Lectures Notes on Chapter 1

In Chapter 1, we will concentrate on the concepts associated with the first law of thermodynamics, that is, that energy must be conserved. While the law itself is fairly intuitive for most people (you can't get something from nothing), how it applies to chemistry and to the quantities of work and heat (which are the two forms of energy what can be transferred in or out of a system) is not obvious.  Note that here, as well as in other lectures to come, I will not necessarily cover things in the same order or the same way as the book.  You are responsible for both.

Let us start with a few definitions:

The above definitions you need to remember and understand. I will use these words on exams and expect you to know what they mean (I may even ask you to define them).

This part of the chapter considers the first law of thermodynamics itself. For this, we will speak of the internal energy of the system, U, which is just the total energy available in the system at any given time. Since we have defined heat (q) and work (w) as the only two possible forms of energy that can enter or leave the system (at least for a closed system where the mass is constant), it follows that any change in the internal energy of the system must be due to heat or work going in or out of the system.

Now we come to an important, but often confusing detail: work and heat are defined in such a way that work and heat are positive if they result in a net increase in internal energy in the system and are negative if they result in a net decrease in internal energy in the system.

Consider an example. If I put my diathermic system in contact with surroundings that are colder than it, heat will pass out of the system. Thus q is negative. If I put the diathermic system in surroundings that are warmer than it, heat will pass into the system and is thus positive. If I do work on the system, such as compression of a gas in the system, then the work is positive because the internal energy of the system will increase. If I let the system do work on the surroundings, such as expansion of a gas in the system, then the work in negative.

Note that this is the sign convention of physical chemists. Engineers sometimes use a different sign convention! Be careful. This brings us to the most important equation you have seen thus far:

In other words, the change in internal energy in the system is just the sum of the work done on (or by) the system and the heat transferred to (or from) the system. DU can either be positive, negative or zero. Remember that it is a change, not an absolute energy.

Now let's consider the mathematical formalism we will use to describe work. In physics, you probably learned that a work is performed if you push against a force, F, for some distance, Dx (assuming the force does not change with distance):

You will notice that I have used a negative sign here. This is in keeping with our sign conventions, as you will see below. Now, what if we have a system which consists of a piston in a cylinder. How would we put work in terms of things we can easily measure (pressure, volume…)?

 

 

Here we can see what is happening. As the piston expands, it is pushing against a force. This force comes from the atmospheric pressure (molecules in the atmosphere hitting the back side of the piston). Remember that pressure was a force per area, so force is just a pressure times an area. To convert to work, we just multiply the force by the change in distance, Dx. This gives Pressure x Area x Distance. But the Area times the distance traveled is just the volume change, so we can say that the work is just the negative of the Pressure times the change in volume.

 

 

 

 

 

If the force (external pressure) is not constant, we can write a more general expression for work:

Above, we considered what would happen is work was against a constant external pressure, like the atmosphere, w = -PexDV, but what if the work occurs reversibly? First, what does reversibly mean? It means that the work is always occurring in a system very, very close to the point where opposing forces are equal. In the case we are considering, this is when the pressures are maintained the same on both sides. This happens, for example, if there is some process going on inside the piston which is slowly increasing the pressure and the piston is continually expanding, keeping the pressure equal on the two sides. This could happen upon heating the gas inside. In this case, we can write Pin = Pex and therefore for a reversible expansion:

In the special case of a reversible expansion where the temperature is constant and the gas is ideal (a reversible, isothermal expansion of an ideal gas), we can substitute the ideal gas law for P, P=nRT/V:

To consider a specific example of pressure volume work, see the tutorial.

In the last section we related changes in two of the things we can measure, pressure and volume, to one way energy can go in or out of a system, work. Here we relate the other obvious observable, temperature, to the other way that energy goes in and out of the system, heat. This turns out to be conceptually and mathematically much simpler. Since temperature is already proportional to the kinetic energy of molecules and heat is just the transfer of random kinetic energy of molecules from one place to another, we should suspect that temperature and heat are directly related:

This is true under conditions where changing the temperature does not cause some chemical change in the system or some change in molecular interactions. It is not true if something happens inside the system which increases or decreases the temperature. C is called the heat capacity, since it is some measure of how much heat is taken up or given off for a given change in temperature. Its value depends on the nature of the molecules under consideration and on the conditions under which the heating and cooling were performed. The simplest case is when we hold the volume of the system constant. In this case there is no PV work to worry about. Only heating is occurring (heat is the only source of energy change in the system). In this case, we use CV for our heat capacity and call this the heat capacity at constant volume. You can look this up in tables for different kinds of compounds .

We know that DU = w + q. At constant volume (assuming only PV type work), w = 0, so for this case DUconst. vol. = q = CV DT. Thus, at constant volume, the internal energy change is just given by the product of the heat capacity and the change in temperature. A more precise way of saying this (without assuming that CV is constant) is:

This expression is a partial derivative. If you have not dealt with such things before, a partial derivative is just the derivative of some function (in this case U) with respect to some variable (in this case T) holding some other parameter constant (in this case V).

The other important case for heating is heating at constant pressure. In this case, heat put into the system not only results in a temperature increase. It also results in work being done do to expansion of the system. As the system heats up, the volume will increase in order to keep the pressure constant. This results in PV work. The change in internal energy is just:

DUconst. P = q - PexDV

but since the pressure is constant and equal inside and outside (such as in an open flask), we can see that

Where P is now the normal P of the system rather than that of the outside world.  We do a great deal of chemistry in open flasks. Thus, we are interested in the kinds of energy changes that can occur under such conditions. Notice that PDV is itself a state function as long as P is constant. In fact, the product PV is a state function whether P is constant or not. This suggests the definition of a new state function called enthalpy.

We can see that

for small changes. At constant pressure this becomes:

Thus, comparing this expression to the expression above for DU at constant pressure, one can see that

 

Remember that this is only true at constant pressure.

So, at constant volume (and no electrical work), the internal energy change is equal to the heat. At constant pressure, the enthalpy change is equal to the heat.

Let us return now to the issue of the change in temperature with heat at constant pressure. We now realize that when heat is put into a system at constant temperature, some of the energy goes into raising the temperature and some of the energy goes into the expansion of the gas. Therefore, in general, it takes more heat to raise the temperature by one degree at constant pressure than it does at constant volume. The heat capacity at constant pressure is called CP and we can write that

Incorporating the idea of enthalpy above (again at constant pressure):

which can be written more exactly as (without assuming that the heat capacity at constant pressure is independent of temperature):

Before we continue, let's say a word about heat capacities in general. Heat capacities (at either constant volume or constant pressure) are normally reported as molar heat capacities. That is, the heat capacity of one mole of material. The expressions we have used thus far are for total heat capacities. That is, the heat capacity of the whole system. We could rewrite these expressions in terms of molar heat capacities by substituting:

Where the bar indicates a molar quantity. Alternatively one could write the whole equation in terms of molar quantities, for example:

This states that the change in the molar enthalpy (the enthalpy per mole of material) with temperature is equal to the molar heat capacity at constant pressure.

An important aside:

Before we leave heat capacities, I would like to consider the internal energy dependence on temperature in an ideal gas. There are some important lessons to be learned which are not explained in the book.

The important point here is that for an ideal gas, the change in internal energy is always, always, always zero if the temperature does not change -- but only for an ideal gas!

You can also get an idea why CP = CV + nR for an ideal gas from the above expression, though this is not a proper proof .

Since DU = w + q, this means that for an ideal gas, if T does not change, w = -q.

In an isothermal ideal gas DU = 0 so w = -q.

In general for an ideal gas DU = CVDT

 

We can make the same kind of argument about enthalpy changes:

So for an isothermal ideal gas, DH = 0

In general for an ideal gas DH = CPDT

Remember these relationships! They come in very handy for problems involving ideal gases!

Now let's consider what happens in an ideal gas system that allows no heat to transfer across the walls (adiabatic). In an adiabatic system, the only change in internal energy of the system that can happen is through work, so DU = w. Thus, the expressions we came up with for the change in internal energy are going to give us the work.

For an adiabatic ideal gas system, w = DU = (CP - nR)DT = CVDT

But this is confusing. We are now using the heat capacity at constant volume to calculate the internal energy change and the work when we are not at constant volume. Let's go through this another way. The only way to change the temperature of an ideal gas in an adiabatic system is through work. This requires a volume change. Since the internal energy change does not depend on path, lets do this in two steps. First we change the volume holding the temperature constant (this requires some heat transfer, but don't worry, it will cancel in a second). The change in the internal energy for this is zero since there is no temperature change. Next, we change the temperature holding the volume constant (which will have just the opposite heat exchange with the environment, zeroing the total amount of heat transferred). The internal energy change here is CVDT. Thus the change in internal energy and the work done in our adiabatic system is CVDT, as above. This, by the way, is also one way to prove that CP = CV + nR. The above argument is general for ideal gases. For all ideal gases the change in internal energy is always the heat capacity at constant volume times the change in temperature. For adiabatic ideal gases, the work is the heat capacity at constant volume times the change in temperature.

All ideal gases:             

Adiabatic ideal gases:               

I also have an on-line treatment of some of the more advanced concepts associated with internal energy and enthalpy, put in mathematically more precise terms.  You are not responsible for this.  It is there to help you understand things at a deeper level if you would like to.

We are going to talk now about reactions and transitions.  Ideal gases do not undergo reactions or transitions because they do not interact.  What we need now to begin to understand is how we can use the concepts derived for ideal gases to discuss non ideal things like reactions and transitions.  For the moment, we will do this for enthalpy.  Later we will extend this to something called free energy.

The point is that what ideal gases tell you about, and what thermodynamics is all about really, is the behavior of numbers of particles.  Therefore, what we have to do when we extend the concepts of ideal gases to other things is to break the problems into pieces.  Typically, this will include a piece that is the reaction or transition itself and then a piece that involves changing the temperature or changing the number of molecules without doing any reactions or transitions.  We can arbitrarily break the system up like this if we are dealing with state functions.  Remember with state functions, it does not matter how we get from state A to state B, so we can do it in any way we like.  We will consider different kinds of reactions and transitions as examples.

Phase Transitions

A phase transition is something you are all familiar with.  It is the conversion from solid to liquid or liquid to gas.  Water is the simplest example.  The tutorials below walk you through phase transitions at the conceptual level.  You might think this is obvious, but it is not.  There is a lot going on in a glass of ice water or when you boil water on the stove.

·        Phase changes in a closed system with no atmosphere present.

·        Phase changes in a closed system with atmospheric gases present.

·        Phase changes in an open system.

The idea that most people do not understand about phase transitions is that they happen (for pure substances) at extremely precise temperatures.  Under standard conditions, glass of ice water (assuming that the system is adiabatic and at equilibrium) has a temperature of 0 C.  Not 0.000001 C.  0 C.  Exactly.  This is because the transition between ice and water is extraordinarily cooperative (we will define that term in a thermodynamic sense later in the course, but for now, just understand that if one molecule leaves an ice crystal it makes it easier for the next to do so, etc.).  It is also because melting ice requires a certain amount of heat energy to be absorbed.  You can heat the ice to 0 C with a certain amount of heat, but it takes much more heat to go from 0 C ice to 0 C water.  In fact the heat capacity of a phase transition is infinite.  You put heat into the system and the temperature does not change.  Instead you just convert some ice to water (or at the boiling point, some water to gas).

In terms of enthalpy, the melting of ice or the evaporation of water is an endothermic process.  It requires the uptake of heat.  It takes 6.01 kJ of heat to melt one mole of water.  This is called the molar enthalpy of fusion.  It takes 44 kJ of heat to evaporate one mole of water at 25 C.  This is called the molar enthalpy of vaporization.  To go the other direction you must remove the equivalent amount of energy.

Now, lets work through a problem.  Let’s say I want to calculate the total change in enthalpy (the total heat required since we will do this at constant pressure) for taking 1 mole of ice at –20 C and converting it into steam at 120 C.  Enthalpy is a state function, so lets break the problem into pieces.  First, we use the fact that DH=CPDT to calculate the enthalpy required to warm the ice from –20 C to 0 C.  CP for ice is 37 J K-1 mole-1.  So the enthalpy change for this part of the process is 740 J/mole.  Now we use the enthalpy of fusion to go from 0 C ice to 0 C water.  We know that for one mole the enthalpy of fusion is 6.01 kJ.  Now we heat the water from 0 C to 100 C (the boiling point).  Again we use the heat capacity equation above for the enthalpy change, but this time the heat capacity is that of water rather than ice.  The heat capacity of water is 75.29 J K-1 mole-1 so for a change of 100 C that is a total enthalpy change of 7529 J/mole.  Now we have to boil (vaporize) the water at 100C to for gas at 100 C.  This is just the enthalpy of vaporization.  For one mole, this is 40.7 kJ/mole (note that this is a little different than the enthalpy of vaporization at 25 C – it depends a bit on temperature).  Finally, we heat the gas up from 100 C to 120 C.  The heat capacity of steam is 33.58 J K-1 mole-1.  So the enthalpy to heat the steam by 20 C is 672 J/mole.  So the total enthalpy is about 55000 J/mole.  Note that almost all of this, 46,000J, comes from the phase transitions.

OK, so you see how to work through an enthalpy calculation but why does this work?  We are using equations for heat capacity developed for ideal gases.  Water in solid or liquid form is certainly not an ideal gas.  This is true, however, what we have done, and what we almost always do, is to use the concepts from ideal gases and add in a fudge factor.  The heat capacity of an ideal gas can be calculated from first principles.  At constant pressure it is about 30 J K-1 mole-1 (note that this is close to the heat capacity of steam).  What we do is to assume that nonideal situations behave similarily, but just use a different constant – we assume that enthalpy is still linearly related to a change in temperature, but we EMPIRICALLY determine a heat capacity instead of calculating it directly from first principles.  In other words, we include an empirical fudge factor to take care of the nonideality and simply list it in a table somewhere.  This is the great dirty little secret of thermodynamics.  Everything is just an ideal gas with a fudge factor added that is determined by experiment!  The enthalpies of fusion and vaporization again are just empirical constants we look up in a table because there is no analog to a phase transition for an ideal gas.  This is an important concept.

Chemical Change and Standard Enthalpies

There are many times in chemistry when we are interested in reactions that occur at constant pressure (like essentially all of biology), and particularly under these conditions, it is useful to consider the enthalpy change (the heat change at constant pressure) that occurs during the reaction. Reactions that produce heat are releasing energy into the environment and are called exothermic. Reactions that remove heat from the surroundings are called endothermic.

We can write down enthalpy changes associated with any process by subtracting the enthalpy in the initial state from the enthalpy in the final state, but before we do this, we need to have some convention about how to report them. This convention is called a standard state and we will use this over and over again from here on out. We normally report enthalpies (and later free energies and entropies) at the standard state. For most of chemistry dealing with simple substances, the standard state is one mole of the pure form of the material at 1 bar of pressure.  We will encounter a slightly different definition when we talk about solution phase chemistry.

This said, we can now write down the enthalpy change for any chemical reaction in this state, and then calculate enthalpy changes at other states as needed (more on this later). For example, I could state the enthalpy change for converting 1 mole of A into 2 moles of B in the chemical reaction Aà 2B. Enthalpy changes also add. If I take two processes, A à B and B à C and add them up chemically, I get A à C. The enthalpy change of B à A is just the negative of the enthalpy change for A à B (this makes sense -- in one direction heat goes in and in the other direction heat comes out).  The enthalpy change for 2A à 2B is double the enthalpy change for A à B.

Chemists have also devised a bookkeeping scheme for calculating the enthalpies of new reactions. You can simply use the enthalpies of formation for each of the products minus the enthalpies of formation for each of the reactants (products are made and reactants are used up) starting from standard materials. This is very well explained in the book and follows directly from the arguments about adding reactions together above. Try this with a few simple reactions just for practice. Here is one:

C6H6(l) + 7.5 O2(g) à 6 CO2(g) + 3H2O(l)

To calculate the reaction enthalpy, we just add the enthalpies of formation of the products and subtract the enthalpies of formation of the reactants, making sure to include the proper stochiometries:

DRH0 = 6DfH0(CO2) + 3DfH0(H2O) - DfH0(C6H6) – 7.5DfH0(O2)

= 6 x (-393.5) + 3 x (-285.83) – (49.0) – 7.5 x (0) =  -3268 kJ/mole

You can look these numbers up in a table.  Note that the enthalpy of formation of O2 is zero.  This is because O2 is the reference state of oxygen and the enthalpy of forming O2 from its elements in their reference state is therefore zero.

Given this kind of thinking, we can talk about reaction enthalpys for pretty much anything.  The book goes through the concept of a bond energy and it uses biological fuels as an example of molecules with different energies of combustion (oxidation with oxygen).  We will discuss this briefly in class if there is time (though this is more a topic for Bch462).

Calculating an enthalpy at a new temperature

If we know the standard enthalpy at one temperature, how do we get it at another? Recall that the change in enthalpy with temperature just depends on the heat capacity at constant pressure. The expression turns out to be:

How did we get this equation?  It is very simple.  We again took advantage of the fact that enthalpy is a state function.  So as long as I get from the initial state to the final state somehow (it does not matter how) and I add up all the enthalpy changes along the way, I must end up with the right total enthalpy change.  Consider then the reaction 2H2(g) + O2(g) à 2H2O(l).  At 25 C this has a standard enthalpy change of –572 kJ/mole (the book gives this as an example).  What would the standard enthalpy change be at 50 C?  We will do this in three steps.  First, we take the reactants at 50 C and we cool them to 25 C.  We can calculate the enthalpy for this from:

DH = CPDT

We are going to assume that hydrogen and oxygen are more or less ideal which means that we can treat them separately and calculate the enthalpy changes due to cooling two moles of hydrogen and due to cooling one mole of oxygen.  For two moles of hydrogen:

DH = CPDT = 2 x 28.8 J K-1mole-1 (-25 K) = -1440 J/mole

For oxygen

DH = CPDT = 29.3J K-1mole-1 (-25K) = -732.5 J/mole

Adding these we get the total enthalpy change for cooling two moles of hydrogen and one mole of oxygen by 25K which is –2172.5 J/mole.  It is negative because heat is going out of the chamber during the cooling (remember at constant pressure, the enthalpy change is just the heat).

Now that we have cooled the reactants to 25 C, we run the reaction at 25 C.  We know the enthalpy change for this reaction at 25 C.  As stated above, it is –572 kJ/mole (a very exothermic reaction).  This is something you would look up in a table of standard reaction enthalpies at 25 C or you would calculate it from enthalpies of formation at 25 C as discussed before.  The point is that most tables are given at 25 C, not 50 C.

Now that we have run the reaction at 25C, we can heat the product, two moles of water, back up to 50 C.

DH = CPDT = 2 x 75.3 J K-1 mole-1 (25K) = 3765 J/mole

This is positive, because we have to put heat into the system in order to raise the temperature by 25K.

OK now we just add all the steps up:

Cool reactants from 50 C to 25 C (-2173 J/mole), Run the reaction at 25 C (-572,000 J/mole), Heat the product from 25 C back to 50 C (3765 J/mole).  The resulting enthalpy change for our reaction run at 50 C is –570,400 J/mole.  This is not very different from the enthalpy change at 25 C because formation of water from hydrogen and oxygen is so exothermic that changing the temperature a little does not much matter.

We are going to use this procedure over and over again.  We will use the rules we learned from ideal gases to put our system in a certain standard condition.  We then use an EMPIRICAL value for the process under that standard condition that we look up in a table somewhere.  We then use our thermodynamic rules to put the system back into the condition of interest.  As long as we are dealing with state functions, this is exactly the same as running the reaction or process under the condition of interest rather than under standard conditions.  This is the essence of thermodynamics and its utility.

So back to our original equation:

Hopefully, you now see where this comes from by comparing it to the example above.

The heat capacity difference is the sum of the product heat capacities minus the sum of the reactant heat capacities weighted by the stochiometric coefficients in the chemical reaction. So for A à 2B it would be

These heat capacities can also be looked up in tables.  This is exactly what we did above, just abbreviated into a single set of equations.