Question

Suppose that a certain biologically important reaction is quite slow at physiological temperature (37 oC) in the absence of a catalyst. Assuming Arrhenius behavior, by how much must an enzyme lower the activation energy of the reaction to achieve a 1 x 105-fold increase in the reaction rate? (Give your answer in kJ)

Answer 1

30 kJ

Step-by-step explanation:

Arrhenius equation is given by:

`k=Aexp(-Ea/RT)\\`

Here, k is rate constant, A is Pre-exponential factor, Ea is activation energy and T is temperature.

taking natural log of both side

ln k = ln A - Ea/RT

In Arrhenius equation, A, R and T are constant.

Therefore,

`ln(k_2)/(k_1) =(Ea_1-Ea_2)/(RT)`

`Ea_1-Ea_2` is the lowering in activation energy by enzyme,

R = 8.314 J/mol.K

T = 37°C + 273.15 = 310 K

`(k_2)/(k_1) =1* 10^5`

`ln 1* 10^5 =(Ea_1-Ea_2)/(RT)\\{Ea_1-Ea_2} = 11.512 * 8.314 * 310\\=29670\ J\\=30\ kJ`