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I am a high school student and I am a little confused in the concept of spontaneity of a reaction and how equilibrium works for a reaction, I got some confusions:

  1. Let's take example of freezing of water: In my textbook its written that freezing point of water at 1 atm pressure is 0 degrees and at this temperature water and ice remains in "dynamic equilibrium". So it means this phase transition from solid to liquid is a "reversible process" and we know at equilibrium the "change in Gibbs free energy is 0". Now if I heat it "∆G would become negative" and what happens is more ice melts but the temperature doesn't changes and stopped heating so a new equilibrium got established, so does it mean here equilibrium constant is changing(as more water is forming and more ice is melting) without increase in temperature? But I used to think it only depends on temperature. So what's going on. please explain it in simple language so that a high school student can understand?

  2. Suppose I have ice at room temperature in a closed beaker. What we observe in our daily life is whole ice melts into water but shouldn't the "equilibrium" still exist because its a reversible reaction ,so just like evaporation in a closed beaker the water reaches an equilibrium state with its vapors at all temperatures, ice should also do the same with water but it doesn't. Why?

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  • $\begingroup$ The amount of ice or water is not important for equilibrium constant. What is important is water activity in both phases is equal. In a very informal layman way, activity is "how much water molecules announces they are there, to do some fun actions". This activity is for pure water or ice at constant T independents on amount. $\endgroup$
    – Poutnik
    Jul 15, 2022 at 12:35
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    $\begingroup$ Room temperature is at 293 K,fyi. $\endgroup$ Jul 15, 2022 at 21:38
  • $\begingroup$ @BruhMoments This is not entirely correct. $\endgroup$
    – andselisk
    Jul 19, 2022 at 13:39
  • $\begingroup$ @andselisk average,I should say. $\endgroup$ Jul 21, 2022 at 2:23

3 Answers 3

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To write down the equilibrium constant expression, you use concentrations, not volumes or mass. To illustrate this with an example you can check in your kitchen, a saturated salt solution does not get saltier by adding more salt (you have pure undissolved solid salt in either case). It does not get less salty if you remove part of the undissolved salt.

The equilibrium between liquid and solid water is a bit special because both species are pure. Adding more ice or adding more water will not change the "concentrations" or the freezing point of water.

  1. [...] more ice melts but the temperature doesn't changes and stopped heating so a new equilibrium got established [...]

No, you still have pure water and pure ice, so the equilibrium constant is still the same, and the concentrations are still the same - it is not a new equilibrium. On the other hand, if you pour salt into the water, the liquid water concentration drops (no longer pure) and the solid water concentration remains constant, so then you are out of equilibrium, and ice will melt and the temperature will decrease. With pure water, when the temperature is higher or lower than the freezing point, ice will melt or water will freeze until we are at the freezing temperature (or we run out of ice or water).

[... from OP's comments] how the temperature will decrease??? suppose if we have ice and water at 100 degrees and I add ideal solute to the water ,the concentration of water would decrease and it will only decrease the no. of water molecules going back to the solid state but the no of solid molecules going to liquid state remains the same so equilibrium will shift forward but here also "the equilibrium constant" should remain the same if we think this way.... but it doesn't if we think of its formula why?

In ice, water has 4 hydrogen bonds. In liquid water, it has less. So transferring a water molecule from ice to water is an endothermic process, and it will cool the system down (thermal energy turns into potential energy).

  1. [...] so just like evaporation in a closed beaker the water reaches an equilibrium state with its vapors at all temperatures, ice should also do the same with water but it doesn't.

The difference between pure liquid water and a gas mixture of water and air ("humid" air) is that the water in humid air is not pure. You can have air with higher or lower humidity. The water in a half-full closed bottle will evaporate until the humidity reaches its equilibrium value. On the other hand, liquid and solid water are both pure, and are at equilibrium (at normal pressure) only if the temperature is equal to the normal freezing point.

What we observe in our daily life is whole ice melts into water but shouldn't the "equilibrium" still exist because its a reversible reaction

Because the ambient temperature in our lives is usually higher than the freezing point of water, and because our containers are not perfectly insulated, we keep transferring heat to the container. By the way, even when ice melts, at the molecular level, some liquid water molecules still attach to the solid. The process of water molecules going from solid to liquid just happens to be faster.

[from OP's comment] does that mean the equilibrium between ice and water exists at all temperatures?? so why do textbooks says and also the phase diagrams says that it exists "only" at freezing point

If water and ice come in contact, there will be a forward and a reverse process (freezing and melting) at the particular level. When the two phases have the same temperature and these two process have the same rate, it is called equilibrium. Because the rates are the same at equilibrium, there is no net change, i.e. the amount of ice will neither increase nor decrease. The temperature where this happens is called the freezing point or melting point, even though at this point, there is not bulk freezing or melting.

[...from OP's comments] can u please explain"dynamically" what's going on when u add "ideal solute" to the water ?? I understand that the concentration of water would decrease and more ice will melt....but if we will think how the activity of the molecules actually affected on adding solute things get difficult? like if i add an ideal solute to the liquid it will not change the bonding because its ideal so what I think is the no. of water molecules going back to solid state should still remain the same??

An ideal solute will not interact different with water than a water molecule, but it can't become part of the ice lattice (it does not fit). So instead of - say - 100 water molecules bumping into the ice surface, now 99 water molecules and one other molecule bump into it, decreasing the rate of attachment of water molecules to the ice.

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  • $\begingroup$ can u please explain"dynamically" what's going on when u add "ideal solute" to the water ?? I understand that the concentration of water would decrease and more ice will melt....but if we will think how the activity of the molecules actually affected on adding solute things get difficult? like if i add an ideal solute to the liquid it will not change the bonding because its ideal so what I think is the no. of water molecules going back to sloid state should still remain the same?? $\endgroup$ Jul 19, 2022 at 6:16
  • $\begingroup$ "Because the ambient temperature in our lives is usually higher than the freezing point of water, and because our containers are not perfectly insulated, we keep transferring heat to the container." does that mean the equilibrium between ice and water exists at all temperatures?? so why do textbooks says and also the phase diagrams says that it exists "only" at freezing point $\endgroup$ Jul 19, 2022 at 6:18
  • $\begingroup$ "so then you are out of equilibrium, and ice will melt and the temperature will decrease. " how the temperature will decrease??? suppose if we have ice and water at 100 degrees and I add ideal solute to the water ,the concetration of water would decrease and it will only decrease the no. of water molecules going back to the solid state but the no of solid molecules going to liquid state remains the same so equilibrium will shift forward but here also "the equilibrium constant" should remain the same if we think this way.... but it doesn't if we think of its formula why? $\endgroup$ Jul 19, 2022 at 6:24
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For the first question, consider a simple reaction for example take Haber's process in which nitrogen and hydrogen combine to form ammonia. Now, at a particular temperature (T), there will be some equilibrium constant and correspondingly, a particular value of $ ∆{{G}_{0}} $. What it simply means is that if you have a mixture of hydrogen, nitrogen and ammonia at a particular temperature, then equilibrium will only be achieved at this specific composition. Some other composition is also possible at the same temperature but it will simply not be at the same temperature but it will not be at equilibrium and will have to tendency to go towards equilibrium and this rate of movement towards equilibrium will depend upon other reaction conditions. So basically in your example,when you provide heat to the mixture, the temperature of mixture suddenly increases leading to change in value of $ ∆{{G}_{0}} $. As a result, melting of ice occurs which decreases temperature. This melting of ice happens till the temperature of mixture reaches the freezing point of water. Talking about equilibrium constant, it is not simple to describe since in high school,liquids and solids have no role in reaction quotient. But gibbs free energy change is the important thing. In real life, the freezing of water or melting of ice is so fast that it almost seems as if the externally supplied heat almost immediately was converted into latent heat

For your second question, as I said, equilibrium constant is a bit difficult to describe but it will definitely be something. So basically, some amount of ice must be left at room temperature but the equilibrium constant is extremely shifted to the right because of which it is in an extremely small amount not percievable practically. (It might be so less that it is less than even a few molecules, but since matter is quantised, it would not be possible to have fractional number of molecules.)

Even when you were talking about water- water vapour equilibrium, if you have taken very small amount of water in very large container, then there will be negligible water left as liquid at equilibrium.

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First some key points that are important to keep in mind when thinking about thermodynamic equilibrium: (1) thermodynamic equilibrium states describe systems that are unchanging, static; (2) an equilibrium constant is relevant only when a system can exist in either of multiple states (as you suggest); (3) the equilibrium constant for the melting of water is just that, a constant equal to the ratio (equal to 1) of the activities of water and ice at the melting point (activities are approximated by concentrations in some cases).

What you are describing (melting of ice) occurs when the system is pushed out of its original equilibrium state by adding heat at a finite rate. Melting, in other words, is not an equilibrium state, it is a transition between different states occurring when you perturb the initial equilibrium state by adding heat.

To explain how melting affects the thermodynamic equilibrium between ice and water, consider the following process: (1) you start with ice and water at equilibrium at the melting point (surroundings and the system are at that point); (2) you isolate the system (by wrapping it with insulation) and transfer some energy into the system as heat; (3) the system returns to equilibrium by redistributing the input energy and forming a new state; (4) you remove the insulation and place the system into contact again with surroundings (at the melting point). Importantly, during steps 2 and 3 the system is not in an equilibrium state. In steps 2/3, if not too much heat is added (so as to melt all of the ice), then a new equilibrium can arise between water and ice, where both are at the melting point. If too much heat is added, then the transition of ice to a liquid will be complete. With sufficient heat the temperature of the system will rise above the melting temperature, with ice no longer present (but then at step 4 the system will be cooled again).

To help wrap your head around the idea that the free energy change after melting some of the ice remains constant ($\Delta G = 0$, $K=1$), consider what would happen to the proportions of ice and water if you should mix together two different samples of ice and water, or simply add water, or ice, to a sample of ice and water, all at the melting point. Nothing! The proportions would remain constant. The point is that at equilibrium (at the melting point) there is no force driving formation of more ice or water.

Superheated ice1 (ice at a temperature above the melting point) is not an equilibrium state relative to water at the same T, because above the freezing point liquid water has a lower free energy than ice. You can in fact create superheated ice, but it will spontaneously melt to water if you remove the constraints that allowed you to create this state.

In the case of a beaker (a diathermal container, allowing heat transfer), there is no equilibrium because heat is constantly being transferred into the container, assuming the surroundings are above the melting point (although if heat is added slowly enough then the system is approximately at equilibrium at any given point). The way of thinking around this is to consider the Gibbs free energy at the initial and final equilibrium states of the system, before and after changes due to addition of heat.

References

  1. Schubert, G. and Lingenfelter, R.E., Superheated Ice Formed by the Freezing of Superheated Water. Science 1970, 168 (3930) pp. 469-470, DOI: 10.1126/science.168.3930.4.
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