The specific reaction rates of a chemical reaction are `2.45xx10^(-5)sec^(-1)` at 273 K and `1.62xx10^(-4)sec^(-1)` at 303 K. Calculate the activation energy.

Explanation

Given data : `k_(1)=2.45xx10^(-5)sec^(-1), " " T_(1)=273 K`
`k_(2)=16.2xx10^(-4)sec^(-1), " " T_(2)=303 K`
`R=8.314 JK^(-1) mol^(-1)`
Formula : `"log"(k_(2))/(k_(1))=(E_(a))/(2.303 R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
Solution :
`"log"(16.2xx10^(-4))/(2.45xx10^(-5))=(E_(a))/(2.303xx8.314)((303-273)/(273xx303))`
`1.8203=(E_(a))/(2.303xx8.314)xx[(30)/(273xx303)]`
`therefore E_(a)=52802.3xx1.8203`
`= 96116J//mol`
`E_(a)=96.116"kJ mol"^(-1)`.