Question

The rate constant is given by Arrhenius equation.
`k=Ae^(-E_(a)//RT)`
Calculate the ratio of the catalysed and uncatalysed rate constants at `25^(@)C` if the energy of activation of a catalysed raction is 162 kJ and for the uncatalysed reaction the value is 350 kJ

Answer 1

Let `K_(ca) and k_(nu)` be the rate contants for catalysed and uncatalysed reactions.
`2.303"log"_(10)k_(ca)=2.303"log"_(10)A-(162xx10^(3))/(RT)" "....(i)`
and `2.303"log"_(10)k_(un)=2.303"log"_(10)A-(350xx10^(3))/(RT)" "....(ii)`
Subtracting eq. (ii) from eq. (i)
`"log"_(10)(k_(ca))/(k_(uc))=(10^(3))/(2.303RT)(350-162)`
`=(188xx10^(3))/(2.303xx8.314xx298)=32.95`
`(k_(ca))/(k_(un))=8.88xx10^(32)`