This question is a little confusing. According to my teacher, the entropy should decrease, but I think that sugar is crystalline first and later it dissolves, so the randomness increases and so the entropy should also increase. What is the correct answer?
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$\begingroup$ Why do you only consider the sugar in your perspective? $\endgroup$– M.A.R.Jun 12, 2015 at 15:52
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2$\begingroup$ @M.A.Ramezani I think that the entropy of both should increase, not only sugar. $\endgroup$– KartikJun 12, 2015 at 15:52
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3 Answers
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While it may seem that randomness always increases when a crystal is dissolved into a liquid phase, it does not have to be that way.
Concerning sugar, the molecule has a large number of hydroxyl groups and is generally rather large when compared with the water molecule around it — much larger than your average sodium or chloride ions. Every hydroxyl will act as both a hydrogen bond donor and an acceptor, likely to two different water molecules, creating a cage of water that is rather ordered and large again when comparing to the sodium or chloride ion’s solvation cell. This will decrease the disorder (and hence entropy) of the free water so much, that it counteracts the increase in entropy of the crystal dissolving. (The effect is more pronounced with sucrose when compared to glucose due to the molecule’s size and the number of hydroxyls.)
On the other hand, the molecules in a sugar crystal are likely less ordered than the ions of a salt crystal. They will form extensive hydrogen-bonding networks, yes; but overall if you shift one row of molecules relative to its neighbour you won’t necessarily break the crystal as you won’t suddenly have ions of the same charge next to each other. And finally, the single salt crystal will break down into much smaller fragments than the sugar molecules.
So summed up, the effects are:
large solvation cells
crystal breaks down into larger (less disordered) fragments upon solvation
sugar crystals themselves are less ordered than salt crystals
Meaning that there is a lot less entropy to be gained.
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$\begingroup$ So that means enough heat is released to compensate for the loss of entropy ? Also I don't understand why sugar has such a high solubility even though each sugar molecule requires a large cage of water molecules ? $\endgroup$ Jun 13, 2015 at 13:58
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$\begingroup$ @olivieradam666 Well yes, if $\Delta G$ were not negative, sugar wouldn’t dissolve. Since $\Delta S$ seems to be negative, $\Delta H$ must be even more negative to outweigh $- T\Delta S$. The high solubility can maybe be explained by considering very high sugar concentrations more like sugar with dissolved water, but I'm not sure if any accepted theory has been established so far on when substances show high solubilities. $\endgroup$– JanJun 13, 2015 at 14:14
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1$\begingroup$ @olivieradam666 See also this answer of mine, where Nicolau states in the comments that sugar's solubility increases linearly with temperature. Remember that $\Delta H$ is also a function of $T$, so likely the contribution of $\Delta H$ always outweighs $- T \Delta S$. $\endgroup$– JanJun 13, 2015 at 14:18
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2$\begingroup$ Hi. This is wrong; sucrose dissolves endothermically. See: webserver.dmt.upm.es/~isidoro/dat1/… books.google.com.co/… pubs.acs.org/doi/abs/10.1021/ja01164a504 When we observe a linear increase in solubility with temperature, the simple explanation is that the process is entropically favorable! $\endgroup$ Jul 24, 2017 at 20:48
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2$\begingroup$ The accepted answer is clearly lacking in emphasizing that no matter what consideration you might have the total entropy of the whole system will always increase (or very rarely remain the same). All questions of the type "does the entropy increase or decrease" in such or such situation should always be answered with yes, unless it is specified that we are only taking into consideration the entropy of a subsystem of any interaction. The answer omits to mention the heat released increasing the temperature (of either the sugar or a heat bath that it is in contact with). Thus entropy still rises. $\endgroup$– KvotheJul 25, 2017 at 8:29
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The second law of thermodynamics states that a process in an isolated system will be spontaneous only if the entropy of the system increases as a result of the said process. Since sugar spontaneously dissolves in water, the total entropy of water and sugar has to increase when sugar dissolves.
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You're the correct one here.
Entropy will go up when a solid becomes aqueous (or dissolved at all) or when liquids mix. This is because there are an increased number of microstates that are indistinguishable from each other when a solid is dissolved or when liquids mix.
(However, do note that if you dissolve a gas, entropy goes down)
answered Jun 12, 2015 at 15:53
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2$\begingroup$ I agree with @Jan that this is too broad a generalization, and it is simply not true in many cases. See Jan's answer for a more nuanced point of view. In short, OP is not correct here. The effect of the solute on the organization of the solvent was not taken into proper consideration. $\endgroup$– MarcoBJun 12, 2015 at 17:29
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$\begingroup$ The organization of the solvent is not sufficient to overcome the entropy of the dissolving sugar. This can be deduced from one simple fact: sugar dissolves faster when the solution is heated. Ergo, this implies first that the dissolving has an enthalpy >0. Second, the dissolving of sugar is a spontaneous process, especially as the temperature goes up. If entropy was less than zero, this would be impossible. (G = H-TS; stick deltas as appropriate). Ergo, entropy of the process MUST be greater than zero. $\endgroup$ Jun 15, 2015 at 13:11
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$\begingroup$ This is so over generalising that it is wrong. While still acceptable in a rough generalisation, it is by reading your comments mixing thermodynamics and kinetics that force me to think that you have confused ideas. $\endgroup$ Feb 2, 2021 at 10:40