Equilibrium

Equilibrium is a state where two opposing processes occur at the same rate, so there is no further net change even though both processes keep happening — this is why it is called dynamic, not static. NCERT builds the idea through three physical examples before moving to chemical reactions.

Solid–liquid: H₂O(s) ⇌ H₂O(l). At the melting point (0 °C, 1 atm) the rate of melting equals the rate of freezing; the mass of ice and water in an insulated vessel stays constant with time, though molecules keep crossing the interface both ways.

Liquid–vapour: H₂O(l) ⇌ H₂O(g). In a closed vessel, evaporation and condensation reach a stage where the rate of evaporation equals the rate of condensation; the pressure exerted by the vapour at this point is the vapour pressure of the liquid at that temperature, and it depends only on temperature, not on the amount of liquid present.

Solid–vapour (sublimation): solid ⇌ vapour, e.g. camphor, iodine, ammonium chloride subliming in a closed container — a saturated vapour pressure is again reached.

Solute–solution (solubility equilibrium): when a saturated solution is in contact with undissolved solute, the rate of dissolution equals the rate of crystallisation, e.g. sugar(s) ⇌ sugar(solution). Solubility of a solid in a liquid is fixed at a given temperature at saturation.

Gas–solution (Henry's law): for a gas dissolving in a liquid under pressure, e.g. soda water, the equilibrium concentration of dissolved gas is proportional to the pressure of the gas above the liquid — this is Henry's law, and it explains why a sealed cold-drink bottle fizzes on opening (pressure drop shifts CO₂ equilibrium toward the gas phase).

A chemical reaction is said to reach equilibrium when the rate of the forward reaction becomes exactly equal to the rate of the reverse reaction, so the concentrations of reactants and products no longer change with time — again dynamic, not static, because both reactions continue at the molecular level. Chemical equilibrium can be approached from either direction: starting with only reactants, or only products, or a mixture — the same equilibrium state (same Kc at a given temperature) is reached, which is a direct consequence of the law of mass action.

NCERT's classic illustration is the reaction of H₂ and I₂ to form HI in a closed vessel at constant temperature: H₂(g) + I₂(g) ⇌ 2HI(g). Whether the vessel is charged with H₂ + I₂, or with pure HI, or with a mixture of all three, the system settles to the same ratio of concentrations at equilibrium at that temperature — this is experimentally how the reversibility and equilibrium nature of the reaction was established.

Chemical equilibria are classified as homogeneous (all species in the same phase, e.g. the HI system above, or N₂(g) + 3H₂(g) ⇌ 2NH₃(g)) and heterogeneous (species in more than one phase, e.g. CaCO₃(s) ⇌ CaO(s) + CO₂(g)).

For a general reversible reaction aA + bB ⇌ cC + dD at a given temperature, the law of chemical equilibrium (law of mass action, given by Guldberg and Waage) states that at equilibrium the ratio of the product of equilibrium concentrations of the products (each raised to its stoichiometric coefficient) to the product of equilibrium concentrations of the reactants (each raised to its coefficient) is a constant, called the equilibrium constant Kc.

Kc = [C]^c[D]^d / [A]^a[B]^b, where square brackets denote equilibrium molar concentrations.

Key facts examiners test: (i) Kc has a fixed value at a given temperature and changes only with temperature, never with initial concentrations or the presence of a catalyst; (ii) pure solids and pure liquids are omitted from the Kc expression because their 'concentration' (density/molar mass) is constant and gets absorbed into K, e.g. for CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc = [CO₂]; (iii) the value of K depends on how the equation is written — if an equation is reversed, K is inverted (K′ = 1/K); if an equation is multiplied by a factor n, K is raised to the power n (K′ = Kⁿ); if two equations are added, the resultant K is the product of the two K values; (iv) a very large Kc (≫10³) means the reaction proceeds nearly to completion (products favoured); a very small Kc (≪10⁻³) means the reaction barely proceeds (reactants favoured); a Kc close to 1 (10⁻³ to 10³) means appreciable amounts of both are present at equilibrium.

For gas-phase reactions it is often more convenient to express the equilibrium constant in terms of partial pressures, Kp, since partial pressure is proportional to concentration (from PV = nRT, P = (n/V)RT = cRT). For the general reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), Kp is related to Kc by:

Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants) = (c + d) − (a + b).

Special cases students must know: if Δn = 0, Kp = Kc (e.g. H₂(g) + I₂(g) ⇌ 2HI(g)); if Δn > 0, Kp > Kc; if Δn < 0, Kp < Kc. R must be in units consistent with P and V — commonly R = 0.0831 L bar K⁻¹ mol⁻¹ when pressure is in bar (NCERT's SI convention) and T in kelvin.

NCERT worked example type: for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Δn = 2 − 4 = −2, so Kp = Kc(RT)⁻², i.e. Kp = Kc / (RT)².

An equilibrium in a system with more than one phase is heterogeneous. The equilibrium constant expression omits pure solids and pure liquids since their activity (effectively their concentration) stays constant. Standard NCERT examples: CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc = [CO₂], Kp = P_CO2; Ag₂O(s) + 2HNO₃(aq) ⇌ 2AgNO₃(aq) + H₂O(l), Kc = [AgNO₃]²/[HNO₃]² (water and Ag₂O omitted).

An important consequence: for CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc depends only on temperature — so at a fixed temperature the equilibrium pressure of CO₂ above CaCO₃/CaO is fixed and does not depend on how much CaCO₃ or CaO is present, only on whether both solid phases are present.

Three standard uses of K, all frequently asked as short-answer questions:

(a) Predicting the extent of a reaction — as noted above, K ≫ 1 means the reaction goes nearly to completion, K ≪ 1 means it barely proceeds, and K close to unity means comparable amounts of reactants and products coexist.

(b) Predicting the direction of a reaction (reaction quotient Qc) — Qc has the same expression as Kc but uses concentrations at any point in time, not necessarily at equilibrium. Comparing Qc to Kc tells which way the reaction will shift to reach equilibrium: if Qc < Kc, the reaction proceeds forward (net formation of more products, since not enough product is present yet); if Qc > Kc, the reaction proceeds in the reverse direction; if Qc = Kc, the reaction is already at equilibrium.

(c) Calculating equilibrium concentrations — using an ICE (Initial, Change, Equilibrium) table: write initial concentrations, express the change in terms of an unknown x (or the degree of dissociation α) using the stoichiometric ratios, add to get equilibrium concentrations, then substitute into the Kc expression and solve for x. This is the standard method for every numerical in this chapter, ionic equilibrium included.

Le Chatelier's principle: if a system at equilibrium is subjected to a change in concentration, pressure, volume or temperature, the equilibrium shifts in the direction that tends to counteract (partially undo) the effect of that change.

Effect of concentration: adding more reactant (or removing product) shifts equilibrium forward; adding more product (or removing reactant) shifts it backward. Example: in N₂ + 3H₂ ⇌ 2NH₃, adding N₂ or H₂, or removing NH₃ as it forms, drives the equilibrium toward more NH₃ — the industrial basis of the Haber process.

Effect of pressure/volume: increasing pressure (decreasing volume) shifts equilibrium toward the side with fewer moles of gas; decreasing pressure shifts it toward the side with more moles of gas. If Δn = 0 (equal moles both sides), pressure change has no effect on the position of equilibrium. Example: N₂ + 3H₂ ⇌ 2NH₃ (4 mol → 2 mol) is favoured by high pressure.

Effect of temperature: this is the one factor that also changes the value of K itself, not just the position of equilibrium. For an exothermic reaction (ΔH negative), increasing temperature shifts equilibrium backward (K decreases); for an endothermic reaction (ΔH positive), increasing temperature shifts equilibrium forward (K increases). Example: N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92.4 kJ mol⁻¹ (exothermic) — industrially run at a moderate ~700 K compromise (rate vs yield), not at low temperature despite low T favouring yield.

Effect of a catalyst: a catalyst speeds up both forward and reverse reactions equally. It has no effect on the position of equilibrium or on the value of K — it only helps the system reach equilibrium faster.

Effect of adding an inert gas: at constant volume, adding an inert gas does not change the partial pressures of the reacting species, so equilibrium is undisturbed. At constant pressure, adding an inert gas increases the total volume, which lowers the partial pressures of reacting gases — the equilibrium then shifts toward the side with more moles of gas (same effect as decreasing pressure).

NCERT gives two applied case studies that combine Le Chatelier's principle with real industrial conditions — both are common in case-based questions.

Haber process for ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92.4 kJ mol⁻¹. High pressure (~200 atm) favours the forward reaction (fewer moles on product side); low temperature would favour yield (exothermic), but industrially a moderate temperature (~700 K) is used with an iron catalyst (with Mo promoter) to get a workable reaction rate, since too low a temperature makes the reaction too slow to be economical. Removing NH₃ as it liquefies also drives the equilibrium forward continuously.

Contact process for sulphuric acid: the key step 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is also exothermic; excess O₂ (from air) and a moderate temperature (~720 K) with a V₂O₅ catalyst are used for a similar rate-vs-yield compromise.

Substances that conduct electricity in molten or aqueous state are electrolytes; strong electrolytes (HCl, NaOH, NaCl) ionise almost completely, weak electrolytes (CH₃COOH, NH₄OH) ionise only partially and set up an ionic equilibrium between unionised molecules and ions.

Arrhenius concept: an acid furnishes H⁺ ions and a base furnishes OH⁻ ions in aqueous solution. Limitation: applies only to aqueous solutions and cannot explain the basicity of substances like NH₃ that have no OH⁻.

Bronsted-Lowry concept: an acid is a proton (H⁺) donor, a base is a proton acceptor. Every acid, on losing a proton, forms its conjugate base, and every base, on accepting a proton, forms its conjugate acid — together an acid and the base formed from it (or vice versa) are called a conjugate acid-base pair, e.g. HCl (acid) / Cl⁻ (conjugate base); NH₃ (base) / NH₄⁺ (conjugate acid). A species that can act as both, like H₂O or HCO₃⁻, is amphoteric. The conjugate base of a strong acid is always weak, and the conjugate acid of a strong base is always weak (inverse strength relationship within a pair).

Lewis concept: an acid is an electron-pair acceptor, a base is an electron-pair donor — this is the broadest definition and explains acidic behaviour of species with no H at all, e.g. BF₃, AlCl₃ (Lewis acids, electron-deficient) and NH₃, H₂O (Lewis bases, lone pair donors).

Water undergoes self-ionisation (autoprotolysis): H₂O + H₂O ⇌ H₃O⁺ + OH⁻, often written simply as H₂O ⇌ H⁺ + OH⁻. The equilibrium constant for this, called the ionic product of water (Kw), is Kw = [H⁺][OH⁻], and at 25 °C, Kw = 1.0 × 10⁻¹⁴. Kw increases with temperature (ionisation is endothermic), so pure water becomes slightly more acidic-seeming (lower pH) at higher T while remaining neutral ([H⁺] = [OH⁻]).

Sørensen's pH scale: pH = −log₁₀[H⁺] (strictly [H₃O⁺]). At 25 °C: neutral solution has pH = 7 ([H⁺] = [OH⁻] = 10⁻⁷ M); acidic solution has pH < 7; basic solution has pH > 7. Similarly pOH = −log₁₀[OH⁻], and since Kw = [H⁺][OH⁻] = 10⁻¹⁴, taking −log of both sides gives pH + pOH = 14 at 25 °C (an examiner-favourite one-liner, valid only at 25 °C).

For strong acids/bases (assumed to dissociate completely), [H⁺] or [OH⁻] is taken directly from the given concentration times the number of ionisable H/OH per formula unit, e.g. for 0.01 M HCl, [H⁺] = 0.01 M, pH = 2; for 0.01 M Ca(OH)₂, [OH⁻] = 0.02 M.

For a weak acid HA ⇌ H⁺ + A⁻, the acid ionisation constant Ka = [H⁺][A⁻]/[HA]. A larger Ka means a stronger acid. Similarly for a weak base BOH ⇌ B⁺ + OH⁻, Kb = [B⁺][OH⁻]/[BOH].

Degree of ionisation (α) is the fraction of the total number of moles that ionises at equilibrium. For a weak acid of initial concentration c, using an ICE table (initial c, 0, 0; change −cα, +cα, +cα; equilibrium c(1−α), cα, cα): Ka = (cα)(cα) / c(1−α) = cα²/(1−α). When α is small (weak acid, α ≪ 1), 1 − α ≈ 1, giving Ostwald's dilution law: Ka ≈ cα², so α ≈ √(Ka/c) — the degree of ionisation increases as the solution is diluted (c decreases). Then [H⁺] = cα = √(Ka·c), and pH = ½(pKa − log c).

pKa and pKb: pKa = −log₁₀Ka, pKb = −log₁₀Kb. A smaller pKa means a stronger acid. For a conjugate acid-base pair, Ka × Kb = Kw, so pKa + pKb = 14 (at 25 °C) — this lets pKb of a conjugate base be found directly from the pKa of the acid, and vice versa.

Relative strength of two acids at the same concentration can be compared using α₁/α₂ = √(Ka₁/Ka₂), derived directly from Ostwald's law.

The common ion effect is the suppression of the ionisation of a weak electrolyte by adding a strong electrolyte that furnishes an ion common to the weak electrolyte's equilibrium. Example: adding CH₃COONa (fully ionised) to a solution of CH₃COOH ⇌ CH₃COO⁻ + H⁺ increases [CH₃COO⁻], and by Le Chatelier's principle the equilibrium shifts backward (left), reducing the ionisation of the weak acid and lowering [H⁺] — this is exactly the mechanism used to prepare a buffer solution.

Similarly, adding NH₄Cl to NH₄OH suppresses the ionisation of the weak base by supplying a common NH₄⁺ ion. Common ion effect is also central to controlling precipitation, e.g. adding H₂S in the presence of dilute HCl (common H⁺) in qualitative salt analysis selectively precipitates only very insoluble sulphides.

Salts formed from the reaction of an acid and a base can be classified by the strength of their parent acid/base, and their aqueous solutions are not always neutral because the ions can react with water (hydrolysis).

Salt of strong acid + strong base (e.g. NaCl): neither ion hydrolyses, solution is neutral, pH = 7.

Salt of strong acid + weak base (e.g. NH₄Cl): the cation hydrolyses (NH₄⁺ + H₂O ⇌ NH₄OH + H⁺), releasing H⁺, so the solution is acidic, pH < 7.

Salt of weak acid + strong base (e.g. CH₃COONa): the anion hydrolyses (CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻), releasing OH⁻, so the solution is basic, pH > 7.

Salt of weak acid + weak base (e.g. CH₃COONH₄): both ions hydrolyse; the resulting pH depends on the relative strengths (Ka vs Kb) of the parent acid and base — solution is close to neutral if Ka ≈ Kb, acidic if Ka > Kb, basic if Kb > Ka.

A buffer solution resists a change in pH on the addition of small amounts of acid or base, or on dilution. An acidic buffer is a mixture of a weak acid and its salt with a strong base (e.g. CH₃COOH + CH₃COONa); a basic buffer is a mixture of a weak base and its salt with a strong acid (e.g. NH₄OH + NH₄Cl).

How it works (mechanism): in an acidic buffer, the large reservoir of CH₃COO⁻ (from the fully ionised salt) neutralises any added H⁺ by combining with it to form unionised CH₃COOH, while the large reservoir of unionised CH₃COOH neutralises any added OH⁻ by releasing H⁺ to replace it (via the equilibrium) — either addition is absorbed by shifting a large existing equilibrium only slightly, so pH barely changes.

The pH of a buffer is given by the Henderson–Hasselbalch equation, derived from the Ka expression of the weak acid: Ka = [H⁺][A⁻]/[HA], so [H⁺] = Ka × [HA]/[A⁻]; taking −log of both sides, pH = pKa + log([salt]/[acid]) = pKa + log([A⁻]/[HA]). For a basic buffer, pOH = pKb + log([salt]/[base]).

Buffers are important physiologically — blood is maintained at pH ≈ 7.4 by the carbonic acid–bicarbonate buffer (H₂CO₃/HCO₃⁻).

For a sparingly soluble salt in a saturated solution, e.g. AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), the equilibrium constant is called the solubility product, Ksp = [Ag⁺][Cl⁻] (the pure solid is omitted, as usual for heterogeneous equilibria). For a general salt A_xB_y(s) ⇌ xA^y+ + yB^x−, Ksp = [A^y+]^x[B^x−]^y.

If solubility (moles dissolved per litre) is S, for a 1:1 salt like AgCl, [Ag⁺] = [Cl⁻] = S, so Ksp = S², S = √Ksp. For a salt like Ag₂CrO₄ (Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻), [Ag⁺] = 2S, [CrO₄²⁻] = S, so Ksp = (2S)²(S) = 4S³.

Predicting precipitation using the ionic product (Qsp): if the ionic product Qsp (calculated from given/mixed concentrations, same expression as Ksp) is greater than Ksp, the solution is supersaturated and precipitation occurs; if Qsp = Ksp, the solution is exactly saturated (equilibrium, no net change); if Qsp < Ksp, the solution is unsaturated and no precipitation occurs (or existing precipitate would dissolve).

Common ion effect on solubility: adding a common ion (e.g. NaCl to a saturated AgCl solution) increases [Cl⁻], and since Ksp is fixed at a given temperature, [Ag⁺] must decrease — i.e. the solubility of AgCl decreases in the presence of a common ion. This is used to purify salts by 'salting out'.