Equilibrium

Equilibrium

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Equilibrium in a reversible reaction is the state where forward and reverse reaction rates become equal, so concentrations of reactants and products stay constant with time — this is called dynamic equilibrium because both reactions continue at the molecular level.

The backbone formulas:

• Equilibrium constant (Kc): Kc = [products]^coeff / [reactants]^coeff (concentrations of only aqueous/gaseous species at equilibrium).
• Relation: Kp = Kc(RT)^Δn_g, where Δn_g = gaseous moles (products − reactants).
• Ionisation product of water: Ka · Kb = Kw = 1.0 × 10⁻¹⁴ at 25 °C.
• Henderson–Hasselbalch: pH = pKa + log([salt]/[acid]) for an acidic buffer.

High-yield JEE pointers: (1) Kc changes only with temperature. (2) Heterogeneous equilibria omit pure solids and pure liquids. (3) Le Chatelier’s principle predicts the direction of shift when concentration, pressure, or temperature is altered — frequently a 1–2 mark MCQ. (4) For a weak monoprotic acid, Ostwald’s dilution law gives α = √(Ka/c).

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Equilibrium Constant and Its Characteristics

For a reversible reaction aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]^c [D]^d / [A]^a [B]^b, where each concentration is measured at equilibrium and raised to its stoichiometric coefficient. Kc is dimensionless when activities are used; with molar concentrations, its value carries the unit (mol L⁻¹)^Δn. The constant depends only on temperature — altering concentration, adding a catalyst, or changing pressure shifts the position of equilibrium but not the numerical value of Kc.

Kp, Kc, and the Δn_g Link

For gas-phase reactions, Kp is written with partial pressures: Kp = (P_C)^c (P_D)^d / (P_A)^a (P_B)^b. The two constants are related by Kp = Kc (RT)^Δn_g, with R = 0.0821 L atm K⁻¹ mol⁻¹. If Δn_g = 0, Kp = Kc. If Δn_g > 0, Kp increases with temperature; if Δn_g < 0, Kp decreases.

Reaction Quotient (Q) and Direction of Shift

Q has the same form as Kc but uses concentrations at any instant:

• Q < Kc ⇒ net reaction proceeds forward.
• Q > Kc ⇒ net reaction proceeds reverse.
• Q = Kc ⇒ system is at equilibrium.

Le Chatelier’s Principle

When a system at equilibrium is disturbed, it shifts in the direction that counteracts the change:

• Adding reactant or removing product ⇒ forward shift.
• Adding product or removing reactant ⇒ backward shift.
• Increasing pressure (or decreasing volume) ⇒ shift toward the side with fewer gaseous moles.
• Raising temperature ⇒ shift in the endothermic direction.
• A catalyst speeds both directions equally, so it does not shift equilibrium.

Heterogeneous Equilibria

Pure solids and pure liquids have unit activity and are excluded from the Kc/Kp expression. For example, for CaCO₃(s) ⇌ CaO(s) + CO₂(g), only Kp = P_CO₂ appears.

Acids, Bases, and Buffers

A buffer resists pH change on adding small amounts of acid/base. An acidic buffer (weak acid + its conjugate base salt) obeys: pH = pKa + log([salt]/[acid]). The relation Ka · Kb = Kw = 1.0 × 10⁻¹⁴ at 25 °C lets you convert between an acid and its conjugate base. Buffer capacity is maximum when pH = pKa (i.e. [salt] = [acid]).

Solubility Product and Common Ion Effect

For a sparingly soluble salt like AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), Ksp = [Ag⁺][Cl⁻]. Adding a common ion (e.g. NaCl supplying Cl⁻) decreases solubility — the dissolution equilibrium shifts backward. This is the common ion effect.

Salt Hydrolysis

• Salt of strong base + weak acid (e.g. CH₃COONa) ⇒ basic solution.
• Salt of weak base + strong acid (e.g. NH₄Cl) ⇒ acidic solution.
• Salt of strong base + strong acid (e.g. NaCl) ⇒ neutral.

Worked Snapshot

For 0.1 M CH₃COOH with Ka = 1.8 × 10⁻⁵, α = √(Ka/c) = √(1.8 × 10⁻⁵/0.1) = √(1.8 × 10⁻⁴) ≈ 1.34 × 10⁻², so [H⁺] ≈ 1.34 × 10⁻³ M and pH ≈ 2.87.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Ostwald’s Dilution Law and α

For a weak monoprotic acid HA ⇌ H⁺ + A⁻ with initial concentration c, Ka = (cα)² / [c(1−α)] ≈ cα² (when α ≪ 1), giving α = √(Ka/c). The degree of dissociation rises on dilution — a textbook application of Le Chatelier’s principle (adding water shifts the equilibrium forward because [H⁺] and [A⁻] decrease).

Polyprotic Acid Hierarchy

For H₃PO₄: Ka1 ≈ 7.1 × 10⁻³, Ka2 ≈ 6.2 × 10⁻⁸, Ka3 ≈ 4.5 × 10⁻¹³. Each successive proton leaves from an anion already carrying negative charge, so dissociation becomes progressively harder; hence Ka1 ≫ Ka2 ≫ Ka3. JEE problems often use only Ka1, but recognizing the cascade explains multi-step titration curves.

Ksp and Precipitation

A precipitate forms only when the ionic product (IP) exceeds Ksp. If IP < Ksp, the solution is unsaturated; if IP = Ksp, it is saturated. Selective precipitation exploits differing Ksp values to separate ions (e.g. Group-I cations in qualitative analysis as chlorides).

Edge Cases That Trip Students Up

• Units of K: Kc has units unless activities are used; Kp likewise. Don’t drop Δn_g when converting Kp ⇌ Kc.
• Heterogeneous rule: omitting solids/liquids is correct only if their activity is unity (pure phase). Dissolved species always count.
• Temperature dependence of K: K changes with T even though the equilibrium “position” changes with concentration/pressure — mixing these is a classic trap.
• Buffer range: a buffer is effective within pH = pKa ± 1; outside this range it loses capacity.
• Henderson–Hasselbalch validity: the equation assumes [salt] and [acid] approximate equilibrium concentrations — fails when one component is too dilute or the acid is too strong.

Connection to Electrochemistry

The Nernst equation E_cell = E°_cell − (RT/nF) ln Q is essentially the equilibrium condition E_cell = 0 ⇒ Q = K, linking ΔG° = −RT ln K with electrochemistry. Recognising this bridge helps in multi-concept JEE problems.

Exam Strategy for JEE Main

Equilibrium contributes about 3% of the Chemistry paper — typically 1 numerical + 1 conceptual MCQ. Highest-weight sub-topics: Le Chatelier shifts, buffer pH calculations, Ksp / common ion effect, and Ka–Kb–pH interconversion. Numerical items usually give Ka and concentration and ask for pH, α, or degree of hydrolysis.

Practice Prompts

1. For N₂O₄(g) ⇌ 2NO₂(g), Kp = 0.36 atm at 100 °C. If pure N₂O₄ is taken at 1 atm total pressure, find the equilibrium partial pressure of NO₂. (Use Δn_g = 1, Kp = Kc(RT)¹.)
2. A buffer is prepared by mixing 0.2 M CH₃COOH and 0.3 M CH₃COONa. Given Ka = 1.8 × 10⁻⁵, calculate the pH and the new pH after adding 0.01 mol HCl to 1 L of the buffer. Verify that the change is small — the buffer’s defining behaviour.

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