Equilibrium

Physical equilibrium involves phase transformations or dissolution processes occurring without changes in chemical composition. When carried out in a closed system at constant temperature, these processes attain a state of dynamic equilibrium:

• Solid-Liquid Equilibrium (Melting / Freezing): $$\text{H}_2\text{O}(s) \rightleftharpoons \text{H}_2\text{O}(l) \quad (\text{at } 0^\circ\text{C}, 1 \text{ atm})$$ At the normal melting point, the rate of melting of ice equals the rate of freezing of water. The mass of ice and water remains constant.
• Liquid-Vapour Equilibrium (Evaporation / Condensation): $$\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_2\text{O}(g) \quad (\text{at constant } T)$$ In a closed vessel, the rate of vaporization equals the rate of condensation. The constant pressure exerted by the vapour is the equilibrium vapour pressure. At normal boiling point ($100^\circ\text{C}$), vapour pressure equals $1 \text{ atm}$ ($1.013 \text{ bar}$).
• Solid-Vapour Equilibrium (Sublimation): $$\text{I}_2(s) \rightleftharpoons \text{I}_2(g), \quad \text{Camphor}(s) \rightleftharpoons \text{Camphor}(g), \quad \text{NH}_4\text{Cl}(s) \rightleftharpoons \text{NH}_4\text{Cl}(g)$$ In a closed vessel, the intensity of violet iodine vapour reaches a constant plateau when dynamic equilibrium is established.
• Dissolution of Solids in Liquids: $$\text{Solute}(s) \rightleftharpoons \text{Solute}(\text{dissolved in solution})$$ At saturation, the rate of dissolution equals the rate of crystallization. The concentration of dissolved solute represents solubility at that temperature.
• Dissolution of Gases in Liquids (Henry's Law):

  William Henry (1803) formulated that at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid surface:

  $$p = K_H \cdot x$$

  where $p$ is the partial pressure of the gas, $x$ is its mole fraction in solution, and $K_H$ is Henry's law constant. As temperature increases, $K_H$ increases and gas solubility decreases (explaining why aquatic life suffers in warm thermal discharge waters).

1. Equilibrium can be established only in a closed system where neither matter enters nor escapes.
2. It is inherently dynamic: the forward and reverse processes continue at identical rates ($r_f = r_b \ne 0$).
3. All measurable macroscopic properties (concentration, total pressure, colour intensity, density, temperature) remain strictly constant.
4. Equilibrium can be approached from either direction (starting from pure reactants or pure products).
5. A dynamic equilibrium represents a state of minimum Gibbs free energy ($G$ is at a minimum, $\Delta G = 0$).

Reversible chemical reactions proceed simultaneously in both forward and reverse directions (denoted by $\rightleftharpoons$). In 1864, Norwegian chemists Cato Maximilian Guldberg and Peter Waage proposed the fundamental Law of Mass Action:

Consider the general reversible reaction:

According to the Law of Mass Action:

where $k_f$ and $k_b$ are the rate constants for forward and reverse reactions. At dynamic chemical equilibrium, the two rates are equal ($r_f = r_b$):

Here, $K_c$ is the equilibrium constant in terms of molar concentration.

For gaseous reactions, concentrations are conveniently replaced by partial pressures. Applying Dalton's law and the ideal gas equation ($P_i = \frac{n_i}{V} RT = C_i RT = [i] RT$):

Substituting $p_i = [i]RT$ for each component:

where $\Delta n_g = (c + d) - (a + b) = \sum n_{\text{gaseous products}} - \sum n_{\text{gaseous reactants}}$.

Three Crucial Cases:

• Case 1: $\Delta n_g = 0$
  Examples: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$, $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$.
  Here $(RT)^0 = 1 \implies K_p = K_c$. $K_p$ and $K_c$ are dimensionless numbers.
• Case 2: $\Delta n_g > 0$
  Examples: $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ ($\Delta n_g = 2 - 1 = +1$).
  Here $K_p = K_c (RT) > K_c$ (assuming $RT > 1$). Unit of $K_p$ is $\text{atm}$ or $\text{bar}$.
• Case 3: $\Delta n_g < 0$
  Examples: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ ($\Delta n_g = 2 - 4 = -2$).
  Here $K_p = K_c (RT)^{-2} < K_c$. Unit of $K_p$ is $\text{atm}^{-2}$ or $\text{bar}^{-2}$.

In a homogeneous equilibrium, all reactants and products reside in a single phase (e.g., all gas or all liquid solution). In a heterogeneous equilibrium, species exist in two or more distinct physical phases.

• $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \implies K_c = [\text{CO}_2], \quad K_p = p_{\text{CO}_2}$
• $\text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \implies K_c = [\text{NH}_3][\text{H}_2\text{S}], \quad K_p = (p_{\text{NH}_3})(p_{\text{H}_2\text{S}})$
• $\text{Fe}_3\text{O}_4(s) + 4\text{H}_2(g) \rightleftharpoons 3\text{Fe}(s) + 4\text{H}_2\text{O}(g) \implies K_p = \frac{(p_{\text{H}_2\text{O}})^4}{(p_{\text{H}_2})^4}$

The Reaction Quotient ($Q$) has the identical mathematical formulation as the equilibrium constant $K$, but the concentrations or partial pressures are measured at any arbitrary non-equilibrium state:

From chemical thermodynamics, the actual Gibbs free energy change $\Delta_r G$ is related to the standard free energy change $\Delta_r G^\circ$ and reaction quotient $Q$ by:

At equilibrium, $\Delta_r G = 0$ and $Q = K$:

• If $\Delta_r G^\circ < 0$: $K > 1$, equilibrium favors products strongly (reaction is exergonic and extensive).
• If $\Delta_r G^\circ = 0$: $K = 1$, reactants and products are equally favored at standard states.
• If $\Delta_r G^\circ > 0$: $K < 1$, equilibrium favors reactants (reaction barely proceeds).

Formulated in 1884 by French chemist Henri Louis Le Chatelier:

1. Effect of Concentration:
   • Increasing the concentration of any reactant causes $Q_c < K_c$, shifting the equilibrium in the forward direction.
   • Increasing the concentration of any product causes $Q_c > K_c$, shifting the equilibrium in the reverse direction.
   • Removing a product as soon as it forms continuously drives the reaction forward (widely exploited industrially to achieve near 100% conversion).
2. Effect of Pressure and Volume:
   • Pressure changes affect only gaseous systems where $\Delta n_g \ne 0$.
   • Increasing pressure (by decreasing volume) forces gas molecules closer together. The system counteracts this by shifting toward the side with fewer moles of gas ($n_g$ decreases).
   • Decreasing pressure (by expanding volume) shifts the equilibrium toward the side with more moles of gas.
   • If $\Delta n_g = 0$ (e.g., $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$), pressure changes have zero effect on the equilibrium composition.
3. Effect of Temperature:
   • Unlike concentration and pressure (which alter $Q_c$ but leave $K$ unchanged), temperature alters the value of the equilibrium constant $K$ itself!
   • Exothermic Reactions ($\Delta H < 0$): Release heat. Increasing temperature shifts equilibrium in the reverse direction (cooling). Hence, $K$ decreases as $T$ increases.
   • Endothermic Reactions ($\Delta H > 0$): Absorb heat. Increasing temperature shifts equilibrium in the forward direction. Hence, $K$ increases as $T$ increases.
   • van 't Hoff Equation: Quantifies the temperature dependence of $K$:
     $$\log_{10}\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)$$
4. Effect of Adding an Inert Gas:
   • At Constant Volume: Total pressure increases, but the partial pressures and molar concentrations of reacting gases ($n_i / V$) remain completely unchanged. Therefore, addition of inert gas at constant volume has no effect on equilibrium.
   • At Constant Pressure: To maintain pressure constant, the total volume must expand. Molar concentrations decrease. The system shifts toward the side with the larger number of gas moles (favors dissociation if $\Delta n_g > 0$).
5. Effect of a Catalyst:

   A catalyst lowers the activation energy equally for both forward and reverse paths. Consequently, it increases $k_f$ and $k_b$ by the exact same factor, leaving $K = k_f / k_b$ unchanged. A catalyst does not alter the position of equilibrium or yield; it merely accelerates the attainment of equilibrium.

A pair of species that differ from each other by only a single proton ($ ext{H}^+$) constitutes a conjugate acid-base pair:

Universal Strength Rule: The stronger the acid, the weaker its conjugate base, and vice versa. For example, $\text{HCl}$ is a very strong acid, so its conjugate base $\text{Cl}^-$ is extraordinarily weak (neutral). Conversely, $\text{HCN}$ is very weak, so $\text{CN}^-$ is a strong Bronsted base.

Amphoteric Substances: Water is amphiprotic (can act as both Bronsted acid and Bronsted base):

Pure water undergoes extremely slight auto-protolysis:

The thermodynamic equilibrium constant is:

Because the degree of ionization is minute, $[\text{H}_2\text{O}]$ is constant ($55.55 \text{ M}$). Combining constants defines the ionic product of water ($K_w$):

In pure water at $25^\circ\text{C}$: $[\text{H}^+] = [\text{OH}^-] = \sqrt{K_w} = 1.0 \times 10^{-7} \text{ M}$.

Temperature Dependence: The auto-ionization of water is endothermic ($\Delta H > 0$). With increase in temperature, $K_w$ increases (at $60^\circ\text{C}$, $K_w \approx 1.0 \times 10^{-13}$, so $[\text{H}^+] = [\text{OH}^-] = 3.16 \times 10^{-7} \text{ M}$, and neutral $\text{pH} = 6.5$).

Søren Peder Lauritz Sørensen (1909) introduced the logarithmic $\text{pH}$ scale to express very small hydronium concentrations concisely:

• Acidic Solution: $[\text{H}^+] > 10^{-7} \text{ M} \implies \text{pH} < 7$
• Neutral Solution: $[\text{H}^+] = [\text{OH}^-] = 10^{-7} \text{ M} \implies \text{pH} = 7$
• Basic Solution: $[\text{H}^+] < 10^{-7} \text{ M} \implies \text{pH} > 7$

For a weak monoprotic acid $\text{HA}$ of molar concentration $C$ with degree of ionization $\alpha$ ($0 < \alpha \ll 1$):

For weak electrolytes, $\alpha \le 0.05$ (less than $5\%$), so $1 - \alpha \approx 1$. Thus, Ostwald's Dilution Law gives:

Polyprotic Acids: For dibasic or tribasic acids (e.g., $\text{H}_3\text{PO}_4$), successive ionization constants follow $K_{a1} \gg K_{a2} \gg K_{a3}$, because removing a positive proton from an increasingly negatively charged anion requires substantially more electrostatic work.

The Common Ion Effect is the suppression of the degree of dissociation of a weak electrolyte by the addition of a strong electrolyte containing a common ion. It is an immediate consequence of Le Chatelier's principle.

• Example: Adding $\text{CH}_3\text{COONa}$ (strong electrolyte) to a solution of $\text{CH}_3\text{COOH}$ (weak acid) floods the solution with $\text{CH}_3\text{COO}^-$, driving the equilibrium $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$ to the left, sharply lowering $[\text{H}^+]$ and raising $\text{pH}$.
• Adding $\text{NH}_4\text{Cl}$ to weak base $\text{NH}_4\text{OH}$ suppresses $[\text{OH}^-]$, an essential condition used in Group III qualitative inorganic analysis to precipitate only $\text{Fe}^{3+}, \text{Al}^{3+}, \text{Cr}^{3+}$ as hydroxides without co-precipitating $\text{Zn}^{2+}, \text{Ni}^{2+}, \text{Mn}^{2+}$.

Salt hydrolysis is the interaction of the cation, anion, or both ions of a salt with water to produce an excess of $\text{H}^+$ or $\text{OH}^-$ ions, altering the solution $\text{pH}$.

A Buffer Solution is an aqueous mixture that possesses the remarkable capability to resist alterations in its $\text{pH}$ upon the addition of small amounts of strong acid, strong base, or upon dilution.

1. Acidic Buffer: Formed by mixing a weak acid and its conjugate salt with a strong base (e.g., $\text{CH}_3\text{COOH} + \text{CH}_3\text{COONa}$). Operates efficiently in the acidic region ($ ext{pH} < 7$).
2. Basic Buffer: Formed by mixing a weak base and its conjugate salt with a strong acid (e.g., $\text{NH}_4\text{OH} + \text{NH}_4\text{Cl}$). Operates in the alkaline region ($ ext{pH} > 7$).

Consider an acidic buffer containing weak acid $\text{HA}$ ($[\text{HA}]$) and salt $\text{NaA}$ ($[\text{A}^-]$). Ionization equilibrium:

Taking the negative logarithm on both sides:

Buffer Capacity and Buffer Range:

• Buffer Capacity ($\beta$): Number of moles of strong acid or strong base required per liter of buffer to change its $\text{pH}$ by one unit: $\beta = \frac{db}{d\text{pH}}$.
• Maximum buffer capacity occurs when $[ ext{Salt}] = [ ext{Acid}]$, which gives $ ext{pH} = ext{pK}_a$.
• Effective buffer action operates over the Buffer Range: $\text{pH} = \text{pK}_a \pm 1$ (corresponding to salt-to-acid ratio between $1:10$ and $10:1$).

When a sparingly soluble salt is stirred in water, a dynamic equilibrium is established between the undissolved solid and its dissolved solvated ions in saturated solution:

Applying the Law of Mass Action (with active mass of pure solid salt $[A_x B_y(s)] = 1$):

If the molar solubility of $A_x B_y$ in water is $S \text{ mol/L}$, then at equilibrium $[A^{y+}] = xS$ and $[B^{x-}] = yS$:

Standard Salt Formulations:

The Ionic Product ($Q_{sp}$) is evaluated identically to $K_{sp}$, but using arbitrary initial ion concentrations present in solution immediately upon mixing:

Adding a soluble salt sharing a common ion dramatically suppresses the molar solubility of a sparingly soluble salt. For example, in a solution containing $C \text{ M } \text{NaCl}$, the solubility $S'$ of $\text{AgCl}$ satisfies:

Because $C \gg S'$, the solubility is reduced by several orders of magnitude compared to pure water ($S = \sqrt{K_{sp}}$).

The systematic separation of basic radicals (cations) into analytical groups depends on controlling $Q_{sp}$ relative to $K_{sp}$ via the common ion effect:

1. Group I ($ ext{Ag}^+, ext{Pb}^{2+}, ext{Hg}_2^{2+}$): Precipitated as insoluble chlorides by adding dilute $\text{HCl}$. ($K_{sp}$ of $\text{AgCl} \sim 1.8 \times 10^{-10}$).
2. Group II ($ ext{Cu}^{2+}, ext{Pb}^{2+}, ext{Bi}^{3+}, ext{Cd}^{2+}, ext{As}^{3+}, ext{Sb}^{3+}, ext{Sn}^{2+}$): Precipitated as sulfides by passing $\text{H}_2\text{S}$ gas in the presence of dilute $\text{HCl}$.
   Mechanism: $\text{HCl}$ provides $\text{H}^+$, which strongly suppresses the ionization of weak diprotic acid $\text{H}_2\text{S}$ (Common Ion Effect). The resulting $[\text{S}^{2-}]$ is very low, yet sufficient to exceed the extremely small $K_{sp}$ of Group II sulfides ($K_{sp} \sim 10^{-30}$ to $10^{-44}$), without precipitating Group IV sulfides ($K_{sp} \sim 10^{-15}$ to $10^{-24}$).
3. Group III ($ ext{Fe}^{3+}, ext{Al}^{3+}, ext{Cr}^{3+}$): Precipitated as hydroxides using $\text{NH}_4\text{OH}$ in the presence of excess $\text{NH}_4\text{Cl}$.
   Mechanism: $\text{NH}_4^+$ from $\text{NH}_4\text{Cl}$ suppresses the ionization of weak base $\text{NH}_4\text{OH}$, keeping $[\text{OH}^-]$ low enough to precipitate only Group III hydroxides ($K_{sp} \sim 10^{-38}$) while preventing premature precipitation of Group IV cations and $\text{Mg}^{2+}$.
4. Group IV ($ ext{Zn}^{2+}, ext{Ni}^{2+}, ext{Co}^{2+}, ext{Mn}^{2+}$): Precipitated as sulfides using $\text{H}_2\text{S}$ in alkaline medium ($ ext{NH}_4\text{OH} + \text{NH}_4\text{Cl}$). $\text{OH}^-$ removes $\text{H}^+$ as water, shifting $\text{H}_2\text{S}$ dissociation forward, generating high $[\text{S}^{2-}]$ to exceed the higher $K_{sp}$ of Group IV sulfides.
5. Group V ($ ext{Ba}^{2+}, ext{Sr}^{2+}, ext{Ca}^{2+}$): Precipitated as carbonates by adding $(\text{NH}_4)_2\text{CO}_3$ in the presence of $\text{NH}_4\text{OH}$ and $\text{NH}_4\text{Cl}$.