Answer to Question #257457 in Physical Chemistry for pawan

Question

Calculate the activation energy for a second-order reaction if its rate constant in- creases from 2.5 mol/(L·s) to 250 mol/(L·s) as the temperature increases from 298 to 340 K. 

Answer 1

We are given that:

When T1

​=298K

Let k1

​=k

When T

=340K

k2

​=2k

Substituting these values the equation:

$Log\frac{2K}{K}$ = $\frac{E_a}{2.303×8.314}$ ($(\frac{340-298}{340×398})$

E_a135320=804.18

E_a=5.94×10^-3J