Freundlich adsorption isotherm model fits well with the experimental data for CO2 adsorption on activated carbon (AC). Can I evaluate the thermodynamic parameters by considering ln K = n (“n” = Freundlich constant) using following expressions? If so, how can it be justified? Could I get few references w r t SOLID-GAS system?
deltaG0 = -RT n
n = -(deltaH0/RT) + (deltaS0/R)
Using the plot of n vs. 1/T , deltaH0 and deltaS0 are estimated from the slope and intercept, respectively.
free energy is related to the concentration of you adsorbed on solid phase and the left over concentration in the liquid. Free energy is simply the energy you arrested of a moving particle in liquid phase on to a solid phase. you decrease the entropy of the system by doing this. So it is just -RTln(equilibirum constant), equilibrium constant will be equivalent K of Freundlich when n = 1. or go for Langmuir isotherm
@ Vasanth Kumar; Thank you for swift response. The equilibrium constant will be equivalent K of Freundlich when n = 1.
But, KF has units ((mmol g-1) (kPa-1)1/n) for F-equation
q (mmol g-1) = KF (PCO2)1/n.
Hence, ln(KF) is irrelevant. Further, I found that "n" is temperature dependant and equal to 3.8, 4.3 and 5.0 at T= 303 K, 323 K, and 348 K, respectively. So, "T" becomes hopeless at n=1.
Bell and Tsezos have considered deltaG = - n x RT (n is the Freundlich isotherm constant, file attached, equation 14) for the Solid-Liquid system.
For a Solid-Gas system, Singh and Kumar have estimated deltaH and delat S from the slope and intercept by plotting ln (b) versus inverse temperature (1/T) using Eq. (9) [File attached].
ln b = (deltaS/R) - (deltaH/RT)
Here again, the same problem, Langmuir parameter "b" has a unit (kPa).
So, I just started wondering about using Freundlich "n" to evaluate all three parameters by considering both equations 14 and 9 as I have shown above in the question.
Just use mol/L to calculate free energy as the K which you get from isotherms correspond to K * molar concentration of standard solution, which we assume as 1 L/mol. So the Free energy always is expressed as work done/mol.
About kPa, you can convert easily in terms of moles in unit volume or to 'number denisty' from the gas density.
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