Asked by Kat

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?
Answered by DrBob222
Use the Arrhenius equation. The problem gives you k1, make k2 twice that.
Answered by Anonymous
-421.41
Answered by Scholar
ln(2)=34900/8.314 - (1/296 - 1/T1)
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