Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110	General Chemistry	Fall 2003
Lecture Notes::Lec 8_12 September	© R. Paselk 2003
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Free Energy and Chemical Reactions

Last time we stopped in the middle of the calculation of DS for the complete combustion of ethane (C₂H₆) with oxygen (O₂). Let's complete that example. Recall that we had just balanced the equation, leaving it in a format for the reaction of one mole of ethane to make thermo calculations easier:

C₂H₆ + 7/2 O₂ Æ 2 CO₂ + 3 H₂O

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

DS° = SnS°_prod - SnS°_rcts

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

DS° = 45.9 JK^-1mol^-1

Note that even though the reaction has 4.5 molecules going to 5, the higher entropies of the reactants make this reaction slightly opposed by the reaction entropy. Of course this reaction is 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 DG problems one of the first things to think about is to guess whether it is spontaneous.

• We have already determined that a (-) DH 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 DS?
  • For physical processes like melting or vaporization we can guess fairly easily: more KE or greater volume indicate a positive DS.
  • 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 Æ 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 DS and DH are in opposition.

So if both of the DG 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 DG.

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

First we need to write the chemical equation:

1/2 N_2(g) + 2 H_2(g) + 1/2 Cl_2(g) Æ NH₄Cl_(s)

From the table of DH_formation, DH° = - 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

Finally:

DG°_f = - 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 DG or DH.

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

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 DT 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.

* I would like to acknowledge 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.

© R A Paselk

Last modified 12 September 2003