Question

What percentage of reactant molecules will crossover the energe barrier t 325 K? Heat of reaction is `0.12` kcal and activation energy of backward reaction is `0.02` kcal.

Answer 1

Activation energy of forward reaction `0.12-0.02 =0.10 "kcal"`
Fraction of moelcuels which are active or which vrossover the energy the barrier `((k)/(A))=e^(-E//RT)`
`log_(3).((k)/(A))=-(E)/(RT)`
`(k)/(A)="antilog"[(-E)/(2.0303RT)]`
`="antilog"[(-0.10xx100)/(2.303xx2xx325)]`
`="antilog"[-0.06680]=0.8574`
`:.` Percentage of reactant molecules crosing over the barrier
`0.8574xx100=85.74`