A. he activation energy of a certain reaction is 46.2 kJ/mol . At 25 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast?
B.Given that the initial rate constant is 0.0190s−1 at an initial temperature of 25 ∘C , what would the rate constant be at a temperature of 140. ∘C for the same reaction described in Part A?
Chemistry Problem?
2018-03-28 12:19 am
回答 (2)
2018-03-28 2:45 am
A.
Eₐ = 46.2 kJ mol⁻¹ = 46200 J mol⁻¹
Gas constant, R = 8.314 J mol⁻¹ K⁻¹
At T₁ = (273 + 25) K = 298 K, k₁ = 0.0190 s⁻¹
At T₂, k₂ = 2 × 0.0190 s⁻¹
Arrhenius equation: ln(k₁/k₂) = (Eₐ/R) [(1/T₂) - (1/T₁)]
ln(1/2) = (46200/8.314) [(1/T₂) - (1/298)]
(1/T₂) - (1/298) = ln(1/2) × 8.314 / 46200
(1/T₂) = {[ln(1/2) × 8.314 / 46200] + (1/298)} K⁻¹
T₂ = {[ln(1/2) × 8.314 / 46200] + (1/298)}⁻¹ K
Temperature at which the reaction goes twice as fast, T₂ = 309.5 K = (309.5 - 273)°C = 36.5°C
B.
A.
Eₐ = 46.2 kJ mol⁻¹ = 46200 J mol⁻¹
Gas constant, R = 8.314 J mol⁻¹ K⁻¹
At T₁ = (273 + 25) K = 298 K, k₁ = 0.0190 s⁻¹
At T₃ = (273 + 140) K = 413 K , k₃ = ? s⁻¹
Arrhenius equation: ln(k₁/k₃) = (Eₐ/R) [(1/T₃) - (1/T₁)]
ln(0.0190/k₃) = (46200/8.314) [(1/413) - (1/298)]
0.0190/k₃ = e^{(46200/8.314) [(1/413) - (1/298)]}
k₃/0.0190 = e^-{(46200/8.314) [(1/413) - (1/298)]}
k₃ = 0.0190 × e^-{(46200/8.314) [(1/413) - (1/298)]} s⁻¹
Rate constant at 140°C = 3.42 s⁻¹
Eₐ = 46.2 kJ mol⁻¹ = 46200 J mol⁻¹
Gas constant, R = 8.314 J mol⁻¹ K⁻¹
At T₁ = (273 + 25) K = 298 K, k₁ = 0.0190 s⁻¹
At T₂, k₂ = 2 × 0.0190 s⁻¹
Arrhenius equation: ln(k₁/k₂) = (Eₐ/R) [(1/T₂) - (1/T₁)]
ln(1/2) = (46200/8.314) [(1/T₂) - (1/298)]
(1/T₂) - (1/298) = ln(1/2) × 8.314 / 46200
(1/T₂) = {[ln(1/2) × 8.314 / 46200] + (1/298)} K⁻¹
T₂ = {[ln(1/2) × 8.314 / 46200] + (1/298)}⁻¹ K
Temperature at which the reaction goes twice as fast, T₂ = 309.5 K = (309.5 - 273)°C = 36.5°C
B.
A.
Eₐ = 46.2 kJ mol⁻¹ = 46200 J mol⁻¹
Gas constant, R = 8.314 J mol⁻¹ K⁻¹
At T₁ = (273 + 25) K = 298 K, k₁ = 0.0190 s⁻¹
At T₃ = (273 + 140) K = 413 K , k₃ = ? s⁻¹
Arrhenius equation: ln(k₁/k₃) = (Eₐ/R) [(1/T₃) - (1/T₁)]
ln(0.0190/k₃) = (46200/8.314) [(1/413) - (1/298)]
0.0190/k₃ = e^{(46200/8.314) [(1/413) - (1/298)]}
k₃/0.0190 = e^-{(46200/8.314) [(1/413) - (1/298)]}
k₃ = 0.0190 × e^-{(46200/8.314) [(1/413) - (1/298)]} s⁻¹
Rate constant at 140°C = 3.42 s⁻¹
2018-03-28 12:35 am
use the following equation (derived from Arrhenius equation
ln(k2/k1) = (Ea/RT)(1/T1 - 1/T2)
in A k2/k1 = 2 solve for T2
in B you know T1, T2, and k1 solve for k2
ln(k2/k1) = (Ea/RT)(1/T1 - 1/T2)
in A k2/k1 = 2 solve for T2
in B you know T1, T2, and k1 solve for k2
收錄日期: 2021-04-24 01:06:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180327161945AA0OtQR