[Judau]

I don't know how to find the difference in activation energy between no catalyst and catalyst added, and the steps to calculate it from Arrhenius equation.
 k = Ae^-E_a/RT rearranged to  ln k = -E_a/RT+ lnA (like y = mx + b)

I need to find E_a-E_ac
                      (NH₂)₂CO₂(aq) + H₂O(l) CO₂(g) + 2NH₃(g)
No catalyst:   4.15 * 10^-5 (s^-1) @ 373K ,first order
With catalyst: 3.4 * 10⁴  (s^-1) @ 294K , first order

Assume the collision factor is the same for both.

I started by converted it to table of ln k vs 1/T

1/T        ln k .00268  -10.1   (No catalyst) .00340   10.4   (with catalyst) so i tried to plug in individually to find the Ea for the slope, referring to y=mx+b, m = -Ea/R Ea= -m (R) Ea = -(-10.1/.00268) (.008314 kJ/mol*K)  =  2.25 *10^-4 kJ/ mol Eac = -(10.4/.00340) (.008314 kJ/mol*K)  = .. and then I m bananaz , answer is 64.0 kJ... how do I calculate it?