ASPEN software for calculating the heat of solution of NaOH

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Recently, in a process calculation, the reaction heat involving NaOH came into play. In order to determine the dissolution heat and reaction heat of NaOH and to provide more detailed data for subsequent calculations related to the process equipment, the dissolution heat of NaOH was calculated first: the software (V7.2) was launched, the appropriate substance was selected, electrolysis was chosen as the method, ELECNRTL was selected as the equation set, the necessary components were entered, and the calculation was carried out. OH, NO!! It’s quite different from what’s stated in the literature (where 1 mol of NaOH is dissolved in 10 mol of water):'( Literature mapping: (from Haichuan), (first calculation value) The calculations showed that NaOH did not ionize; it merely formed hydrates, and the heat released was only about half of the dissolution heat (at this point, I realized I had returned the physical chemistry materials to the teacher). And I don’t know how to adjust it. Then I began to go back and review the various data provided by ASPEN, trying to understand that the difference between the standard enthalpy of formation of a solid and the enthalpy of formation at infinite dilution should also be the heat of solution of NaOH. The difference was calculated to be 47 KJ/mol, but the question arose as to how this value should be incorporated into the unit calculations. After conducting two more reaction simulations to calculate the dissolution of NaOH, it was found that the resulting value was very close to the values reported in the literature, and complete ionization also occurred. At this point, I believe it is now possible to roughly simulate and calculate the heat of solution of NaOH. There may be misunderstandings during this process; please **correct me if necessary, or let me know if there is a simpler way to do it.**