[化學]反應速率公式點樣In ?

2008-02-29 3:10 pm

it is given that, in first order reaction,
rate = -d[A]/dt=k[A]
How can we get the equation, that
[A]t = [A]0e^-kt ,
where [A]t is the molarity when time is t & [A]0 is the molarity when the reaction start.
And how can we get
ln[A]t=ln[A]0-kt

係一級反應入面，
反應速率= -d[A]/dt=k[A]
我地點樣推導下面的公式：
[A]t = [A]0e^-kt ,
其中[A]t 是時間t的濃度，[A]0是初始濃度。
而且，點樣推下一條式：
ln[A]t=ln[A]0-kt

回答 (1)

Audrey Hepburn

2008-03-01 8:50 pm

✔ 最佳答案

In the first orderreaction,
Rate = -d[A]/dt = k[A]
With the initialmolarity [A]0

-d[At]/dt =k[A]
∫-1/[At] d[A] =k∫dt
∫1/[At] d[A] = -k∫dt
ln[At] =-kt + C, where C is a constant

With the condition whent = 0, [A] = [A0]
i.e. ln[A0] =C

∴ ln[At] = -kt + ln[A0]
ln[At] – ln[A0]= -kt
ln{[At] / [A0]}= -kt
[At] / [A0]= e-kt
∴ [At] = [A0]e-kt

You can derive from [At]= [A0]e-kt
Take natural log inboth sides
ln[At] =ln[A0]e-kt
ln[At] =ln[A0] + lne-kt
ln[At] =ln[A0] – ktlne
∴ ln[At] = ln[A0] - kt

參考: Myself~~~

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