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
Therefore we have for the un-catalyzed reaction
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
since A, R and T remain constant
The above relation becomes
A ×1.171 ×10⁻¹³ s⁻¹ = which is the same as
1.171 ×10⁻¹³ s⁻¹ = that is ㏑(1.171 ×10⁻¹³ ) =
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.
