Question

The rate constant of a chemical reaction increased from 0.100 s−1 to 3.00 s−1 upon raising the temperature from 25.0 ∘C to 37.0 ∘C .
Calculate the value of ln(k1/k2) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined in part A.

Answer 1

$k_1=0.1 \ s^{-1};k_2=3 \ s^{-1}$

So, $ln(\frac{k_1}{k_2})=ln(\frac{0.1}{3})=-3.4$