The rate constant of a reaction is `1.2xx10^(-3)sec^(-1)" at "30^(@)C" and "2.1xx10^(-3)sec^(-1)" at" 40^(@)C`.
Calculate the energy of activation of the reaction.

Explanation

`{:("Here, we are given that :",,,k_(1)=1.2xx10^(-3)sec^(-1)",",,,T_(1)=30+273=303" K"),(,,,k_(2)=2.1xx10^(-3)sec^(-1)",",,,T_(2)=40+273=313" K"):}`
Substituting these values in the equation :
`log""(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))," we get"`
`log""(2.1)/(1.2)=(E_(a))/(2.303xx8.314)xx(10)/(303xx313("J mol"^(-1)))`
This on solving gives `E_(a)=44126.3" J mol"^(-1)=44.13" kJ mol"^(-1)`