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Spontaneity of a solvation process question. Solvation of sodium decahydrate.?
So I did an experiment of solvation of sodium decahydrate. It is an endothermic process, but at the same time, the process is spontaneous. How could it be? What is the reason behind this?
When I try to calculate the change of entropy, I got a negative value, which suggest that it was not a spontaneous process. I used the delta S = Cv ln (T2/T1), where T2 < T1.
How can I explain it then? Thanks in advance
My bad. It is sodium sulfate decahydrate. I missed the sulfate bit lol. Thanks
1 Answer
- hfshawLv 79 years agoFavorite Answer
There is no such thing as "sodium decahydrate". I assume you are using a decahydrate salt of sodium in which the anion is carbonate, sulfate, borate, or some other anion. Let's assume it's sodium sulfate decahydrate (Na2 SO4 * 10H2O).
Your calculation of the entropy change only takes into account the change in entropy due to the change in temperature of the solution, and does not account for the entropy change of the dissolution reaction:
Na2SO4*10H2O(s) -> 2Na+(aq) + SO4--(aq) + 10H2O(liq)
The entropy change for this reaction, which produces 10 moles of liquid water for every mole of solid reactant should have a large positive value. This will dominate over what I assume was a rather small negative entropy change due to the drop in temperature.
