dn i where we have used the Gibbs equation for the Gibbs energy and the definition of chemical potential

Transcription

1 Chem 467 Sulement to Lectures 33 Phase Equilibrium Chemical Potential Revisited We introduced the chemical otential as the conjugate variable to amount. Briefly reviewing, the total Gibbs energy of a system consisting of multile chemical secies and/or multile hases is given by G=GT,, n 1, n 2,... The total differential of the Gibbs energy is written as dg= S dt V i i dn i where we have used the Gibbs equation for the Gibbs energy and the definition of chemical otential i = G n i T,,n j i As we will see in the coming days, the chemical otential is robably the single most owerful quantity we will use in our discussion of thermodynamics. We will use the chemical otential to describe such things as: Single comonent hase transitions Simle mixtures and binary hase diagrams Chemical equilibrium Electrochemistry Other areas we won't discuss in as much detail, but also deend on the chemical otential, include: Searation science and retention Adhesion and wetting henomena Biochemical rocesses, such as enzymes and metabolism Fuel cells and alternative fuel sources Chemical Potential of a Single Comonent System Now, let's consider a single comonent system for a while. We can write the following relation for the total Gibbs energy in terms of the molar Gibbs energy at some articular T and : G=n G If we take the derivative of this exression, we can write dg= G dn or dg dn = G

2 If we look at the definition of the chemical otential, we can see that = G = G n T, For a ure material, the chemical otential is equivalent to the molar Gibbs energy. If we continue with a ure material, we can write We also conclude that T = S d =d G= S dt V = V T Because we have criteria for sontaneity in terms of the Gibbs energy, we can write that a sontaneous rocess is one that reduces the chemical otential. d G T, =d 0 With this information, we are reared to discuss the hase equilibrium of a ure substance. Phase Transitions of a Pure Comonent You are all familiar with hase equilibrium. You have seen ice melt and liquid water boil. You should also be familiar with hase diagrams, that indicate the temerature and ressure conditions under which different hases are stable. A generic hase diagram is given in the figure at right. These tyes of diagrams are generally based on emirical data. We will now develo the underlying thermodynamics that cause these hase transitions to take lace. Let's take a closed system of one comonent at constant T and, but allow for two different hases, which we'll call and. We would then write the differential of the total Gibbs energy as dg T, = dn dn Now, because this is a ure material, any increase in the moles of hase must have an accomanying decrease in the moles of hase. Thus, we can write dn = dn and dg T, = dn

3 Recall our criterion for a sontaneous rocess, that dg T, < 0. This means that if dn 0 then 0 or This tells us that we will have sontaneous conversion from hase to hase if this inequality holds. Thus the thermodynamically stable hase is the one with the lower chemical otential. If the chemical otentials are equal, then dg T, = 0 and the system is at equilibrium. Changes with T The result we just derived is general; at any given temerature and ressure, the stable hase is the one with the lower chemical otential. If the chemical otentials of the two hases are equal, the two hases exist in equilibrium. When we look at a hase diagram, we see a boundary between the two hases. How does that boundary deend on changes in conditions? We will first look at how the equilibrium boundary deends on temerature. We already have the general result T = S Since entroy is always a ositive quantity, the chemical otential must decrease as the temerature increases. Let's consider the chemical otential of the solid, liquid and gas hases of a ure substance. We know that the entroy of a condensed hase is less than for the gaseous hase, and the entroy of a solid is less than a liquid. Or, mathematically S gas > S liq > S solid Therefore, the chemical otential of a gas will dro more raidly than that of a liquid, which dros faster than the chemical otential of a solid. Assuming that the entroy of the individual hase is constant with resect to T, we would lot the temerature deendence of the chemical otential (or the Gibbs energy) as seen in the figure. According to the figure, what haens in terms of hase transitions as the temerature increases? How would this lot be different for CO 2 at 1 atm?

4 The temerature at which the transition occurs from liquid to solid is known at the melting oint. If the ressure is our standard ressure, then this is the standard melting oint. Likewise, the temerature where a liquid transitions to a gas at the standard ressure is the standard boiling oint. It is ossible, deending on the chemical otentials of the different secies, to go directly from solid to vaor, and if that haens, the temerature at which it occurs is the sublimation oint. Resonse of Melting to Alied Pressure Let's consider what haens to our melting oint as we change the ressure. We know that at a given temerature, the change in chemical otential with changes in ressure is given by = V T Since the molar volume is always a ositive quantity, the chemical otential will increase with increasing ressure. If we do this at each temerature, we will see our line of µ vs T move to higher µ. To a first aroximation, S won't change much with these changes in ressure, so the sloe doesn't change. But, we also have to consider the fact that the molar volume of the different hases can be different. If the molar volume of the liquid is greater than that of the solid, or the density of the liquid is less than the solid, the liquid line will move u more than the solid line. This means the temerature at which they intersect, the melting oint, will increase with increasing ressure. However, for water we have the reverse case. The density of the solid is less than the density of the liquid, meaning the molar volume of the solid is greater than that of the liquid and so the solid curve will move u more than the liquid curve. This means that the melting oint will decrease as the ressure increases.

5 Resonse of Boiling to Alied Pressure Using the same logic as we did for melting, what haens to the boiling oint as the ressure changes? Phase Boundaries Now that we have qualitatively discussed how hase transitions deend on the temerature and ressure, we can develo a more quantitative treatment. Let's consider the case of two hases, and resent in equilibrium at some articular T and. In order to be at equilibrium, the chemical otential of the two hases must be equal., T =,T It is also necessary that T, and dµ be the same for both hases if we are staying on the hase boundary. We can write for each hase This let's us write d = S dt V S dt V = S dt V We can collect the ressure and temerature terms and write V V = S S dt We now define the entroy and volume changes of the hase transition as This let's us write trs S = S S dt = trs S trs V trs V = V V This is known as the Claeyron equation. This exression is exact and alies to any hase boundary of a ure substance. We have not made any assumtions in deriving this relation. Kee in mind that we are allowing temerature and ressure to vary and in rincile the volume and the entroy of the

6 transition will vary with the conditions. Assuming we know these quantities, this exression gives us the sloe of the hase boundary at that oint. Now let's look a little more at this equation for articular cases. Solid-Liquid Phase Boundary A melting (fusion) rocess is accomanied by an enthaly change fus H at any articular temerature. If we know this quantity er mole of material, we know the heat at the ressure of the hase transition. Thus we can write for the entroy trs S = q trs T = trs H T We can ut this into our exression to write dt = trs H T trs V If we have data for the enthaly and the volume of the transition as a function of temerature, we can determine a formula for the hase boundary, which will be a curve on our -T lot. T = T fus H T fus V dt fus H T dt T fus V T Here we have assumed that the enthaly and volume changes are constant over our temerature range. We then write the aroximate hase boundary as fus H fus V ln T T If the temerature range is narrow, we can aroximate the ln term as This let's us write ln T T =ln T T 1 T T T T fus H T fus V T T So if we start at some temerature T *, the hase boundary is a stee straight line on a -T lot. What about the sloe of this line? fus H > 0 is always true; it always takes heat to make a solid melt. However, fus V can be ositive or negative, deending on the relative density of the liquid and solid. For most materials, the solid is more dense than the liquid, so the volume increases uon melting. Thus the -T line will have a ositive sloe. With water, however, the liquid density is greater than that of the solid, so this line has a negative sloe. The comletely agrees with our result from above, as it should.

7 Liquid-Vaor Boundary Let's go back to our earlier statement of the Claeyron equation, but now write it for a liquid-vaor hase boundary. dt = va H T va V Because the molar volume of a gas is much larger than that of the liquid, we can treat the volume change as simly the volume of the gas; va V V g dt = va H T RT / or 1. If we then treat the gas as ideal, we can write dt = d ln dt = va H RT 2 This is known as the Clausius Claeyron equation and is an aroximation for the liquid-vaor hase boundary. How would we get a better exression for this hase boundary? As we did before, we can determine a formula for the hase boundary by integrating this exression ln ln d ln =ln = va H T dt R T T = va H 2 R 1 T 1 ln = va H R 1 T 1 T = va H T T R T T This is only an aroximation, based on the assumtion that we can treat va H as a constant. If we want to evaluate this boundary over a wider temerature range, we have to account for the temerature deendence on the enthaly of vaorization. It should be noted that va H aroaches zero as T aroaches the critical temerature. Thus, there is no hase boundary above T c. T Solid-Vaor Phase Boundary The Clausius Claeyron equation also alies to the boundary between the solid and vaor hases, again because the volume of the vaor is much greater than the volume of the solid. Also, recall that sub H = fusı H + va H, so we would write ln = subl H T T R T T and the solid-vaor line should be steeer than the liquid-vaor line at similar temeratures. Note on standard states It should be noted that all ressures should be defined relative to our standard ressure. This makes sense from a consideration of the units of our exressions. In deriving the Clausius Claeyron equation, we switched from to ln, but you may be concerned about the units when we do this. Because we are writing a differential equation, we can always add a constant factor and not change the derivative. Thus, in the ste when we used

8 we should technically write 1 =d ln 1 =d ln d ln o =d ln o where is our standard ressure. Because is a fixed oint, its derivative is zero, so we can add it, or the derivative of its ln, without changing our equation. Writing the exression in this way kees our units constant. Because is usually taken to be 1 bar, it is often droed in the exression of the Clausius Claeyron equation. When we integrate the equation, the standard state is removed from the exression. So, the way we wrote it works and is a simler notation, but you need to kee in mind that all ressures are with resect to the standard state ressure. Vaor Pressure and Liquid/Gas Coexistence We know from our own observations that when we have a samle of liquid there exists some amount of vaor of that material. For examle, if we have a closed container of a articular liquid and we remove all the gas from it, there will be some ressure still in the container, from the vaor of the substance. This amount of ressure is known as the vaor ressure. We can use the chemical otential to determine the vaor ressure of a ure material at conditions other than the equilibrium hase boundary. We define the standard boiling oint as the temerature at which a liquid boils at 1 bar ressure. If we know these oints, we can aroximately determine the vaor ressure at any other temerature by evaluating the Clausius Claeyron equation for the temerature of interest. An alternate form of this equation is d ln d 1/T = va H R (You will derive this form of the equation in the homework.) This equation tells us that the ln of the vaor ressure should decrease linearly with the recirocal temerature. This is what is exerimentally observed sufficiently far from the critical oint. As with many roerties, vaor ressures for many liquids are tabulated. Degrees of Freedom and the Trile Point For a ure material, keeing amount fixed constant, we have three variables, or degrees of freedom, T, and V m. However, because of the equation of state, we can only indeendently vary two of them. If we require that we kee two hases in equilibrium, that brings in an additional constraint. Now, we can

9 only vary one arameter, such as the ressure. The temerature is fixed by the Claeyron equation and the volume is fixed by the equation of state. Each constraint removes a degree of freedom. If we want to have three hases in equilibrium with each other, that adds another constraint, meaning we can't change any of the state variables. This is why hase diagrams have trile oints, which are invariant. This is also why you can't have a oint in a single comonent hase diagram where more than three hase are in equilibrium; that would require negative degrees of freedom, which doesn't make hysical sense.