Govt. Exams
Entrance Exams
For weak acid: α = √(Ka/C) = √(1×10⁻⁵/0.1) = √(1×10⁻⁴) = 0.01 or 1%. Using exact formula: α ≈ 0.0316 (3.16%)
Initial: 2 atm of A. At equilibrium with α = 0.5: A dissociates to (1-0.5)×2 = 1 atm, producing B and C each at 0.5 atm. Total = 1 + 0.5 + 0.5 = 2 atm × (1 + α) = 2 × 1.5 = 3 atm
At higher dilutions, weak electrolytes ionize more completely, increasing degree of ionization and thus molar conductivity
ΔG = ΔH - TΔS; For spontaneity ΔG < 0. Exothermic (ΔH < 0) with ΔS > 0 always gives ΔG < 0
New rate = k(2[A])²(0.5[B]) = k × 4[A]² × 0.5[B] = 2k[A]²[B] = 2 × original rate
Benzene and toluene are both aromatic hydrocarbons with similar molecular structures, forming nearly ideal solutions
ΔG° = -nFE°; For bivalent ion transfer: ΔG° = -2 × 96485 × 0.92 ≈ -177.6 kJ/mol
In presence of 0.1 M Cl⁻: Ksp = [Ag⁺][Cl⁻]; 1.8 × 10⁻¹⁰ = s × 0.1; s = 1.8 × 10⁻⁹ M
Using Henderson-Hasselbalch: pH = pKa + log([salt]/[acid]) = 4.74 + log(0.15/0.1) = 4.74 + 0.48 = 5.22
After 3 half-lives (30 min): (1/2)³ = 1/8 = 12.5% remains