The activation energy for the reaction:
`2AB rarr A_(2)+B_(2)(g)`
is `159.7 kJ mol^(-1)` at `500 K`. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
(Given: `2.3 xx 8.314 J K^(-1) mol^(-1) xx 500 K = 9561.1 J mol^(-1)`)

Explanation

Fraction of molecules having energy equal to or greater than activation energy is:
`E_(a) = (n)/(N) = x = e^(-E_(a)//RT)`
`:. x = e^(-E_(a)//RT)`
`ln x = (-E_(a))/(RT) rArr 2.3 log x = (-E_(a))/(RT)`
or `log x = - (159.7 xx 10^(3) J mol^(-1))/(2.3 xx 8.314 J K^(-1) mol^(-1) xx 500 K)`
`= -(159.7 xx 10^(3) J mol^(-1))/(9561.1 J mol^(-1))`
`= -16.7`
`:. x = "Antilog"(-16.7)`
`= "Antilog"(-16-0.7 + 1-1)`
`= "Antilog"(bar(17).3)`
`= 2 xx 10^(-17)`