Calculate how long a hydrogen atom will remain on the surface of a solid at 298 K if its desorption activatino energy is (a) 15 kJmol ^(-1) (b) 150 kJ mol^(-1). Assume that tau_(0)=10^(-13) s. Also calculate the results at 1000 K. What do you conclude from your results ?

Calculate how long a hydrogen atom will remain on the surface of a sol

1.	Calculate how long a hydrogen atom will remain on the surface of a solid at 298 K if its desorption activatino energy is (a) 15 kJmol ^(-1) (b) 150 kJ mol^(-1). Assume that tau_(0)=10^(-13) s. Also calculate the results at 1000 K. What do you conclude from your results ?
Answer» Solution :`tau=tau_(0)e^(E_(a)//RT)` In `tau=ln tau_(0)+(E_(a))/(2.303RT)` At 298 K (a) `log tau =log10^(-3)+(15000)/(2.303xx8.314xx298)=-13+2.6289=bar(11).6289` `therefore tau=` Autilog `bar (11).6289=4.255xx10^(-11)` (b) `logtau log 10^(-13)+(150000)/(2.303xx8.314xx298)=-13+26.289=13.289` `tau=` Antilog `=13.289=1.909xx10^(3)s~~600.000` years At 1000 K (a) `log tau =log 10^(-13)""+(15000)/(2.303xx8.314xx1000)=-13+0.7834=bar(13).7834` `tau=` Antilog `bar (13).7834 =6.073xx10^(-13)~~6.1xx10^(-13)s` (b) `logtau=-13+7.834=bar 6. 834` `tau=` Antilog `bar6.834=6.823xx10^(-6)s~~6.8xx10^(-6)s` Conclusion. (i) The time for a hydrogen atom to remain on the surface of the adsorbent depends upon desorption activation energy and temperature. (ii) Greater the desorption activation energy `(E_(a)).` greater teh time. Greater the temperature, lesser the time.