Collision theory question.. help?
2017-12-26 7:39 pm
A reaction is found to have a rate constant of 8.60 x 10-1sec-1 at 523 K and an activation energy of 120.8 kJ/mol. What is the value of the rate constant at 270 K?
回答 (2)
2017-12-27 1:06 am
Eₐ = 120.8 kJ mol⁻¹ = 120.8 × 1000 J mol⁻¹ = 120800 J mol⁻¹
R = 8.314 J mol⁻¹ K⁻¹
When T₁ = 523 K, k₁ = 8.60 × 10⁻¹ sec⁻¹
When T₂ = 270 K, k₂ = ?
One of the forms for Arrhenius equation : ln(k₁/k₂) = (Eₐ/R) [(1/T₂) - (1/T₁)]
ln[(8.6 × 10⁻¹) / k₂] = (120800 / 8.314) × [(1/270) - (1/523)]
(8.6 × 10⁻¹) / k₂ = e^{(120800 / 8.314) × [(1/270) - (1/523)]}
k₂ = (8.6 × 10⁻¹) / e^{(120800 / 8.314) × [(1/270) - (1/523)]}
Rate constant at 270 K, k₂ = 4.25 × 10⁻¹² sec⁻¹
R = 8.314 J mol⁻¹ K⁻¹
When T₁ = 523 K, k₁ = 8.60 × 10⁻¹ sec⁻¹
When T₂ = 270 K, k₂ = ?
One of the forms for Arrhenius equation : ln(k₁/k₂) = (Eₐ/R) [(1/T₂) - (1/T₁)]
ln[(8.6 × 10⁻¹) / k₂] = (120800 / 8.314) × [(1/270) - (1/523)]
(8.6 × 10⁻¹) / k₂ = e^{(120800 / 8.314) × [(1/270) - (1/523)]}
k₂ = (8.6 × 10⁻¹) / e^{(120800 / 8.314) × [(1/270) - (1/523)]}
Rate constant at 270 K, k₂ = 4.25 × 10⁻¹² sec⁻¹
2017-12-26 8:33 pm
ln (k1/k2) = Ea/R * (1/T2 – 1/T1)
ln(k1/0.86) = 120.8*10^3((1/523) - (1/270)) = -26.03
k1/0.86 = 4.3*10^-12 (s^-1)
ln(k1/0.86) = 120.8*10^3((1/523) - (1/270)) = -26.03
k1/0.86 = 4.3*10^-12 (s^-1)
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https://hk.answers.yahoo.com/question/index?qid=20171226113912AAA8rfg