a) The rate of a particular reaction doubles when the temperature changes from 300 K to 310 K. Calculate the energy of activation of the reaction. [Given : `R="8.314 JK"^(-1)" mol"^(-1)`].

Explanation

a) `log""(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`log""(2k_(1))/(k_(1))=(E_(a))/(2.303xx8.314)[(300-310)/(300xx310)]`
`log2=(E_(a))/(19.1471)[(10)/(93000)]`
`E_(a)=53598J or 53.598kJ`