Entropy
Microstates
1
Statistics of coin flips:
# molecules
10
100
1000
10000
1 × 106
45%–55%
0.246
0.632
0.998
1
1
49%–51%
49.9%–50.1%
0.452
0.953
1
0.151
0.954
General statements about entropy, often (but not always) true:
1. Solids are typically low-entropy: molecules are locked down
into well-defined positions and (often) well-defined orientations.
2. Liquids have higher entropy: once the solid is allowed to melt,
molecules can now translate and rotate.
3. Gases have much higher entropy: each location any molecule
can find itself in is a different configuration, and there are a
near-infinite (but not infinite!) number of of locations.
4. Solvation often increases entropy, but not always: introducing ions into aqueous solution can sometimes create more
order as the water molecules lock into position around ions.
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5. Larger molecules have more internal motions possible, and
have more entropy than smaller molecules and atoms.
Entropy and qrev
∆S = Sfinal − Sinitial
Relating heat to entropy requires a reversible process:
A reversible process is one where the system remains at
equilibrium throughout. Because of this, it can change
direction at any time.
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We can melt an ice cube reversibly by putting it in an environment that is infinitesimally warmer than 0 ◦ C. When we do
this, heat is transferred very slowly from the surroundings into the
system, and when ∆Hfus has been transferred, the ice is melted.
Lowering the surroundings infinitesimally below 0 ◦ C will freeze the
ice cube again. We will use
qrev = T ∆S
or
∆S =
qrev
T
Here,
qrev = ∆Hfus
and so
∆Sfus
=
=
=
=
qrev
T
∆Hfus
T
6.01 kJ/mol
273 K
22.0 J mol−1 K−1
The Second Law of Thermodynamics
The entropy of the universe (system and surroundings)
remains constant in a reversible process, and it increases
in an irreversible (spontaneous) process.
In the previous example, if the surroundings are not at 0 ◦ C, the
process will be irreversible. Let’s say
Tcold = 273 K
∆Ssurr
and
qsys
T
∆Hfus
= −
T
= −
4
Thot = 363 K
6.01 kJ/mol
363 K
= −16.6 J mol−1 K−1
= −
Now,
∆Suniverse = ∆Ssurr + ∆Ssys = 22.0 − 16.6 = 5.4 J mol−1 K−1
Standard Molar Entropy
Standard molar entropy, S ◦ :
1. S ◦ is the entropy change to take 1 mole of a substance from
crystalline form at absolute zero (0 K) to a standard temperature of 298 K. (This is slightly simplified.)
2. S ◦ is always positive.
3. Solids tend to have lower molar entropies than liquids, which
tend to have lower entropies than gases.
4. Heavy atoms and floppy bonds increase molar entropy; light
atoms and rigid bonds decrease it.
Entropy of a substance always increases with temperature, and we
can plot an idealized curve of this:
To calculate entropy changes for chemical reactions, we just subtract the entropy of the products from the entropy of reactants.
For
H2 (g) + Cl2 (g) −→ 2HCl(g)
we have
S ◦ (H2 )
=
130.7 J mol−1 K−1
S ◦ (Cl2 )
=
223.1 J mol−1 K−1
=
186.9 J mol−1 K−1
◦
S (HCl)
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◦
Srxn
◦
◦
= Sprod
− Sreact
=
2(186.9) − (130.7 + 223.1)
=
20 J mol−1 K−1
In most cases, producing gases (or increasing the number of moles
of gas) will increase entropy. Take
CaCO3 (s) −→ CaO(s) + CO2 (g)
with
S ◦ (CaCO3 )
=
88.0 J mol−1 K−1
S ◦ (CO2 )
=
213.8 J mol−1 K−1
S ◦ (CaO)
=
38.1 J mol−1 K−1
which gives us
◦
Srxn
◦
◦
= Sprod
− Sreact
=
(213.8 + 38.1) − (88.0)
=
163.9 J mol−1 K−1
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Thermodynamic Cycles
When a system changes temperature, its entropy increases (higher
T ) or decreases (lower T ). The amount of entropy change is related
to the heat capacity of the material, like so:
∆S
=
∆S
=
Tfinal
Tinitial
Tfinal
nCV ln
Tinitial
nCP ln
(constant pressure)
(constant volume)
These equations make it possible to calculate the change in entropy
to
1. heat something,
2. do a phase change or reaction, and
3. cool it back down.
For example, the standard molar entropies of liquid and gaseous
water—remember, this is at 298 K—are 70.0 and 188.8 J mol−1 K−1 ,
respectively. The molar heat capacities are 0.232 and 0.115 J mol−1 K−1 .
We can calculate the entropy of vaporization at room temperature
very simply:
∆Svap = 188.8 − 70.0 = 118.8 J mol−1 K−1
at 373 K, we do the following cycle:
and calculate ∆S for each step:
1. Cool liquid water at 373 K to 298 K:
∆S1
Tfinal
Tinitial
298
= 0.232 ln
J mol−1 K−1
373
= −0.052 J mol−1 K−1
= nCP ln
2. Vaporize:
∆S2 = 118.8 J mol−1 K−1
3. Heat the vapor back to 373 K:
∆S3
Tfinal
Tinitial
373
= 0.115 ln
J mol−1 K−1
298
= 0.026 J mol−1 K−1
= nCP ln
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The entropy of vaporization is
∆Svap @ 373 K = ∆S1 +∆S2 +∆S3 = −0.052+118.8+0.026 = 118.78 J mol−1 K−1
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