Question

# The rate constant of a chemical reaction increased from 0.100 s−1 to 2.60 s−1 upon raising...

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?

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!

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

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