When a biochemical reaction is carried out in laboratory from outside of human body in the absence of enzyme, the rate of reaction obtained is 10-6 times, then activation energy of the reaction in the presence of enzyme is

Question

When a biochemical reaction is carried out in laboratory from outside of human body in the absence of enzyme, the rate of reaction obtained is 10-6 times, then activation energy of the reaction in the presence of enzyme is (A) (6/RT) (B) P is required (C) different from Ea obtained in laboratory (D) cannot say any things

Tardigrade Admin · Accepted Answer

When a biochemical reaction is carried out laboratory from outside of human body in absence of enzyme, then rate of reaction obtain 106 times than activation energy of reaction presence of enzyme. It is different from Ea ob in laboratory because for a given chemical reac k=Ae-Ea/RT (Arrhenius equation)

Q. When a biochemical reaction is carried out in laboratory from outside of human body in the absence of enzyme, the rate of reaction obtained is $1 0^{- 6}$ times, then activation energy of the reaction in the presence of enzyme is

$\frac{6}{RT}$

P is required

different from $E_{a}$ obtained in laboratory

cannot say any things

Q. When a biochemical reaction is carried out in laboratory from outside of human body in the absence of enzyme, the rate of reaction obtained is 10−6 times, then activation energy of the reaction in the presence of enzyme is

RT6​

P is required

different from Ea​ obtained in laboratory

cannot say any things

Q. When a biochemical reaction is carried out in laboratory from outside of human body in the absence of enzyme, the rate of reaction obtained is $1 0^{- 6}$ times, then activation energy of the reaction in the presence of enzyme is

$\frac{6}{RT}$

different from $E_{a}$ obtained in laboratory