Initial: PA = 2, PB = 1. Change: -1.5 for A, -1.5 for B, +1.5 for C and D. Equilibrium: PA = 0.5, PB = -0.5 (invalid). Using Kp: 4 = (PC × PD)/(0.5 × PB). Since stoichiometry is 1:1:1:1, PC = PD = 1.5 atm.
At equilibrium, the chemical potential of a substance is the same in all phases present. This is the condition for phase equilibrium.
For first-order: ln([A]₀/[A]t) = kt. If 75% complete, [A]t = 0.25[A]₀. ln(4) = k × 45. Also, t₁/₂ = 0.693/k. From ln(4) = k × 45, k = ln(4)/45. t₁/₂ = 0.693 × 45/ln(4) ≈ 22.5 min.
At 25°C (298 K), (2.303RT/F) = (2.303 × 8.314 × 298)/96485 ≈ 0.0592 V. This is commonly used in half-cell potential calculations.
Although H₂O has strong hydrogen bonding, the question asks for lowest. H₂S has the lowest boiling point (-60°C) among stable hydrides due to weak intermolecular forces. Note: H₂O has anomalously high bp due to H-bonding.
π = MRT, where π = 10 atm, T = 300 K, R = 0.0821 L·atm/(mol·K). M = π/(RT) = 10/(0.0821 × 300) ≈ 0.41 M.
ΔTf = Kf × m, where m = molality = 0.5 mol/1 kg (approximately). ΔTf = 0.93°C, so Kf = 0.93/0.5 = 1.86 K·kg/mol.
All colligative properties (boiling point elevation, freezing point depression, osmotic pressure, and vapor pressure lowering) depend only on the number of solute particles, not their nature.
ΔG° = -RT ln K. With ΔG° = -40 kJ/mol = -40000 J/mol, ln K = 40000/(8.314 × 298) ≈ 16.2, so K = e^16.2 >> 1 (very large).
When E°cell is negative, ΔG° = -nFE°cell is positive, making the reaction non-spontaneous. Also, negative E°cell indicates Kequilibrium < 1.