Equilibrium — NEET Notes

Equilibrium in physical processes

The cleanest way to meet equilibrium is to watch it in a phase change. Place ice and water in a perfectly insulated thermos at 273 K and 1 atm: the temperature stays fixed, the mass of ice and water never changes, yet at the boundary an intense traffic continues — molecules of ice escape into liquid water at exactly the same rate that liquid molecules adhere to the ice surface. This is the hallmark of dynamic equilibrium: macroscopic stillness, microscopic activity. The same pattern appears whenever two phases of the same substance, or solute and solvent, are confined in a closed container.

Three signatures show up in every physical equilibrium NCERT lists. For a solid–liquid system at one atmosphere there is exactly one temperature — the normal melting point — at which both phases coexist. For a liquid–vapour system at a given temperature there is a unique equilibrium vapour pressure; this is why water boils at 100 °C only at 1.013 bar and at lower temperatures up in the mountains. For a saturated solution there is a unique solubility at each temperature. Carbon-dioxide-in-water (the soda-bottle equilibrium) obeys Henry's law: the dissolved gas is proportional to its partial pressure above the liquid, which is why opening a bottle releases the fizz.

Equilibrium in chemical processes — dynamic equilibrium

When reactants in a closed vessel at constant temperature react, the concentrations of reactants fall and those of products rise — but only for a while. Eventually the concentrations stop changing. At this point, the forward rate has fallen to meet the reverse rate, and the reverse rate has risen to meet the forward rate; the two have become equal. The mixture sitting in the vessel is an equilibrium mixture. NCERT classifies real chemical equilibria into three families: reactions that go almost to completion (negligible reactants left), reactions that barely move (negligible products formed), and reactions where reactants and products coexist in comparable amounts. Only the third category is informative for equilibrium calculations; the other two are limits.

The double-headed arrow ⇌ marks every equilibrium. It says explicitly that both processes are still happening — they have simply tied. Radioactive labelling proves this: if a saturated sugar solution is in equilibrium with solid sugar and one drops in some radioactive sugar, radioactivity appears in both phases within minutes. There is no "still" molecule at equilibrium; only a balanced exchange.

At equilibrium the rate of the forward reaction equals the rate of the reverse — concentrations are constant in time but molecules keep moving.

The defining property of dynamic equilibrium

Law of mass action and the equilibrium constant

For a general reversible reaction a A + b B ⇌ c C + d D at constant temperature, Cato Guldberg and Peter Waage's law of mass action (1864) states that the ratio of product concentrations to reactant concentrations, each raised to its stoichiometric coefficient, is a constant — the equilibrium constant Kc:

K_c  =  [C]^c[D]^d / [A]^a[B]^b

Concentration-based equilibrium constant

For reactions involving gases it is more natural to use partial pressures. The pressure-based constant is

K_p  =  (p_C)^c(p_D)^d / (p_A)^a(p_B)^b

Partial-pressure-based equilibrium constant

Using the ideal-gas relation p = (n/V)RT = [gas]RT, Kp and Kc are linked by

K_p  =  K_c (RT)^Δn

Where Δn = (moles of gaseous products) − (moles of gaseous reactants)

If Δn = 0 — equal moles of gas on both sides, as in H₂ + I₂ ⇌ 2 HI — then Kp = Kc numerically. For N₂ + 3 H₂ ⇌ 2 NH₃, Δn = 2 − 4 = −2, so Kp = Kc(RT)⁻². R must be used in units consistent with the pressure unit; with p in atmospheres and concentration in mol L⁻¹, R = 0.0821 L atm K⁻¹ mol⁻¹.

Four properties of K are tested directly. First, if a reaction is reversed, the new constant is 1/K. Second, if a reaction is multiplied by n, the new constant is Kⁿ. Third, if two reactions are added, the resulting K is the product of the individual constants — NEET 2017 tested this exact rule with the ammonia–oxidation network. Fourth, K is a function of temperature alone; changing concentration, pressure or adding a catalyst does not change K.

Homogeneous vs heterogeneous equilibria

A homogeneous equilibrium has all species in a single phase — gaseous (N₂O₄ ⇌ 2 NO₂) or aqueous (CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺). Every species appears in the K expression with its stoichiometric exponent. A heterogeneous equilibrium involves more than one phase — typically a solid or a pure liquid alongside a gas or solution. For these, the activity of any pure solid or pure liquid is taken as 1, so they drop out of the K expression entirely.

The thermal decomposition of strontium carbonate appeared in NEET 2017: SrCO₃(s) ⇌ SrO(s) + CO₂(g). The two solids drop out, so K_p = p_CO₂. At equilibrium the pressure of CO₂ over the solid is fixed at 1.6 atm regardless of how much SrCO₃ is present — a hallmark of heterogeneous equilibrium. Compressing the container only matters until p_CO₂ reaches 1.6 atm; beyond that, more CO₂ converts back to SrCO₃.

Applications of the equilibrium constant

Three questions can be answered the moment K is known: how far the reaction proceeds, in which direction it currently runs, and what the equilibrium concentrations will be. The extent of reaction is governed by the magnitude of K. A very large K (≫ 10³) means the equilibrium lies overwhelmingly to the right — the reaction is essentially complete. A very small K (≪ 10⁻³) means the reverse dominates. K close to unity means reactants and products coexist in comparable amounts.

To predict direction, we use the reaction quotient Q. Q is written with the same algebraic form as K but evaluated at any non-equilibrium concentration:

Q_c  =  [C]^c[D]^d / [A]^a[B]^b   at any instant

Reaction quotient — the moving sibling of K

For a NEET-style numerical, given Kc and starting concentrations, one writes an ICE table (Initial, Change, Equilibrium), substitutes into the K expression, and solves for x. The 3 O₂ ⇌ 2 O₃ problem in NEET 2022 was a direct K rearrangement: with Kc = 3 × 10⁻⁵⁹ and [O₂] = 0.04 M, [O₃]² = Kc[O₂]³ gives [O₃] ≈ 4.38 × 10⁻³².

K, Q and the Gibbs free-energy connection

The equilibrium constant is not an isolated quantity — it is the thermodynamic fingerprint of the reaction, linked to the standard Gibbs free-energy change by

ΔG°  =  − RT ln K

The bridge between thermodynamics and equilibrium

Reading this expression three ways tells you everything. When ΔG° is negative the exponent is positive, K > 1, and products are favoured — the reaction is spontaneous as written. When ΔG° is positive the exponent is negative, K < 1, and reactants are favoured — the reverse direction is spontaneous. When ΔG° = 0 the system is already at equilibrium at standard concentrations; K = 1.

Under non-standard conditions, the running quantity is ΔG = ΔG° + RT ln Q. At equilibrium ΔG = 0 and Q = K, which immediately gives back ΔG° = − RT ln K. NEET 2020 tested this directly: hydrolysis of sucrose has Kc = 2 × 10¹³ at 300 K, and the answer is ΔG° = − (8.314 J K⁻¹ mol⁻¹)(300 K) ln(2 × 10¹³) — sign and substitution are the entire problem.

Le Chatelier's principle

Le Chatelier's principle — proposed by Henry Louis Le Chatelier in 1888 — states that if a system at equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium shifts in the direction that partially counteracts the change. It is the qualitative compass for every "which way does it shift?" question NEET asks. The principle does not change K; it only changes where the system sits on the K = constant curve.

Effect of concentration

Adding a reactant raises Q below K and the system shifts forward to consume the added species; removing a product also lowers Q and shifts forward. Removing a reactant or adding a product raises Q above K and the system shifts backward. The shift is always opposite to the change.

Effect of pressure

Pressure changes matter only when Δn_gas ≠ 0. Increasing pressure (decreasing volume) shifts the equilibrium toward the side with fewer moles of gas; decreasing pressure shifts it toward the side with more moles. For N₂ + 3 H₂ ⇌ 2 NH₃ (4 mol gas → 2 mol gas), high pressure favours ammonia. For reactions with Δn = 0 (H₂ + I₂ ⇌ 2 HI), pressure has no effect on the position of equilibrium.

Effect of inert gas addition

Adding an inert gas at constant volume does not change partial pressures or concentrations — so no shift. Adding it at constant pressure forces the container to expand, lowering the partial pressures of all reactive species, and the system shifts toward the side with more moles of gas.

Effect of temperature

Temperature is the only variable that changes K itself. For an exothermic reaction (ΔH < 0), heat is a product, so raising T shifts the system backward and lowers K. For an endothermic reaction (ΔH > 0), heat is a reactant, so raising T shifts it forward and raises K. NEET 2018 asked precisely this: A₂(g) + B₂(g) ⇌ X₂(g), ΔH = − X kJ, Δn = −1 — to maximise product, use low temperature (exothermic favoured) and high pressure (fewer moles favoured). Answer: low T, high P.

Effect of catalyst

A catalyst lowers the activation energy of both forward and reverse reactions by the same amount. It speeds up the arrival at equilibrium but does not move the destination. The position of equilibrium and K are unchanged.

Acid–base theories — Arrhenius, Brønsted-Lowry, Lewis

Three concepts of acids and bases run in parallel through NCERT and NIOS, each broader than the last. The Arrhenius definition, the oldest, requires water as solvent: an acid releases H⁺ and a base releases OH⁻. The Brønsted-Lowry definition removes the water constraint and works in any solvent: an acid is a proton donor and a base is a proton acceptor. The Lewis definition removes the proton itself: an acid is an electron-pair acceptor and a base is an electron-pair donor. Every Arrhenius acid is a Brønsted acid; every Brønsted acid is a Lewis acid; the reverse is not true.

Conjugate-pair reasoning lives inside the Brønsted-Lowry framework: when HCl + H₂O → H₃O⁺ + Cl⁻, HCl is the acid and Cl⁻ is its conjugate base; H₂O is the base and H₃O⁺ is its conjugate acid. A strong acid has a weak conjugate base and vice versa — strength flips when you cross the arrow. Water itself appears on both sides of the acid/base register; in pure water two molecules briefly act as one acid and one base in autoionisation, the source of Kw.

pH, ionisation constants and the ionic product of water

For a weak acid HA ⇌ H⁺ + A⁻ in water, the ionisation constant is Ka = [H⁺][A⁻]/[HA]; for a weak base B + H₂O ⇌ BH⁺ + OH⁻, the corresponding constant is Kb = [BH⁺][OH⁻]/[B]. Both depend only on temperature. Because Ka and Kb often span many orders of magnitude (CH₃COOH: Ka ≈ 1.8 × 10⁻⁵; HCN: Ka ≈ 6.2 × 10⁻¹⁰), it is conventional to use their logarithms: pKa = − log Ka and pKb = − log Kb. A small Ka means a large pKa, which means a weaker acid.

The pH scale compresses hydrogen-ion concentrations into a workable range: pH = − log [H⁺]. Pure water at 298 K has [H⁺] = 10⁻⁷ M, giving pH = 7. Acidic solutions have pH < 7, basic solutions pH > 7. The temperature dependence is real but small enough that 7 is the standard reference at 25 °C.

Water itself is feebly ionised: H₂O ⇌ H⁺ + OH⁻, with the equilibrium constant for the ionised products called the ionic product of water:

For conjugate acid–base pairs, multiplying Ka × Kb gives Kw — equivalently pKa + pKb = pKw = 14. This single identity is the backbone of every hydrolysis-pH calculation. NEET 2021 used it inside the pH formula for the salt of a weak acid and weak base: pH = 7 + ½(pKa − pKb). For dimethylammonium acetate with pKa(CH₃COOH) = 4.77 and pKb(dimethylamine) = 3.27, pH = 7 + ½(4.77 − 3.27) = 7.75.

For strong acids/bases (HCl, HNO₃, NaOH, KOH), ionisation is essentially complete, so the H⁺ or OH⁻ concentration equals the formal concentration of the acid or base. For weak acids, [H⁺] = √(Ka·C) is the simplifying approximation (degree of ionisation small). For weak bases the analogous result is [OH⁻] = √(Kb·C).

Buffer solutions and the Henderson–Hasselbalch equation

A buffer solution resists changes in pH when small amounts of acid or base are added. Two recipes give buffers: a weak acid plus its salt with a strong base (acidic buffer, e.g. CH₃COOH + CH₃COONa), and a weak base plus its salt with a strong acid (basic buffer, e.g. NH₄OH + NH₄Cl). The pH of an acidic buffer is given by the Henderson–Hasselbalch equation:

pH  =  pK_a  +  log ([salt] / [acid])

Henderson–Hasselbalch equation, acidic buffer

For a basic buffer the analogue is pOH = pKb + log([salt]/[base]). When [salt] = [acid], log(1) = 0 and pH = pKa — this is the buffer's central operating point, where buffering capacity is maximum. A buffer holds pH within about ±1 unit of pKa.

NEET 2022 walked the equation step by step: 50 mL of 0.10 M CH₃COONa plus 50 mL of 0.01 M CH₃COOH, pKa = 4.57. This is a weak acid plus the salt of its conjugate base — an acidic buffer. The salt-to-acid ratio is 0.10 / 0.01 = 10, so pH = 4.57 + log 10 = 4.57 + 1 = 5.57.

Hydrolysis of salts and pH of salt solutions

When a salt dissolves in water, its ions may react with water in a process called hydrolysis. Whether the resulting solution is acidic, basic or neutral depends on the parentage of the cation and anion. Strong-acid–strong-base salts (NaCl, KNO₃) do not hydrolyse — neither ion has any inclination to react with water — and the solution is neutral. Strong-acid–weak-base salts (NH₄Cl, AlCl₃) produce acidic solutions because the cation, the conjugate acid of a weak base, donates a proton to water. Weak-acid–strong-base salts (CH₃COONa, KCN) produce basic solutions because the anion, the conjugate base of a weak acid, abstracts a proton from water. Weak-acid–weak-base salts (CH₃COONH₄) give a pH that depends on the relative magnitudes of Ka and Kb.

For a salt of a weak acid and a weak base — NEET 2021's example, dimethylammonium acetate — the cation and anion both hydrolyse. The pH is independent of concentration to a first approximation:

pH  =  7  +  ½ (pK_a − pK_b)

Salt of weak acid + weak base

The result tilts in whichever direction the parent is stronger — if Ka > Kb (acid is stronger than base), pH < 7; if Kb > Ka, pH > 7; if Ka = Kb the salt solution is exactly neutral.

Solubility equilibria — Ksp and the common-ion effect

For a sparingly soluble salt M_pX_q in water, dissolution is itself an equilibrium: M_pX_q(s) ⇌ p M^q+(aq) + q X^p−(aq). The solid is a pure phase, so its activity is 1; only the ions appear in the constant. This constant is the solubility product:

K_sp  =  [M^q+]^p[X^p−]^q

Solubility product — equilibrium constant for dissolution

Two routine NEET operations follow. Ksp from solubility: for BaSO₄ (1:1 salt) with solubility s, Ksp = s². NEET 2018 used solubility = 2.42 × 10⁻³ g L⁻¹ ÷ 233 g mol⁻¹ = 1.03 × 10⁻⁵ M, giving Ksp = (1.03 × 10⁻⁵)² ≈ 1.08 × 10⁻¹⁰. For a 1:2 salt like Ag₂C₂O₄ (NEET 2017), Ksp = (2s)²(s) = 4s³. For a 1:3 salt like NY₃ (NEET 2016), Ksp = (s)(3s)³ = 27 s⁴. The general rule: Ksp = (p)^p(q)^q·s^p+q.

Solubility from Ksp reverses the calculation. Comparing two salts with equal Ksp but different stoichiometry — the NEET 2016 MY vs NY₃ trap — gives different solubilities: s(MY) = √Ksp ≈ 8 × 10⁻⁷ M, s(NY₃) = (Ksp/27)^1/4 ≈ 3.89 × 10⁻⁴ M. The 1:3 salt is more soluble despite the identical Ksp because the fourth-root flattens the comparison.

The common-ion effect is the most important consequence of Ksp. Adding an ion already present in the solubility equilibrium suppresses dissociation — the system shifts back toward solid, lowering the molar solubility. NEET 2020 used Ni(OH)₂ in 0.1 M NaOH: Ksp = s(0.1 + 2s)² ≈ s(0.1)² = 2 × 10⁻¹⁵, so s = 2 × 10⁻¹³ M — about eight orders of magnitude lower than the solubility in pure water. The dramatic suppression makes the common-ion effect the chemistry behind selective precipitation in qualitative analysis.