Lucas Lima
Your case is similar to the thermodynamics of the dissolution of hydrocarbons in water. This is due to quantum phenomena in the process. In some cases, it is impossible to formulate a theory of thermodynamics without the concept of a microstate. Microstates are best understood in terms of quantum states S = − Tr ρ log ρ , where Tr -trace, ρ is the density matrix. In linear algebra trace square matrix A, denoted tr(A). You calculated thermodynamics within the framework of classical thermodynamics. It is necessary to take into account quantum phenomena.
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Lucas Lima
ΔG = ΔH - T.ΔS
If I take your enthalpy and entropy values and a temperature of 298K I calculate ΔG = 10.50 kJ which is close to your value of 12.77 (note that entropy is usually given in J, while enthalpy and Gibbs energy are usually given in kJ)
So, if - T.ΔS is more positive than ΔH is negative, then the result for ΔG is positive. The justification is build in in the formula ΔG = ΔH - T.ΔS
Jan Dolfing
You don’t understand Lucas Lima question. Let me explain. If the adsorption process occurs, then it is a spontaneous process. The change in Gibbs energy must have a negative sign. Lucas Lima results are positive. So he is surprised and asks a question.
In addition, ΔG has units of kJ/mol, and ΔS J/mol K.
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