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:
[tex]k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}[/tex]
Rate factor in the presence of catalyst:
[tex]k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}[/tex]
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:
[tex]\dfrac{k_2}{k_1}={ \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}[/tex]
[tex]\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}[/tex]
Thus;
[tex]Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}[/tex]
Let say the assumed temperature = 25° C
= (25+ 273)K
= 298 K
Then ;
[tex]Ea_1-Ea_2 = 8.314 \ J/mol/K * 298 \ K * In (10^6)[/tex]
[tex]Ea_1-Ea_2 = 34228.92 \ J/mol[/tex]
[tex]\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}[/tex]
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
