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

activation energy is given as:

x=e-Ea/RT
ln x=-Ea/RT
log x=-Ea203.0 RT
logx=209500 j mol-12.303×8.314 jk-1mol-1×581
Now , x =Anti log (18.82323)
=Anti log19.¯1677
=1.471 ×10-19