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

Answer:

30 kJ

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,

$ln\frac{k_2}{k_1} =\frac{Ea_1-Ea_2}{RT}$

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

R = 8.314 J/mol.K

T = 37°C + 273.15 = 310 K

$\frac{k_2}{k_1} =1\times 10^5$

$ln 1\times 10^5 =\frac{Ea_1-Ea_2}{RT}\\{Ea_1-Ea_2} = 11.512 \times 8.314 \times 310\\=29670\ J\\=30\ kJ$