# Dissolution of sugar in water and its temperature dependence

As far as I can remember, I was taught in 6th grade that warm water would dissolve sugar with greater ease than cool water. In 11th grade, I was taught that Hydrogen Bonding is the reason behind compounds dissolving in water, and was later taught that at higher temperatures hydrogen bonds tend to break. Now, I realize that the two kind-of contradict each other as if sugar were to dissolve easily at a higher temperature, then it should form hydrogen bonds more easily at higher temperatures, however, higher temperature weakens the hydrogen bonding between sugar (say sucrose) and water and that must inhibit dissolution. So what actually is going on?

We can start by watching two videos showing how sugar and salt are dissolved in water: https://www.youtube.com/watch?v=fwjvwoFHTbg and https://www.youtube.com/watch?v=14iv71ZTRTo. Particularly, the second video is fun to watch, where molecular dynamics simulation is used to show how water molecule solvates an ion at the corner of a NaCl particle.

To reconcile the contradiction, you need to consider two points here, one being thermodynamics and the other being kinetics. Thermodynamics is about the change of state functions of a system, whereas kinetics is about the rate of the change with time involved.

You can first consider thermodynamics. Sugar crystals or granules put into water are the initial state and the sugar molecules dissolved in water forming a homogeneous solution are the final state. Since sugar can be dissolved in water, it must be true that the final state is more stable than the initial state, even at elevated temperature. You can do a simple experiment at home to verify this is the case. Using thermodynamics terminologies, $$\Delta G = \Delta H - T\Delta S$$, where G is Gibbs free energy, H is enthalpy, S is entropy, T is temperature, and the symbol $$\Delta$$ means changes, the greater stability of the final state than the initial state means $$\Delta G$$ is negative. This means that the dissolution of sugar in water is thermodynamically favorable. The hydrogen bonding between sugar molecules and water molecules may be weakened with increasing temperature, and therefore $$\Delta H$$ may be less negative at a higher temperature. However, due to the loss of hydrogen bonding the final state could be more disordered, and therefore $$\Delta S$$ may be more positive with increasing temperature, which can contribute to make the $$\Delta G$$ more negative favoring the dissolution of sugar at a higher temperature.

Then you need to also consider kinetics. Your statement that "warm water would dissolve sugar with greater ease than cool water" has to do with kinetics, where water molecules have higher energies to move around to dissolve sugar molecules more quickly from the initial crystal/granule state into the final homogeneous solution state. More specifically, water molecules at the solid/liquid interface carry sugar molecules from the solid surface into the liquid phase, especially from the crystal corners and edges where the chemical potentials of sugar molecules are higher than at basal planes. Additionally, at a higher temperature, water molecules have higher diffusion coefficients to move from the liquid bulk to the solid/liquid interface to dissolve sugar molecules and then the dissolved sugar molecules also have higher diffusion coefficients to move from the solid/liquid interface to the liquid bulk. Generally speaking, at higher temperature molecules have higher energies and move faster to speed up the dissolution process.

Therefore, you need to consider both thermodynamics and kinetics to get a more comprehensive understanding of this problem. This is my simple answer with only some handwaving arguments, which may not be complete and others may provide corrections and/or additional explanations.

• Thank you! Pretty comprehensive, I was previously looking at it from a very narrow perspective. Jan 5 at 16:00
• Your statement " ΔS is positive, which can contribute to make the ΔG more negative favoring the dissolution of sugar at a higher temperature." is incorrect. The entropy change does not contribute in any way to the temperature dependence of the equilibrium. Only the enthalpy change matters. See the derivation of the van't Hoff equation for validation. Jan 5 at 22:50
• I believe it does. Please do not just look at the differential form of the van't Hoff equation. You need to look at the linear form of the van't Hoff equation. If you plot lnK as a function of 1/T, the slope is affected by $\Delta H$ but the intercept is affected by $\Delta S$, so both $\Delta H$ and $\Delta S$ contribute to affect the equilibrium as temperature is changed. Jan 5 at 23:12
• @Josiah_H Andrew is correct. The temperature dependence of the equilibrium constant is determined only by $\Delta H$. It should also be noted that your picture is oversimplified, since both $\Delta H$ and $\Delta S$ are temperature-dependent (though their temperature-dependencies cancel out, so we can ignore them when it comes to determining the temperature dependence of $\ln K_{eq}$). Thus, for the purposes of addressing your picture, let's assume $\Delta H$ and $\Delta S$ are constant. What determnes the temperature dependence is the slope ($\Delta H$), not the intercept ($\Delta S$). Jan 7 at 4:27
• @Josiah_H I.e., think about two different simple linear equations: $y = 3 x + 1$ and $y = 3x +2$. The amount by which $y$ changes when we increase $x$ from, say, $5$ to $6$, is the same for both equations. The difference in intercept between the two only changes the vertical offset. The $x$-dependence (analogous to the $T$-dependence) is the same in both Jan 7 at 4:36

As far as I can remember, I was taught in 6th grade that warm water would dissolve sugar with greater ease than cool water.

That is true. Sometimes, the sugar in honey crystallizes. If you carefully heat up the honey, you can get it back to a liquid. So there is experimental evidence that sugar is more soluble as you heat up the solution.

In 11th grade, I was taught that Hydrogen Bonding is the reason behind compounds dissolving in water, and was later taught that at higher temperatures hydrogen bonds tend to break.

Sometimes, water interacts with solutes through hydrogen bonds (in the case of sugar). Sometimes, it interacts through ion-dipole interactions (in the case of ionic solutes). If you stick with sugar, you should know that sugar molecules in a sugar crystal also interact through hydrogen bonds.

It is true that hydrogen bonds tend to break at higher temperatures because the hydrogen bonds restrict the movement of particles, and particles move "more" at higher temperature. However, because all the molecules interact via hydrogen bonds in the water/sugar system (i.e. water with water, sugar with sugar, water with sugar) this does not let you predict how solubility will change with temperature.

Now, I realize that the two kind-of contradict each other as if sugar were to dissolve easily at a higher temperature, then it should form hydrogen bonds more easily at higher temperatures, however, higher temperature weakens the hydrogen bonding between sugar (say sucrose) and water and that must inhibit dissolution.

There is a misconception there. The hydrogen bonds do not tend to break at higher temperature because they get weaker. They break because the particles move faster (have more kinetic energy).

So what actually is going on?

The dissolution of sugar is an endothermic process (the temperature drops as you mix in the sugar). If you increase the temperature in an endothermic process, you favor the product. Hydrogen-bonded or not, some substances dissolve in an endothermic process and others in an exothermic process. The former will be more soluble in warmer water, while the latter will be less soluble in warmer water. There is no good explanation for this at the level of this answer.

... dissolve sugar with greater ease ...

The answer focuses on solubility. "Ease" might also refer to how fast the sugar dissolves. The higher the temperature the faster the process, and the finer the solid sugar (i.e. confectioners sugar instead of rock candy), the faster the process.