Arrhenius equation. The activation energy of a certain reaction is 34.6kJ/mol . At 22 ∘C , the rate constant?

The activation energy of a certain reaction is 34.6kJ/mol . At 22 ∘C , the rate constant is 0.0170s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

Answer 1

ln(k2/k1) = (Ea / R)(1/T1 – 1/T2)
ln(2) = (3.46e4 J/mol / 8.314 J/mol-K)(1/(22 + 273K) – 1/T2)
1.6656e-4 = 1/295 – 1/T2
T2 = 310K —> 37°C

ln(k2/0.0170 s^-1) = (3.46e4 J/mol-K / 8.314 J/mol-K)(1/295K – 1/(180 + 273K))
ln(k2/0.0170 s^-1) = 4.9204
k2 = (0.0170 s^-1)e^(4.9204)
k2 = 2.33 s^-1

Leave a Comment