The rate constant of the chemical reaction doubled for an increase of 10 k in absolute temperature from 295 k. Calculate the (activation energy),`E_(a).`

Options

A. `51.8kJ" mol"^(-1)`
B. `82.1kJ" mol"^(-1)`
C. `23.8kJ" mol"^(-1)`
D. `62.1kJ" mol"^(-1)`

Answer

A

Explanation

According to Arrhenius equation,
`"log"(k_(2))/(k_(1))E_(a)/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
`k_(2)/(k_(1))=2,T_(1)=295k`
`T_(2)=305k`
`R=8.314Jk^(-1)mol^(-1)`
`:." "log2=(E_(a))/(2.303xx(8.314Jk^(-1)mol^(-1)))[(1)/(295k)-(1)/(305k)]`
`E_(a)=(0.3010xx2.303xx8.314xx295xx305)/(10)(Jmol^(-1))`
`=51855Jmol^(-1)=51.855kJmol^(-1)`