| Chem 110 |
General Chemistry |
Fall 2003 |
| Lecture Notes::Lec 9_15 September |
© R. Paselk 2003 |
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More on Entropy, cont.
Like enthalpy, DH, DS
is a State Function, and thus DS
is pathway independent. So for a chemical reaction we can
write:
DS°rxn =
SnpDS°(prod)
- SnrDS°(rct)
Example: Find the entropy change in the reduction of
aluminum oxide to aluminum metal by hydrogen gas.
Al2O3(s) + 3 H2(g)
Æ 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:
DS°rxn =
2DS°Al(s)
+ 3 DS°H2O(g)
- [DS°Al2O3(s)
+ 3 DS°H2(g)]
DS°rxn =
2 (28 J/molK) + 3 (189 J/molK) - [ (51 J/molK) + 3 (131 J/molK)
DS°rxn
= 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 DS°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:
- 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.
DG Calculations
We are going to look at four different ways to calculate DG:
- By the formula for free energy: DG°
= DH° -TDS°
- In this case we calculate DH°
and DS° from tabulated values.
- Then use these values and the specified temperature to calculate
DG°.
- By the Hess's Law approach - this is based on the fact that
DG 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 DG
for the overall reaction.
- We can simply look up and add together values of free energies
of formation (if they are available): DG°
= SnpDG°f(prod)
- SnrDG°f(rct).
- Calculate from the equilibrium constant. (We will look at
this later.)
So let's look at examples of each method.
1. Let's look at another example of the first method, using
the equation: DG° = DH°
-TDS°.
Example: Find the free energy of the reaction of sulfur
dioxide and oxygen to give sulfur trioxide under standard conditions
(25°C and 1 atm)
2 SO2(g) + O2(g) Æ
2 SO3(g)
From Appendix 4 of your text we find:
| |
DH°f
(kJ/mol) |
S° (J/molK) |
| SO2(g) |
-297 |
248 |
| O2(g) |
0 |
205 |
| SO3(g) |
-396 |
257 |
|
We need to calculate DH° and
DS° separately.
DH° = SnpDH°f(prod) -
SnrDH°f(rct)
DH° = 2 DH°f(SO3(g))
- {2 DH°f(SO2(g))
+ DH°f(O2(g))}
DH° = 2(-396 kJ/mol)
- {2 (-297 kJ/mol) + 0 kJ/mol)}
DH° = -198 kJ/mol.
DS° = SnpS°(prod)
- SnrS°f(rct)
DS° = 2 S°(SO3(g))
- {2 S°(SO2(g)) + S°(O2(g))}
DS° = 2 (257 J/molK)
- {2 (248 J/molK) + 205 J/molK}
DS° = -187 J/molK
DG° = DH°
-TDS°
DG° = -198 kJ/mol -
(298)(-187 J/molK)
DG° = -198 x 103J/mol
+ 55.73 x 103J/mol
DG° = -142 x 103J/mol
= -142 kJ/mol
© R A Paselk
Last modified 16 September 2003
