Thermodynamics behind Deliquescence? The phenomenon of deliquescence wherein a water soluble substance absorbs water from atmosphere and dissolve in it forming a saturated solution, has thermodynamic roots. I have read the fact that the vapour pressure of the saturated solution should be lower than the the vapour pressure of water at that temperature. Why is this so?  

Moreover, since deliquescence is a spontaneous process and therefore thermodynamically favourable, there must be a negative $\Delta G$ for the absorption of water vapour and conversion of solid substance into its saturated solution? How can we derive, understand or quantify various thermodynamic parameters ($\Delta G, \Delta S, \Delta H$ etc) for the process of deliquescence?
I am comfortable with basic thermodynamic manipulations, binary phase diagrams, multiphase systems and equilibria.
The vapour pressure lowering is a consequence of Raoult's law with a non-volatile solute, i.e. a solvent that has zero vapour pressure (or at least, lower vapour pressure than the solvent).

Regarding the second part (“How can we derive, understand or quantify various thermodynamic parameters (ΔG,ΔS,ΔH etc) for the process of deliquescence?”), I am not sure exactly what you are looking for. This is a complex phenomenon, and there is a lot of research literature of the topic. To give only one example, see


  “Water-Solids Interactions: Deliquescence”, Lisa J. Mauer and Lynne S. Taylor, Ann. Rev. of Food Sci. Technol., 2010, 1, 41-63.