If the activation energy for a given compound is found to be 42.0 kJ/mol, with a frequency factor of #8.0 x 10^10 s^(-1)#, what is the rate constant for this reaction at 298 K?

1 Answer
Jan 26, 2018

Answer:

#3.48 × 10^3#

Explanation:

Rate constant can be calculated from Arrhenius equation

#k = A xx e^(-E_a/(RT))#

Where,

k = Rate constant

A = Frequency factor

#E_a# = Activation energy

R = Gas constant

T = Absolute Temperature

#k = 8.0 xx 10^10 s^-1 × e^-("42.0 kJ/mol × 1000 J/kJ"/("8.314 J/(mol K) × 298 K")) = 3.48 × 10^3#