JEE Chemistry Equilibrium Notes Set 02

Chemical Equilibrium

A reversible reaction reaches equilibrium when the rate of the forward reaction equals the rate of the backward reaction. Equilibrium is dynamic: both reactions continue, but the concentrations of reactants and products stay constant at a given temperature.
Law of mass action: the rate of a reaction is proportional to the product of the active masses (molar concentrations) of the reactants.

Equilibrium Constant

For \( aA + bB \rightleftharpoons cC + dD \): \( K_c = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \). For gases \( K_p = \dfrac{p_C^{\,c}\,p_D^{\,d}}{p_A^{\,a}\,p_B^{\,b}} \).
Relation: \( K_p = K_c\,(RT)^{\Delta n} \), where \( \Delta n \) = moles of gaseous products - moles of gaseous reactants. Kp = Kc when \( \Delta n = 0 \).
Unit of Kc = (mol/L)\(^{\Delta n}\). Pure solids and pure liquids do not appear in K.
K depends only on temperature, not on initial concentrations, pressure or catalyst.
Manipulating K: reversing a reaction gives 1/K; multiplying coefficients by n gives \( K^{n} \); adding two reactions multiplies their K values; \( k_f/k_b = K_c \).

Reaction Quotient and Direction

Q has the same form as K but uses the concentrations at any instant. If Q < K the reaction goes forward, if Q > K it goes backward, and if Q = K the system is at equilibrium.

Le Chatelier's Principle

If a system at equilibrium is disturbed, it shifts so as to oppose the change.
Concentration: adding a reactant (or removing a product) shifts the equilibrium forward.
Pressure: raising the pressure favours the side with fewer moles of gas; if \( \Delta n = 0 \) there is no effect. Adding an inert gas at constant volume has no effect.
Temperature: heating favours the endothermic direction and changes K. A catalyst speeds up both directions equally and does not change K or the position of equilibrium.
Examples: Haber process \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \) (exothermic, favoured by high pressure and moderate temperature); contact process \( 2SO_2 + O_2 \rightleftharpoons 2SO_3 \).

Degree of Dissociation

For \( PCl_5 \rightleftharpoons PCl_3 + Cl_2 \) starting with 1 mol and degree of dissociation \( \alpha \): total moles = \( 1+\alpha \) and \( K_p = \dfrac{\alpha^{2}P}{1-\alpha^{2}} \). For \( N_2O_4 \rightleftharpoons 2NO_2 \): \( K_p = \dfrac{4\alpha^{2}P}{1-\alpha^{2}} \).

Ionic Equilibrium: Electrolytes and Ionisation

Strong electrolytes (NaCl, HCl, NaOH) ionise almost completely; weak electrolytes (CH3COOH, NH4OH) ionise partially. Non-electrolytes such as sugar and urea give no ions.
Ostwald's dilution law for a weak electrolyte: \( K = \dfrac{C\alpha^{2}}{1-\alpha} \approx C\alpha^{2} \), so \( \alpha = \sqrt{K/C} \) and \( [H^{+}] = \sqrt{K_aC} \). Ionisation increases on dilution and tends to 100% at infinite dilution.
Common-ion effect: adding an ion common to a weak electrolyte suppresses its ionisation.

Acids and Bases

Arrhenius: acids give H+, bases give OH- in water. Bronsted-Lowry: an acid is a proton donor and a base is a proton acceptor; every acid has a conjugate base and the weaker the acid, the stronger its conjugate base. Lewis: an acid is an electron pair acceptor (BF3, AlCl3, Zn2+) and a base an electron pair donor (NH3, OH-).
Amphiprotic species such as H2O and HCO3- can act both as acid and as base.

pH, pOH and Water

\( pH = -\log[H^{+}] \), \( pOH = -\log[OH^{-}] \), and \( K_w = [H^{+}][OH^{-}] = 10^{-14} \) at 25 °C, so pH + pOH = 14. Kw increases with temperature (10\(^{-12}\) at 90 °C, where neutral water has pH 6).
Mixing acid and base: find the excess milliequivalents and divide by the total volume to get the concentration of H+ or OH-.

Hydrolysis of Salts

Strong acid + strong base: neutral (NaCl). Strong base + weak acid: basic (CH3COONa, NaCN). Weak base + strong acid: acidic (NH4Cl, FeCl3). \( K_h = K_w/K_a \) for an anion and \( K_w/K_b \) for a cation, and \( h = \sqrt{K_h/C} \).

Buffer Solutions

A buffer resists changes in pH on adding small amounts of acid or base. Acidic buffer: weak acid and its salt; basic buffer: weak base and its salt. Henderson equation: \( pH = pK_a + \log\dfrac{[salt]}{[acid]} \) and \( pOH = pK_b + \log\dfrac{[salt]}{[base]} \). The maximum buffer capacity is when [salt] = [acid]. Blood is buffered by H2CO3 / HCO3-.

Solubility Product

For \( A_pB_q \rightleftharpoons pA^{q+} + qB^{p-} \): \( K_{sp} = p^{p}q^{q}S^{p+q} \). For AB, \( K_{sp} = S^2 \); for \( AB_2 \), \( K_{sp} = 4S^3 \).
Precipitation occurs when the ionic product exceeds Ksp. A common ion lowers solubility; a complexing agent (for example NH3 for AgBr) raises it.

Worked Examples

Example 1. At equilibrium [N2] = 0.5, [H2] = 1.0 and [NH3] = 0.2 mol/L. Find Kc for N2 + 3H2 <=> 2NH3.
Answer:
\( K_c = \dfrac{(0.2)^2}{0.5\times(1.0)^3} = 0.08 \)

Example 2. Find the pH of 0.1 M acetic acid (\( K_a = 1.8\times10^{-5} \)).
Answer:
\( [H^{+}] = \sqrt{1.8\times10^{-5}\times0.1} = 1.34\times10^{-3} \) M, so pH = 2.87.

Example 3. Find the solubility of AgCl (\( K_{sp} = 1.8\times10^{-10} \)) in pure water.
Answer:
\( S = \sqrt{1.8\times10^{-10}} = 1.34\times10^{-5} \) mol/L.