Question 4.9:The activation energy for the reaction
2HI(g) → H2 + I2(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?

In the given case:

Ea=209.5kjmol-1=209500kmol-1
T=581K
R=8.314jk-1
Now, the fraction of molecules of reactants having energy equal to or greater than

activation energy is given as:

x=e-Ea/RT
lnx=-Ea/RT
logx=-Ea203.0RT
logx=209500jmol-12.303×8.314jk-1mol-1×581
Now,x=Antilog(18.82323)
=Antilog19.¯1677
=1.471×10-19