Hello Mujahid,

One of the advantages of RGibbs reactor in Aspen is that you do not need to know the reaction stoichiometry, only the chemical components present in the process. The RGibbs is capable to determine the outlet composition assuming the reacting system reached the equilibrium.

For more information about how to calculate Gibbs free energy I recommend you check this text: Chemical thermodynamicss for process simulation by Jurgen Gmehling et al. Specifically section 12.2 (Chemical equilibrium). In regard to RGibbs reactor in Aspen you can check Process analysis and simulation in Chemical Engineering by Ivan Dario Gil Chaves et al. (see Chapter 5, Chemical Reactors).

Hope it can help you.

Best regards,

Junior

Junior Lorenzo Llanes

Thankyou very much. i have simulated a few chemicals however in case of DBT it says that it cannot evaluate the properties DHFORM which is the enthalpy of formation of that compound. the chemical DBT is user defined and i have input its structure. moreover i was trying to manually input enthalpy of formation and gibbs free energy but i am unable to calculate either.

thankyou for the books. ill review them.

Thanks again.

You can get an estimate of the enthalpy of formation using Benson’s method. Several applications implement this method including CHETAH available from ASTM.

The so-called "entropy " doesn't exist at all.

During the process of deriving the so-called entropy, in fact, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.

The so-called entropy was such a concept that was derived by mistake in history.

It is well known that calculus has a definition,

any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.

Based on the definition of calculus, we know:

to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.

As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,

∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.

1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.

2) On the other hand, In fact, Q=f(P, V, T), then

∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).

We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.

that is, the so-called "entropy " doesn't exist at all.

.

.

.

Why did the wrong "entropy" appear ?

In summary , this was due to the following two reasons:

1) Physically, people didn't know Q=f(P, V, T).

2) Mathematically, people didn't know AΔB couldn‘t become AdB directely .

If people knew any one of them, the mistake of entropy would not happen in history.

Please read my paper and those answers of the questions related to my paper in my Projects.

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

1. Logic of the Second Law of Thermodynamics: Subjectivism, Logical Jump, Interdisciplinary Argumentation.

2. New thermodynamics pursues universality, two theoretical cornerstones:

2.1 Boltzmann formula: ro=A*exp(-Mgh/RT) - Isotope centrifugal separation experiments show that it is suitable for gases and liquids.

2.2. Hydrostatic equilibrium: applicable to gases and liquids.

3. The second and third sonic virial coefficients of R143a derived from the new thermodynamics are in agreement with the experimental results.

3.1. The third velocity Virial coefficient derived is in agreement with the experimental data, which shows that the theory is still correct when the critical density is reached.

4. See Appendix Pictures and Documents for details.