The rate constant k for the first-order reaction A→3B is 0.527 s−1. After how many seconds will 45.0% of the initial amount of A remain?
回答 (1)
2020-11-23 3:54 pm
✔ 最佳答案
Method 1:ln(A/Aₒ) = -kt
ln(45%) = -(0.527 s⁻¹)t
Time taken, t = -ln(0.45)/0.527 s = 1.52 s
====
Method 2:
Let n be the number of half-lives taken.
45% = (1/2)ⁿ
ln(0.45) = ln(0.5)ⁿ
n ln(0.5) = ln(0.45)
n = ln(0.45)/ln(0.5)
Time taken = [ln(0.45)/ln(0.5)] × [ln(2)/0.527 s] = 1.52 s
收錄日期: 2021-04-23 23:06:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20201123054319AA3dZVa