Question

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the rate of the hydrolysis reaction by a factor of 1 million, how much lower does the activation barrier have to be when sucrose is in the active site of the enzyme

Answer 1

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

$k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}$

Rate factor in the presence of catalyst:

$k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}$

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

$\dfrac{k_2}{k_1}={ \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}$

$\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}$

Thus;

$Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}$

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

$Ea_1-Ea_2 = 8.314 \ J/mol/K * 298 \ K * In (10^6)$

$Ea_1-Ea_2 = 34228.92 \ J/mol$

$\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}$

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme