Calculate the ratio of catalysed and uncatalysed rate constant
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We are asked to calculate kcatk for a reaction at a fixed temperature. We can start from the Arrhenius equation:
k=Ae−Ea/RT
Having a fixed T and two different activation energies Ea gives rise to two different rate constants ki:
kcat=Ae−Ea,cat/RT
k=Ae−Ea/RT
Assuming the frequency of collisions is the same in the different mechanism:
kcatk=e−Ea,cat/RTe−Ea/RT
=e−Ea,catRT+EaRT
=e−Ea,cat−EaRT
As usual, we must have temperature in K...
20∘C=293.15 K
k=Ae−Ea/RT
Having a fixed T and two different activation energies Ea gives rise to two different rate constants ki:
kcat=Ae−Ea,cat/RT
k=Ae−Ea/RT
Assuming the frequency of collisions is the same in the different mechanism:
kcatk=e−Ea,cat/RTe−Ea/RT
=e−Ea,catRT+EaRT
=e−Ea,cat−EaRT
As usual, we must have temperature in K...
20∘C=293.15 K
Answered by
2
k
c
a
t
k
=
6.31
×
10
9
We are asked to calculate
k
c
a
t
k
for a reaction at a fixed temperature. We can start from the Arrhenius equation:
k
=
A
e
−
E
a
/
R
T
Having a fixed
T
and two different activation energies
E
a
gives rise to two different rate constants
k
i
:
k
c
a
t
=
A
e
−
E
a
,
c
a
t
/
R
T
k
=
A
e
−
E
a
/
R
T
Assuming the frequency of collisions is the same in the different mechanism:
k
c
a
t
k
=
e
−
E
a
,
c
a
t
/
R
T
e
−
E
a
/
R
T
=
e
−
E
a
,
c
a
t
R
T
+
E
a
R
T
=
e
−
E
a
,
c
a
t
−
E
a
R
T
As usual, we must have temperature in
K
...
20
∘
C
=
293.15 K
Thus:
k
c
a
t
k
=
e
−
20 kJ/mol
−
75 kJ/mol
0.008314472 kJ/mol
⋅
K
⋅
293.15 K
=
6.31
×
10
9
This means your reaction is over a billion times faster due to the catalysis.
Why must we have
R
in
kJ/mol
⋅
K
?
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