The Total solution for NCERT class 6-12
The rate constant for thedecomposition of hydrocarbons is 2.418 ├Ч 10тИТ5┬аsтИТ1┬аat546 K. If the energy of activation is 179.9 kJ/mol, what will be the value ofpre-exponential factor.
k┬а=2.418 ├Ч 10тИТ5┬аsтИТ1
T┬а=546 K
Ea┬а=179.9 kJ molтИТ1┬а= 179.9 ├Ч 103┬аJ molтИТ1
According to the Arrheniusequation,
= (0.3835 тИТ 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 ├Ч 1012┬аsтИТ1┬а(approximately)