Question

A reaction is followed and found to have a rate constant of 3.36 × 104 M-1s-1 at 344 K and a rate constant of 7.69 M-1s-1 at 219 K. Determine the activation energy for this reaction.23.8 kJ/mol11.5 kJ/mol12.5 kJ/mol42.0 kJ/mol58.2 kJ/mol

Answer 1

Ea = 42,0 kJ/mol

Step-by-step explanation:

It is possible to solve this question using Arrhenius formula:

`ln(k2)/(k1) = (-Ea)/(R) ((1)/(T2) -(1)/(T1) )`

Where:

k1: 3,36x10⁴ M⁻¹ s⁻¹

T1: 344K

Ea = ???

R = 8,314472x10⁻³ kJ/molK

k2 : 7,69 M ⁻¹ s⁻¹

T2: 219K

Solving:

`-8,382 = (-Ea)/(8,314472x10^(-3)kJ/molK) (1,659x10^(-3)K^(-1))`

`-8,382 = -Ea*0,19956mol/kJ`

`-8,382 = -Ea*0,19956mol/kJ`

`-42,0 kJ/mol = -Ea`

Ea = 42,0 kJ/mol

I hope it helps!