question_answer 32)

The rate constant for the decomposition of a hydrocarbon is 2.418 x 10–5 s–1 at 546 K. If the energy of activation is 179.9 kJ mol–1, what will be the value of pre-exponential factor?  

Answer:

According to Arrhenius equation, log k = log A ?   k = 2.418 x 10?5 s?1 ; Ea = 179900 J mol?1 R = 8.314J K?1 mol?1 ; T = 546 K = ? 4.6184 + 17.21 = 12.5916 A = antilog (12.5919) = 3.9 × 10?12.