What is the equilibrium constant for the reaction below at 298 K? 2C(s)+ 3H₂(g)→ C₂H₆(g) Given: Δ_rH° = -84.68 kJ; Δ_rS° = -173.8 J/K at 298 K.(R = 8.314 J/K⋅mol) A) 5.8 × 10⁵ B) 1.0 × 10−5 C) 8.6 × 10^-10 D) 1.7 × 10^-6 E) 7.0 × 10¹⁴

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The equilibrium constant (K) can be calculated using the Gibbs free energy change (ΔG°) at 298 K:ΔG° = -RTlnKΔG° = ΔH° - TΔS°ΔG° = (-84.68 kJ/mol) - (298 K)(-173.8 J/K⋅mol) / 1000 J/kJΔG° = -33.5 kJ/molK = e^(-ΔG°/RT)K = e^(-(-33.5 kJ/mol) / (8.314 J/K⋅mol)(298 K))K = 5.8 × 10^5

What Is the Equilibrium Constant for the Reaction Below at 298