Dr Cooper's comments:

JCrishi

**Helmholz Energy and the Notion of Maximum Work.**

We saw in Lecture 8 that, given constant temperature and volume, the (infinitesimal) maximum work extractable from a system is

For a measureable change in a system the relation becomes

(1)

This equation tells us that the maximum work available from the system is equal to the Internal Energy of that system minus (Temperature times) the change in entropy *of that system*. One is tempted to believe that the work extractable may never exceed the change in Internal Energy as we are always subtracting a negative value for entropy ( ) as heat flows out of the system. Subtracting a negative from a negative, which is adding a positive, thereby ensuring that the right side of the equation is always smaller than alone. Thus making Wmax less than . Recall however that the entropy may increase or decrease for a system depending upon its environment. The surroundings may be much hotter than the system in which case heat would flow into the system raising its entropy value. These two cases are illustrated below.

In figure 3-15 above we see the case in which heat flows out of the system, raising the entropy of the surroundings. As is negative for the system the effect will be to make less negative (see eqn.1) hence lessening the work extractable.

In figure 3-16 above we see the case in which heat flows into the system, decreasing the entropy of the surroundings while raising the entropy of the system. As is positive for the system the subtraction (see eqn. 1) will make the right hand side more negative hence increasing the work extractable to a value greater than .

The following problem demonstrates the calculation of the heat and work that can be extracted from the oxidation of glucose at 25 °C. It is also worked and explained in the book.

Problem?

For the heat at constant p:

Since the number of moles of gas don't change from products to reactants we have -

The enthalpy change is equal to the internal energy change as given so -

Since the work extractable is equal to the Helmholz energy we have from eqn 1 above:

Substituting gives -

And finally -

**Gibbs Energy and the Spontaneity of a Reaction**

From the Clausius inequality we know that

Rearranging we arrive at

Recall that at constant p

Therefore

(2)

For eqn. 2 at constant enthalpy (dH = 0) we have TdS >= 0 or

dS >= 0. (3)

For eqn. 2 at constant entropy (dS = 0) we have

0 >= dH. (4)

Eqn. 3 states that for a spontaneous change, at constant p, the entropy of the system must increase if the enthalpy remains constant. At constant pressure the volume of the *system* will increase giving rise to greater entropy there.

Eqn. 4 states that for a spontaneous change the enthalpy will decrease if the entropy remains constant. This can only happen if heat flows into the *surroundings*, increasing the random motion in the surroundings which constitutes increasing the entropy there.

Rearranging eqn. 2 gives

As with the Helmholz energy define a new thermodynamic quantity G such that

(5)

Or macroscopically

where G is the Gibbs Energy.

Since the Gibbs Energy is defined at constant temperature and pressure and constant temperature and pressure are the conditions found in the lab, the Gibbs Energy is of principal interest to chemists. The constant volume conditions used for the Helmholz Energy rarely pertain to reactions of interest.

Recalling equation 2 above we see that . Keeping that relationship in mind as we examine equation 5 we see that means that any spontaneous change in a system (at constant T and p) must occur such that .

Reactions can be endothermic, , or exothermic, .

Regardless, for spontaneous endothermic or exothermic reactions it is required that .

For endothermic reactions this occurs via a drop in entropy of the surroundings as heat flows into the system, increasing the entropy of the system.

For exothermic reactions this occurs via a drop in entropy of the system as heat flows out of the system and into the surroundings, raising the entropy there.

PChadwell

The information above shows us how internal energy, entropy, and temperature relate to the idea of maximum work in a system. However, there is another concept that is related inexorably to the amount of work a system can do but differs from the above in that it is non-expansive rather than expansive. The difference in these two equations is due to what is held constant. For Hemholtz the temperature and volume are held constant, for Gibbs the temperature and pressure are held constant. This difference is why the Gibbs energy is used to calculate the maximum non expansive work because the pressure is subtracted from the Gibbs energy in the equation to make it such. Proof of this and exactly how to use the Gibbs free energy to find the non expansive work of a system is as follows.

Any way to reformat the following? A little confusing.

A1: Take the dirivative

D1:

A2: Plug in D1

C1:

A3: Take the dirivative

D2: Plug in C1

C2: Its isothermal therfore dt=o

C3:

A4: Because the change is reversable we can say A4 and A5

A5: The subsript r here stands for reversable, the computer didnt want to play nice

C4: Plug in A4 and A5

A6: Plug into C4

C5: Because Gibbs energy is done at constant pressure dp=0

T1: The subscript a here stands for additional and reversable

This proof shows that for a system at constant pressure and temperature that the maximum additional non-expansive work of a system is equal to the Gibbs energy of that system. This is extremely useful not only to find electrical work but for the practical application that it is much easier to work at constant pressure in the lab then it is in to work at constant volume.

Gibbs energy is also useful because it is the energy that a chemical system is trying to decrease when the system undergoes a spontaneous process at constant temperature and pressure. This makes it an obvious choice to be used as the criteria for use as a standard for the energy of formation because when the Gibbs energy of a compound is zero that element is in the state it is trying to get to via spontaneity. This means essentially that the Gibbs free energy of a compound can be used as the standard for energies of formation because the Gibbs energy is what drives the spontaneity of reactions and; therefore, whether or not something will happen by chance and lower energy or whether work will have to be pumped in to for the compound to change. Which means that the amount of work, which is what we are ultimately looking for, can be found via the Gibbs energy because weather or not the compounds will change depends on it as shown in the previous proof. The change in Gibbs energy of a reaction then will show how much energy is released or required to perform a reaction. These quantities are weighted with respect to stoichiometric coefficients.

With this equation in mind it is also useful to know the energies associated with disassociated ions. As a standard chemists have designated the H^{+}ion to be zero and all other ions judged by it. Therefore, with the above equation and some laboratory data about the change in Gibbs energy we can find the energy associated with disassociated ions other than H^{+} by using the reaction of disassociation and the above equation.

(Dustin Chu-qchu1)

**Standard Reaction Gibbs Energies**

The Standard Reaction Gibbs Energy is the energy associated with a chemical reaction to do work. It is defined as the change in enthalpy of the system minus the product of temperature and the change in entropy of the system at the constant temperature and pressure is:

**Standard Reaction Gibbs Energies of Formation **

The difference between the sum of the standard Gibbs energy of formation of the products and the sum of the standard Gibbs energy of formation of the reactants:

- The Standard Gibbs energy of formation describes the energy that is created either released or consumed when a compound is formed from its elements at their reference states.
- The Reference state of an element is known to be most stable at the specific temperature and the pressure of 1 bar.
- The standard Gibbs energy of formation of an element is zero at its reference state.

For example: Nitrogen when is at its reference state which is N_{2} gas, Mercury as in liquid Hg, Carbon as graphite all have the standard Gibbs energy of formation of zero.

__Example:__ Calculate the Standard Gibbs energy of the reaction at 25^{0}C

As we know :

(The standard Gibbs energies of formation values (at 298K) can be found in Table 3.4)

= -394.4 kJmol^{-1 }- [(-137.2) +(1/2)(0)] kJmol^{-1}

= -257.2 kJmol^{-1}

- The standard Gibbs enrgies of formation of an ion cannot be determined by the characteristics of the ion itself but also the accompanying ion since a cation cannot be prepared without the simultaneous formation of the accompanying anion.
- Conventionally, the standard of Gibbs energy of formation of Hydrogen ion is zero at all temperature. (at all temperatures)

Similarly,

(at all temperatures)

__Example:__ Calculate the from from the reactions

^{ }

From (R1)

^{ }

Therefore,

From (R2)

So

**Standard Gibbs Eneries of Solvation**

The calculation of the Gibbs energy of formation of an ion in solution can be broken down into smaller steps as in a thermodynamic cycle, and is illustrated as showed below:

The Thermodynamic cycles for the discussion of the Gibbs energies of solvation and formation of Cl^{-} (left) and I^{- }(right) in aqueous solution.

- The Gibbs energies of solvation of individual ions can be estimated through the Born's equation:

For water at 25^{0}C, the equation becomes:

- From the Born's equation, since z, r, p, m are all positive values, is always negative.
- As the radius decreases, the increases, and as the charge number of the ion increases, the increses as well. Thus, Gibbs energies of solvation of smaller and highly charged ions are greater.

**Combining First and Second Laws **

First Law of thermodynamics states: dU= dq + dw

- In a closed system for a reversible change without any addition work, we have dw
_{rev}= -pdV where p is the pressure of the system.
- Second Law dq
_{rev}= TdS where T is the temperature of the system.

Combine first and second law, we then have the fundamental equation: dU=TdS - pdV

From the fundamental equation, dU is proportional with dS and dV. Thus, we can express U as a function of S and V:

Where

and

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