Given statement: Enthalpy of adsorption is the strength at which molecule (say $\ce{H2}$) is bound to the adsorbent (say some metal center). For a free $\ce{H2}$ molecule, the entropy contribution is from translational, rotational and vibrational motion. After the adsorption, the translational and rotational motions will not contribute to the entropy change $\mathrm{\Delta S}$. Thus the only contribution to $\mathrm{\Delta S}$ is from the vibrational motion of $\ce{H2}$ which will decrease for increase in adsorption strength.

As per the given statement, increase in enthalpy should decrease the entropy. However, Enthalpy-Entropy correlation states that increase in enthalpy leads to increase in entropy. How this is true?


closed as unclear what you're asking by Jon Custer, Pritt says Reinstate Monica, M.A.R., NotEvans., Buttonwood Jul 21 '17 at 19:38

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  • $\begingroup$ Could you explain why you think that the enthalpy increases? Isn't the absorption exothermic as, for instance, stated here (--> decrease of enthalpy) ? $\endgroup$ – user45298 Jul 20 '17 at 3:35
  • $\begingroup$ Why, recalling the change of Free / Gibbs enthalpy expressed as $\Delta{}_R G = \Delta_RH - T\Delta_RS$, both changes of enthalpy and of entropy (multiplied by temperature) are taken into consideration? Currently your question reads like one were enough, which I doubt. $\endgroup$ – Buttonwood Jul 21 '17 at 19:38