Equilibrium

Equilibrium is a state reached in a closed system at constant temperature where the rate of the forward process exactly equals the rate of the reverse process. It is called a dynamic equilibrium because both processes keep occurring at the molecular level, but their opposite effects cancel out, so measurable bulk properties (pressure, colour, concentration) stay constant with time.

Physical equilibria involve a change of physical state only, no new substance is formed:

• Solid-liquid: ice and water coexist at 273 K and 1 atm; rate of melting equals rate of freezing.
• Liquid-vapour: a liquid in a closed vessel evaporates until rate of evaporation equals rate of condensation, giving a constant saturated vapour pressure at that temperature.
• Solid-vapour: sublimable solids such as iodine or camphor set up a constant vapour pressure through direct solid-vapour interconversion.
• Dissolution of a solid in a liquid: in a saturated solution the rate of dissolution equals the rate of crystallisation of the solute.
• Dissolution of a gas in a liquid: governed by Henry's law, the mass/amount of gas dissolved in a given quantity of liquid at a given temperature is proportional to the pressure of the gas in equilibrium with the liquid (e.g. CO₂ in a sealed soda bottle).

Common characteristics of any equilibrium (physical or chemical): (i) it is attained only in a closed system at constant temperature; (ii) it is dynamic, not static; (iii) all measurable properties become constant; (iv) it can be approached from either direction (starting from pure reactants or pure products gives the same final equilibrium mixture under identical conditions); (v) a catalyst helps the system reach equilibrium faster but does not alter the equilibrium state itself.

Worked example: A stoppered bottle is half filled with water and left undisturbed at 25 °C. Initially, evaporation dominates and the amount of water vapour above the liquid increases. As vapour molecules accumulate, some start condensing back. Eventually the rate of evaporation becomes equal to the rate of condensation, and the space above the water contains a fixed amount of water vapour exerting the saturated vapour pressure of water at 25 °C, which remains unchanged as long as temperature and the closed condition are maintained -- this is liquid-vapour equilibrium.

For a general reversible reaction aA + bB ⇌ cC + dD at a given temperature, the law of chemical equilibrium (law of mass action) states that the equilibrium constant is:

K_c = [C]^c[D]^d / ([A]^a[B]^b)

where square brackets denote equilibrium molar concentrations. K_c depends only on temperature, not on the initial amounts taken or on the presence of a catalyst. When all reactants and products are gases, it is often more convenient to express the constant in terms of partial pressures, giving K_p, defined the same way but using partial pressures instead of concentrations.

Using the ideal gas equation, the partial pressure of a gas is p = (n/V)RT = [gas]RT. Substituting this into the K_p expression and simplifying gives the relation:

K_p = K_c(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants).

If Δn = 0, K_p = K_c. The reaction quotient, Q, has the same algebraic form as K but can be evaluated at any stage of the reaction, not just at equilibrium. Comparing Q with K_c predicts the direction a reaction will proceed: if Q < K_c, the forward reaction is favoured; if Q > K_c, the reverse reaction is favoured; if Q = K_c, the system is already at equilibrium.

Worked example: In a 1 L closed vessel at constant temperature, the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) reaches equilibrium with 0.6 mol PCl₅, 0.4 mol PCl₃ and 0.4 mol Cl₂ present. Since the volume is 1 L, molar concentrations equal the number of moles: [PCl₅] = 0.6 M, [PCl₃] = 0.4 M, [Cl₂] = 0.4 M. Therefore K_c = ([PCl₃][Cl₂]) / [PCl₅] = (0.4 × 0.4) / 0.6 = 0.16 / 0.6 ≈ 0.267 mol L^-1. Since Δn = (1 + 1) − 1 = 1, K_p = K_c(RT) at that temperature.

Le Chatelier's principle states that 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.

• Concentration: adding more reactant shifts equilibrium forward (towards products); removing a product (e.g. by letting a gas escape) also shifts it forward.
• Pressure/volume: increasing pressure (decreasing volume) shifts equilibrium towards the side with fewer moles of gas; decreasing pressure shifts it towards the side with more moles of gas. If moles of gas are equal on both sides, pressure has no effect on the position of equilibrium.
• Temperature: increasing temperature shifts equilibrium in the endothermic direction (the direction that absorbs the added heat); decreasing temperature shifts it in the exothermic direction. Unlike concentration and pressure changes, a temperature change also changes the numerical value of the equilibrium constant.
• Catalyst: a catalyst speeds up the forward and reverse reactions equally, so it helps equilibrium to be reached faster but does not shift its position or change the value of K.
• Inert gas addition: at constant volume, adding an inert gas does not change partial pressures of reacting species, so there is no shift; at constant total pressure, adding an inert gas increases the volume, which shifts equilibrium towards the side with more moles of gas.

Worked example: In the industrial synthesis of ammonia, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92.4 kJ mol^-1 (exothermic). (i) Increasing pressure shifts equilibrium forward, since 2 moles of gaseous product form from 4 moles of gaseous reactants, increasing the yield of NH₃. (ii) Increasing temperature shifts equilibrium backward (since the forward reaction is exothermic), lowering the yield of NH₃ even though it speeds up how fast equilibrium is reached; industrially, a moderate temperature (about 700 K) with a catalyst is used as a compromise between yield and rate. (iii) Continuously removing NH₃ from the mixture shifts equilibrium further forward, improving conversion.

Arrhenius concept: an acid is a substance that increases the concentration of H⁺ (H₃O⁺) ions when dissolved in water; a base is a substance that increases the concentration of OH^- ions in water. This concept is limited to aqueous solutions.

Bronsted-Lowry concept: an acid is a proton (H⁺) donor; a base is a proton acceptor. When an acid donates a proton, it forms its conjugate base; when a base accepts a proton, it forms its conjugate acid. Together, an acid and the base formed from it (or vice versa) are called a conjugate acid-base pair, differing by exactly one proton. Species such as H₂O, HCO₃^- and HSO₄^- can act as either an acid or a base depending on the other species present, and are called amphoteric (or amphiprotic).

Lewis concept: an acid is a species that can accept a pair of electrons (an electron-deficient species or electrophile), while a base is a species that can donate a pair of electrons (a species with a lone pair, a nucleophile). This is the broadest definition: it includes all Bronsted acids and bases, but also covers species like BF₃ or AlCl₃ that have no proton to donate yet still behave as acids by accepting an electron pair.

Worked example (mechanism): Consider the reaction of boron trifluoride with ammonia: BF₃ + :NH₃ → F₃B←NH₃. Boron in BF₃ has only six electrons around it (an incomplete octet), so it can accept the lone pair on nitrogen in NH₃, forming a coordinate (dative) bond. Here BF₃ is the Lewis acid (electron-pair acceptor) and NH₃ is the Lewis base (electron-pair donor), even though no proton is transferred, so this reaction cannot be classified using the Bronsted-Lowry concept at all.

Worked example (Bronsted-Lowry): NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq). Here H₂O donates a proton to NH₃, so H₂O is the acid and NH₃ is the base; NH₄⁺ is the conjugate acid of NH₃, and OH^- is the conjugate base of H₂O.

Water undergoes slight self-ionisation: 2H₂O(l) ⇌ H₃O⁺(aq) + OH^-(aq), often written simply as H₂O ⇌ H⁺ + OH^-. The equilibrium constant for this, the ionic product of water, K_w, equals [H₃O⁺][OH^-] = 1.0 × 10^-14 at 298 K. In pure water [H⁺] = [OH^-] = 1.0 × 10^-7 M, so pure water is neutral.

The pH scale expresses acidity conveniently: pH = −log[H⁺], and similarly pOH = −log[OH^-]. Taking the log of the K_w expression gives pH + pOH = 14 at 298 K. Solutions with pH < 7 are acidic, pH = 7 neutral, pH > 7 basic.

For a weak acid HA, HA(aq) ⇌ H⁺(aq) + A^-(aq), with ionisation constant K_a = [H⁺][A^-]/[HA]. If C is the initial concentration and α the degree of dissociation, Ostwald's dilution law gives K_a = Cα²/(1 − α); when α << 1, this simplifies to K_a ≈ Cα², so α ≈ √(K_a/C). An analogous K_b equation applies to weak bases. For any conjugate acid-base pair (e.g. NH₄⁺ and NH₃), the two ionisation constants are linked through the ionic product of water: K_a × K_b = K_w.

A buffer solution resists a change in pH when small amounts of acid or base are added, or on dilution. An acidic buffer (weak acid + its salt with a strong base, e.g. CH₃COOH + CH₃COONa) maintains a pH below 7; a basic buffer (weak base + its salt with a strong acid, e.g. NH₄OH + NH₄Cl) maintains a pH above 7. The pH of an acidic buffer is given by the Henderson-Hasselbalch equation: pH = pK_a + log([salt]/[acid]); similarly pOH = pK_b + log([salt]/[base]) for a basic buffer.

Worked example: Calculate the pH of a buffer containing 0.1 M CH₃COOH and 0.1 M CH₃COONa, given K_a = 1.8 × 10^-5. pK_a = −log(1.8 × 10^-5) = 4.74. Since [salt] = [acid] here, log([salt]/[acid]) = log(1) = 0, so pH = pK_a + 0 = 4.74.

For a sparingly soluble ionic salt A_xB_y, dissolution sets up the equilibrium: A_xB_y(s) ⇌ xA^y+(aq) + yB^x-(aq). The equilibrium constant for this process is the solubility product, K_sp = [A^y+]^x[B^x-]^y, which is constant at a given temperature (the solid is not included, as its activity is taken as 1). If the molar solubility of the salt is s, then [A^y+] = xs and [B^x-] = ys, giving K_sp = x^xy^ys^(x+y).

For a 1:1 salt like AgCl, K_sp = s², so s = √K_sp. For a salt like Ag₂CrO₄ (Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄^2-), K_sp = (2s)²(s) = 4s³.

The ionic product, Q_sp, has the same form as K_sp but can be calculated for any concentrations, not only at saturation. Comparing Q_sp to K_sp predicts whether a precipitate forms: if Q_sp > K_sp, the solution is supersaturated and precipitation occurs until Q_sp = K_sp; if Q_sp < K_sp, the solution is unsaturated and no precipitate forms (more solid can dissolve); if Q_sp = K_sp, the solution is exactly saturated (at equilibrium).

The common ion effect lowers the solubility of a sparingly soluble salt: adding a soluble salt that supplies an ion already present in the equilibrium (e.g. adding NaCl to a saturated AgCl solution, which supplies extra Cl^-) shifts the dissolution equilibrium backward by Le Chatelier's principle, decreasing the solubility of AgCl. This effect is used, for example, to purify common salt by passing HCl gas through a saturated impure brine solution, which precipitates purer NaCl.

Worked example: The K_sp of AgCl is 1.8 × 10^-10. Will a precipitate form when equal volumes of 0.002 M AgNO₃ and 0.002 M NaCl are mixed? On mixing equal volumes, each concentration is halved: [Ag⁺] = [Cl^-] = 1.0 × 10^-3 M. So Q_sp = [Ag⁺][Cl^-] = (1.0 × 10^-3)(1.0 × 10^-3) = 1.0 × 10^-6. Since Q_sp (1.0 × 10^-6) > K_sp (1.8 × 10^-10), a precipitate of AgCl will form.