ΔG° = -nFE° = -2 × 96500 × 1.10 = -212,300 J/mol = -212.3 kJ/mol.
For a reversible adiabatic process, q = 0 and ΔS = 0 because there is no heat exchange and the process is reversible, making it isentropic.
pH = -log[H⁺], so [H⁺] = 10⁻ᵖᴴ = 10⁻⁴ = 1 × 10⁻⁴ M.
E°cell is directly related to the maximum electrical work (w_max = -nFE°cell) that can be obtained from a galvanic cell.
Hydration enthalpy depends on the charge density of the ion (charge/size ratio). Higher charge and smaller size lead to greater hydration enthalpy.
If 30% of PCl₅ dissociates, [PCl₃] = 1 × 0.30 = 0.3 M and [Cl₂] = 0.3 M.
In endothermic reactions, ΔH is positive because heat is absorbed from surroundings. The system absorbs energy.
For first-order reactions, t₁/₂ = 0.693/k = 0.693/0.693 = 1.0 s.
E°cell = E°cathode - E°anode = 0.80 - (-0.40) = 1.20 V. A positive E°cell indicates a spontaneous galvanic cell.
Using ln(k₂/k₁) = (Ea/R)(T₂-T₁)/(T₁T₂), ln(2) = (Ea/8.314)(10)/(93000). Solving: Ea ≈ 52.8 kJ/mol.