'Equilibrium'

Equilibrium in Chemical Processes

Equilibrium is a state where opposing processes (forward and reverse reactions) occur at equal rates, resulting in no net change.

Physical Equilibrium

• Solid-liquid equilibrium: Rate of melting = Rate of freezing.
• Liquid-vapour equilibrium: Rate of evaporation = Rate of condensation.
• Solid-vapour equilibrium: Sublimation equilibrium.

Chemical Equilibrium

Dynamic equilibrium where rates of forward and reverse reactions are equal.

Law of Mass Action and Equilibrium Constant

For a reversible reaction: aA + bB <=> cC + dD

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

Where concentrations are at equilibrium.

Kp (Equilibrium Constant in Terms of Pressure)

For gaseous reactions: K_p = (P_C^c * P_D^d)/(P_A^a * P_B^b)

Relation Between Kp and Kc

K_p = K_c (RT)^(Delta n_g)

Where Delta n_g = (c + d) - (a + b) (change in moles of gas).

Characteristics of K

• K depends only on temperature.
• K is dimensionless (concentrations relative to standard state).
• If K > 1: Products favoured at equilibrium.
• If K < 1: Reactants favoured.
• If K = 1: Equal concentrations.

Le Chatelier's Principle

If a system at equilibrium is subjected to a change (concentration, temperature, pressure), the equilibrium shifts in the direction that tends to counteract the change.

Effect of Concentration

• Increasing reactant concentration: Equilibrium shifts to the right (more products).
• Increasing product concentration: Equilibrium shifts to the left.

Effect of Pressure

• Increasing pressure: Shifts towards fewer gas molecules.
• No effect if equal gas molecules on both sides.
• No effect on reactions with no gases.

Effect of Temperature

• Endothermic reaction (Delta H > 0): Increasing temperature shifts equilibrium to the right.
• Exothermic reaction (Delta H < 0): Increasing temperature shifts equilibrium to the left.

Effect of Catalyst

Catalyst does NOT affect equilibrium position. It only speeds up attainment of equilibrium.

Ionic Equilibrium

Strong and Weak Electrolytes

• Strong electrolytes: Completely ionised (HCl, NaOH, NaCl).
• Weak electrolytes: Partially ionised (CH3COOH, NH4OH).

Ostwald's Dilution Law

For weak electrolyte AB: alpha = sqrt(K_a/C), where alpha = degree of dissociation.

Ionisation of Water

H2O <=> H+ + OH- K_w = [H+][OH-] = 1.0 x 10^(-14) at 25 C.

pH Scale

pH = -log[H+] pOH = -log[OH-] pH + pOH = 14

Ionisation Constants (Ka and Kb)

• Ka = [H+][A-]/[HA] for weak acid HA.
• Kb = [BH+][OH-]/[B] for weak base B.
• Stronger acid = higher Ka, lower pKa.
• Kw = Ka * Kb (for conjugate acid-base pair).

Buffer Solutions

Solutions that resist change in pH when small amounts of acid or base are added.

Types

• Acidic buffer: Weak acid + its salt with strong base (CH3COOH + CH3COONa).
• Basic buffer: Weak base + its salt with strong acid (NH4OH + NH4Cl).

Henderson-Hasselbalch Equation

For acidic buffer: pH = pKa + log([Salt]/[Acid]) For basic buffer: pOH = pKb + log([Salt]/[Base])

Buffer Capacity

phi = (number of moles of acid/base added)/(change in pH)

Maximum buffer capacity when [Salt] = [Acid] (or [Salt] = [Base]), i.e., pH = pKa.

Solubility Product (Ksp)

For sparingly soluble salt A_x B_y: A_x B_y(s) <=> xA^(y+) + yB^(x-) Ksp = [A^(y+)]^x [B^(x-)]^y

Relation with Solubility (s)

• AB type: Ksp = s^2
• AB2 or A2B type: Ksp = 4s^3
• AB3 type: Ksp = 27s^4

Common Ion Effect

The degree of dissociation of a weak electrolyte is suppressed in presence of a common ion. Used in buffer preparation and qualitative analysis.

Worked Examples

Example 1: For N2 + 3H2 <=> 2NH3, Kc = 0.5 at 400 C. If [N2] = 2 M, [H2] = 4 M, [NH3] = 4 M at equilibrium, verify Kc. Solution: Kc = [NH3]^2/([N2][H2]^3) = 16/(2*64) = 16/128 = 0.125 != 0.5. Not at equilibrium.

Example 2: Find pH of 0.01 M HCl. Solution: [H+] = 0.01 M. pH = -log(0.01) = 2.

Example 3: Calculate pH of 0.1 M acetic acid (Ka = 1.8 x 10^(-5)). Solution: [H+] = sqrt(Ka*C) = sqrt(1.8x10^(-5)*0.1) = sqrt(1.8x10^(-6)) = 1.34 x 10^(-3) M. pH = -log(1.34x10^(-3)) = 2.87.

Common Mistakes

1. Kc and Kp relation: Kp = Kc(RT)^(Delta n_g). Remember the exponent is Delta n_g.
2. Square brackets mean equilibrium concentration: Not initial concentration.
3. pH of strong acids: For very dilute acids (< 10^(-6) M), consider H+ from water.
4. Buffer range: Buffers work best within pKa +/- 1 pH unit.

ISC Exam Focus

• Theory (70%): Kc and Kp expressions, Le Chatelier's principle, pH, buffer solutions, Ksp.
• Application (30%): Numerical problems on Kc, pH, buffer, solubility product.
• ISC frequently asks: "State Le Chatelier's principle and apply to ..." and "Calculate pH of ...".
• Buffer solution problems using Henderson-Hasselbalch equation are common.

Self-Test Questions

Q1: State Le Chatelier's principle. Answer: If a system at equilibrium is disturbed, it shifts in the direction that counteracts the disturbance.

Q2: Find pH of 0.001 M NaOH. Answer: [OH-] = 0.001 M. pOH = 3. pH = 14 - 3 = 11.

Q3: For 2SO2 + O2 <=> 2SO3, write Kc and Kp expressions. Answer: Kc = [SO3]^2/([SO2]^2[O2]). Kp = P_SO3^2/(P_SO2^2 * P_O2).

Q4: Calculate Kp from Kc for the above reaction at 300 K (Delta n_g = -1). Kc = 4. Answer: Kp = Kc(RT)^(Delta n_g) = 4*(0.0821*300)^(-1) = 4/24.63 = 0.162.

Q5: Define buffer solution. Give an example of an acidic buffer. Answer: A solution that resists pH change. Acidic buffer: CH3COOH + CH3COONa.

Q6: Find the pH of a buffer containing 0.1 M CH3COOH and 0.1 M CH3COONa (Ka = 1.8 x 10^(-5)). Answer: pH = pKa + log([Salt]/[Acid]) = -log(1.8x10^(-5)) + log(1) = 4.74 + 0 = 4.74.