In the Arrhenius plot of ln k vs `(1)/(T)` , a linear plot is obtained with a slope of `-2 xx 10^(4)` K . The energy of activation of the reaction (in kJ `"mole"^(-1)`) is (R value is 8.3 J`K^(-1) mol^(-1)`)

Options

A. 83
B. 166
C. 249
D. 332

Answer

B

Explanation

`k=Ae^(-E_(a)//RT)`
`lnK=lnA-E_(a)//RT`
For `lnk vs 1//T`
`lnA=` intercept
`-E_(a)//R=` slope =`-2xx10^(4)K`
`E_(a)=Rxx2xx10^(4)K=8.3xx2xx10^(4)Jmol^(-1)`
`=16.6xx10^(4)Jmol^(-1)` or `166kJ mol^(-1)`