Question

If the activation energy for a given compound is found to be 103 kJ/mol, with a frequency factor of 4.0 × 1013 s-1, what is the rate constant for this reaction at 398 K?
A) 1.2 s-1
B) 8.2 s-1
C) 3.9 × 1010 s-1
D) 1.7 × 1010 s-1
E) 2.5 × 107 s-1
Answer: A

Answer 1

1.) k=4.0*1013e^(10300)/(8.314)(398)= about 1.2