Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the reaction equation
2H2O2(aq)⟶2H2O(l)+O2(g)
The activation energy for this reaction is 75 kJ⋅mol−1. The enzyme catalase, found in blood, lowers the activation energy to 8.0 kJ⋅mol−1.
At what temperature would the non‑catalyzed reaction need to be run to have a rate equal to that of the enzyme‑catalyzed reaction at 25 ∘C?
Chemistry Help Please! t=____K?
2020-06-22 4:08 pm
回答 (1)
2020-06-22 5:07 pm
Arrhenius equation:
k = A e^[-Eₐ/(RT)]
ln(k) = [-Eₐ/(RT)] + ln(A)
Non-catalyzed reaction at T K:
ln(k₁) = [-75 × 1000 / (RT)] + ln(A)
Catalyzed reaction at 25°C (298 K):
ln(k₂) = [-8.0 × 1000 / (R × 298)] + ln(A)
When the two rates are equal, k₁ = k₂:
[-75 × 1000 / (RT)] + ln(A) = [-8.0 × 1000 / (R × 298)] + ln(A)
[-75 × 1000 / (RT)] = [-8.0 × 1000 / (R × 298)]
75 / T = 8.0 / 298
T = 75 × 298 / 8.0
T = 2794
The required temperature = 2794 K = 2521°C
k = A e^[-Eₐ/(RT)]
ln(k) = [-Eₐ/(RT)] + ln(A)
Non-catalyzed reaction at T K:
ln(k₁) = [-75 × 1000 / (RT)] + ln(A)
Catalyzed reaction at 25°C (298 K):
ln(k₂) = [-8.0 × 1000 / (R × 298)] + ln(A)
When the two rates are equal, k₁ = k₂:
[-75 × 1000 / (RT)] + ln(A) = [-8.0 × 1000 / (R × 298)] + ln(A)
[-75 × 1000 / (RT)] = [-8.0 × 1000 / (R × 298)]
75 / T = 8.0 / 298
T = 75 × 298 / 8.0
T = 2794
The required temperature = 2794 K = 2521°C
收錄日期: 2021-04-18 18:32:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200622080812AAPBbab