The activation energy for the reaction `2HI(g)rarrH_(2)+I_(2)(g) " is " 209.5"kJ.mol"^(-1)` at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation. energy ?

Explanation

From Boltzmann's equation , fraction of molecules having energy equal to or greater than activation energy is,
`x=(n)/(N) = e^(-E_(a)//RT) therefore" In"x=-(E_(a))/(RT)`
[n = no .of molecules with energy `ge209.5"kJ.mol"^(-1) and ` N = total no . of molecules ]
or, `logx=-(E_(a))/(2.303RT)=-(209.5xx10^(3))/(2.303xx8.314xx581)=-18.8323`
`thereforelog ="antilog"(-18.8323)=1.471xx10^(-19)`