The Total solution for NCERT class 6-12
The rate constant for thedecomposition of hydrocarbons is 2.418 × 10−5 s−1 at546 K. If the energy of activation is 179.9 kJ/mol, what will be the value ofpre-exponential factor.
k =2.418 × 10−5 s−1
T =546 K
Ea =179.9 kJ mol−1 = 179.9 × 103 J mol−1
According to the Arrheniusequation,
= (0.3835 − 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 × 1012 s−1 (approximately)