Question

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kj/mol. 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? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical and the temperature is 298 K .)

Answer 1

Answer:

The answer to the question is

If an enzyme increases the rate of the hydrolysis reaction by a factor of 1 million the activation barrier has to be 34.229 kJ lower when sucrose is in the active site of the enzyme

Explanation:

To solve the question we note that Arrhenius's equation for the rate constant is given by

$k = Ae^{\frac{-E_{a} }{RT} }$

Therefore we have for the un-catalyzed reaction

$k = Ae^{\frac{-108000 }{(8.314)(298)} } =$ A ×1.171 ×10⁻¹⁹ s⁻¹

When an enzyme is added the rate of hydrolysis increases by a factor of 1 million therefore

new rate = initial rate × 1000000 =  A ×1.171 ×10⁻¹⁹ s⁻¹ ×100000 = A ×1.171 ×10⁻¹³ s⁻¹

Hence we have the new given by

$k_{new} = Ae^{\frac{-E_{a(new)} }{RT} }$ since A,  R and T remain constant

The above relation becomes

A ×1.171 ×10⁻¹³ s⁻¹  = $Ae^{\frac{-E_{a(new)} }{(8.314)(298)} }$   which is the same as

1.171 ×10⁻¹³ s⁻¹  = $e^{\frac{-E_{a(new)} }{(8.314)(298)} }$ that is ㏑(1.171 ×10⁻¹³ ) = ${\frac{-E_{a(new)} }{(8.314)(298)} }$

or  Eₐ = 29.8 × 8.314 × 298 = 73771.1 J = 73.7711 kJ

This means that the activation energy barrier needs to be 73.7711 kJ or (108000 J - 73771.1 J), 34228.9 J lower

The activation energy needs to be 34.229 kJ lower.