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

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

Step-by-step 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^{^{^{ \frac {- Ea_1}{RT}}`

Rate factor in the presence of catalyst:

`k_2= A*e^{^{^{ \frac {- 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:

`(k_2)/(k_1)={ \frac {e^{ \frac {- Ea_2}{RT} }} { e^{ \frac {- Ea_1}{RT} }}`

`(k_2)/(k_1)={ \frac {e^([ Ea_1 - Ea_2 ] )}{RT} }}`

Thus;

`Ea_1-Ea_2 = RT In (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