ΔG = ΔG°+ RT ln K
How to derive this equation ? I searched it a lot but couldn't find a clear explanation.
The equation, ΔG = ΔG°+ RT ln Q, is derived on Wikipedia, under the subsection Thermodynamics.
https://en.wikipedia.org/wiki/Chemical_equilibrium
Remember, Q is the reaction quotient, which at equilibrium is equal to the equilibrium constant, K. Then you have your equation ΔG = ΔG°+ RT ln K. Since, at equilibrum, ΔG = 0, the expression can be rearranged to the familiar ΔG°= -RT ln K.
The calculation of thermodynamic parameters is wrong. Equilibrium constants Kd= qe/Ce or KC= (Co-Ce)/(Co), or Kd= (V/m)*(Co-Ce)/Co, or Kc= Cs/Ce., or K= Ln(qe/Ce), or Kd= qexdensity/ce. Although there are several papers reported in the literature using these relationships as equilibrium constant for obtaining the thermodynamic parameters such as: enthalpy changes, entropy changes, free Gibbs energy, the thermodynamic parameter obtained by these Kc or KD are not correct. Please read the following papers: Separation and Purification Technology 61 (2008) 229-242; Biochemical Engineering Journal 35 (2007) 174-182; J. Chem. Eng. Data 2009, 54, 1981-1985; Journal of Industrial and Engineering Chemistry, 70 (2019) 346-354.
Hong-Tao Fan , please, could you elaborate a bit more your statement?
The question was about a basic relationship between the standard Gibbs energy change, dGo, and the equilibrium constant, K.
However, you just say that "the calculation of thermodynamic parameters is wrong", indicate some equations, and list several papers regarding adsorption thermodynamics...
Dear colleague
I can send you a few pages regarding equilibrium constants in correlation with the form of chemical potential for different kinds of systems, from my textbook : St. Perişanu, "General Chemistry", Ed. Politehnica Press, Bucuresti, 2004.
I need your e-mail address therefore.
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