Question

In a decomposition reaction N₂O₅ --> products, a plot of the on [N₂O₅] vs time yields...

In a decomposition reaction N₂O₅ --> products, a plot of the on [N₂O₅] vs time yields a straight line with a slope of -0.0363/s.
If the initial concentration of [N₂O₅] is 4.0 M, what is the concentration after 155 seconds?
What is the half-life of the reaction?
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Homework Answers

Answer #1

1)
ln[N2O5] vs t is straight line
So, it is 1st order reaction
for 1st order reaction,
slope = -k
so,
k = 3.63*10^-2 s-1
[N2O5]o = 4.0 M
t = 155.0 s

use integrated rate law for 1st order reaction
ln[N2O5] = ln[N2O5]o - k*t
ln[N2O5] = ln(4) - 3.63*10^-2*155
ln[N2O5] = 1.3863 - 3.63*10^-2*155
ln[N2O5] = -4.2402
[N2O5] = 1.44*10^-2 M
Answer: 1.44*10^-2 M

2)
Given:
k = 0.0363 s-1
use relation between rate constant and half life of 1st order reaction

t1/2 = (ln 2) / k
= 0.693/(k)
= 0.693/(3.63*10^-2)
= 19.09 s
Answer: 19.1 s

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