The activation energy of a certain reaction is 34.9kJ/mol . At 23 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast?
3 answers
Use the Arrhenius equation. The problem gives you k1, make k2 twice that.
-421.41
ln(2)=34900/8.314 - (1/296 - 1/T1)
1 answer