chem question can someone answer please?

Due to the same extent of rotting, k and directly proportional to (1/time). k is represented by (1/time).

At room temp:  T₁ = (273.2 + 24.0) K = 297.2 K,  k₁ = (1/3) /day
In refrigerator:  T₂ = (273.2 + 0.5) K = 273.7 K,  k₂ = (1/11) /day
Gas constant, R = 8.314 J /(mol K)

Arrhenius equation: ln(k₁/k₂) = (Eₐ/R) [(1/T₂) - (1/T₁)
ln[(1/3)/(1/11)] = (Eₐ/8.314) × [(1/273.7) - (1/297.2)]
ln(11/3) = (Eₐ/8.314) × [(1/273.7) - (1/297.2)]
Eₐ = 8.314 × ln(11/3) / [(1/273.7) - (1/297.2)]

Activation energy, Eₐ = 37400 J/mol = 37.4 kJ/mol

from the arrhenius equation
.. ln(k1 / k2) = (Ea/R) * (1/T2 - 1/T1)
.. Ea = R * ln(k1/k2) / (1/T2 - 1/T1)

and you are given TIME to rot and temperatures.. (not "k" and temperatures) and asked to find Ea.  so we need to modify that equation a bit

********
the general form of a rate equation is
.. rate = k * [A]^n
for 2 different values of k (because we're running the experiment at 2 different temperatures) where the initial concentration is the same
.. rate1 = k1 * [A]^n
.. rate2 = k2 * [A]^n
dividing
.. rate1 / rate2 = k1 / k2

AND we need to know that rate = amount / time.  if we assume the amount of rot is the same for both apples.. let's call that A.. then
.. rate1 / rate2 = k1 / k2 = (A / t1) / (A / t2) = (t2 / t1)
i.e
.. k1 / k2 = t2 / t1

*******
so that our equation becomes
.. Ea = R * ln(t2 / t1) / (1/T2 - 1/T1)

solving
... .... ... ... ... ... ... ... .. ... ... ln(11 days / 3.0 days).. .. . 1 kJ
.. Ea = 8.314 J/molK * ----- ---- ----- ----- ---- ----- ----- x ---- ---- = 37.4 kJ/mol 
... ... ... ... ... ... ... ... .. . (1 / 273.65K) - (1 / 297.15K).. 1000J