Govt. Exams
Entrance Exams
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
ΔS = nR ln(V₂/V₁) = 1 × R × ln(2V/V) = R ln(2) for one mole
Using Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.745 + log(0.1/0.1) = 4.745 + 0 ≈ 4.74
Molar conductivity Λm = κ × (1000/C) = 0.0141 × (1000/0.01) = 0.0141 × 100000 = 1410. Wait, correct is 1410 S·cm²/mol, but checking: 0.0141/(0.01) × 1000 = 1.41 × 1000 = 1410. Answer should be B.