Question

The light emitting decay of excited mercury atoms (Hg vapor lamps) is a first order reaction: Hg_(excited) --> Hg_{(ground state)}​. The half life of the reaction is 4.2 x 10^-7 s. An initial sample of excited mercury atoms has a concentration of 4.5 x 10^-6 M.
1. How much time passes to reach a concentration of excited Hg atoms of 4.5 x 10^-7 M?
​a) 5.5 x 10⁶ s
b) 3.4 x 10³ s
c) 7.2 x 10^-3 s
​d) 1.4 x 10^-6 s

2. What is the [Hg(excited)] after 2.5 x 10^-6 s (assuming the same initial conditions)?
a) 7.3 x 10^-8 M
​b) 4.5 x 10^-6 M
​c) 2.4 x 10^-6 M
d) 2.8 x 10^-4 M

Answer 1

1)
Given:
Half life = 4.2*10^-7 s
use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k
= 0.693/(half life)
= 0.693/(4.2*10^-7)
= 1.65*10^6 s-1
we have:
[Hgs-excited]o = 4.5*10^-6 M
[Hgs-excited] = 4.2*10^-7 M
k = 1.65*10^6 s-1

use integrated rate law for 1st order reaction
ln[Hgs-excited] = ln[Hgs-excited]o - k*t
ln( 4.2*10^-7) = ln(4.5*10^-6) - 1.65*10^6*t
-14.683 = -12.3114 - 1.65*10^6*t
1.65*10^6*t = 2.3716
t = 1.4*10^-6 s

Answer: d

2)
we have:
[Hgs-excited]o = 4.5*10^-6 M
t = 2.5*10^-6 s
k = 1.65*10^6 s-1
Given:
[Hgs-excited]o = 4.5000E-6 M

use integrated rate law for 1st order reaction
ln[Hgs-excited] = ln[Hgs-excited]o - k*t
ln[Hgs-excited] = ln(4.5000*10^-6) - 1.65*10^6*2.5*10^-6
ln[Hgs-excited] = -1.2311*10^1 - 1.65*10^6*2.5*10^-6
ln[Hgs-excited] = -1.6436*10^1
[Hgs-excited] = 7.3*10^-8 M

Answer: a