The first-order rate constant of the reaction of CH3Cl with water to produce CH3OH and HCI is 3.32x 10^-10 s^-1 at 25 °C. Calculate the rate constant at 40 °C if the activation energy is 116 kJ/mol.

Answer 1

For this type of problem, you are looking to use the Arrhenius equation, which is...

ln(k2/k1) = - Ea/R (1/T2 - 1/T1)

k1 = 3.32x10^-10

k2 = ?

Ea = 115 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol

T1 = 25º + 273 = 298K

T2 = 40º + 273 = 313K

Plugging these values into the equation and solving, we have...

ln(k2/3.32x10^-10) = -115 / 0.008314 (1/313 - 1/298)

ln(k2/3.32x10^-10) = -13832 (0.00319 - 0.00336)

ln(k2/3.32x10^-10) = 2.35

k2 / 3.32x10^-10 = 12.7

k2 = 4.21x10^-9 s^-1