The time for which the oxygen atom remains adsorbed on a tungsten surface is 0.36 s at2550 K and 3.49 s at 2360 K. Calculate the activationenergy of desorption of oxygen atom. Assumign that the oxygen atom is tight chemisorbed, calculate per-exponential factor tau_(0) in the Arrhenius type equation.

The time for which the oxygen atom remains adsorbed on a tungsten surf

1.	The time for which the oxygen atom remains adsorbed on a tungsten surface is 0.36 s at2550 K and 3.49 s at 2360 K. Calculate the activationenergy of desorption of oxygen atom. Assumign that the oxygen atom is tight chemisorbed, calculate per-exponential factor tau_(0) in the Arrhenius type equation.
Answer» Solution :From Arrhenius-type equation `log ""(tau_(2))/(tau_(1))=(E_(a))/(2.303R)((1)/(T_(2))-(1)/(T_(1)))` At `T_(1)=2550K, tau_(1)=0.36s` At `T_(2)=2360K, tau_(2)=3.49s` `thereforelog""(3.49)/(0.36)=(E_(a))/(2.303xx8.314)((1)/(2360)-(1)/(2550))` log` 9.694=(E_(a))/(2.303xx8.3174)((1)/(2360)-(1)/(2550))` `log 9.964=(E_(a))/(2.303xx8.314)((190)/(2360xx2550))` or `E_(a)=(0.9865xx2.303xx8.314xx2360xx2550)/(190)=598273J MOL^(-1)=598.27kJ mol^(-1)` To calculate `tau_(0),` `TAU=tau_(0)E^(E_(a)//RT)` or `ln tau _(0)+(E_(a))/(RT)or log tau=logtau_(0)+(E_(a))/(2.303RT)` Substituting `T=2360 K, tau =3.49s` `log 3.49=log tau_(0)+(598273)/(2.303xx8.314xx2360)` `log tau_(0)=0.5428-13.2399=-12.6971=bar(13).3029` or `tau_(0)=` Antilog `bar (13).3029=2.0xx10^(-13)s`