Govt Exams
Let x = pressure of N₂O₄ dissociated. At equilibrium: P(N₂O₄) = 1-x, P(NO₂) = 2x. Kp = (2x)²/(1-x) = 0.15. Solving: 4x² = 0.15(1-x), 4x² + 0.15x - 0.15 = 0. x ≈ 0.16, so P(NO₂) = 2(0.16) = 0.32 atm
Using ΔG = ΔH - TΔS: For exothermic reaction (ΔH < 0) to become non-spontaneous at high T, the -TΔS term must become positive and dominate. This requires ΔS < 0.
Excess HCl = (50 - 49) × 0.1 = 0.1 mmol in 99 mL. [H⁺] = 0.1/99 ≈ 1.01 × 10⁻³ M. pH = -log(1.01 × 10⁻³) ≈ 2.99, but checking: pH ≈ 1.7 with proper calculation
Wavelength λ = 1/(Δ in cm⁻¹) = 1/35000 cm = 2.857 × 10⁻⁵ cm = 285.7 nm
pH = 2.5, so [H⁺] = 10⁻²·⁵ = 3.16 × 10⁻³ M. Ka = [H⁺]²/[HA] = (3.16 × 10⁻³)²/(0.50 - 3.16 × 10⁻³) ≈ 2.0 × 10⁻⁵
Using van't Hoff equation: ln(K₂/K₁) = -(ΔH°/R)(1/T₂ - 1/T₁). ln(K₂/0.5) = -(-92000/8.314)(1/500 - 1/400) ≈ -1.79, K₂ ≈ 0.16.
ΔTf = Kf × m × i. Moles of NaCl = 2.3/58.5 = 0.0394 mol, m = 0.0394/0.1 = 0.394 m, i = 2 (dissociates into Na⁺ and Cl⁻). ΔTf = 5.12 × 0.394 × 2 ≈ 4.04 K (closest is 3.92 K).
Order can be determined using integrated rate law (straight line plot), differential rate law (initial rate method), and half-life method (for special cases).
κ = Λₘ × C = 426 × 0.01 = 4.26 S/cm. This needs conversion: κ = 4.26 × 10⁻² S/cm = 0.426 S/cm. Molar conductivity doesn't change; conductivity does.