Answer:
t
(
2
)
1/2
=
85.25 s
Notice how you're given the half-life (for one temperature), a second temperature, and the activation energy. The key to doing this problem is recognizing that:
the half-life for a first-order reaction is related to its rate constant.
the rate constant changes at different temperatures.
Go here for a derivation of the half-life of a first-order reaction. You should find that:
t
1/2
=
ln
2
k
Therefore, if we label each rate constant, we have:
k
1
=
ln
2
t
(
1
)
1/2
k
2
=
ln
2
t
(
2
)
1/2
Recall that the activation energy can be found in the Arrhenius equation:
k
=
A
e
−
E
a
/
R
T
where:
A
is the frequency factor, i.e. it is proportional to the number of collisions occurring over time.
E
a
is the activation energy in
kJ/mol
.
R
=
0.008314472 kJ/mol
⋅
K
is the universal gas constant. Make sure you get the units correct on this!
T
is the temperature in
K
(not
∘
C
).
Now, we can derive the Arrhenius equation in its two-point form. Given:
k
2
=
A
e
−
E
a
/
R
T
2
k
1
=
A
e
−
E
a
/
R
T
1
we can divide these:
k
2
k
1
=
e
−
E
a
/
R
T
2
e
−
E
a
/
R
T
1
Take the
ln
of both sides:
ln
(
k
2
k
1
)
=
ln
(
e
−
E
a
/
R
T
2
e
−
E
a
/
R
T
1
)
=
ln
(
e
−
E
a
/
R
T
2
)
−
ln
(
e
−
E
a
/
R
T
1
)
=
−
E
a
R
T
2
−
(
−
E
a
R
T
1
)
=
−
E
a
R
[
1
T
2
−
1
T
1
]
Now if we plug in the rate constants in terms of the half-lives, we have:
ln
⎛
⎜
⎝
ln
2
/
t
(
2
)
1/2
ln
2
/
t
(
1
)
1/2
⎞
⎟
⎠
=
−
E
a
R
[
1
T
2
−
1
T
1
]
This gives us a new expression relating the half-lives to the temperature:
⇒
ln
⎛
⎜
⎝
t
(
1
)
1/2
t
(
2
)
1/2
⎞
⎟
⎠
=
−
E
a
R
[
1
T
2
−
1
T
1
]
Now, we can solve for the new half-life,
t
(
2
)
1/2
, at the new temperature,
40
∘
C
. First, convert the temperatures to
K
:
T
1
=
25
+
273.15
=
298.15 K
T
2
=
40
+
273.15
=
313.15 K
Finally, plug in and solve. We should recall that
ln
(
a
b
)
=
−
ln
(
b
a
)
, so the negative cancels out if we flip the
ln
argument.
⇒
ln
⎛
⎜
⎝
t
(
2
)
1/2
t
(
1
)
1/2
⎞
⎟
⎠
=
E
a
R
[
1
T
2
−
1
T
1
]
⇒
ln
⎛
⎜
⎝
t
(
2
)
1/2
400 s
⎞
⎟
⎠
=
80 kJ/mol
0.008314472 kJ/mol
⋅
K
[
1
313.15 K
−
1
298.15 K
]
=
(
9621.78 K
)
(
−
1.607
×
10
−
4
K
−
1
)
=
−
1.546
Now, exponentiate both sides to get:
t
(
2
)
1/2
400 s
=
e
−
1.546
⇒
t
(
2
)
1/2
=
(
400 s
)
(
e
−
1.546
)
=
85.25 s
This should make sense, physically. From the Arrhenius equation, the higher
T
2
is, the more negative the
[
1
T
2
−
1
T
1
]
term, which means the larger the right hand side of the equation is.
The larger the right hand side gets, the larger
k
2
is, relative to
k
1
(i.e. if
ln
(
k
2
k
1
)
is very large,
k
2
>>
k
1
). Therefore, higher temperatures means larger rate constants.
Furthermore, the rate constant is proportional to the rate of reaction
r
(
t
)
in the rate law. Therefore...
The higher the rate constant, the faster the reaction, and thus the shorter its half-life should be.
Explanation:
Sorry just go here https://socratic.org/questions/588d14f211ef6b4912374c92#370588
