilibrium values into K and solve for x.
K says nothing about how fast equilibrium is reached. That is kinetics.
K, Q and Gibbs energy
Gibbs energy decides if a change can happen by itself. For a reaction mixture:
ΔG = ΔG° + RT ln Q
At equilibrium ΔG = 0 and Q = K, so ΔG° = −RT ln K = −2.303 RT log K. Also ΔG = RT ln(Q/K).
- ΔG° negative → K > 1 → products favoured.
- ΔG° positive → K < 1 → reactants favoured.
- ΔG° = 0 → K = 1.
Use R = 8.314 J K⁻¹ mol⁻¹ here, and T in kelvin. Try it: at step 5 the readout shows ΔG turning to 0 as Q comes back to K.
Exam corner
The Equilibrium unit carries about 7 marks in CBSE Class 11. Common questions: write Kc/Kp expressions, Kp–Kc relation, find K from ICE tables, predict direction using Q, and ΔG° = −RT ln K numericals. Always write the unit of K when asked and state the temperature.
Key formulas and definitions
- Kc = [C]^c[D]^d / [A]^a[B]^b
- Kp = Kc (RT)^Δn, Δn = gaseous product moles − gaseous reactant moles
- Reverse reaction: K' = 1/K; multiply by n: K' = Kⁿ
- Q < K → forward; Q > K → backward; Q = K → equilibrium
- ΔG = ΔG° + RT ln Q; ΔG° = −RT ln K = −2.303RT log K
Worked examples
1. Write Kc for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).
Products over reactants, each raised to its coefficient: Kc = [SO₃]² / ([SO₂]²[O₂]).
2. For H₂ + I₂ ⇌ 2HI, Kc = 50 at 700 K. Find Kc for 2HI ⇌ H₂ + I₂ and for HI ⇌ ½H₂ + ½I₂.
Reversed: K = 1/50 = 0.02. Halved: K = √0.02 = 0.141.
3. 1.0 mol N₂O₄ is kept in a 1 L flask. At equilibrium 0.52 mol NO₂ is present. Find Kc for N₂O₄ ⇌ 2NO₂.
NO₂ formed = 0.52 mol, so N₂O₄ used = 0.26 mol. [N₂O₄] = 0.74 M, [NO₂] = 0.52 M. Kc = 0.52² / 0.74 = 0.2704/0.74 ≈ 0.37 mol/L.
4. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = 0.5 L² mol⁻² at 500 K. Find Kp (R = 0.0831 L bar K⁻¹ mol⁻¹).
Δn = 2 − 4 = −2. RT = 0.0831 × 500 = 41.55. Kp = 0.5 × (41.55)⁻² = 0.5 / 1726.4 ≈ 2.9 × 10⁻⁴ bar⁻².
5. For A ⇌ B, K = 4. A mixture has [A] = 0.5 M and [B] = 1.0 M. Which way will it go?
Q = [B]/[A] = 1.0/0.5 = 2. Q (2) < K (4), so the reaction goes forward, making more B.
6. Find ΔG° at 300 K for a reaction with K = 10 (R = 8.314 J K⁻¹ mol⁻¹).
ΔG° = −2.303RT log K = −2.303 × 8.314 × 300 × 1 = −5744 J/mol ≈ −5.74 kJ/mol. Negative, so products are favoured.
7. For H₂ + I₂ ⇌ 2HI, Kc = 64. We start with 1 mol H₂ and 1 mol I₂ in 1 L. Find the equilibrium amounts.
ICE: H₂ = 1 − x, I₂ = 1 − x, HI = 2x. Kc = (2x)² / (1 − x)² = 64. Square root: 2x/(1 − x) = 8 → 2x = 8 − 8x → x = 0.8. So H₂ = I₂ = 0.2 mol, HI = 1.6 mol.
8. ΔG° for a reaction at 298 K is +11.4 kJ/mol. Find K.
log K = −ΔG°/(2.303RT) = −11400/(2.303 × 8.314 × 298) = −11400/5705.8 ≈ −2.0. K = 10⁻² = 0.01. Reactants are favoured.
Common mistakes
- Writing reactants on top. K is always products over reactants.
- Putting pure solids or liquids (like CaCO₃(s) or H₂O(l) as solvent) inside K. Leave them out.
- Counting all moles in Δn. Only gaseous moles count.
- Thinking that at equilibrium the reaction stops or that reactant and product amounts are equal. Only the rates are equal; amounts are just constant.