For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

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## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Properties of the Helmholtz & Gibbs Energies

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Properties of the Helmholtz & Gibbs Energies
- Equating the Differential Coefficients
- An Increase in T; a Decrease in A
- An Increase in V; a Decrease in A
- We Do the Same Thing for G
- Increase in T; Decrease in G
- Increase in P; Decrease in G
- Gibbs Energy of a Pure Substance at a Constant Temperature from 1 atm to any Other Pressure.
- If the Substance is a Liquid or a Solid, then Volume can be Treated as a Constant
- For an Ideal Gas
- Special Note
- Temperature Dependence of Gibbs Energy

- Intro 0:00
- Properties of the Helmholtz & Gibbs Energies 0:10
- Equating the Differential Coefficients
- An Increase in T; a Decrease in A
- An Increase in V; a Decrease in A
- We Do the Same Thing for G
- Increase in T; Decrease in G
- Increase in P; Decrease in G
- Gibbs Energy of a Pure Substance at a Constant Temperature from 1 atm to any Other Pressure.
- If the Substance is a Liquid or a Solid, then Volume can be Treated as a Constant
- For an Ideal Gas
- Special Note
- Temperature Dependence of Gibbs Energy 27:02
- Temperature Dependence of Gibbs Energy #1
- Temperature Dependence of Gibbs Energy #2
- Temperature Dependence of Gibbs Energy #3
- Temperature Dependence of Gibbs Energy #4

### Physical Chemistry Online Course

### Transcription: Properties of the Helmholtz & Gibbs Energies

*Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.*0000

*Today, we are going to talk about the properties of the Helmholtz and Gibbs energies.*0004

*Let us jump right on in.*0009

*For the Helmholtz energy, one of the fundamental equations that we have is the following.*0013

*We have DA = - S DT – P DV.*0018

*In this particular case, the temperature and the volume are the natural variables for the Helmholtz energy.*0030

*Essentially, what we are saying is that the Helmholtz energy is a function of temperature and volume.*0043

*We can go ahead and express the total differential this way.*0052

*DA is the general total differential, it is going to be DA DT we already know this, holding the other variable constant not pressure,*0058

*this is volume, DT + the partial derivative with respect to the other variable DA DV holding that one constant DV.*0069

*This is a general total differential.*0080

*This is the one that we actually extracted from the fundamental equation of thermodynamics.*0082

*This is one of the 4 fundamental equations of thermodynamics.*0087

*If we equate this and this, and this and this, here is what we get.*0090

*Equating the differential coefficients, we get DA DT with respect to V = - S and DA DV holding T constant = -P.*0100

*Basically, what this says is that for every unit change in temperature,*0134

*a change in the Helmholtz energy of that system is going to equal negative of the entropy.*0140

*What these negative signs tell us, this is a rate of change, the rate at which*0146

*the Helmholtz energy changes as you change the temperature is going to be a negative of the entropy at that moment.*0151

*That is all this is saying.*0161

*If we make a change in the volume of the system, the Helmholtz energy of that system*0163

*is going to equal the negative of the pressure of the system at that point.*0169

*We are holding V constant, here we are holding T constant.*0175

*Because of these negative signs, this is what it means.*0180

*Because of the negative signs these equations they say the following.*0187

*Number 1, an increase in the temperature implies a decrease in the Helmholtz energy.*0206

*That is what the negative sign means, if you increase this, this is dependent on this.*0222

*If this goes up, this negative sign means that this goes down.*0229

*An increase in temperature implies a decrease in the Helmholtz energy.*0234

*The higher the entropy, the higher this value is, the more quickly the Helmholtz energy decreases.*0242

*It is important when you see equations like this, you are equating some thing, some value.*0251

*You are equating it with a rate, this is a rate.*0259

*It is very easy to forget that we are talking about a rate of change.*0261

*An increase in temperature implies a decrease in A, the higher the entropy, in other words, if S itself is a bigger number,*0266

*the higher the entropy of the substance the greater the rate of change at which it is changing when you change T.*0282

*Let us say I’m at a particular temperature and let us say that my system has an entropy of 50 units.*0315

*If I raise the temperature by a certain number of degrees, the Helmholtz energy of the system is going to decrease at a rate of 50 J/°K.*0322

*At that point, if my entropy was actually higher, if it was 100 then at that point it is going to decrease by 100 J for every °K.*0345

*That is what this means.*0356

*The rate, the higher S is, the fact that there is something going to drop the negative sign.*0358

*The second one says, exactly what you think based on what we did for the first one.*0370

*An increase in the volume implies a decrease in the Helmholtz energy.*0375

*Increase in V implies a decrease in A.*0386

*I’m just making sure that I'm keeping all of my variables clear here.*0396

*What this says, the higher the pressure at that moment, the higher the P the greater the rate of change.*0406

*In other words, the greater the rate of change or decrease.*0425

*If I’m at a particular temperature and if I'm at a particular pressure,*0436

*if I change the temperature the Helmholtz energy is going to change by whatever the pressure happens to be.*0439

*It is going to be a rate of change.*0446

*If the pressure is higher, the rate of change is going to be bigger, the decrease is going to be bigger if I'm increasing the temperature*0447

*and it is the other way around.*0454

*If I’m decreasing the temperature it is going to go the other correction, the Helmholtz energy is going to increase.*0456

*It is very important.*0461

*We are talking about, what we want to do here is we want to cover every single base possible.*0464

*As far as chemistry is concerned, we are going to concern ourselves with Gibbs energy,*0471

*we are not going to concern ourselves with the Helmholtz energy but what we need to know that it is there.*0474

*If for any reason we decide to run an experiment where we are actually holding temperature and*0479

*volume constant, then we have to use the Helmholtz energy and not that Gibbs energy,*0485

*because the Gibbs energy is for conditions of constant temperature and pressure.*0489

*Constant temperature and pressure is what 99% of all chemical reactions are run under so*0494

*that is the standard laboratory conditions for chemistry, not for other things.*0501

*We would do the same thing for G.*0509

*We do the same thing for G, the Gibbs energy.*0517

*I think I would go back to blue.*0523

*Once again, we take a look at our fundamental equations for G so we have DG = - S DT.*0526

*This is going to be + V DP.*0534

*In the case of free energy, the Gibbs energy it is the temperature and pressure that are the natural variables.*0537

*The total differential DG is going to be DG DT constant P × DT + DG DP at constant T × DP.*0545

*When we equate the differential coefficients, this and this, we have the following relationships.*0562

*The rate of change of the Gibbs free energy per unit change in temperature holding pressure constant is – S,*0570

*the same thing as we saw before the Helmholtz.*0579

*The other relationship is the rate of change of the free energy as you change the pressure of the system*0584

*while you hold the temperature constant = V.*0591

*This is very interesting.*0595

*What this says, I’m going to write down in just a second.*0596

*I will go ahead and tell you.*0598

*As you change the temperature, if you increase the temperature, the free energy, the Gibbs energy of the system is going to decrease.*0600

*It is going to decrease as a rate, at a rate of whatever entropy that happens to be at that moment.*0608

*Here, if you have a system and if you increase the pressure because this is positive,*0615

*if you increase the pressure you increase the free energy of the system and the rate at which you are increasing the free energy*0620

*is going to equal whatever the volume of the system happens to be.*0628

*That is all these says, these equations are not mysterious, just take them for what they are.*0631

*Completely a phase value.*0636

*As you change the pressure, the free energy of the system is going to change the rate*0638

*at which it changes is going to equal the volume at that particular moment.*0644

*That is all this is saying.*0650

*The first one, it says that an increase in T implies a decrease in G.*0653

*A decrease in G is –S.*0667

*Once again, the higher the entropy, in other words, the higher S is the higher the entropy S, the greater the rate of decrease.*0671

*I will just say the rate of change because it might be an increase and decrease in temperature.*0689

*The second one says, now this is positive so an increase in the pressure implies an increase in G.*0698

*The bigger the volume is at that particular moment, the greater the V or volume the, the greater the rate of change of G.*0718

*That is all that is going on here.*0753

*These are very important relationships and again what is important to know is that,*0756

*it is the 4 fundamental equations of thermodynamics.*0765

*They are not really 4 fundamental equations, there is only 1 fundamental equation of thermodynamics.*0769

*These other ones were derived simply based on the definitions of enthalpy, the Helmholtz energy, and Gibbs energy.*0774

*Those are the composite functions.*0783

*Remember what we did, the only real thermodynamics functions are temperature, energy, and entropy.*0786

*The composite thermodynamic functions, the ones that are accounting devices that take other things into consideration,*0793

*that was the enthalpy, the Helmholtz energy, and the Gibbs energy.*0800

*Again, all of these equations they come from only one fundamental equation.*0804

*These are just different variations of it and it is a way of expressing some particular property of the system, some state property in this case.*0809

*G in terms of other properties of the system that is all we are doing.*0816

*At entropy volume, we are expressing G as a function of temperature and pressure but it is also going to be volume and entropy.*0821

*That is all we are doing, we are expressing one state property in terms of other state properties.*0832

*Things that are reasonably easily measurable.*0837

*Let us go ahead and take this particular equation right here and we can actually integrate this.*0843

*Let us go ahead and go to the next page.*0850

*We can integrate the DG/ DP at constant T = V to obtain an expression for the Gibbs energy of a pure substance at a constant temperature.*0853

*We took an expression for the Gibbs energy of a pure substance at constant temperature*0912

*from standard pressure which is equal to 1 atm to any other pressure P.*0917

*Let us go ahead and do that here.*0933

*We have our DG/ DP at constant T = V temperature is being held constant so essentially what this is, this is going to end up being just DG/ DP.*0935

*When we integrate this, let us go ahead and write this way.*0949

*DG = V DP again, we are holding the temperature constant.*0954

*For all practical purposes, the G is just a function of the pressure so T is held constant.*0965

*That is all it says, the function of two variables.*0974

*If we hold one constant it essentially just becomes a function of one variable.*0977

*In this case, the one variable is P so I can go ahead and use regular differential notation.*0980

*I can go ahead and integrate both of these expressions and end up getting the following.*0985

*I end up with δ G = the integral of V DP from P0 which is a 1 atm to any other value of P.*0990

*Well, δ G is just G - G0, the G0 on top is a standard free energy, free energy at 1 atm pressure.*1003

*Remember, this little degree sign it does not mean standard temperature and pressure,*1016

*it only refers to standard pressure of 1 atm because temperature can change.*1020

*It is equal this P0/ P V DP and then when we end up moving this over, we end up with the following.*1026

*G a function of temperature and pressure is going to equal G0 which is just a function of temperature.*1037

*I will explain this in just a minute, 0 to P V DP.*1046

*If you are going to find free energy at a specific pressure, you are fixing the pressure, say we fix the pressure 1 atm that is what P0 is.*1060

*The 0 on top of the degree sign, that is standard conditions, standard condition is 1 atm.*1071

*If you fix the pressure, now the pressure is no longer variable.*1076

*The free energy becomes only a function of the temperature that is why we have G °of T.*1080

*However, over here when we are actually calculating what the new G is at a new temperature and at a new pressure,*1087

*This G that you are looking for is a function of the temperature and pressure.*1095

*That is why this is G (TP) it is a function of temperature and pressure but this one is only a function of temperature because of that degree sign.*1101

*That degree sign says we are fixing the pressure at 1 atm, we are locking it in.*1110

*There is our standard, therefore, I no longer have to consider the pressure.*1114

*If you happen to know what the standard free energy is at 1 atm and*1121

*if you want to know what the free energy is off the system at another pressure, this is the equation that you use.*1125

*Let us go ahead and say a little bit more.*1138

*If the substance is a liquid or a solid, because all of these equations that we are developing they actually apply across the board,*1140

*to any substance, any system, and anytime.*1156

*These are universal, it does not apply just to gases or liquid or solids.*1160

*If the substance is a liquid or a solid then the volume can be treated as a constant.*1165

*We know already that for a liquid and a solid, unless you are applying a lot of pressure and by a lot of pressure I’m talking thousands of atmospheres,*1180

*the volume of liquid or solid is not going to change very much.*1191

*In fact, it is going to be very little.*1194

*You remember K, the compressibility coefficient is very tiny on the order of 10⁻⁵ or 10⁻⁶, things like that.*1196

*For all practical purposes, the volume is constant so it actually ends up becoming independent of the pressure.*1204

*If the substance is a liquid or a solid, then V can be treated as a constant and we can pull that V out from under the integral sign.*1212

*You will end up with the following once you integrate it.*1223

*For a liquid or solid it is still going to be a function of temperature and pressure but you do not have to worry about the integration,*1226

*It is going to be + the volume × just the δ P which we will write as P - P0, something like that.*1232

*Unless, P is really high like the difference between 1 and let us say 3000 atm, P is extremely high.*1246

*This term V × P - P0 is going to be very small which means we can usually ignore it.*1263

*Therefore, our G is actually just going to equal G(T).*1287

*Basically, what we are saying is if you are dealing with a liquid or solid this is the general equation for it right here.*1293

*Because V is essentially constant, you can pull it out from under the integral sign of the equation*1299

*that we just had at the previous slide and you end up with just the integral of TP which is δ P.*1303

*That is what this is right here.*1311

*But unless, the pressure is really high this can pretty much drop out.*1313

*When this drops out, you are just left with the free energy of a system of a liquid or solid,*1320

*it is pretty much just equal to free energy under standard conditions.*1325

*For liquids or solids, the free energy of the system is not going to change that much under pressure, that is all this is saying.*1330

*For an ideal gas, let us go ahead and write our expression again.*1340

*We had G as a function of T and P is going to equal G (T) + the integral of V DP from there to there.*1350

*For an ideal gas, V = nRT/ P, just rearranging the ideal gas law.*1364

*We go ahead and put this in here and we get the following.*1377

*We will get G (TP) = G standard function of T + nRT integral DP/ P when P is 0 to P.*1382

*We get our final equations.*1401

*This is for an ideal gas.*1403

*G0 at T + nRT LN (P)/ P0.*1406

*We said that P0 is 1 atm so this just becomes LN P.*1415

*We will go ahead and use this one as my basic equation.*1421

*That is fine, I will just go ahead and write it down.*1434

*Since P is 0 = 1 atm we end up with G of TP = G0 T + nRT × log of the pressure,*1436

*if you are looking for the free energy of a particular substance, of an ideal gas.*1460

*I’m sorry this is an ideal gas under a certain temperature and pressure conditions, you find the standard free energy*1465

*which is under 1 atm conditions and you just take the number of moles you are dealing with × R × T × log of the new pressure.*1472

*That is all that is going on here.*1480

*This is just an equation to find the free energy of an ideal gas at a particular temperature and a particular pressure.*1486

*Special note, the free energy is a very special thing.*1497

*If the function G (TP) is known, if you happen to know the functional form, this is free energy is some function of temperature and pressure.*1507

*If you happen to know what that function is, it is known, then all other thermodynamic functions can be derived from it.*1523

*Can be derived not just from it, all to say all the other thermodynamic functions can be derived*1545

*using the relations that we have + the definitions of the composite functions.*1560

*We are actually could be doing one of these problems when we do the whole problem sets for free energy.*1564

*At a specific temperature, they should be obvious from the equation above but not a problem, we will go ahead and write it down anyway.*1577

*At a specific temperature, the G of an ideal gas, the free energy of an ideal gas is a function of pressure only.*1588

*Of course, if you hold one of these constant at a certain temperature, that temperature it is just going to be a function of P.*1606

*Let me go ahead and talk about the temperature dependence of the Gibbs energy.*1618

*We already have a relation for the temperature dependence of the Gibbs energy.*1636

*Depending on a particular problem that you are working on or a problem that you run across,*1643

*we have developed different relations to express the relationship,*1651

*that is temperature dependence of Gibbs energy to what extent does Gibbs energy rely on temperature.*1657

*It is not just one equation, we actually have several equations.*1663

*We will go ahead and derive those now and they are actually quite simple.*1666

*Let us go ahead and start with the first one, the one that we already know.*1670

*We know that DG DT under constant pressure = -S.*1673

*We know this already, as the temperature changes the free energy of the system is going to change according to –S.*1682

*This is the relationship, this expresses the temperature dependents of G on T at constant pressure.*1689

*I’m going to call this equation number 1.*1697

*We know the G = H - TS that is the definition of G.*1700

*Let us go ahead and move this around a little bit.*1709

* G - H = - TS and let us go ahead and divide by T so we get G - H ÷ T = - S.*1712

*-S = G - H/ T - S = DG DT at constant P.*1729

*We are going to go ahead and put this into here and we are going to get our second equation.*1735

*DG DT at constant P = G - H/ T.*1745

*This is just another way of expressing so I will call this equation number 2.*1754

*This is just another way of expressing the temperature dependence of the Gibbs energy.*1759

* For every unit change in temperature the Gibbs energy is going to change according to what the G is*1763

*at the time - the enthalpy that quantity ÷ the temperature of the system.*1769

*That is all we are doing, let us say I do not have the entropy but let us say I have the G at the time, let us say I know the temperature and enthalpy.*1775

*This just gives me another way of expressing the change in the Gibbs energy as a function of temperature.*1783

*Sometimes, in this case I’m going to say do not worry about Y.*1792

*Sometimes, we will want the derivative of the function G/ T with respect to T.*1798

*In this particular case, we get the following.*1815

*D (G/T DT) is equal to, we have a ratio.*1820

*When you take the derivative of a ratio you get this × the derivative of that - that × the derivative of this all over that squared.*1831

*1 - S let me go ahead put P here.*1853

*We will go ahead and substitute the equation.*1860

*We already have the equation DG, always variables floating around.*1860

*T at constant P = - S so this is our first equation, equation number 1.*1869

*We know the DG DT P = - S so we can go ahead and put that - S in for this DG DT P and we end up with the following.*1876

*The partial derivative of the function G/ T with respect to T under constant pressure is going to equal*1888

*- TS - G/ T² that = - S/ T - G/ T² or it also = -ST + G ST + G/ T².*1898

*We know that G = H - TS so H = G + TS, G + TS which is ST + G that is actually equal to H.*1929

*I can go ahead and put that in there and I have my final form which it can end up being equation number 3.*1947

*Let me write it again.*1956

*The derivative of the function G/ T with respect to T constant P is going to equal - H/ T².*1958

*This is equation 3 and this is usually the equation let us call the Gibbs Helmholtz equation.*1970

*This is the form that is called the Gibbs Helmholtz equation.*1978

*Actually, any one of these equations I’m actually going to write one more you can consider as the Gibbs Helmholtz equation.*1983

*This is just variations of the same equation, pretty much this one.*1988

*We just took it and we are expressing the temperature dependence of the free energy in different ways depending on the particular problem we want.*1991

*We can go ahead and stick with this and at a particular problem, on a particular situation,*1999

*we can derive all of these which is how it actually happened.*2004

*What they decided to do was they decided they were talking about free energy so when we present this material to you as students,*2009

*we are just going to present all of these variations of the equation for you.*2015

*That is all we are doing here.*2020

*This is called the Gibbs Helmholtz equation and this is usually the form that you see it in.*2021

*The derivative of G/ T with respect to T holding P constant = -H/ T².*2027

*If you want to know how the free energy ÷ temperature changes as you change the temperature,*2033

*it is equal to the -enthalpy/ the temperature².*2041

*It is just equations.*2045

*A lot of times in thermodynamics, especially when it comes to things like free energy, when we are dealing with G,*2051

*I do not worry about that, I am actually going to explain what G is and what Helmholtz is in relation to the system.*2059

*I’m going to give a physical meaning to it, particularly when I start doing the problems.*2065

*A lot of this is just mathematical so just create it mathematically, sort of pull yourself away a little bit.*2069

*And if you do not completely understand what is happening physically, that is not a problem, that is natural.*2075

*Just go ahead and treat it mathematically.*2082

*Let us go ahead and give one more version of the Gibbs Helmholtz equation.*2085

*Let us go ahead, if I take the derivative with respect to T of 1/ T that is going to equal -1/ T².*2091

*We have D of 1/ T = -1/ T² DT.*2110

*I'm going to move the T² over so I have - T² D of 1/ T = DT.*2121

*I can do this, I can go ahead and create this thing D of G/ T with respect to D of 1/ T.*2134

*I can go ahead and rewrite this as DT = - T² D of 1/ T.*2160

*I have D of GV/ DT, I have this thing.*2170

*I have this, I'm going to go ahead and put this in the denominator so I end up with.*2177

*Let me go ahead write this whole thing out, things are confusing enough as it is.*2186

*We have the D of the G/ T DT = - H / T², that is the Gibbs Helmholtz equation, that is equation number 3.*2193

*I went ahead and I did a little bit more manipulation with this D of 1/ T.*2204

*I expressed it this way so DT = -T² D 1/ T.*2209

*I went ahead and converted to a partial derivative and now wherever I see the DT which is right here, I can go ahead and replace it with this.*2214

*I end up with the following.*2222

*I end up with D of G/ T - T² D of 1/ T = - H / T².*2224

*The - T² cancels and I end up with a partial derivative of G/ T with respect to 1/ T = H.*2238

*And this is going to be equation number 4 for the Helmholtz equation.*2253

*And again, these are all just 4 different variations of the same equation.*2256

*The temperature dependence of the free energy, when I change the temperature how does the free energy change?*2262

*This is just a different way of expressing it.*2271

*Personally, I do not like this form, it is confusing and you probably not going to run across it*2272

*altogether that much except maybe in a couple of problems.*2278

*The most problems are really only going to require some mathematical manipulation.*2281

*The way we put those problems in these problem sets with the free energy section of your books*2286

*simply to get you familiar with the mathematics of the equations.*2291

*You have noticed, in this free energy section when we dropped all of those fundamental equations*2295

*at Maxwell relations there is a lot of mathematics going on.*2302

*One of the problem sets in thermodynamics books for the free energy section*2305

*tend to involve mathematical manipulation a little more so than with energy and entropy.*2311

*That is all that is going on here, it is not altogether important that you totally grasp this.*2318

*Again, if there is one equation that you should probably remember more than any other it is the original one, the DG DT.*2324

*This one that was the equation number 1 that gave rise to the equation number 2 and equation number 3.*2333

*This one is the Gibbs Helmholtz equation.*2341

*They are all Gibbs Helmholtz equations, they are all just variations of this one.*2344

*In any case, I will go ahead and leave it that.*2348

*We will see some more of this when we actually do the problem sets and probably make more sense.*2351

*Thank you so much for joining us here at www.educator.com.*2355

*We will see you next time, bye.*2357

1 answer

Last reply by: Professor Hovasapian

Wed Nov 4, 2015 9:57 PM

Post by Manish Shinde on November 4, 2015

How can we show the dG = VdP - S dT for pure phases where the super bars refer to molar quantities. (Note:- the above equation has a bar)

1 answer

Last reply by: Professor Hovasapian

Wed Feb 4, 2015 4:16 PM

Post by matt kruk on February 4, 2015

hi professor i dont understand how you came up with the d(g/t)/dt = [T(dg/dt)p - G] /T^2 expression