68.
b) 212 kJ
c) 35 kj
d) 12110
The E, of a non-catalyzed reaction at 27 C is 300 kcal/mol and E of the same reaction catalyzed
by an enzyme is 10 kcal mol.Calculate the ratio of rate constants for catalyzed and non-catalyzed
reaction.
Answers
Answer:
The Arrhenius equation looks like this
∣
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
k
=
A
⋅
exp
(
−
E
a
R
T
)
a
a
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−
, where
k
- the rate constant for a given reaction
A
- the pre-exponential factor, specific to a given reaction
E
a
- the activation energy of the reaction
T
- the absolute temperature at which the reaction takes place
In essence, the Arrhenius equation establishes a relationship between the rate constant of a reaction and the absolute temperature at which the reaction takes place.
In other words, this equation allows you to figure out how a change in temperature will ultimately affect the rate of the reaction.
The two temperatures at which the reaction takes place can be calculated using the conversion factor
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
T
[
K
]
=
t
[
∘
C
]
+
273.15
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−
In your case, you will have
T
1
=
100
∘
C
+
273.15
=
373.15 K
T
2
=
120
∘
C
+
273.15
=
393.15 K
If you take
k
1
to be the rate constant of the reaction at
T
1
, you can say that
k
1
=
A
⋅
exp
(
−
E
a
R
⋅
T
1
)
(
1
)
Similarly, if you take
k
2
to be the rate constant of the reaction at
T
2
, you will have
k
2
=
A
⋅
exp
(
−
E
a
R
⋅
T
2
)
(
2
)
Now, let's assume that your reaction is
n
order with respect to a reactant
A
n
A
→
products
The differential rate law for this generic reaction would look like this
rate
=
k
⋅
[
A
]
n
Assuming that you'll perform the reaction at
T
1
and at
T
2
using the same concentration for the reactant, you can say that you have
rate
1
=
k
1
⋅
[
A
]
n
and
rate
2
=
k
2
⋅
[
A
]
n
Your goal here will be to find the ratio that exists between the rate of the reaction at
T
2
and the rate of the reaction at
T
1
. This comes down to finding
rate
2
rate
1
=
k
2
⋅
[
A
]
n
k
1
⋅
[
A
]
n
=
k
2
k
1
=
?
Now, divide equations
(
2
)
and
(
1
)
to get
k
2
k
1
=
A
⋅
exp
(
−
E
a
R
⋅
T
2
)
A
⋅
exp
(
−
E
a
R
⋅
T
1
)
This will be equivalent to
k
2
k
1
=
exp
[
E
a
R
⋅
(
1
T
1
−
1
T
2
)
]
Before plugging in your values, make sure that you do not forget to convert the activation energy from kcal per mole to cal per mole by using the conversion factor
1 kcal
=
10
3
cal
You will have
k
2
k
1
=
exp
[
15
⋅
10
3
cal
mol
−
1
1.987
cal
mol
−
1
K
−
1
⋅
(
1
373.15
−
1
393.15
)
K
−
1
]
k
2
k
1
=
2.799
I'll leave the answer rounded to two sig figs
rate
2
rate
1
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
2.8
a
a
∣
∣
−−−−−−−
Therefore, the reaction will proceed
2.8
times faster if the temperature is increased by
20
∘
C