(ii)
Firstorder reactions (145):
Rate depends on
reactant concentration
1
[A]
d[A]
=
−
k
d
t
Now integrate:
1
[A]
[
A
]
o
[
A
]
∫
d[A]
=
−
k
d
t
0
t
∫
Get: ln [A]  ln [A]
o
= –
k t
OR:
How do you know that the data is for 1st order reaction ?
ln[A]
Intercept : ln [A]
o
slope = 
k
Time (s)
Plot of
ln[A] vs t
is
linear
for firstorder reactions.
ln [A] = ln [A]
o
–
k t
Integrated rate law
for 1
st
order reactions
1.8
Halflife for first order reactions :
ln
[A]
[A]
o
=
−
k
t
ln
1/2[A]
o
[A]
o
=
−
k
t
1/2
ln 1 / 2
( ) =
−
k
t
1/2
ln 2
( ) =
k
t
1/2
t
1/2
=
ln 2
k
=
0.693
k
About natural logarithms: ln (say “lawn”)
log x = y, so x = 10
y
ln x = y, so x = e
y
(e base of ln = 2.718…)
Rate
=
−
d
A
[ ]
dt
=
k
A
[ ]
1
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: Radioactive technetium (
99m
Tc) is used clinically because of its
short halflife [a metastable
99m
Tc, decays to
99
Tc emitting
γ
radiation].
(1) If the decay rate constant is 0.116 h
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 Spring '09
 DrK
 Reaction, #, 1 K, Rate equation, 6 %, 0.116 hr, 5.97 hours

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