### Richard A. Paselk

Chem 110

General Chemistry

Summer 2006

Lecture Notes::Lec 6_6 June

© R. Paselk 2006

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# Free Energy and Chemical Reactions

We very frequently want to find deltaG for a chemical process. Last semester we saw that we can determine the values for deltaH for chemical reactions using tabulated heats (enthalpies) of formation (deltaH°). Similarly we can find deltaS from tabulations of Standard Molar Entropy values, deltaS°.

Example: Calculate deltaS for the complete combustion of ethane (C2H6) with oxygen (O2).

First we need to write the chemical equation:

C2H6 + O2 CO2 + H2O

Balancing by inspection:

C2H6 + 7/2 O2 2 CO2 + 3 H2O

(Notice that we left the coefficients in fractional form. This is acceptable for thermo problems because we are attempting to find the molar values of deltaH, deltaS, deltaG, etc. and giving whole number coefficients will add another step and thus another chance for error.)

From Tables we can get values of S° in the vapor state:

 Vapor S° JK-1mol-1 C2H6 229.5 O2 205.1 CO2 213.6 H2O 188.7

deltaS° = Sum(nS°)prod - Sum(nS°)rcts

deltaS° = [2(213.6) + 3(188.7)] - [229.5 + 7/2(205.1)]

deltaS° = 45.9 JK-1mol-1

Note that the reaction has 4.5 molecules going to 5, making this reaction slightly favored by the reaction entropy. Of course this reaction is also highly exothermic so the entropy of the Universe increases significantly and the in fact the reaction is highly favored.

Something I like to emphasize to students are various "tools for thought." What I mean by this is the use of various approachs and techniques to help you determine outcomes and solve problems. An important use for these mental tools is determining if an answer is reasonable. (Sometimes its just a matter of paying attention. For example an answer of 2 x 10-4 atoms is obviously ridiculous since we can't have fractional atoms - you either have an atom or you don't.)

For deltaG problems one of the first things to think about is to guess whether it is spontaneous.

• We have already determined that a (-) deltaH favors a process. So if a reaction gives off heat and/or light it is likely to be spontaneous.
• However, how do we guess about deltaS?
• For physical processes like melting or vaporization we can guess fairly easily: more KE or greater volume indicate a positive deltaS.
• What about a chemical reaction? Here things can get a bit more complicated because the number and types of particles changes. Let's look at an example. Consider the entropy change for converting hydrogen into atomic hydrogen: H2 2 H. If we assume a container divided into a million cells which can be occupied by up to two particle each and we start with one molecule of hydrogen then:
• For H2 we have one million (106) possibilties.
• For 2 hydrogens consider the first hydrogen occupies cell number one, then the second can occupy any of the one million cells. Thus the total number of possibilities is (106)(106) = 1012. Obviously the entropy has increased considerably.
• Finally, we also have to take the temperature into account if deltaS and deltaH are in opposition.

So if both of the deltaG terms are favorable, then easy to predict, otherwise we need to be a bit more thoughtful.

With this in mind let's look as some calculations involving deltaG.

Example: Calculate the deltaG of formation for ammonium chloride (NH4Cl) @ 25°C.

First we need to write the chemical equation with the corfficient of unity for ammonium chloride:

1/2 N2(g) + 2 H2(g) + 1/2 Cl2(g) NH4Cl(s)

From the table of deltaHformation, deltaH° = - 314.4 kJmol-1

From the table of Standard Molar Entropy we find:

 S° JK-1mol-1 N2(g) 191.5 H2(g) 130.6 Cl2(g) 223.0 NH4Cl(s) 94.6

deltaS° = SnS°prod - SnS°rcts

deltaS° = S°NH4Cl(s) -(1/2S°N2(g) + 2S°H2(g) + 1/2S°Cl2(g))

= 94.6 - [1/2(191.5) + 2(130.6) + 1/2(223.0)]

deltaS° = -373.8 JK-1mol-1

Finally:

deltaf = deltaf - Tdeltaf

= - 314.4 kJK-1mol - (298 K)(-373.8 JK-1mol-1)

deltaf = - 2.030 x 105 Jmol-1 = -203.0 kJmol-1

# More on Entropy*

Recall that earlier I said that we can get an absolute value for S, since entropy, like temperature has a zero value. Thus for a pure substance a perfect crystal at absolute zero has an entropy of zero. This is quite different than deltaG or deltaH.

Though its beyond the level of this course, it can be shown that:

deltaS = (c/T)deltaT at temperature above 0 K

where c = the heat capacity (JK-1mol-1)

The point is, we can actually find the entropy of a substance etc. via direct measurements of c and T. If we then plot c/T vs. T the area under the curve (S(c/T x deltaT from 0-T) is S! Note that we cannot actually get data for any system all the way to absolute zero, and for many systems we can only get within a few degrees of 0 K. However, the last bit can be extrapolated via calculation from theory, or extrapolated as on the curve below, with little error for normal temperatures, to give an excellent estimate of the true value of the standard molar entropy. A table of Standard Molar Entropy (sometimes called Third Law Entropy) values can be found in Appendix 4 of your text. These values are determined for 25°C and 1 atm. Let's look at a few values as illustrations.

• For liquid water, S° = 70 J/molK
• For water vapor, S° = 189 J/molK. Note that although water is most stable as a liquid at 25°C, some does exist as vapor!
• Solid carbon (diamond) S° = 2 J/molK. Note the incredibly small value of S. This is because in a perfect diamond every carbon atom in the bulk phase is connected to four others in a very rigid geometry by very strong covalent bonds, so very little motion is allowed, even at room temperature.
• Solid carbon (graphite) S° = 6 J/molK. Though still a very low value of S, graphite's values is significantly higher than diamond since the carbons are arranged in planes linked to three other carbons. The lower rigidity allows greater motional freedom than we see in diamond.

Like enthalpy, deltaH, deltaS is a State Function, and thus deltaS is pathway independent. So for a chemical reaction we can write:

deltarxn = Sum(nS°)prod - Sum(nS°)rcts

Example: Find the entropy change in the reduction of aluminum oxide to aluminum metal by hydrogen gas.

Al2O3(s) + 3 H2(g) 2 Al(s) + 3 H2O(g)

Note that we have both solid and gaseous forms of the reactants and products, in particular water which is generally listed as both a gas and a solid, so pay attention to which values you pick!

 S° (J/molK) Al2O3(s) 51 H2(g) 131 Al(s) 28 H2O(g) 189

Plugging in the stoichiometric coefficients and values of S:

deltarxn = 2 Al(s) + 3 S°H2O(g) - [S°Al2O3(s) + 3 S°H2(g)]

deltarxn = 2 (28 J/molK) + 3 (189 J/molK) - [ (51 J/molK) + 3 (131 J/molK)

deltarxn = 179 J/molK

This is a somewhat surprising result given our discussions so far. After all we've gone from one mole of solid and three moles of gas to one mole of solid and three moles of gas. We should expect deltarxn to be about zero, but in fact its quite large: 179. So what's going on?

Let's look at the change between hydrogen and water, since gases are simpler, as an example. There are equal numbers of each, at the same pressure and temperature, so why the large difference in S? It turns out the water molecule has more degrees of freedom than hydrogen.

If we look at the structures of these two molecules: We can see that:

• Water has three atoms, and in a Cartesian (x,y,z) coordinates system as shown on the diagram, each atoms movement can be described as having three positional degrees of freedom: along the x-axis, along the y-axis, and along the z-axis.
• The atoms in the molecule then have 3n = 3 + 3 + 3 = 9 degrees of freedom.
• In addition the water molecule as a whole can translate along the three axis (3 degrees of freedom) and rotate around the three axis (3 degrees of freedom) giving 6 df.
• If we subtract the translational and rotational df from the atomic df we get 9 - 6 = 3 df for water.
• Hydrogen has two atoms, and by the same reasoning:
• The atoms in the molecule then have 3n = 3 + 3 = 6 degrees of freedom.
• In addition the hydrogen molecule as a whole can translate along the three axis (3 degrees of freedom), however it can rotate around only two axis (2 degrees of freedom) giving 5 df. (For hydrogen the reduction in rotational freedom is most easily seen by arranging the x-axis along the bond axis. The molecule can then rotate around the z- and y-axes, but rotations around the x-axis is meaningless, as the atoms have no solidity.)
• If we subtract the translational and rotational df from the atomic df we get 6 - 5 = 1 df for hydrogen.

Thus water has more entropy because it has more degrees of freedom - more microstates are available to it.

In general we can say the more complex the molecule, the higher the standard molar entropy.

# deltaG Calculations

We are going to look at four different ways to calculate deltaG:

1. By the formula for free energy: deltaG° = deltaH° -Tdelta
1. In this case we calculate deltaH° and deltaS° from tabulated values.
2. Then use these values and the specified temperature to calculate deltaG°.
2. By the Hess's Law approach - this is based on the fact that deltaG is a State Function, so it is pathway independent. Thus we can look at any combination of reactions which will add up to the desired reaction, regardless of how impractical in practice, and find the value of deltaG for the overall reaction.
3. We can simply look up and add together values of free energies of formation (if they are available): deltaG° = Snpdeltaf(prod) - Snrdeltaf(rct).
4. Calculate from the equilibrium constant. (We will look at this later.)

* I would like to thank my colleague, Dr. William Golden, for allowing me to use his notes as the basis for my own discussion on Thermodynamics over the next few days. Any errors in these notes are, however, my own responsibility.

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