# Is adsorption exothermic, and if so, why?

The explanation given in my textbook is:

• For a process to be spontaneous, the thermodynamic requirement is that, at constant temperature and pressure, $\Delta G < 0$.
• As the entropy change of adsorption is negative ($\Delta S < 0$), the enthalpy change must therefore be negative ($\Delta H < 0$).
• If $\Delta H$ were to be positive, then $\Delta G = \Delta H - T\Delta S$ would always be positive.

It basically says adsorption is exothermic because $\Delta G$ has to be negative. Isn't there any other reason?

• This is just an educated guess, but the adsorption of atoms on a surface is akin to creating weak bonds; the creation of bonds is exothermic (since it stabilizes the system), hence adsorption is exothermic too. – Hippalectryon Mar 20 '16 at 10:54

There are two common arguments presented as for why $\Delta H < 0$:

Argument (1): Well, indeed I see nothing wrong with the argument presented by the textbook. If adsorption takes place spontaneously, then one can conclude that the change in Gibbs free energy of the process is indeed negative. Since, the entropy change associated with process is necessarily negative (if we assume the entropy of the adsorbent is necessarily greater in the gaseous or liquid state than it is in the adsorbed state), we need a sufficiently large negative value for the change in enthalpy to ensure spontaneity.

Argument (2): Now, adsorption as a phenomenon is associated with "surface energy" (not, unlike surface tension). The surface of the adsorbent molecule attracts and "attaches" adsorbates either via weak van der waals forces (physisorption) or stronger chemical interactions (chemisorption)--in either cases, the surface energy of the system is minimised due to the formation of these new attractions. Thus, one would say adsorption would be exothermic in nature.

The key difference in physisorption and chemisorption is that the electronic structures of atoms/molecules of the adsorbent and adsorbate remain largely unperturbed in the first case (i.e no chemical reaction takes place so to speak).

Anyway, these seem to be the oft cited reasons. Both these arguments are generally sound and hold true under most, but not all conditions. These arguments are often introduced in most introductory level courses, but they do have some holes.

Food for thought - what if a positive entropy change occurs? Or, the chemical interactions that take place during chemisorption are endothermic in nature? So is adsorption always exothermic, no exceptions? Well, short answer: No.

The following discussion is inspired (largely) by ref 1.

If we just look physisorption exclusively, since this is essentially a condensation process analogous to liquefaction on the absorbent. This is so because the number of degrees of freedom of the adsorbed species is less than the number it possessed prior to adsorption, and because the entropy of the absorbent is unaltered, since there is no chemical interaction. So argument (1) can be applied.

The same need not hold true for chemisorption, since a chemical reaction is indeed taking place. In argument (2), the assumption was that whatever chemical reaction that is taking place at the surface is exothermic, however this need not be the case. If an endothermic reaction takes place then for the process to remain spontaneous the entropy change should not only be positive, but $T\Delta S$ must exceed, numerically, $\Delta H$ so that the net Gibbs free energy change as a whole is negative

Recall: $\Delta G= \Delta H-T\Delta S$

This can, indeed be realised. For instructional purposes consider the following hypothetical system: A molecule, $A_2$, is dissociatively chemisorbed on the surface of a solid, M; consider strength of the M-A bond equal to half that of the A-A bond. This would give us a net enthalpy change of zero. If the adsorbed atoms, A, have complete two-dimensional mobility, it would then follow that a positive entropy change, would result, corresponding to a net gain of one degree of freedom, and the associated free energy change is a) negative and numerically equal to $T\Delta S$.

Similarly, one can envisage a system where the net positive entropy change is caused not because of increased mobility of the adsorbate molecules; the entropy of the adsorbate may very well decrease, but an associated large positive entropy change of the adsorbent (consequence of the chemical reaction taking place at the surface) can make up for it (possible, again). Again, one can expect endothermic adsorption in such a scenario.

I'll quote the conclusion from the paper:

Summarising, it may be said that in all adsorptions entropy changes due to surface-structural changes in the adsorbent itself must be considered along with the entropy changes of the adsorbate. In physical adsorption, the contribution of the first of these two entropy changes to the total entropy change is always likely to be insignificant compared to the contribution of the second (though further investigations alone can establish this unambiguously). In chemisorption, however, there is ample evidence already available to show that the entropy change due to surface structural changes in the adsorbent contributes significantly to the total entropy of adsorption. Whenever the positive entropy change due to adsorbent surface changes exceeds numerically the negative entropy change due to loss of freedom of the adsorbate, the net entropy of adsorption is positive. Failure to recognise this has led to serious error when using equation (1) to arrive at the sign of the heat of chemisorption. This, in essence, is why endothermic chemisorption has been discounted until very recently, though its existence should be considered as normal as the existence of endothermic solution.

You can look up specific examples, which are themselves referenced within ref 1. One such example is in ref 2, which provides some detailed potential energy diagrams illustrating endothermic chemisorption.

References

(1) Thomas, J. M. The existence of endothermic adsorption. J. Chem. Educ. 1961, 38 (3), 138. DOI: 10.1021/ed038p138.

(2) Dowden, D. A.; Mackenzie, N.; Trapnell, B. M. W. 9 Hydrogen-Deuterium Exchange on the Oxides of Transition Metals. Adv. Catal. 1957, Vol. 9, 65–69. DOI: 10.1016/S0360-0564(08)60154-6.

• "If the adsorbed atoms, A, have complete two-dimensional mobility" I can't envision what you mean here by two-dimensional mobility. Do you mean stretching? – Mockingbird Jul 4 '17 at 16:59