課題
n次反応 –d [A]/dt = k [A]n (n≠1) の半減期を求めよ
∫ xp dx = 1 xp+1
p+1
n次反応(n≠1)の半減期
∫ xp dx = 1 xp+1
p+1
–d [A]/dt = k [A]n より、 - [A]-nd [A] = k dt
t: 0 → t, [A]: [A]0 → [A]t で積分すると
(左辺) = - [1/(-n+1) [A]-n+1] = {1/(n -1)} ([A]t-n+1 – [A]0-n+1)
(右辺) = k t
t = t1/2 のとき、[A]t = [A]0 / 2 だから、上式に代入して
t1/2 = [1/{k(n -1)}] {([A]0/2)-n+1 – [A]0-n+1}
= [1/{k(n - 1)}] [A]0-n+1(2n-1 -1)
検算
n = 2 のとき
t1/2 = [1/{k(2-1)}] [A]0-2+1 (22-1 -1)
= (1/ k) [A]0-1 (2-1)
= 1/ (k [A]0)
(a) n = 3 より、 t1/2 = [1/{k(3-1)}] [A]0-3+1(23-1 - 1) = 3/(2k [A]02)
(b) n = -1より、 t1/2 = [1/{k(-1-1)}] [A]0-(-1)+1(2(-1)-1 - 1) = {3/(8k)} [A]02
(c) n = 1/2 より、 t1/2 = [1/{k(1/2 -1)}] [A]0-(1/2)+1(21/2 -1 - 1)
-1/2
(1/√2) – 1
= {(2 - √2)/k} [A]01/2
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