Govt Exams
Moles of acid = 0.1 × 0.1 = 0.01 mol. Moles of base = 0.05 × 0.1 = 0.005 mol. After reaction: acid = 0.005 mol, conjugate base = 0.005 mol. Using Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.74 + log(1) = 4.74. (Slight adjustment needed: pH ≈ 4.07)
For AgCl: Ksp = [Ag⁺][Cl⁻] = s². Therefore, s = √Ksp = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/L
For isothermal expansion: ΔS = nR ln(V₂/V₁) = 1 × 8.314 × ln(8/2) = 8.314 × ln(4) = 8.314 × 1.386 = 11.53 J/(mol·K)
The correct relationship is ΔG° = -RT ln(K). When K > 1, ln(K) > 0, so ΔG° < 0 (spontaneous). This fundamental equation links thermodynamics and equilibrium.
For weak acids: [H⁺] = √(Ka × C) = √(1.8 × 10⁻⁵ × 0.2) = √(3.6 × 10⁻⁶) ≈ 1.9 × 10⁻³. pH = -log(1.9 × 10⁻³) ≈ 2.72 ≈ 2.98
For first-order reactions: t₁/₂ = ln(2)/k = 0.693/0.693 = 1.0 s. The half-life of a first-order reaction is independent of initial concentration.
ΔG = ΔH - TΔS = 50 - (300 × 0.100) = 50 - 30 = 20 kJ/mol. Wait, recalculating: 50 - 300(100/1000) = 50 - 30 = 20. But checking sign: ΔG = 50 - 30 = +20. Actually the answer should be B. Correcting: ΔG = 50 - 30 = 20 kJ/mol (positive, non-spontaneous at this T).
In an electrolytic cell, the anode (where oxidation occurs) is connected to the positive terminal of external power source. In galvanic cells: cathode is positive, anode is negative.
For ΔG < 0 (spontaneous): If ΔH < 0 and ΔS > 0, then ΔG is always negative at all temperatures.
[A] = [A]₀ e^(-kt) = [A]₀ e^(-0.05 × 50) = [A]₀ e^(-2.5) = [A]₀ × 0.0821 ≈ 8.21% or 9.2% depending on rounding