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In the gas phase, formic acid forms a dimer, 2HCOOH(g) (HCOOH)2(g).?
For this reaction, ΔH° = –60.1 kJ/mol and ΔG° = –13.9 kJ/mol at 25°C. Find the equilibrium constant (Kp) for this reaction at 75 °C. why is the answer Kp = 8.33?
1 Answer
- 冷眼旁觀Lv 72 years agoFavorite Answer
2 HCOOH(g) ⇌ (COOH)₂(g) …… Kp
ΔG° = -R T ln(Kp)
At 25°C:
-13.9 × 1000 J/mol = {-8.314 J/(mol K)} × {(273 + 25) K} × ln(Kp)
ln(Kp) = 13.9 × 1000 / (8.314 × 298)
Kp = e^{13.9 × 1000 / (8.314 × 298)} = 273.2
At 25°C: T₁ = (273 + 25) K = 298 K, Kp₁ = 273.2
At 75°C: T₂ = (273 + 75) K = 348 K, Kp₂ = ?
ln(Kp₁/Kp₂) = (ΔH/R){(1/T₂) - (1/T₁)}
ln(273.2/Kp₂) = {(-60.1 × 1000 J/mol) / [8.314 J/ (mol K)]} × {[(1/348) - (1/298)] 1/K}
273.2/Kp₂ = e^{[(-60.1 × 1000) / 8.314] × [(1/348) - (1/298)]} = 32.63
Kp at 75°C, Kp₂ = 273.2/32.63 = 8.37
