Consider the following reaction: 2C(s)+ 2H₂(g)$\to$ C₂H₄(g); $\Delta$_rH° = 52.47 kJ; $\Delta$_rS° = -53.5 J/K at 298 K What is the equilibrium constant at 298 K for this reaction? (R = 8.314 J/K.mol)

Question

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The equilibrium constant can be calculated using the Gibbs free energy change (ΔG°) equation: ΔG° = ΔH° - TΔS°. At 298 K, ΔG° = 52.47 kJ - (298 K)(-53.5 J/K) = 52.47 kJ + 15.943 kJ = 68.413 kJ = 68413 J. The equilibrium constant K is then calculated using ΔG° = -RT ln(K), giving K = e^(-ΔG°/RT) = e^(-68413/(8.314*298)) ≈ 1.0 × 10^-12.

Consider the Following Reaction: 2C(s)+ 2H2(g) →\to→ C2H4(g); Δ\DeltaΔ rH° = 52 Δ\DeltaΔ

Consider the Following Reaction:
2C(s)+ 2H₂(g) $\to$ C₂H₄(g); $\Delta$ _rH° = 52 $\Delta$