The activation energy for a reaction when the temperature is raised from 300K to 310K is

Options

A. `50.6kJ mol^(-1)`
B. `53.6kJmol^(-1)`
C. `56.6kJ mol^(-1)`
D. `59.6 kJ mol^(-1)`

Answer

B

Explanation

b) According to Arrhenius equation,
`log (K_(2))/(K_(1)) = E_(a)/2.303[1/T_(1)-1/T_(2)]`
`log(2K_(1))/(K_(1))=(E_(a))/(2.303R) [1/T_(1)-1/T_(2)]`
`log2 = (E_(a))/(2.303 xx 8.314 Jk^(-1)mol^(-1))[1/(300K-1/310K)]`
`0.30 = (E_(a))/(2.303 xx (8.314 J mol^(-1)) xx (9300)`
`=53.6 kj mol^(-1)`