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