a first order of reaction has activation energy and a rate constant equal to 45.0KJmol^-1 and 5.0*10^-4s^-1 at 37°C. to lower the energy needed for the reaction an enzyme is added as a catalyst and the rate id 2.0*10^2S^-1. if the activation energy is the factor affected by the presence of enzymes by how much must the enzymes lower the activation energy(take R = 8.314Jmol^-1K^-1?

Marvelous_1

3 years ago

mole

K= Rate constant without catalyst =5.0×10
−4
s
−1


K
′
= Rate constant with catalyst =2.0×10
−2
s
−1


K=Ae
−Ea/RT
& K
′
=Ae
−Ea
/RT

K
K
′

​
=
e
−Ea/RT

e
−Ea/RT

​


2.303×log(
K
K
′

​
)=
RT
1
​
(Ea−Ea
′
)

(Ea−Ea
′
)=2.303RTlog(
K
K
′

​
)

(Ea−Ea
′
)=2.303×8.314×10
−3
×310log
5×10
−4

2×10
−2

​


(Ea−Ea
′
)=5.9356×log
5
200
​
=