17.2 Reversible Reactions, Equilibrium Constants (Kc, Kp), Le Chatelier & Ksp

Reversible Reactions, Equilibrium Constants (Kc, Kp), Le Chatelier & Ksp

In the Intermediate Chemistry Part I syllabus and the University of Engineering and Technology (UET) ECAT entry test, chemical and ionic equilibria form the core of physical chemistry problem-solving. ECAT questions frequently challenge candidates to derive and convert between $K_c$ and $K_p$, evaluate reaction quotient ($Q_c$) shifts, determine optimal operating parameters for industrial syntheses, calculate the pH of conjugate acid-base buffer solutions using the Henderson-Hasselbalch equation, and predict selective precipitation using solubility products ($K_{sp}$) in the presence of common ions. Rapidly identifying algebraic relationships without relying on electronic calculators is essential for achieving a top percentile score.

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1. Dynamic Equilibrium & The Law of Mass Action

Chemical reactions are fundamentally divided into two categories based on completeness:

1. Irreversible Reactions: Reactants convert completely into products, proceeding in a single forward direction until at least one reactant is entirely exhausted. Such reactions occur primarily in open vessels or involve the escape of a gaseous product, precipitation of an insoluble solid, or vigorous neutralization (e.g., $2\text{KClO}_3(s) \xrightarrow{\Delta} 2\text{KCl}(s) + 3\text{O}_2(g)$, or $\text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)$).
2. Reversible Reactions: Products react together to regenerate the original reactants simultaneously. These reactions never proceed to absolute completion and can establish an equilibrium state if maintained within a closed system where matter cannot enter or escape (e.g., $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$).

The Nature of Dynamic Chemical Equilibrium

When reactants are initially mixed, the forward reaction rate is at its maximum while the reverse reaction rate is zero. As reactants are consumed, their concentrations decrease, causing the forward rate ($R_f$) to decline. Concurrently, as products accumulate, the reverse reaction rate ($R_r$) increases. Eventually, a point is reached where both rates become exactly equal:

$R_{\text{forward}} = R_{\text{reverse}}$

This condition is defined as dynamic chemical equilibrium:

• Dynamic: Chemical reactions have not ceased. Molecules continue to react in both directions at equal, non-zero velocities.
• Macroscopic Invariance: The observable macroscopic properties of the system—such as total pressure, individual concentrations, color intensity, density, and temperature—remain perfectly constant over time.
• Approachability: Equilibrium can be established starting from either direction (pure reactants or pure products), provided temperature and pressure remain fixed.

The Law of Mass Action

Formulated in 1864 by Norwegian chemists Cato Maximilian Guldberg and Peter Waage, the Law of Mass Action states:

The rate at which a substance reacts is directly proportional to its active mass (molar concentration), and the rate of a chemical reaction is directly proportional to the product of the active masses of the reacting substances, each raised to a power equal to its stoichiometric coefficient in the balanced chemical equation.

For a general reversible elementary reaction:

$aA + bB \rightleftharpoons cC + dD$

The forward and reverse rates are expressed as:

$R_f = k_f [A]^a [B]^b \quad \text{and} \quad R_r = k_r [C]^c [D]^d$

where $k_f$ and $k_r$ are the forward and reverse specific rate constants, and square brackets denote equilibrium molar concentrations in $\text{mol}\cdot\text{dm}^{-3}$. At dynamic equilibrium, equating the rates gives:

$k_f [A]^a [B]^b = k_r [C]^c [D]^d$

Rearranging the ratio of specific rate constants defines the equilibrium constant in terms of concentration ($K_c$):

$K_c = \frac{k_f}{k_r} = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

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2. Expressions for Equilibrium Constants: $K_c$, $K_p$, $K_x$, and $K_n$

Depending on the physical state of the reacting species and the measurement units employed, equilibrium constants can be expressed in multiple thermodynamic frameworks:

1. Concentration Constant ($K_c$): Utilizes molar concentrations ($[i] = n_i / V$ in $\text{mol}\cdot\text{dm}^{-3}$): $K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$
2. Partial Pressure Constant ($K_p$): Utilizes equilibrium partial pressures ($P_i$ in $\text{atm}$ or $\text{bar}$) for gaseous reactions: $K_p = \frac{P_C^c P_D^d}{P_A^a P_B^b}$
3. Mole Fraction Constant ($K_x$): Utilizes equilibrium mole fractions ($X_i = n_i / n_{\text{total}}$): $K_x = \frac{X_C^c X_D^d}{X_A^a X_B^b}$
4. Mole Constant ($K_n$): Utilizes the actual stoichiometric mole quantities at equilibrium ($n_i$): $K_n = \frac{n_C^c n_D^d}{n_A^a n_B^b}$

Heterogeneous Equilibria Rule

In heterogeneous systems involving multiple physical phases (solids, pure liquids, gases, solutions), the molar concentration or active mass of any pure solid or pure liquid is constant because density and molar mass are intrinsic properties independent of sample size ($[\text{pure solid}] = 1$, $[\text{pure liquid}] = 1$). Consequently, pure solids and pure solvents are omitted from equilibrium constant expressions:

$\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \implies K_c = [\text{CO}_2] \quad \text{and} \quad K_p = P_{\text{CO}_2}$

Mathematical Derivation Relating $K_p$ and $K_c$

Assuming all reacting gases obey the ideal gas equation $P_i V = n_i R T$, the partial pressure of component $i$ is:

$P_i = \left(\frac{n_i}{V}\right) R T = [i] R T$

Substituting this relation into the expression for $K_p$ for the reaction $aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)$:

$K_p = \frac{([C]RT)^c ([D]RT)^d}{([A]RT)^a ([B]RT)^b} = \frac{[C]^c [D]^d}{[A]^a [B]^b} \cdot \frac{(RT)^{c+d}}{(RT)^{a+b}}$

Defining the change in stoichiometric moles of gas as $\Delta n_g = (c + d) - (a + b)$:

$K_p = K_c (R T)^{\Delta n_g}$

where $R = 0.0821\text{ dm}^3\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ and $T$ is temperature in Kelvin.

Crucial Cases Governing $K_p$ and $K_c$

• Case 1: $\Delta n_g = 0$ (No change in moles of gas): $(RT)^0 = 1 \implies K_p = K_c$ Both $K_p$ and $K_c$ are dimensionless. Changes in total pressure or system volume have zero effect on the equilibrium position. Examples: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$, and $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$.
• Case 2: $\Delta n_g > 0$ (Increase in moles of gas): $K_p = K_c(RT)^{\Delta n_g} > K_c \quad (\text{provided } RT > 1)$ Units: $K_c$ has units of $(\text{mol}\cdot\text{dm}^{-3})^{\Delta n_g}$; $K_p$ has units of $(\text{atm})^{\Delta n_g}$. Example: $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ ($\Delta n_g = 2 - 1 = +1$).
• Case 3: $\Delta n_g < 0$ (Decrease in moles of gas): $K_p = K_c(RT)^{\Delta n_g} < K_c \quad (\text{provided } RT > 1)$ Example: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ ($\Delta n_g = 2 - (1+3) = -2$).

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3. Reaction Quotient ($Q_c$) and Predicting Reaction Direction

The reaction quotient ($Q_c$) carries the exact same mathematical formulation as the equilibrium constant $K_c$, but the concentrations inserted are the instantaneous, non-equilibrium concentrations measured at any arbitrary point in time:

$Q_c = \frac{[C]_t^c [D]_t^d}{[A]_t^a [B]_t^b}$

Comparing the numerical value of $Q_c$ against the true equilibrium constant $K_c$ at a given temperature dictates the spontaneous directional drive of the system:

1. $Q_c < K_c$: Product concentrations are deficient relative to reactants. The reaction must proceed spontaneously from left to right (forward direction) to generate more products until $Q_c$ increases to equal $K_c$.
2. $Q_c = K_c$: The system is at dynamic chemical equilibrium. Forward and reverse reaction rates are identical; no net concentration changes occur.
3. $Q_c > K_c$: Product concentrations are in excess relative to reactants. The reaction must proceed spontaneously from right to left (reverse direction) to consume products until $Q_c$ decreases to match $K_c$.

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4. Le Chatelier's Principle: System Response to External Stresses

Formulated in 1884 by French chemist Henri Louis Le Chatelier, this governing principle predicts how an equilibrium system shifts when disturbed:

Statement: If a chemical system at dynamic equilibrium is subjected to an external stress or disturbance (such as a change in concentration, pressure, volume, or temperature), the system adjusts its equilibrium position in such a direction as to counteract and minimize the applied stress.

1. Effect of Concentration Variations

• Adding Reactants / Removing Products: Increases reactant concentration or decreases product concentration, causing $Q_c < K_c$. The system shifts forward (to the right) to consume added reactants or replace removed products.
• Adding Products / Removing Reactants: Causes $Q_c > K_c$. The system shifts in reverse (to the left).
• Note: Changing concentrations alters the equilibrium position (equilibrium concentrations), but does not change the numerical value of $K_c$.

2. Effect of Pressure and Volume Changes (Gaseous Equilibria)

According to Boyle's law, pressure is inversely proportional to volume ($P \propto 1/V$):

• Increasing Pressure (Decreasing Volume): Compresses the gas mixture, increasing the number of molecules per unit volume. The system counteracts this stress by shifting toward the side possessing fewer moles of gas (lower $\sum n_{\text{gas}}$).
• Decreasing Pressure (Increasing Volume): The system shifts toward the side possessing greater moles of gas.
• Systems with $\Delta n_g = 0$: If moles of gaseous reactants equal moles of gaseous products (e.g., $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$), pressure changes compress both sides equally. Total pressure changes have no effect on the equilibrium position.

Effect of Adding an Inert Gas (e.g., Argon, Helium)

• At Constant Volume ($V = \text{constant}$): Adding an inert gas increases total system pressure, but the partial pressures and molar concentrations ($n_i / V$) of the reacting gases remain completely unchanged. Therefore, inert gas addition at constant volume exerts zero effect on equilibrium position.
• At Constant Pressure ($P = \text{constant}$): To keep total pressure constant when adding inert gas, system volume must expand. Expansion dilutes all reacting species, shifting the equilibrium toward the side with more moles of gas (higher $\Delta n_g$).

3. Effect of Temperature Variations

Temperature is the only external parameter that changes the numerical value of the equilibrium constant ($K_c$ and $K_p$):

• Exothermic Reactions ($\Delta H < 0$): Heat behaves as a product ($A + B \rightleftharpoons C + \text{Heat}$).
  • Increasing temperature ($T \uparrow$) shifts equilibrium to the left (reverse), causing $K_c$ and $K_p$ to decrease.
  • Decreasing temperature ($T \downarrow$) shifts equilibrium to the right (forward), causing $K_c$ and $K_p$ to increase.
• Endothermic Reactions ($\Delta H > 0$): Heat behaves as a reactant ($A + B + \text{Heat} \rightleftharpoons C$).
  • Increasing temperature ($T \uparrow$) shifts equilibrium to the right (forward), causing $K_c$ and $K_p$ to increase.
  • Decreasing temperature ($T \downarrow$) shifts equilibrium to the left (reverse), causing $K_c$ and $K_p$ to decrease.
• The quantitative shift is governed by the van 't Hoff isochore equation: $\ln\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

4. Effect of a Catalyst

A catalyst lowers the activation energy ($E_a$) of both the forward and reverse reactions by providing an alternative reaction pathway. Because $E_{a,f}$ and $E_{a,r}$ are lowered by the exact same numerical quantity:

• The forward rate constant $k_f$ and reverse rate constant $k_r$ increase by the identical multiplicative factor.
• The equilibrium constant $K_c = k_f / k_r$ remains completely unaltered.
• A catalyst does not shift the equilibrium position nor does it affect product yield. It merely accelerates the speed at which equilibrium is attained.

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5. Key Industrial Applications: Haber & Contact Processes

1. Haber-Bosch Synthesis of Ammonia

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92.4\text{ kJ/mol} \quad (\Delta n_g = 2 - 4 = -2)$

• Pressure: Because $\Delta n_g < 0$, high pressure drives the equilibrium to the right. Industrially, an operating pressure of $200\text{ to }300\text{ atm}$ is utilized (limited by mechanical vessel engineering and compressor costs).
• Temperature: Because the reaction is exothermic, low temperature thermodynamically favors ammonia yield. However, at low temperatures, the reaction rate is impractically sluggish. An optimum compromise temperature of $400^\circ\text{C to }450^\circ\text{C}$ is selected to provide acceptable kinetic conversion rates.
• Catalyst: Finely divided iron ($\text{Fe}$) enriched with metal oxide promoters ($4\%\text{ Al}_2\text{O}_3$, $1\%\text{ K}_2\text{O}$, and $\text{SiO}_2$) to increase surface roughness and prevent catalytic sintering.
• Continuous Removal: Ammonia gas is condensed and liquefied out of the cycle at $-33^\circ\text{C}$, keeping $[\text{NH}_3]$ low and continuously driving the reaction forward.

2. Contact Process for Sulfuric Acid ($2\text{nd}$ Stage Oxidation)

$2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \quad \Delta H = -196\text{ kJ/mol} \quad (\Delta n_g = 2 - 3 = -1)$

• Pressure: Moderate pressure of $1\text{ to }2\text{ atm}$ is sufficient because conversion reaches $>98\%$ even at ambient pressure, avoiding expensive high-pressure gas compressors.
• Temperature: Exothermic reaction operated at an optimum $400^\circ\text{C to }450^\circ\text{C}$.
• Catalyst: Vanadium pentoxide ($\text{V}_2\text{O}_5$) pellets (previously platinum, which was abandoned due to arsenic poisoning).

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6. Ionic Equilibria: Water Autoionization, pH, pOH & Buffers

Pure liquid water undergoes self-ionization (autoionization) to a minute extent:

$\text{H}_2\text{O}(l) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$

The thermodynamic equilibrium constant for this process is the ionic product of water ($K_w$):

$K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\text{ mol}^2\cdot\text{dm}^{-6} \quad (\text{at } 25^\circ\text{C})$

Because autoionization is an endothermic bond cleavage ($\Delta H > 0$):

• As temperature rises, $K_w$ increases (e.g., at $100^\circ\text{C}$, $K_w \approx 5.5 \times 10^{-13}$).
• At $25^\circ\text{C}$ in neutral water: $[\text{H}^+] = [\text{OH}^-] = \sqrt{K_w} = 1.0 \times 10^{-7}\text{ M}$.

The pH and pOH Scales

Søren Sørensen defined pH in 1909 as the negative logarithm (base 10) of hydrogen ion molarity:

$\text{pH} = -\log_{10}[\text{H}^+] \quad \text{and} \quad \text{pOH} = -\log_{10}[\text{OH}^-]$

Taking the negative logarithm of $K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}$ yields:

$\text{pH} + \text{pOH} = \text{p}K_w = 14.00 \quad (\text{at } 25^\circ\text{C})$

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

Buffer Solutions and the Henderson-Hasselbalch Equation

A buffer solution is an aqueous solution that resists drastic changes in its hydrogen ion concentration (pH) upon the addition of moderate quantities of strong acid or strong base:

1. Acidic Buffer: Prepared by mixing a weak acid with its conjugate salt formed with a strong base (e.g., acetic acid $\text{CH}_3\text{COOH}$ and sodium acetate $\text{CH}_3\text{COONa}$).
   • Dissociation equilibria: $\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq)$ $\text{CH}_3\text{COONa}(aq) \rightarrow \text{CH}_3\text{COO}^-(aq) + \text{Na}^+(aq)$
   • The acid dissociation constant is $K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$.
   • Taking negative logarithms yields the Henderson-Hasselbalch Equation: $\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{salt}]}{[\text{acid}]}\right)$
2. Basic Buffer: Prepared by mixing a weak base with its conjugate salt formed with a strong acid (e.g., ammonium hydroxide $\text{NH}_4\text{OH}$ and ammonium chloride $\text{NH}_4\text{Cl}$): $\text{pOH} = \text{p}K_b + \log_{10}\left(\frac{[\text{salt}]}{[\text{base}]}\right) \quad \text{and} \quad \text{pH} = 14 - \text{pOH}$

Buffer Capacity & Biological Buffering

• Maximum Buffer Capacity: Occurs when $[\text{salt}] = [\text{acid}]$, making $\log_{10}(1) = 0$, so $\text{pH} = \text{p}K_a$. A buffer functions effectively within the range $\text{pH} = \text{p}K_a \pm 1$.
• Blood Buffering System: Human arterial blood is maintained within a narrow physiological pH window of $7.35\text{ to }7.45$ by the carbonic acid-bicarbonate buffer system ($\text{H}_2\text{CO}_3 / \text{HCO}_3^-$): $\text{H}^+(aq) + \text{HCO}_3^-(aq) \rightleftharpoons \text{H}_2\text{CO}_3(aq) \rightleftharpoons \text{H}_2\text{O}(l) + \text{CO}_2(g)$

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7. Solubility Product ($K_{sp}$) & Common Ion Effect

When a sparingly soluble ionic salt is placed in water, a dynamic equilibrium is rapidly established between the undissolved crystalline solid and its dissolved solvated ions:

$A_x B_y(s) \rightleftharpoons x A^{y+}(aq) + y B^{x-}(aq)$

Because the concentration of undissolved solid remains constant ($[A_x B_y(s)] = 1$), the equilibrium constant is formulated as the Solubility Product Constant ($K_{sp}$):

$K_{sp} = [A^{y+}]^x [B^{x-}]^y$

Mathematical Relations between Molar Solubility ($S$) and $K_{sp}$

Let $S$ represent the molar solubility of the salt in $\text{mol}\cdot\text{dm}^{-3}$:

1. $1:1$ Electrolyte (e.g., $\text{AgCl}, \text{BaSO}_4$): $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq) \implies K_{sp} = (S)(S) = S^2 \implies S = \sqrt{K_{sp}}$
2. $1:2$ or $2:1$ Electrolyte (e.g., $\text{PbCl}_2, \text{Ca(OH)}_2, \text{Ag}_2\text{CrO}_4$): $\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq) \implies K_{sp} = (S)(2S)^2 = 4S^3 \implies S = \sqrt[3]{\frac{K_{sp}}{4}}$
3. $1:3$ Electrolyte (e.g., $\text{Al(OH)}_3, \text{Fe(OH)}_3$): $\text{Fe(OH)}_3(s) \rightleftharpoons \text{Fe}^{3+}(aq) + 3\text{OH}^-(aq) \implies K_{sp} = (S)(3S)^3 = 27S^4 \implies S = \sqrt[4]{\frac{K_{sp}}{27}}$
4. $2:3$ Electrolyte (e.g., $\text{Ca}_3(\text{PO}_4)_2, \text{Bi}_2\text{S}_3$): $K_{sp} = (2S)^2 (3S)^3 = 108S^5 \implies S = \sqrt[5]{\frac{K_{sp}}{108}}$

Criterion for Precipitation & Selective Separation

Comparing the ionic product quotient ($Q_{sp}$) against $K_{sp}$ dictates precipitation:

• $Q_{sp} < K_{sp}$ (Unsaturated): No precipitate forms; additional solid solute can dissolve.
• $Q_{sp} = K_{sp}$ (Saturated): System is in dynamic dissolution-crystallization equilibrium.
• $Q_{sp} > K_{sp}$ (Supersaturated): Solid crystalline precipitate forms instantaneously until ion concentrations drop to satisfy $Q_{sp} = K_{sp}$.

The Common Ion Effect in Qualitative Analysis

The Common Ion Effect is the suppression of the degree of dissociation of a weak electrolyte or sparingly soluble salt caused by the addition of a strong electrolyte that provides a shared ionic species.

• In salt analysis, Group II basic radicals ($\text{Cu}^{2+}, \text{Cd}^{2+}, \text{Pb}^{2+}$) are precipitated as sulfides by passing $\text{H}_2\text{S}$ gas in the presence of dilute $\text{HCl}$. The high concentration of $\text{H}^+$ from fully dissociated $\text{HCl}$ suppresses $\text{H}_2\text{S}$ ionization via the common ion effect, suppressing $[\text{S}^{2-}]$ so that only the very low $K_{sp}$ sulfides of Group II precipitate, leaving Group IIIB cations ($\text{Zn}^{2+}, \text{Ni}^{2+}, \text{Mn}^{2+}$) in solution.
• Conversely, Group IIIB radicals are precipitated in alkaline medium using $\text{H}_2\text{S} + \text{NH}_4\text{OH} + \text{NH}_4\text{Cl}$, where hydroxide ions remove $\text{H}^+$ to yield sufficiently high $[\text{S}^{2-}]$ to exceed the larger $K_{sp}$ thresholds of Group IIIB sulfides.

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8. Le Chatelier Effects and Equilibrium Expressions Table