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=logA-(E_(a))/(2.303 RT)(y=c+mx)`
Slope `=(-E_(a))/(2.303R)=(1)/(2.303)(given)(tan theta=(1)/(2.303))`
`-E_(a)= 2.303Rxxslope`
`=2.303xx(R )/(2.303)=R=2 cal`