7. Calculate the fraction of the starting quantity of A that will be used up after...
7. Calculate the fraction of the starting quantity of A that will be used up after 60 s. Given the reaction below which is found to be the first order in A and t1/2=40s A→B+C
Homework Answers
Given:
Half life = 40 s
use relation between rate constant and half life of 1st order reaction
k = (ln 2) / k
= 0.693/(half life)
= 0.693/(40)
= 1.733*10^-2 s-1
we have:
[A]o = 100 [Let initial concentration be 100]
t = 60.0 s
k = 1.733*10^-2 s-1
use integrated rate law for 1st order reaction
ln[A] = ln[A]o - k*t
ln[A] = ln(1*10^2) - 1.733*10^-2*60
ln[A] = 4.605 - 1.733*10^-2*60
ln[A] = 3.566
[A] = e^(3.566)
[A] = 35.4
35.4 of 100 is remaining.
Amount used up is 100 - 35.4 = 64.6 which is 64.6 %
Answer: 64.6 %
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