That  DELG = DELH- T  DEL S    ın your case you have 

DELG = _ ABS DELH +  T  ABSDELS   the sign could be positive or  negative depends on the relative absolute values of DEL  H and TDELS.  But at the equilibrium it should be ZERO. Sınce in your case  DELG> 0 that means absorption and desorption reactions are not in mutual thermodynamic equilibrium.  Rather  the reverse reaction, which is  desorption would be the natural process and that takes place in reality. This way  the  direction of the reaction proceeds along the decreasing path  of the Gibbs free energy (towards the direction of equilibrium).. 

Which kind of system are you using, and which procedure?

i done batch adsorption studies with adsorbent to remove metal ions from aqueous solutions

How, and where, do you calculate the thermodynamical values?

Sorry for the questions, but such a situation is very interesting and worth to be investigated.   

Performed adsorption studies at different temperatures and thermodynamic parameters (del H and Del S) were calculated by using Gibb's energy equation.

del G = del H - (T * del S)

For reference follow thermodynamics portion in attached publication

In nature processes always seek the lowest energy state, then when does the adsorption, for its to be spontaneous at the end of the process energy must be less than the initial matter by contacting the adsorbent and the material to be adsorbed, the relationship spontaneity is known in thermodynamics and correlated by the Gibbs energy, the most famous, or entropy; so that its value is negative, because the end is less the initial.

What will happen if the adsorption process is non-spontaneous. means del G value is negative

Yes, is possible that is necessary heating (>40-50°C) or presence of some substances can be non-spontaneous.

As i found from literature, the negative values of ΔG° confirm the feasibility of the process and the spontaneous nature of adsorption. The decrease in the negative value of ΔG° with an increase in temperature indicates that the adsorption process becomes more favorable at higher temperatures.

as per above statement, what will happen if ΔG° value is positive?

 I got this explanation from one literature, please check it once and let me know me whether that statement is true or not. 

positive ΔG° values indicates the non-spontaneous nature of the adsorption process. The positive value of ΔG° also shows that there is an energy barrier. When the ions enter from the solution into the particle surface, some of the water molecules forming hydration shell of ions are stripped off and simultaneously the degree of freedom of ions declines.

With the thermodynamic data on ions in aqueous solution, one has to consider that the standard gibbs free energy of formation of H+(aq) is arbitrarily set to zero, and this affects the values for other ions in solution. 

That  DELG = DELH- T  DEL S    ın your case you have 

DELG = _ ABS DELH +  T  ABSDELS   the sign could be positive or  negative depends on the relative absolute values of DEL  H and TDELS.  But at the equilibrium it should be ZERO. Sınce in your case  DELG> 0 that means absorption and desorption reactions are not in mutual thermodynamic equilibrium.  Rather  the reverse reaction, which is  desorption would be the natural process and that takes place in reality. This way  the  direction of the reaction proceeds along the decreasing path  of the Gibbs free energy (towards the direction of equilibrium).. 

Dear Bas, That wouldn't change any thing since the energy of any kind is  relative quantity, and further more only the differences of energies (i.e., the chemical potential or the electrochemical potentials of charged  species)  enter  into  the calculations for every physico-chemical reactions.

it would be interesting to know how Mr.Nagireddy arrived at his positive DG values. 

fDear Srinu;  This is what you have written 'As per literature's if Gibb's free energy values are negative the adsorption process is spontaneous and if it is positive process is non-spontaneous.'  This a partially wrong statement about if you could recalled the general definition of  Gibbs free energy of reaction or adsorption as follows:

Del G =  SUMNProduct  ChemPot Product  -  SUM NReactant  ChemPotReactant If   it is negative the   Reaction proceed along the forward direction,  if  it is positive proceed in the opposite direction.  Of course both reactions proceed spontaneously because of the positive internal entropy production.  When Del G =0   the reaction takes place in both directions in equal rate with only fluctuation of  the molar densities of product and reactant species, that is called non-equilibrium stationary state according to Prigogine:  

In general one defined the Affinity of a chemical reaction as ; Affinity = -  Del G

In solutions the reference state (standard state) is  c=1 mol/L. What is the reference state in the adsorped phase ??

Srinu, you are confused and your statements are ambiguous letting out the confusion. You did the experiments of adsorption studies of a liquid, say, L, on a solid, say, S. The process may be represented as: S + L = sl. Such a process may either be spontaneous or non spontaneous according to thermodynamics. Since you carried out the experiments of the adsorption process we expect you to get negative values for the ΔG of the process. However, you find you get negative values for ΔH and ΔS and positive values for ΔG. The positive values of ΔG you got , if correct, indicate that adsorption did not occur. But your experimental results, if correct, indicate the process did occur! Both can not be true afterall! You must check your calculations again. Or you must analyse your experimental results again to see what your reactants and products are. Incidentally how do you check for your product, that is, the liquid adsorbed on solid?

You may do some trial experiments with known systems such as nonstick stuff used in making cookware. See if you get positive values for ΔG, You may also consider systems of immissible liquids - This is a physical and not a chemical process - and see what values you get for ΔG. These are only suggestions, if you don' like them, don't worry!

Dear Radhakrishnamurty;  Unfortunately,  without an exception all the thermodynamics calculations done in metallurgical and chemical engineering & sciences rely on the equilibrium thermodynamics, which assumes that for the isobaric and isothermal reactions   Del G= 0 (See: Physical Chemistry of Metals by Darken & Gurry;  The Principles of Chemical Equilibrium by K. Denbigh, The computation of Chemical Equilibria by F. van Zeggeren and  S Storey etc.). Then they try to compute the followings:

DEL G(standard) = products  SUMk (SNk x  Standard ChemPotentK ) -  Reactants  SUMJ (SNJ x standard ChemPotentJ)    = -  RT Ln { ProK Activity CoefficientKSNK /  PropJ  Activity CoeffJ  SNJ }

where  SN  stoichometric numbers.  If he finds positive Gibbs free energy  as you and I stated previously that means this is an unnatural process, which can never take place in reality. Then there is some thing wrong in the calculations.  Of course there are some  short of  dynamics non equilibrium processes such as the lattice cooling by adiabatic nuclear spin demagnetization where there is a heat flow from  low temperature (nuclear spin assemble) to high temperature (lattice phonon). 

I don't understand what Tarik Ömer Oğurtani is saying. Without doubt, ΔGT,P = 0 for any process at equilibrium.

There is a correction in the eqn for ΔG0 he wrote. The set of activity terms must correspond to the set of equilibrium values and not any other set of values!

The last sentence appears to be irrelevant for the discussion of the issue (chemical reactions) at hand. Note that during chemical reactions, nuclei are assumed to remain unchanged. Therefore, bringing in irrelevant aspects such as nuclear spins will only help unnecessarily complicate the discussion and serves no purpose in my opinion.

Dear Radhakrishnamurty, There is no mistake in that expression.  Better you should check your class notes first. No body told you that those terms belong to the non=equilibrium case.

All these arguments belong to the sophomore year thermodynamics. My last remark was intended to show that the limitation of the second law of thermodynamics for the multi dimensional spaces.During the above mentioned dynamic process temporal negative absolute temperature is realized in the nuclear spin space. 

Entropy: A concept that is not a physical quantity

https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity