I need help with biochem

Question

Staphylococcal nuclease accelerates phosphodiester bond hydrolysis by a factor of 5.6x1014 over the spontaneous rate (1.7x10-13). How much does staphylococcal nuclease decrease the activation energy of its reaction at 25?C?

Answer 1

ln k₂/k₁ = - Ea/R (1/T₂ - 1/T₁) ... Arrhenius equation

Since temperature is constant at 25ºC we have ...

ln k₂/k₁ = - Ea/R

Given k₂/k₁ = 5.6x10¹⁴, the ln k2/k1 = 33.96

Thus, 33.96 = -Ea/R = -Ea/8.314

Ea = -282 Joules = change in activation energy