Homework Answers
Given:
Half life = 8.04 days
use relation between rate constant and half life of 1st order reaction
k = (ln 2) / k
= 0.693/(half life)
= 0.693/(8.04)
= 8.619*10^-2 days-1
we have:
[A]o = 100 [Let initial concentration be 100]
68 % has decayed.
So, remaining is 32 %
[A] = 32
k = 8.619*10^-2 days-1
use integrated rate law for 1st order reaction
ln[A] = ln[A]o - k*t
ln(32) = ln(100) - 8.619*10^-2*t
3.466 = 4.605 - 8.619*10^-2*t
8.619*10^-2*t = 1.139
t = 13 days
Answer: a
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