The activation energy and enthalpy of chemisorption of oxygen on a metal surface is 37.3kJ mol^(-1) and -72.1kJ mol^(-1) At a certain pressure, the rateconstant for chemisorption is 1.2xx10^(-3) at 3198 K . What will be the value of the rate constant at 308K ?

The activation energy and enthalpy of chemisorption of oxygen on a met

1.	The activation energy and enthalpy of chemisorption of oxygen on a metal surface is 37.3kJ mol^(-1) and -72.1kJ mol^(-1) At a certain pressure, the rateconstant for chemisorption is 1.2xx10^(-3) at 3198 K . What will be the value of the rate constant at 308K ?
Answer» `7.6 xx 10^(-4)s^(-1)` `1.6xx10^(-6)s^(-1)` `7.6xx10^(-2)s^(-1)` `1.6xx10^(-5)s^(-1)` Solution :`k(` RATE CONSTANT `)=Ae^(En//RT)` `ln.(K_(2))/(K_(1))=(E_(a))/(R)((1)/(T_(1))-(1)/(T_(2)))`or `lnk_(2)+(E_(a))/(R)((1)/(T_(1))-(1)/(T_(2)))` Now,`T_(1)=318K,E_(a)=37.3kJmol^(-1)` and `T_(2)=308K` `logk_(2)=log(1.2xx10^(-3))+(37300)/(8.31xx2.303)((1)/(318)-(1)/(308))=-2.9208-(37300xx10)/(8.31xx2.303xx318xx308)` `=-2.9208-0.1989` `=k_(2)=7.6xx10^(-4)s^(-1)`