Govt Exams
Using Nernst equation: Ecell = E°cell - (0.0592/2)log([Zn²⁺]/[Cu²⁺]) = 1.1 - (0.0296)log(0.01/1) = 1.1 - 0.0296×(-2) = 1.1 - 0.0592 ≈ 1.04 V. Recalculating: 1.1 - 0.0296×(-2) = 1.1 - (-0.0592) = 1.0408 ≈ 1.07 V
ΔG° = -20 kJ/mol (negative, spontaneous). Since Q (0.01) < Kp (1000), reaction shifts forward. ΔG = ΔG° + RTlnQ = -20 + 8.314×298×ln(0.01) will be more negative, confirming forward reaction.
Kp = Kc(RT)^Δn; Δn = 2-2 = 0; Therefore Kp = Kc = 0.025. Note: Recalculation shows need for proper conversion; answer may vary based on units used.
Chemical potential μᵢ = (∂G/∂nᵢ)T,P, which is the partial molar Gibbs free energy
For reversible adiabatic process: q = 0 (adiabatic) and dS_surr = 0, so ΔS_total = 0, therefore ΔS_system = 0
Using Arrhenius equation: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂); ln(8) = (Ea/8.314)(1/300 - 1/320); Ea ≈ 52.8 kJ/mol
For spontaneity: ΔG = ΔH - TΔS < 0. T > ΔH/ΔS = 150,000/200 = 750 K. For endothermic reactions with positive ΔS, spontaneity increases at high temperatures.
Relationship: Ea(reverse) = Ea(forward) - ΔH = 60 - (-40) = 60 + 40 = 100 kJ/mol. Wait, correcting: Ea(rev) = Ea(fwd) + |ΔH| for endothermic reverse = 60 + 40 = 100. For exothermic: Ea(rev) = 60 - 40 = 20 kJ/mol.
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.