|Lecture Notes::Lec 6_6 June||
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.
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:
deltaG°f = - 2.030 x 105 Jmol-1 = -203.0 kJmol-1
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:
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.
Like enthalpy, deltaH, deltaS is a State Function, and thus deltaS is pathway independent. So for a chemical reaction we can write:
Example: Find the entropy change in the reduction of aluminum oxide to aluminum metal by hydrogen gas.
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!
Plugging in the stoichiometric coefficients and values of S:
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 deltaS°rxn 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:
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.
We are going to look at four different ways to calculate deltaG:
* 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.
© R A Paselk
Last modified 6 June 2006