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.
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Expert's answer
2020-06-10T07:40:59-0400
"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"
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