Question

The rate constant of a chemical reaction increased from 0.100 s−1 to 2.60 s−1 upon raising the temperature from 25.0 ∘C to 45.0 ∘C .

I solved

(1T2−1T1) =

−2.11×10−4

K−1

In (k1/k2) = -3.26

But, I'm having problems on this question:

What is the activation energy of the reaction?

Express your answer numerically in kilojoules per mole.

It would be great if you could show all of your work. I've been trying to figure out this problem for a while now and really need help!

Answer 1

We use Arrhenius equation to solve this problem.

Ln (k2/k1) = ( - Ea / R ) x ( 1 / T2 - 1 / T1)

Here k2 is the rate constant at Temperature T2

And k1 is at T1

Ea is activation energy in J / mol and R is gas constant = 8.314 J / ( K mol )

Let k1 = 0.100 per s T1 = 25.0 ⁰C = 25.0 + 273.15 = 298.15 K

k 2 = 2.60 per s , T2 = 45.0⁰ C = 45.0 + 273.15 = 318.15 K

Lets plug these value in order to get Ea.

Ln ( 2.60 / 0.100 ) = (- Ea / 8.314 ) ( 1 / 318.15 – 1 / 298.15 )

Ea = 128472.7 J /mol

Ea in kJ = 128.5 kJ /mol