Half-life equation for first-order reactions: t 1/2=0.693 k where t 1/2 is the h
ID: 1005493 • Letter: H
Question
Half-life equation for first-order reactions: t1/2=0.693k
where t1/2 is the half-life in seconds (s), and k is the rate constant in inverse seconds (s1).
A.) What is the half-life of a first-order reaction with a rate constant of 1.90×104 s1?
B.) What is the rate constant of a first-order reaction that takes 369 seconds for the reactant concentration to drop to half of its initial value?
C.) A certain first-order reaction has a rate constant of 2.70×103 s1. How long will it take for the reactant concentration to drop to 18 of its initial value? Thank you!
Explanation / Answer
t1/2=0.693k
a) t 1/2 = 0.693 / 1.9 x 10^-4 = 3647 second
b) ln([A]/[A]o) = -kt
Given = [A]/[A]o = 1/2
ln 0.5 = - k x 369
K= -0.693/369= 1.8 * 10^-3 s-1
C) Given = [A]/[A]o = 1/8 = 0.125
ln 0.125= - 2.7*10-3 x t
t= -2.07/- 2.7*10-3 = 766.6 second
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