Question

The activation energy for the reaction, `2Hi(g) to H_(2)(g) + I_(2)(g)` is 209.5 kJ `moli^(-1)` at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

Answer 1

The fraction of the molecules (x) having energy equal to or more than activation energy may be calcualted as follows.
`x=n//N=e^(-E_(a)//Rt)`
In x = `E_(a)/(2.303RT)`
log x = `-(209.5 xx 10^(3)J mol^(-1))/(2.303 xx (8.314 J K^(-1) mol^(-1)) xx 581 K)=-18.8323`
`x="Antilog"(-18.8324) = " Antilog " bar19.1677`
`=1.471 xx 10^(-19)`.