Chemistry XI • Ch 6 Equilibrium
Beginner - Story Mode
1. Equilibrium - Dynamic BalancingPart 1
Imagine a closed plastic bottle half-filled with water left on your dining table. Inside, water is constantly evaporating into vapour, and vapour is condensing back into water. After some time, the water level looks constant. Does that mean everything has stopped? Not at all.

It means the rate of evaporation has become exactly equal to the rate of condensation. The same substance exists in two states at the same time, but in a balanced way. This is Physical Equilibrium. A classic example is ice at 0°C — melting rate = freezing rate — so solid and liquid co-exist peacefully.

Now think of a reversible chemical reaction like N₂ + 3H₂ ⇌ 2NH₃ in a closed box. In the beginning, mostly forward reaction happens, but slowly the reverse also starts. At equilibrium, the forward rate becomes equal to the reverse rate. Concentrations become constant, but not necessarily equal. It is dynamic — like two opposite escalators moving at the same speed. The number of people on the floor stays constant, but people are continuously moving both ways. Or like two teams, one filling and one emptying a water tank at the same speed — water level is constant, but water is still flowing.
Revision: Physical equilibrium = same substance in two phases in closed system, evaporation rate = condensation rate. Chemical equilibrium = reversible reaction where forward rate = backward rate, concentrations constant (not equal), dynamic in nature — like two teams filling & emptying tank at same speed.
Equilibrium

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  • Physical: H2O(l)⇌H2O(g) closed evap=cond, H2O(s)⇌H2O(l)
  • Chemical: N2+3H2⇌2NH3 forward=reverse conc constant not equal dynamic
  • Graph flat after equilibrium
2. Kc & Kp - RatioPart 2
Think of a cricket player's batting average — runs scored divided by times out. At equilibrium, chemistry also looks for a ratio of products to reactants.

This is the Law of Mass Action. For a general reaction aA + bB ⇌ cC + dD, at a fixed temperature, the ratio Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ becomes constant. Here [ ] means molar concentration. If gases are involved, we can also write the ratio in terms of partial pressures as Kp — Kp = (pC)ᶜ(pD)ᵈ / (pA)ᵃ(pB)ᵇ.

They are related by Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) – (moles of gaseous reactants). For example, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) has Δn = 2 – 4 = -2, so Kp = Kc/(RT)². Just like a good average tells you who dominates the game, a large K means products dominate at equilibrium.
Revision: Kc = (products)^coeff / (reactants)^coeff = [C]ᶜ[D]ᵈ/[A]ᵃ[B]ᵇ, Kp uses partial pressures. Relation Kp = Kc(RT)^Δn where Δn = gaseous products – reactants; e.g., N₂+3H₂⇌2NH₃, Δn=-2 so Kp=Kc/(RT)².
KcKp

Quick Revision Box

  • Kc=[products]^c/[reactants]^c
  • Kp=(P prod)^c/(P react)^c
  • Kp=Kc(RT)^Δn
  • N2+3H2⇌2NH3 Δn=-2
3. Types & ApplicationsPart 3
In your kitchen, sometimes you mix everything in one kadhai, sometimes you have solid rice, liquid water and steam together. Equilibrium is classified similarly.

If all reactants and products are in the same phase, like all gases in N₂(g) + 3H₂(g) ⇌ 2NH₃(g), we call it Homogeneous Equilibrium. If they are in different phases, like CaCO₃(s) ⇌ CaO(s) + CO₂(g) where solids and gas co-exist, it is Heterogeneous Equilibrium. For heterogeneous cases, we ignore pure solids and pure liquids while writing K, because their concentration is constant. So for this example, Kc = [CO₂] only.

Why is K useful? If K >> 1, products are highly favoured; if K << 1, reactants dominate; if K ≈ 1, both are present in comparable amounts. To know the position at any time, we use Reaction Quotient Qc — same formula as Kc but with current concentrations. If Qc < Kc, reaction goes forward; if Qc > Kc, it goes backward; if Qc = Kc, system is already at equilibrium — like checking scoreboard before deciding to attack.
Revision: Homogeneous = same phase (all gases); Heterogeneous = different phases, pure solids/liquids omitted — e.g., CaCO₃(s)⇌CaO(s)+CO₂(g), Kc=[CO₂]. K>>1 products favoured, K<<1 reactants; QcKc→backward, Qc=Kc→equilibrium.
Types

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  • Homogeneous same phase heterogeneous different solids omitted
  • CaCO3(s)⇌CaO(s)+CO2(g) Kc=[CO2]
  • K>>1 products K<<1 reactants
  • QcKc backward
4. Gibbs & Le Chatelier - SeesawPart 4
Every reaction maintains an energy account called Gibbs free energy (G). At any moment, ΔG = ΔG° + RT ln Q. At equilibrium the account is settled, so ΔG = 0 and Q = K, which gives the very important relation ΔG° = -RT ln K = -2.303 RT log K. This links energy and equilibrium directly.

If ΔG° is negative, then K > 1 — products are favoured and forward reaction is spontaneous. If ΔG° is positive, then K < 1 — reactants are favoured. And then comes Le Chatelier's Principle — like a village seesaw. If you put extra weight on one side, the seesaw tilts and tries to balance again. Similarly, if a system at equilibrium is disturbed by changing concentration, pressure or temperature, it shifts in a direction that counteracts the disturbance to restore balance. It is nature's way of resisting stress.
Revision: ΔG = ΔG°+RT lnQ; at eqm ΔG=0, Q=K so ΔG°=-RT lnK. ΔG° -ve → K>1 products spontaneous, +ve → K<1 reactants. Le Chatelier: disturbance → system shifts to counteract it, like a seesaw restoring balance.
Gibbs

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  • ΔG=ΔG°+RT lnQ ΔG°=-RT lnK
  • ΔG° -ve K>1 products +ve K<1 reactants
  • Le Chatelier disturbance shifts to counteract
5. Le Chatelier Concentration PressurePart 5
Let us see Le Chatelier in daily life. Suppose you are making chai. Add more milk (reactant), you get more chai — reaction shifts forward. Add more ready chai (product) to the pan, system tries to use it and shifts backward. In short: adding reactant → forward, adding product → backward; removing reactant → backward, removing product → forward.

Pressure effect matters only for gases. Think of a crowded general compartment in a train. If you reduce space (increase pressure), people will try to go where there is more space — fewer moles. So increasing pressure shifts equilibrium to side with fewer moles of gas. For N₂ + 3H₂ ⇌ 2NH₃, 4 moles → 2 moles, so increasing P favours more NH₃. Adding inert gas at constant volume increases total pressure but partial pressures remain same, so no effect. Adding inert gas at constant total pressure increases volume (like expanding the compartment), so it shifts to side with more moles.
Revision: Add reactant→forward, add product→backward, remove reactant→backward, remove product→forward. For gases, ↑P shifts to fewer moles side (N₂+3H₂⇌2NH₃, 4→2, ↑P gives more NH₃). Inert gas at const V = no effect, at const P → shifts to more moles side.
Conc Pressure

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  • Add reactant forward product backward
  • ↑P fewer moles side N2+3H2⇌2NH3 4→2 ↑P forward
  • Inert const V no effect const P more moles
6. Temperature CatalystPart 6
Temperature is unique because it treats heat as an ingredient. In an exothermic reaction, heat is released, we can write A + B ⇌ C + Heat. If you increase temperature, you are adding heat, so Le Chatelier says equilibrium will go backward, K decreases. It likes cold. So decreasing T shifts forward and K increases.

In an endothermic reaction, heat is required — A + B + Heat ⇌ C. Increasing temperature adds reactant, so it shifts forward, K increases, it likes hot. Decreasing temperature shifts backward and K decreases. A catalyst is like a shortcut road in your mohalla — it lowers activation energy (Ea) for both forward and backward equally, so equilibrium is reached faster but it does not change position or K. Remember: among concentration, pressure and catalyst, only temperature changes the value of K.
Revision: Exothermic A+B⇌C+Heat: ↑T→backward K↓ (likes cold), ↓T→forward K↑. Endothermic A+B+Heat⇌C: ↑T→forward K↑ (likes hot), ↓T→backward K↓. Catalyst = shortcut, lowers Ea both ways, no shift, no change in K; only T changes K.
Temp Catalyst

Quick Revision Box

  • Exothermic ↑T backward K↓ likes cold
  • Endothermic ↑T forward K↑ likes hot
  • Catalyst no shift no K faster only
  • Only T affects K
7. Acids Bases Ka Kb pHPart 7
There are three popular definitions of acids and bases, like three ways to identify a shopkeeper in your market.

Arrhenius said acid gives H⁺ in water and base gives OH⁻ — simple but works only in water. Bronsted-Lowry is more general, a proton game — acid donates H⁺, base accepts H⁺. So CH₃COOH / CH₃COO⁻ are a conjugate acid-base pair differing by one proton. Then Lewis looked at electrons — Lewis acid accepts an electron pair (electron-deficient like BF₃), Lewis base donates an electron pair (has lone pair like :NH₃).

Strength is measured by dissociation constant Ka = [H⁺][A⁻]/[HA] — larger Ka means stronger acid. For convenience, pKa = -log Ka, so smaller pKa = stronger acid. For water itself, Kw = [H⁺][OH⁻] = 1.0×10⁻¹⁴ at 25°C. To handle small numbers, we use pH = -log[H⁺] and pOH = -log[OH⁻] with relation pH + pOH = 14. Lower pH means more acidic, like nimbu pani.
Revision: Arrhenius: acid H⁺, base OH⁻; Bronsted: acid donates H⁺, base accepts H⁺ → conjugate pair differs by 1 H⁺; Lewis: acid accepts e⁻ pair (BF₃), base donates e⁻ pair (:NH₃). Ka larger = stronger, pKa=-logKa, Kw=1e-14, pH=-log[H⁺], pH+pOH=14.
Acids

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  • Arrhenius H+ acid OH- base
  • Bronsted acid donates H+ base accepts
  • Lewis acid accepts e- pair BF3 base donates NH3
  • Ka larger stronger pKa=-logKa Kw=1e-14 pH=-log[H+]
8. Common Ion HydrolysisPart 8
Have you noticed that adding extra common salt to your nimbu paani makes it less sour? That is Common Ion Effect in action.

Take weak acid CH₃COOH ⇌ H⁺ + CH₃COO⁻. If you add CH₃COONa which gives the same CH₃COO⁻ ion, the common ion pushes equilibrium to left by Le Chatelier, so dissociation is suppressed and less H⁺ is formed. This effect is widely used to control pH.

Salt hydrolysis is when an ion of the salt reacts with water and changes pH. Think of family background: salt of weak acid + strong base like CH₃COONa — anion takes H⁺ from water → solution basic, pH > 7, Kh = Kw/Ka. Salt of strong acid + weak base like NH₄Cl → cation gives H⁺ → acidic, pH < 7, Kh = Kw/Kb. Salt of weak acid + weak base → pH depends on Ka vs Kb. Salt of strong acid + strong base like NaCl does not hydrolyse, stays neutral pH = 7.
Revision: Common ion (same ion added) shifts left, suppresses dissociation (e.g., CH₃COOH + CH₃COONa → less H⁺). Hydrolysis: weak+strong → basic >7 (CH₃COONa, Kh=Kw/Ka), strong+weak → acidic <7 (NH₄Cl, Kh=Kw/Kb), weak+weak → depends on Ka/Kb, strong+strong → neutral.
Common Ion

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  • Common ion shifts left suppresses
  • Weak+Strong basic >7 CH3COONa
  • Strong+Weak acidic <7 NH4Cl
  • Strong+Strong neutral NaCl
9. Buffer Shock AbsorberPart 9
A buffer solution resists pH change — just like a shock absorber on your bicycle that resists jerks on a bumpy road. Even if you add a small amount of acid or base, its pH hardly changes.

There are two types. Acidic buffer — weak acid + its salt, e.g., CH₃COOH + CH₃COONa, works around pH < 7. Basic buffer — weak base + its salt, e.g., NH₄OH + NH₄Cl, works around pH > 7. How does it work? Added H⁺ is caught by conjugate base: H⁺ + A⁻ → HA. Added OH⁻ is neutralised by weak acid: OH⁻ + HA → A⁻ + H₂O.

Its pH is given by Henderson-Hasselbalch equation: pH = pKa + log([Salt]/[Acid]) for acidic buffer. Buffer capacity is maximum when [Salt] = [Acid] then pH = pKa. The best natural example is our blood — H₂CO₃ / HCO₃⁻ system maintains pH around 7.4, essential for life.
Revision: Buffer = resists pH change (shock absorber). Acidic: CH₃COOH+CH₃COONa, basic: NH₄OH+NH₄Cl. Actions: H⁺+A⁻→HA, OH⁻+HA→A⁻+H₂O. Henderson: pH=pKa+log([Salt]/[Acid]), max when [Salt]=[Acid]→pH=pKa; blood H₂CO₃/HCO₃⁻ pH 7.4.
Buffer

Quick Revision Box

  • Buffer resists pH shock absorber
  • Acidic CH3COOH+CH3COONa Basic NH4OH+NH4Cl
  • Henderson pH=pKa+log(Salt/Acid)
  • Max [Salt]=[Acid] pH=pKa blood 7.4
10. Ksp Ostwald DilutionPart 10
Some salts are so sparingly soluble that only a tiny amount dissolves, like trying to dissolve haldi in cold water. For such salts we define Solubility Product Ksp. For simple salt AB(s) ⇌ A⁺ + B⁻, Ksp = [A⁺][B⁻]. In general for AxBy(s) ⇌ xAʸ⁺ + yBˣ⁻, Ksp = [A]ˣ[B]ʸ = xˣ·yʸ·s^(x+y) where s = molar solubility. So for AB, s = √Ksp; for AB₂ type like CaF₂, Ksp = 4s³.

To predict whether precipitation will occur, we calculate ionic product Qsp from present concentrations: if Qsp > Ksp → precipitation occurs, if Qsp < Ksp → unsaturated, no ppt, if Qsp = Ksp → saturated solution. For weak electrolytes, degree of dissociation α is less than 1 and is given by Ostwald's Dilution Law: α = √(Ka/c) and Ka = cα²/(1-α) ≈ cα². So on dilution (decreasing c), α increases — just like adding more water to sharbat makes sugar spread more completely.
Revision: Sparingly soluble: AB, Ksp=[A⁺][B⁻]; general Ksp=xˣyʸs^(x+y) → AB s=√Ksp, AB₂ 4s³. Qsp>Ksp→ppt, Qsp
Ksp

Quick Revision Box

  • AB Ksp=[A+][B-] AxBy Ksp=x^x y^y s^(x+y)
  • AB s=√Ksp AB2 4s³
  • Qsp>Ksp precipitation Qsp
  • Ostwald α=√(Ka/c) Ka=cα²/(1-α)