Graph between log k and `1//T` [where K is rate constant in `s^(-1)` and T is the temperature (in K) is a straight line with Hence, `E_(a)` will be

Options

A. `2.303 xx 2` cal
B. `2/(2.303) cal`
C. 2 cal
D. None of these

Answer

C

Explanation

log k = log A`-(E_(a))/(2.303 R) = tan phi= 1/(2.303)` (Given)
`E_(a) = 2.303 R xx` Slope
`=2.303 R xx 1/(2.303) = R=2 Cal`