Question

At what temperature would the non-catalyzed reaction need to be run to have a rate equal to that of the metal-catalyzed reaction at 25°c?

Answer 1

Fixing the balanced equation:

2 H2O2 ----> 2 H2O + 1 O2

not

2 H2O ----> 2 H2O + 1 O2

Arrhenius equation

k = Ao exp (-Ea/RT)

Two runs:
k1 = Ao exp (-Ea1 / RT1)
k2 = Ao exp (-Ea2 / RT2)

Rearrangement:
k1 / exp (-Ea1 / RT1) = Ao
k2 / exp (-Ea2 / RT2) = Ao

since they both = Ao, they must = each other and ;
k1 / exp (-Ea1 / RT1) = Ao = k2 / exp (-Ea2 / RT2)

Rearrangement:
(k1 / k2) = exp(-Ea1 / RT1) / exp(-Ea2 / RT2)

Rearrangement: since n^a / n^b = n^(a-b)..
(k1 / k2) = exp (-Ea1 / RT1) - (-Ea2 / RT2) )

Rearrangement:
(k1 / k2) = exp ( (1/R) x (( Ea2 / T2) - (Ea1 / T2)) )

taking "ln" of both sides
ln(k1 / k2) = (1/R) x (Ea2/T2 - Ea1/T1)

rate1 = k1 x [H2O2]^n
rate2 = k2 x [H2O2]^n

rate1 / rate2 = k1 / k2

ln(k1 / k2) = (1/R) x (Ea2/T2 - Ea1/T1)

ln(rate1 / rate2) = (1/R) x (Ea2/T2 - Ea1/T1)

rate1 = rate2
T1 = ?
Ea1 = 75 kJ/mol
T2 = 25°C = 298K
Ea2 = 49 kJ/mol

ln(rate2 / rate2) = (1/R) x (Ea2/T2 - Ea1/T1)
ln(1) = (1/R) x (Ea2/T2 - Ea1/T1)
0 = (1/R) x (Ea2/T2 - Ea1/T1)
0 = Ea2/T2 - Ea1/T1

Ea2/T2 = Ea1/T1
T1 = T2 x (Ea1 / Ea2) = 298K x (75 kJ/mol / 49kJ/mol) = 456K = 183°C