2
$\begingroup$

Why is it that chemisorption decreases with higher temperatures?

The explanations I found online were quite vague and one of them cited Le Chatelier's principle, but I thought chemisorption is an irreversible process (because covalent bonds are formed between the adsorbate and adsorbent) so there is no reversible equilibrium.

Explanations I found online:

  1. "Further increase will increase the energy of the molecules adsorbed and will also increase the rate of desorption. Therefore, the extent of adsorption decreases." - why does the rate of desoprtion increase?

  2. "The initial increase is due to the fact that heat supplied acts as activation energy. The decrease afterward is due to the exothermic nature of adsorption equilibrium."

$\endgroup$
3
  • 1
    $\begingroup$ The chemisorption barrier depend on the electronic states involved which form a barrier to dissociation, i..e an activation energy. Adding energy , i.e increasing the temperature, brings the molecules energy closer to the top of the barrier and so it can dissociate more easily. This is true of chemical reactions in general. $\endgroup$
    – porphyrin
    Feb 13, 2021 at 17:42
  • $\begingroup$ @porphyrin Oh, I see. Thank you for your answer! Also, is it that dissociation occurs with higher temperatures because the increase in kinetic energy breaks the covalent bonds? Just wanted to clarify because I'm a high school biology student, but I need to explain some chemistry concepts in my report. $\endgroup$
    – Emma_K
    Feb 14, 2021 at 9:24
  • $\begingroup$ Not just kinetic, molecules have vibrational and rotational energy also but eventually a molecule will find that it has enough energy to surmount the energy barrier and break a bond. The fraction of energy at any given temperature is given by the Boltzmann distribution. $\endgroup$
    – porphyrin
    Feb 14, 2021 at 13:22

1 Answer 1

2
$\begingroup$

Let's answer this question with the help of some thermodynamics and kinetics. Chemisorption is a chemical process with $\Delta S<0$ and $\Delta H<0$. Thus, in simpler words it is an enthalpy driven exothermic equilibrium

Every chemical process is an equilibrium with varying extent of completion. So your idea of chemisorption not being an equilibrium is wrong. Every chemical reaction is theoretically reversible in nature. What we call irreversible reactions in general are such reactions which are heavily favoured on one side of the equilibrium.

Now,

$\ln K = -\frac{\Delta H}{RT} + \frac{\Delta S}{R}$

It is evident from this above expression that the $K$ will decrease when we increase $T$ when $\Delta H<0$ and thus the aforementioned conclusion.

But this is only half the story. Like all other chemical reactions, chemisorption also needs some activation energy to overcome the transition state. At lower temperatures there are very few molecules which have the kinetic energy enough to overcome the activation energy. The fraction of these molecules increases with increase in temperature and thus increasing the extent of chemisorption.

Conclusion

Due to the combination of thermodynamic and kinetic effects the rate of chemisorption first increases with the increase in temperature but after a certain saturation point the thermodynamic effects start dominating and thus the extent of chemisorption starts to decrease.

The graph below might help you understand how the extent of chemisorption changes with change in temperature.

Chemisorption extent vs Temperature

Here,

$x$ = Mass of adsorbate
$m$ = Mass of adsorbent

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.