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Author Topic: Differences between (g), (l), (s) and (aq) in molecules when predicting entropy  (Read 606 times)

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Hello Chemical Forums, I've got some homework for the finals and one of the topics that I could never understand is if there is any difference between the three states of matter and aqueous when they are next to a molecule and you have to predict the sign of the entropy change.
One of the exercises say: NaNO3(s) :rarrow: NaNO3(aq)
As you can see, there's no change in the molecule but the state of it (going from solid to aqueous), but other exercises say CH4(g)+2O2(g) :rarrow: CO2(g)+2H2O(l)

Am I overthinking all of this or is there an actual difference when trying to calculate the same molecules but different states of matter?


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Definitely change of state means change in the entropy.

Imagine sublimating iodine, I2(s) :rarrow: I2(g) - which form has a higher entropy?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation,,, PZWT_s1

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