Answer:
k ≈ 9,56x10³ s⁻¹
Explanation:
It is possible to solve this question using Arrhenius formula:
[tex]ln\frac{k2}{k1} = \frac{-Ea}{R} (\frac{1}{T2} -\frac{1}{T1} )[/tex]
Where:
k1: 1,35x10² s⁻¹
T1: 25,0°C + 273,15 = 298,15K
Ea = 55,5 kJ/mol
R = 8,314472x10⁻³ kJ/molK
k2 : ???
T2: 95,0°C+ 273,15K = 368,15K
Solving:
[tex]ln\frac{k2}{k1} = 4,257[/tex]
[tex]\frac{k2}{k1} = 70,593[/tex]
[tex]{k2} = 9,53x10^3 s^{-1}[/tex]
k ≈ 9,56x10³ s⁻¹
I hope it helps!
