A plot of `"Ink "v//s " 1/T"` for a reaction gives the slope `-1xx10^(4)k.` The energy of activation for the reaction is `(Given,R=8.314k^(-1)" mol"^(-1))`

Options

A. `8314kJ" mol"^(-1)`
B. `1.202kJ" mol"^(-1)`
C. `1202kJ" mol"^(-1)`
D. `83.14kJ" mol"^(-1)`

Answer

D

Explanation

If a plot is drawn between in k and `(1)/(T),` then according to Arrhenius equation, `"In " k="In " A -(E_(a))/(RT)`
Here, `slope=-(E_(a))/(R)rArr-1xx10^(4)k=-(E_(a))/(8.314)`
`therefore" "E_(a)=8.314xx10^(4)J"mol"^(-1)=83.14kJ"mol"^(-1)`