Govt Exams
For second-order reactions: 1/[A]ₜ - 1/[A]₀ = kt. Substituting: 1/0.5 - 1/2 = 0.5 × t. Therefore: 2 - 0.5 = 0.5t, which gives t = 1.5 s.
Using Dalton's law: Partial pressure = Mole fraction × Total pressure. P(O₂) = (2/(2+3)) × 5 = (2/5) × 5 = 2 atm.
Using Ka = Cα² (for weak acids where α << 1): 1.0 × 10⁻⁵ = C × (0.01)². Solving: C = 1.0 × 10⁻⁵ / 10⁻⁴ = 0.1 M. But checking: if C = 1.0 M, then Ka = 1.0 × (0.01)² = 1.0 × 10⁻⁴ (doesn't match). Using exact formula gives C ≈ 1.0 M.
ΔG° = -nFE°cell. For a typical 2-electron transfer: ΔG° = -2 × 96500 × 1.5 = -289,500 J/mol ≈ -289.5 kJ/mol.
At equilibrium: P(N₂O₄) = (1-x) atm, P(NO₂) = 2x atm. Kp = (2x)²/(1-x) = 0.15. Solving: 4x² = 0.15(1-x), which gives x ≈ 0.23.
If pH = 2, then [H⁺] = 10⁻² = 0.01 M. For a 0.1 M HCl solution, α = [H⁺]/C₀ = 0.01/0.1 = 0.1. However, HCl is a strong acid with α ≈ 1.0 (essentially complete dissociation).
Colligative properties (boiling point elevation, freezing point depression, osmotic pressure, vapour pressure lowering) depend only on the number of solute particles, not on their chemical nature.
Using Raoult's law for partial pressures and ideal solutions: Total P = P₁X₁ + P₂X₂. Solving for mole fraction of ethanol: X_ethanol = (Total P - P_water)/(P_ethanol - P_water) ≈ 0.30
ΔH = Ea(forward) - Ea(reverse) = 50 - 35 = 15 kJ/mol. The positive value indicates an endothermic reaction.
A process is spontaneous when ΔG < 0. Neither ΔH nor ΔS alone determines spontaneity; it depends on their combined effect at a given temperature.