3A(g)+4B(g)↽−−⇀2C(g)+2D(g)
were found to be 𝑃A=4.80 atm, 𝑃B=5.25 atm, 𝑃C=4.07 atm, and 𝑃D=5.42 atm.
What is the standard change in Gibbs free energy of this reaction at 25 ∘C ?
At 25 ∘C , the equilibrium partial pressures for the reaction?
2020-11-10 2:23 pm
回答 (1)
2020-11-10 2:55 pm
✔ 最佳答案
3A(g) + 4B(g) ⇌ 2C(g) + 2D(g) KpKp = 𝑃C² × 𝑃D² / (𝑃A³ × 𝑃B⁴) = 4.07² × 5.42² / (4.80³ × 5.25⁴) = 0.00579
Standard change in Gibbs free energy, ΔG°
= -R T ln(Kp)
= - (8.314 × 10⁻³ kJ mol⁻¹ K⁻¹) × [(273 + 25) K] × ln(0.00579)
= +12.8 kJ mol⁻¹
收錄日期: 2021-04-26 12:54:14
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20201110062347AAP1zQ5