Calculate the rate constant, 𝑘 , for a reaction at 52.0 °C that has an activation energy of 79.6 kJ/mol and a frequency factor of 1.05×1011 s−1.

Question

K=

Answer 1

ln k = ln A + (-Ea/RT)

k = ?

A = 1.05x10¹¹s^-1

Ea = 79.6 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol

T = 52ºC + 273 = 325K

ln k = ln 1.05x10¹¹ + (-79.6/0.008314 x 325)

ln k = ln 1.05x10¹¹ + (-20.46)

ln k = 25.38 - 20.46

ln k = 4.92

k = 137s^-1

(be sure to check the math)