# Change in enthalpy and entropy when sugar dissolves in water

We know that dissolution of sugar in water is a spontaneous process.

So, change in Gibbs free energy $$(ΔG)$$ must be negative for the overall process:

$$ΔG = ΔH - TΔS < 0$$

Hence either the enthalpy or entropy must drive the reaction.

Now, after searching for the values, most websites suggest the process is endothermic (i.e. $$ΔH > 0).$$

But Jan's answer has me convinced that the overall entropy of a system should decrease (i.e. increase in entropy of sugar molecules is outweighed by decrease in entropy of water molecules).

So, what's the real reason for reaction to be spontaneous if neither entropy and enthalpy are driving the reaction?

• This doesn't answer your question, but serves as an important point of information: You can have a reaction where the inherent enthalpy and entropy are both unfavorable, yet it will still proceed spontaneously if you are away from equilibrium, because of the entropy of mixing. See: chemistry.stackexchange.com/questions/129589/… Feb 2, 2021 at 7:25
• How can you say that neither energy nor enthalpy drive the reaction? Feb 2, 2021 at 8:40
• @ultralegend5385 For a reaction to be driven in a direction, ΔG must be negative in that direction. For ΔG to be negative, either ΔH must be -ve or ΔS must be +ve. (Neither of which is happening in this situation) Feb 2, 2021 at 8:52
• I think that no matter what, since the process is spontaneous, $\Delta G<0$ . As you said, $\Delta H>0$ is true for certain. So, that implies $\Delta S$ must be positive. I think that @Jan's reasoning has some flaws. Read the comments and other answers in the link that you mentioned, they do point out that it is wrong.
– V.G
Feb 2, 2021 at 9:30
• What about the entropy change of mixing? Feb 2, 2021 at 12:32

The entropy change for ideal mixing is $$\Delta S=-R(x_A\ln{x_A}+x_B\ln{x_B})$$