The activation energy for the reaction,`2HI(g)toH_2(g)+I_2(g)`is `209.5kj mol^(-1)`at 581 K.Calculate the fraction of molecules of reactant having energy equal to or greater than activation energy.

Explanation

Let the fraction be x
`e^(E//RT)` i.e.,x=`e^(E//RT)`
In `x= -E_a/(RT)log x=-E_a/(2.303RT)=(-209.5xx10^3J)/(2.303xx8.314xx581)=-18.8323`
`thereforex=Antilog` of -18.8323=Antilog of `bar19 .1677=1.471xx10^(-19)`