4.5 Chemical Equilibrium, Le Chatelier's Principle & Acid-Base Theories

4.5 Chemical Equilibrium, Le Chatelier's Principle & Acid-Base Theories

Chemical equilibrium governs reversible reactions in physical and analytical chemistry. This section covers equilibrium constant expressions, Le Chatelier's principle, industrial synthesis optimizations, classical and modern acid-base theories, buffer solutions, and solubility products.

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Dynamic Equilibrium & Law of Mass Action

A reaction reaches dynamic equilibrium when forward and reverse reaction rates become equal while reactant and product concentrations remain constant.

Law of Mass Action & Equilibrium Constants ($K_c, K_p$)

For a general reversible reaction: $aA + bB \rightleftharpoons cC + dD$

• Equilibrium Constant in Concentration ($K_c$): $K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$
• Equilibrium Constant in Partial Pressures ($K_p$): $K_p = \frac{P_C^c P_D^d}{P_A^a P_B^b}$
• Relationship Between $K_p$ and $K_c$: $K_p = K_c (R T)^{\Delta n_g}$ where $\Delta n_g = (c+d) - (a+b)$ gaseous moles.
  • If $\Delta n_g = 0 \implies K_p = K_c$ (e.g., $\text{H}{2(g)} + \text{I}{2(g)} \rightleftharpoons 2\text{HI}_{(g)}$).
  • If $\Delta n_g > 0 \implies K_p > K_c$ (e.g., $\text{PCl}{5(g)} \rightleftharpoons \text{PCl}{3(g)} + \text{Cl}_{2(g)}$).
  • If $\Delta n_g < 0 \implies K_p < K_c$ (e.g., $\text{N}{2(g)} + 3\text{H}{2(g)} \rightleftharpoons 2\text{NH}_{3(g)}$).

Reaction Quotient ($Q_c$)

• $Q_c < K_c$: Reaction proceeds forward (to the right).

• $Q_c = K_c$: System is at equilibrium.

• $Q_c > K_c$: Reaction proceeds reverse (to the left).

• CRITICAL RULE: $K_c$ and $K_p$ values depend ONLY on temperature. They are completely independent of initial concentrations, pressure, volume, or presence of a catalyst.

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Le Chatelier's Principle & Industrial Applications

Le Chatelier's Principle: If a stress (change in concentration, pressure, volume, or temperature) is applied to a system at dynamic equilibrium, the system shifts its equilibrium position to relieve the stress.

Summary of Disturbances

Industrial Synthesis Case Studies

1. Haber Process for Ammonia: $\text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} \quad \Delta H = -92.4 \text{ kJ/mol} \quad (\Delta n_g = -2)$
   • Optimal Yield Conditions: High Pressure ($200 \text{ atm}$), Compromise Temperature ($400-450^\circ\text{C}$), Iron catalyst with $\text{Al}_2\text{O}_3 / \text{K}_2\text{O}$ promoter, continuous removal of liquefied $\text{NH}_3$.
2. Contact Process for Sulfur Trioxide: $2\text{SO}_{2(g)} + \text{O}_{2(g)} \rightleftharpoons 2\text{SO}_{3(g)} \quad \Delta H = -198 \text{ kJ/mol}$
   • Optimal Conditions: $400-450^\circ\text{C}$, $1-2 \text{ atm}$, Vanadium Pentoxide ($\text{V}_2\text{O}_5$) catalyst.

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Acid-Base Theories

Brønsted-Lowry Conjugate Acid-Base Pairs

When an acid donates a proton, it forms a Conjugate Base. When a base accepts a proton, it forms a Conjugate Acid: $\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+$

• Acid: $\text{CH}_3\text{COOH} \rightarrow$ Conjugate Base: $\text{CH}_3\text{COO}^-$
• Base: $\text{H}_2\text{O} \rightarrow$ Conjugate Acid: $\text{H}_3\text{O}^+$
• Rule: Strong acids have weak conjugate bases; weak acids have strong conjugate bases.
• Amphoteric Species: Can act as either an acid or a base (e.g., $\text{H}_2\text{O}, \text{HCO}_3^-, \text{HSO}_4^-$).

Lewis Acid and Base Examples

• Lewis Acids: Electron-deficient molecules ($\text{BF}_3, \text{AlCl}_3, \text{SO}_3$) or cations ($\text{H}^+, \text{Fe}^{3+}, \text{Cu}^{2+}$).
• Lewis Bases: Species with lone pairs ($\text{NH}_3, \text{H}_2\text{O}, \text{R-OH}$) or anions ($\text{F}^-, \text{OH}^-, \text{CN}^-$).

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pH Scale, Weak Acids & Buffer Solutions

• Autoionization of Water: $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ at $25^\circ\text{C}$. $\text{pH} + \text{pOH} = 14$
• Dissociation Constants: $\text{p}K_a = -\log_{10} K_a$. Stronger acid $\rightarrow$ larger $K_a \rightarrow$ smaller $\text{p}K_a$. $K_a \times K_b = K_w = 10^{-14} \implies \text{p}K_a + \text{p}K_b = 14$

Buffer Solutions

Buffers resist sharp changes in pH upon addition of small amounts of strong acid or base.

1. Acidic Buffer: Weak acid + Salt of weak acid with strong base (e.g., $\text{CH}_3\text{COOH} + \text{CH}_3\text{COONa}$).
   • Henderson-Hasselbalch Equation: $\text{pH} = \text{p}K_a + \log_{10}\left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$
2. Basic Buffer: Weak base + Salt of weak base with strong acid (e.g., $\text{NH}_4\text{OH} + \text{NH}_4\text{Cl}$).
   • Henderson-Hasselbalch Equation: $\text{pOH} = \text{p}K_b + \log_{10}\left( \frac{[\text{Salt}]}{[\text{Base}]} \right) \quad \text{and} \quad \text{pH} = 14 - \text{pOH}$

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

For a sparingly soluble salt $A_x B_y \rightleftharpoons x A^{y+} + y B^{x-}$ with molar solubility $S$: $K_{sp} = [A^{y+}]^x [B^{x-}]^y = (x S)^x (y S)^y = x^x y^y S^{x+y}$

• Example for $\text{AgCl}$: $K_{sp} = [\text{Ag}^+][\text{Cl}^-] = S^2 \implies S = \sqrt{K_{sp}}$.
• Precipitation Condition: Precipitation occurs when Ionic Product ($Q_{sp}$) $> K_{sp}$.

Common Ion Effect

The suppression of ionization of a weak electrolyte by adding a strong electrolyte containing a common ion.

• Application: Used in qualitative inorganic analysis to selectively precipitate Group III cations ($\text{Fe}^{3+}, \text{Al}^{3+}, \text{Cr}^{3+}$) as hydroxides using $\text{NH}_4\text{OH}$ in the presence of $\text{NH}_4\text{Cl}$.

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Worked Numerical Examples

Example 1: Buffer Solution pH Calculation

Problem: Calculate the pH of a buffer solution containing $0.10 \text{ M} \text{ CH}_3\text{COOH}$ and $0.20 \text{ M} \text{ CH}_3\text{COONa}$ given $\text{p}K_a(\text{CH}_3\text{COOH}) = 4.74$.

Solution: Using Henderson-Hasselbalch equation: $\text{pH} = \text{p}K_a + \log_{10}\left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$ $\text{pH} = 4.74 + \log_{10}\left( \frac{0.20}{0.10} \right) = 4.74 + \log_{10}(2) = 4.74 + 0.301 = 5.041$

Example 2: $K_p$ from $K_c$ Calculation

Problem: For $\text{N}{2(g)} + 3\text{H}{2(g)} \rightleftharpoons 2\text{NH}_{3(g)}$, if $K_c = 0.50 \text{ M}^{-2}$ at $500 \text{ K}$, calculate $K_p$.

Solution:

1. $\Delta n_g = 2 - (1 + 3) = -2$.
2. $R = 0.0821 \text{ atm}\cdot\text{dm}^3\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$.
3. $R T = 0.0821 \times 500 = 41.05$.
4. $K_p = K_c (R T)^{\Delta n_g} = 0.50 \times (41.05)^{-2} = \frac{0.50}{1685.1} = 2.97 \times 10^{-4} \text{ atm}^{-2}$.

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Environmental Chemistry Links (Acid Rain, Smog, Ozone)

Link acid–base and redox ideas to the short FSc environmental chapter:

• Acid rain: primarily from SO₂ and NOₓ dissolving to form H₂SO₄ / HNO₃; damages marble (CaCO₃) and aquatic systems.
• Photochemical smog: NO₂ + hydrocarbons + sunlight → ozone and PAN at ground level.
• Stratospheric ozone depletion: catalytic chlorine cycles from CFCs (older FSc wording still appears in banks).
• Greenhouse gases: CO₂, CH₄, N₂O—know qualitative warming contribution, not contested policy debates.

These items are usually one-fact recalls; connect them to Le Chatelier only when an equilibrium shift is explicitly asked.