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 rate constant for chemisorption is `1.2xx10^(-3)` at `3198 K` . What will be the value of the rate constant at `308K` ?

Options

A. `7.6 xx 10^(-4)s^(-1)`
B. `1.6xx10^(-6)s^(-1)`
C. `7.6xx10^(-2)s^(-1)`
D. `1.6xx10^(-5)s^(-1)`

Answer

1

Explanation

`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)`