Question

From the formulae:
( mathrm{k}=mathrm{Ae}^{wedge}(-mathrm{Ea} / mathrm{RT}) )
( log mathrm{k}=log mathrm{A}-2.303 mathrm{Ea} / mathrm{RT} ldots ldots ldots(1) )
Given eqn. is:
( log k=5.4-212 / T ldots ldots ldots ldots ldots ldots .(2) )
Compariing both the eqns. We get,
( 2.303 mathrm{Ea} / mathrm{R}=212 )
( mathrm{Ea}=>212^{*} 25 /left(3^{*} 2.303right)=767.11 mathrm{J} )

# The rate constant of a certain reaction is given by: log, ok = 5.4.212 Calculate the activation energy at 127°C.

Solution