How do different types of mixtures affect phase equilibria?

[PhaseShifter]

The first is related to statistics more than to any particular physical mechanism. Think of how relative standard deviation is related to sample size and consider what that means when considering a system of 1000 atoms vs a system of 100 000 000 000 000 000 000 000 atoms.

The seconds means exactly what it says--if one part of your system is in one phase, and another part is in a different phase, then you generally expect a surface or interface to separate the two phases.

What's so mysterious about having three phases in equilibrium? It simply means that $${\Delta~G_fusion}={\Delta~G_vaporization}=0$$

 

[PhaseShifter]

How can we use this paragraph to explain the equation : F = C − P + 2 ,
and explain how Gibbs reasoned this out?

Keep in mind that's not exactly correct. Chemical equiibrium poses additional constraints, so it's actually F=C-P-R+2.

Let's start out considering the simplest system, which would have one component and only one phase.

The chemical potential is a function of temperature and pressure, regardless of what the equation of state is for that phase.

Now consider what happens when you increase the number of phases: The chemical potential of the substance has to be the same at equilibrium, so we have the constraint $${\mu_a}(T,P)={\mu_b}(T,P)$$, or $$\Delta\mu_{ab}(T,P)={\mu_a}(T,P)-{\mu_b}(T,P)=0$$. We can choose either T or P as an independent variable, but we can only choose one of them--the new constraint makes the other a dependent variable by default.

Now if we make a third phase

Likewise adding a third phase gives a second delta:
$${\mu_a}(T,P)={\mu_c}(T,P)$$ gives us $$\Delta\mu_{ac}(T,P)={\mu_a}(T,P)-{\mu_c}(T,P)=0$$. This gives us a different curve on a T,P graph, which will intersect the first curve at a single point if all three phases coexist at once. (or won't intersect at all if the three phases can't coexist)

At this point you're probably wondering why I don't use a third delta:
$${\mu_b}(T,P)={\mu_c}(T,P)$$ gives us $$\Delta\mu_{bc}(T,P)={\mu_b}(T,P)-{\mu_c}(T,P)=0$$. But notice that $$\Delta\mu_{bc}(T,P)={\mu_b}(T,P)-{\mu_a}(T,P)-{\mu_c}(T,P)+{\mu_a}(T,P)=\Delta\mu_{ac}(T,P)-\Delta\mu_{ab}(T,P)$$--so this third delta is simply a combination of the other two.
Mathematically it will always be zero when both of the others are.

Now let's go back to the one-phase system, but add a second component.
Now we have the chemical potential of each substance as a function of three independent variables: temperature, pressure, and mole fraction of component #1. We could also use the mole fraction of component #2, but $${X_1}+{X_2}=1$$ so only one mole fraction is an independent variable.

 

[Urmi Roy]

Basically,does this mean that since chemical potential is a function of temperature and pressure,we can make any two components have the same chemical potential by fixing them at certain tempertatures and pressures (which must be specific for the pair,no?) and taking another anyone of them,if we equate its chemical potential at a certain temperature and pressure with yet another substance,and if that value of chem. potential is the same as the previous pair's,we get all three of them in equilibrium..right?

According to this,any 2 substances may have the same chemical potential at many values of temperature and pressure (like along the solid-liquid equilibrium curve of water),so we may even have many points at which all three substances stay in equlibrium--all we need to do is match the value of chemical potential of all the components.,after all,we can match up the chemical potentiala os the substances merely by adjusting their temperatures and pressures.

Also,it is not exactly clear why a 2 phase system has a degree of greedom of 2,and how the number degrees of freedom vary with increasing number of phases.

 

[PhaseShifter]

Having two components with the same chemical potential is possible, but not particularly useful or even meaningful.

What is important is having the chemical potential of any given component equal in each phase--the system will shift towards the state of lowest free energy, so a component in a phase where it has a high chemical potential will move to a phase where it has a lower chemical potential. In each phase, the chemical potential of a given component will be a different function of temperature, pressure, and composition.

Also, if you add the chemical potentials of all components multiplied by their stoichiometric coefficients for a given reaction, you get $$\Delta~G$$ for that particular reaction.
Since $$\Delta~G={{\partial G}\over{\partial\xi}}$$, the system will again minimize the Gibbs free energy by reacting until $$\Delta~G=0$$. Because of this, a chemical reaction places an additional constraint just like an extra phase does.

Any time you add a new component to the system, you need an additional degree of freedom to describe the composition of the system.
Any time you add an additional phase or reaction, you need to use one of those degrees of freedom to describe "distance from (phase or chemical) equilibrium", which of course is no longer a degree of freedom under equilibrium conditions, since by definition it must be equal to zero.

 

[Urmi Roy]

Any time you add an additional phase or reaction, you need to use one of those degrees of freedom to describe "distance from (phase or chemical) equilibrium", which of course is no longer a degree of freedom under equilibrium conditions, since by definition it must be equal to zero.

I think I get it now. Whenever I add a new phase to the system,initially,it has its own temperature and pressure and hence chemical potential.However,it has to gradually adjust its chemical potential to attain $$\Delta~G=0$$. Since temperature and pressure of a system are dependant-a change in one will prompt the other to change,the newly added phase must modify any of these factors, thus changing its chemical potential to attain minimum chemical potential condition.Right?

 

[Urmi Roy]

"At alloy compositions and temperatures between the start of solidification and the point at which it becomes fully solid (the eutectic temperature) a mushy mix of either alpha( a mixture of mainly A metal with some less amount of B) or beta (mixture containing mainly B with some small amount of A)will exist as solid lumps with a liquid mixture of A and B. These partially solid regions are marked on the phase diagram."

Alpha and beta are also mixtures, but weren't we discussing that only one metal ,say A,(lead in our original discussion) separates out leaving liquid B (tin) in the mixture?

Do we define the compositions of the alpha or beta states at the eutectic point,or do we consider alpha as pure A,and beta as pure B?

 

[PhaseShifter]

Whenever I add a new phase to the system,initially,it has its own temperature and pressure and hence chemical potential.However,it has to gradually adjust its chemical potential to attain $$\Delta~G=0$$. Since temperature and pressure of a system are dependant-a change in one will prompt the other to change,the newly added phase must modify any of these factors, thus changing its chemical potential to attain minimum chemical potential condition.Right?

Right.

"At alloy compositions and temperatures between the start of solidification and the point at which it becomes fully solid (the eutectic temperature) a mushy mix of either alpha( a mixture of mainly A metal with some less amount of B) or beta (mixture containing mainly B with some small amount of A)will exist as solid lumps with a liquid mixture of A and B. These partially solid regions are marked on the phase diagram."

Alpha and beta are also mixtures, but weren't we discussing that only one metal ,say A,(lead in our original discussion) separates out leaving liquid B (tin) in the mixture?

This is correct above the eutectic temperature, The alpha phase can exist in an equilibrium mixture with the melt and the beta phase can also coexist in a mixture with the melt--but the alpha phase and beta phase can't coexist at equilibrium unless the temperature is lowered below the eutectic temperature.

Below the eutectic temperature, the mixture spontaneously separates into alpha and beta phases, but the melt is no longer thermodynamically stable.

The alpha and beta phases are themselves mixtures of two components, but they are each a homogeneous solid solution of one metal in the other. In the case of lead and tin, you can think of the two solid phases as "lead contaminated by tin" and "tin contaminated with lead".

 

[Urmi Roy]

Cool,it seems I'm getting somewhere at last!

I know its getting kind of long,but I can't help squeezing some things in!

1.When we are cooling the mixture,how can small amounts of lead separate out at a time,since we are lowering the temperature of the entire mixture --the entire lead should freeze all at once,shouldn't it?

2.Why does the tin suddenly freeze all at once at the eutectic point,which is not even the proper temperature for it to do so?

3. Why does lead’s rate of cooling decrease when it has started freezing?

 

[PhaseShifter]

1.When we are cooling the mixture,how can small amounts of lead separate out at a time,since we are lowering the temperature of the entire mixture --the entire lead should freeze all at once,shouldn't it?

When lead crystallizes out of solution, the composition of the remaining melt changes. This means the temperature has to be lowered still further to freeze out more lead, which will shift the melt composition even more...

2.Why does the tin suddenly freeze all at once at the eutectic point,which is not even the proper temperature for it to do so?

But it is the proper temperature for the tin to freeze out. This is a sort of triple point where solid lead, solid tin, and melt can coexist. Below this temperature the melt is no longer thermodynamically stable at any composition. At higher tin concentrations, there will be no solid lead above the eitectic temperature because that is not thermodynamically stable. At higher lead concentrations there will be no solid tin above the eutectic temperature because that phase is not thermodynamically stable.

3. Why does lead’s rate of cooling decrease when it has started freezing?

This is mainly due to the latent heat of fusion.

 

[Urmi Roy]

When lead crystallizes out of solution, the composition of the remaining melt changes. This means the temperature has to be lowered still further to freeze out more lead, which will shift the melt composition even more...

This means that say at a particular lowered temperature, the lead composition is getting prepared to freeze--there may be some parts of it which due to certain factors (unnevenness of heating etc.) may start loosing heat first---as soon as that part has separated out,the composition has changed. This lowers the freezing point and ther same process is repeated--am I okay?

But it is the proper temperature for the tin to freeze out. This is a sort of triple point where solid lead, solid tin, and melt can coexist. ...thermodynamically stable..

To tell you the truth,this still isn't very clear.
I think what it means is that its only at the eutectic point that freezing of tin is thermodynamically favoured...

but there seems to be some blurred aspects of it...1.inspite of the gradual lowering of temperature of the mixture,the tin does not cool at all...2.the eutectic temperature is lower than even the freezing point of tin,so does it in a way influence the freezing point of tin too?...3after the eutectic composition is reached,can we say that the alloy has become homogeneous and the grains of lead and tin are fully incorporated in a single stable structure,accounting for the fact that at this composition,the cooling curve resembles that of a pure,homogeneous substance.

This is mainly due to the latent heat of fusion.

If I'm not wrong,this represents the fact that the heat loss during cooling is partially offset due to the heat evolved during the solidification process.

 

[PhaseShifter]

To tell you the truth,this still isn't very clear.
I think what it means is that its only at the eutectic point that freezing of tin is thermodynamically favoured...

Freezing of tin can be thermodynamically favored above the eutectic temperature as long as the tin concentration is higher than at the eutectic composition. Below the eutectic temperature, both solid lead and solid tin are more stable than the melt, so the melt does not exist.

but there seems to be some blurred aspects of it...1.inspite of the gradual lowering of temperature of the mixture,the tin does not cool at all...

I'm not sure what you're getting at here...at equilibrium the entire system will be the same temperature. The phase diagram only shows the system at equilibrium, so the tin will be no warmer or cooler than any other part of the mixture.

2.the eutectic temperature is lower than even the freezing point of tin,so does it in a way influence the freezing point of tin too?...

As one substance crystallizes out of the melt, the composition of the melt shifts away from the composition of the solid. This gives you one curve for lead crystallizing out of the melt, and another curve for tin crystallizing out of the melt. The eutectic point is where the two curves intersect, and both the lead and tin can coexist with the melt. If you lower the temperature any further, the melt ceases to be stable leaving you with only solid tin and solid lead.

The situation is not unique to metals...a salt solution has a lower freezing point than either pure salt or pure water.

3after the eutectic composition is reached,can we say that the alloy has become homogeneous and the grains of lead and tin are fully incorporated in a single stable structure,accounting for the fact that at this composition,the cooling curve resembles that of a pure,homogeneous substance.

If there are grains of lead and grains of tin, the mixture is not homogeneous.

If I'm not wrong,this represents the fact that the heat loss during cooling is partially offset due to the heat evolved during the solidification process.

This part is correct. When cooling below the eutectic point, the melt will freeze completely, and both the lead and the tin have latent heats of fusion.

 

[Urmi Roy]

As one substance crystallizes out of the melt, the composition of the melt shifts away from the composition of the solid. This gives you one curve for lead crystallizing out of the melt, and another curve for tin crystallizing out of the melt. The eutectic point is where the two curves intersect, and both the lead and tin can coexist with the melt. If you lower the temperature any further, the melt ceases to be stable leaving you with only solid tin and solid lead..

Right,so basically, what happens is that the prescence of either of the metals lowers the freezing point (F.P) of the other. Since the lowering of freezing point depends upon the amount of impurity, the lead gets it F.p lowered to 183 deg. at 38% of tin and tin has its F.p lowered to 183 deg at 62 % of lead. This is dependant on the nature of the metals--how much of the other it requires to lower the temp. Since by 183 deg. tin has solidified,there is no more liquid tin to depress the F.p of lead anymore,(similar argument can be applied to lead),so at that temperature,both the metals have solidified.
Is that okay?

If there are grains of lead and grains of tin, the mixture is not homogeneous..

From what I understand now, below the eutectic point,since both the metals are frozen,they cool individually,so the cooling curve for the mixture resembles that of a pure element.

 

[PhaseShifter]

That seems to be correct.

 

[Urmi Roy]

That's a great relief.

I think I've got my foundations pretty strong,thanks to Phaseshifter,though I'll need some books to read.

I really appreciate the wonderful and more importantly,the continuous help that Phaseshifter has provided.

Thankyou very very much. :)

 

[Urmi Roy]

Hi,I'm sorry to be back with all my questions after almost 2 months,but inspite of all my books and websites, I can't find any complete description of phase diagrams of solid solutions and of peritectic mixtures.
I have my exams next week,so please try and help me within that period...as soon as my exams begin I'll be completely tied up with my other subjects also.

Here are all the doubts I have in regard to solid solution phase diagrams...I have only one more set of questions for peritectic systems...

(In reference to http://csmres.jmu.edu/geollab/Fichter/IgnRx/SolidSol.html )

We consider the phase diagram of Anorthrite and Albite solid solution...I'll refer to albite as Na and anorthrite to Ca, as these are the major metals involved (They've done the same thing on the website also)..

1. Why does calcium freeze so drastically at the beginning (of the freezing process of the liquid melt)?

2. While initially the rate of crystallization of calcium is greater, why does sodium incorporation (into the forming crystal) start increasing as the crystallisation continues?

3.Why will the first crystal formed due to solidification of calcium consist of any sodium at all -sodium has a much lower freezing point?

4.Is the liquidus curve in the albite-anorthrite curve in any way similar to that of the simple two component eutectic mixture (Like lead and tin)?

5. Why are there two curves-solidus and liquidus--(in the diagrams for the simple eutectic systems,the liquid phase and forming cryatal phase were represented on the same diagram)?

6.What does it mean to move down a vertical line crossing the liquidus line, solid + liquid region and the solidus line?

 

[Urmi Roy]

I found that this thread isn't being displayed in the chemistry forum..can anybody see it?

 

[Borek]

Hardly surprising - it is in Other Sciences, not in Chemistry.

 

[PhaseShifter]

1. Why does calcium freeze so drastically at the beginning (of the freezing process of the liquid melt)?

Anorthite has a higher melting point than albite, and will freeze first as the temperature is lowered.

2. While initially the rate of crystallization of calcium is greater, why does sodium incorporation (into the forming crystal) start increasing as the crystallisation continues?

According to the phase diagram, anorthite and albite are soluble in one another. Only one solid phase is formed as the mixture cools.
The freezing point decreases as the sodium ion concentration increases in the solid--therefore in order to be in equilibrium with the melt, the solid must incorporate more sodium as the temperature is lowered.

3.Why will the first crystal formed due to solidification of calcium consist of any sodium at all -sodium has a much lower freezing point?

Again, albite and anorthite don't form two separate solid phases. Rather, they represent the two extremes in possible compositions of a solid solution. If any sodium exists in the melt, it will also exist in the solid.

4.Is the liquidus curve in the albite-anorthrite curve in any way similar to that of the simple two component eutectic mixture (Like lead and tin)?

It's actually much simpler than the eutectic mixture, since there is only one solid phase. (and only two curves here, where the eutectic phase diagram has seven)

In this case you have two compounds that are completely miscible in both the solid and liquid phases, where in the eutectic mixture a miscibility gap exists between two solid phases.

5. Why are there two curves-solidus and liquidus--(in the diagrams for the simple eutectic systems,the liquid phase and forming cryatal phase were represented on the same diagram)?

The area between the liquidus and solidus curves is thermodynamically unstable.
A mixture inside this region will spontaneously separate out into a liquid phase on the liquidus curve) and a solid phase (on the solidus curve). Given the overall composition of the mixture along with the compositions indicated by the liquidus and solidus curves at a given temperature, you can apply the lever rule to determine the relative amounts of solid and melt.

Incidentally, this is related to the technique of recrystallization used for purification in chemistry. Since the liquid phase has a higher-than-average concentration one compound and the solid phase has a higher-than-average concentration of the other compound, you can separate the phases and then cool the liquid (or heat the solid) and repeat until you reach a desired level of purity.

6.What does it mean to move down a vertical line crossing the liquidus line, solid + liquid region and the solidus line?

Above the liquidus curve and below the solidus curve, moving down a vertical line just reduces the temperature of the existing phase.

The interesting part is when the system enters the thermodynamically unstable region. When the mixture cools below the liquidus curve, it begins to form a solid phase.
As the mixture cools through the thermodynamically unstable region, the composition of the solid phase will follow the solidus curve while the composition of the liquid will follow the liquidus curve. The relative amounts of each phase present can be determined by the lever rule.

 

[Urmi Roy]

Thanks PhaseShifter. As you can understand,my exams have already begun,so I'll take some time to go through the post,sometime in between my exams.

Thanks again.

 

[sp105]

How we can learn from the given transition if it is first order or second order phase transition? e.g. paramagnetic to ferromagnetic phase transitions are ...?

 

[Urmi Roy]

Well, I'm encountering something very much related to the eutectic mixtures stuff that I did in freshman year now in 3rd year...phaseshifter did a lot of hard work in explaining what goes on in an eutectic mixture and how we get at the eutectic composition etc etc...but is there any similar theory or explanation relating to eutectoid, peritectic and peritectoid mixtures..? In most texts, I find these terms just explained as simple definitions...can anyone suggest any website where they are explained thoroughly?

Thanks,
Urmi.