Graph between `log k` and `1//T` [`k` rate constant `(s^(-1))` and `T` and the temperature `(K)`] is a straight line with `OX =5, theta = tan^(-1) (1//2.303)`. Hence `-E_(a)` will be

Options

A. `2.303 xx 2` cal
B. `2 // 2.303` cal
C. 2 cal
D. none

Answer

C

Explanation

`logk=log A-(E_(a))/(2.303 RT)` (y= c + mx)
`"Slope"=(-E_(a))/(2.303 R)=(1)/(2.303)` (given)(`tan theta=(1)/(2.303)`)
`-E_(a)=2.303R xx "Slope"`
`=2.303 xx (R)/(2.303)=R=2` cal