The rate constant of a reaction increases by 5% when its temperature is raised from `27^@c` to `28^@c.` The activation energy of the reaction is

Options

A. 36 kJ/mol
B. 16.6 kJ/mol
C. 46.6 kJ/mol
D. 26.6 kJ/mol

Explanation

`log. (k_2)/(k_1)= E_(a)/(2.303R)[(T_2-T_1)/(T_1T_2)]`
`log. (150)/(100)= E_a/(2.303 xx 8.314)[1/(300 xx 301)]`
or `E_a = (log 1.05) xx 2.303 xx 8.314 xx 300 xx 301`
`=36.65 kJ`