The Third Law of Thermodynamics TL-1homepages.gac.edu/~anienow/CHE-371/Lectures/ThirdLawStudent...
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TL-1 The Third Law of Thermodynamics 0 → Δ S r 0 → T Walther Nernst Nernst suggested that the change in entropy for chemical reactions approached 0 as the temperature approached 0 K. as I think that the entropy of a pure substance approaches 0 at 0 K! Max Planck The Third Law: Every substance has a finite positive entropy, but at 0 K the entropy may become 0, and does so in the case of a perfectly crystalline substance.
Transcript of The Third Law of Thermodynamics TL-1homepages.gac.edu/~anienow/CHE-371/Lectures/ThirdLawStudent...
TL-1The Third Law of Thermodynamics
0→Δ Sr 0→T
Walther Nernst
Nernst suggested that the change in entropy for chemical reactions approached 0 as the temperature approached 0 K.
as
I think that the entropy of a pure substance approaches 0 at 0 K!
Max Planck
The Third Law: Every substance has a finite positive entropy, but at 0 K the entropy may become 0, and does so in the case of a perfectly crystalline substance.
TL-2Statistical Mechanics and the 3rd Law
WkS B ln=
jj
jB ppkS ln∑−=
The third law was formulated before the full development of quantum theory. However, statistical thermodynamics gives us molecular insight to the third law.
At 0 K, we expect that the system will be in its lowest energy state and therefore W = 1, S = 0.
p0 = 1 and all other pj ’s = 0. S = 0.
TL-31st and 2nd Law vs 3rd Law
revrev qwdU δδ +=
TdSqrev =δ
PdV−
The 1st and 2nd Law of Thermodynamics introduced new state functions. The 3rd Law of Thermodynamics simply provides an absolute scale for entropy.
EX-TL1
PdVTdSdU −=
VdPPdVdUPVUddH ++=+= )(VdPTdSdH +=
Law 1Law 2
1st and 2nd Law
1st and 2nd Law
TL-4Results of EX-TL1
TC
TS V
V
=⎟⎠⎞
⎜⎝⎛∂∂
TC
TS P
P
=⎟⎠⎞
⎜⎝⎛∂∂
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛∂∂
+=⎟⎠⎞
⎜⎝⎛∂∂
TT VUP
TVS 1
⎥⎦
⎤⎢⎣
⎡−⎟
⎠⎞
⎜⎝⎛∂∂
=⎟⎠⎞
⎜⎝⎛∂∂ V
PH
TPS
TT
1
We work at constant pressure most of the time…
TC
TS P
P
=⎟⎠⎞
⎜⎝⎛∂∂ Integrate with respect to T at constant P
∫=−=Δ 2
1
)()()( 12
T
TP dTT
TCTSTSS
∫==Δ 2
02)()(
T P dTT
TCTSSIf T1 = 0 K
TL-5Phase Transitions!!
∫==Δ 2
02)()(
T P dTT
TCTSS
trs
revtrs T
qS =Δtrs
trstrs T
HS Δ=Δ
If we know Cp(T) we can find ΔS
What happens at phase transitions?
At constant P:
TL-6Practical Absolute Entropies
Values of entropies for gases given in the literature are standard entropies. These are by convention corrected for the non-ideality of gases (to be discussed in detail in Ch 22).
N2
Table 21.1Figure 21.1
TL-7Low T and Debye Theory
3)( TTC sP →
0→T
Peter Debye
34
512
⎟⎟⎠
⎞⎜⎜⎝
⎛Θ
=D
PTRC π
lowTT ≤<0
3)(
512)(
0
23
4 TCdTTRTS PT
D
=Θ
= ∫′π
as Less than 15 K
TL-8Partition Functions and 3rd Law
VNBB T
QTkQkS,
lnln ⎟⎠⎞
⎜⎝⎛
∂∂
+=
∑ ∑
∑−
−
− +=j
j
TkEj
TkEj
TkEB Bj
Bj
Bj
e
eE
TekS /
/
/ 1ln
Remember:
How does S behave as the temperature goes to 0?
Is this consistent with the 3rd Law of Thermodynamics?
See page 861-862 for proofs. 862 and 863 also discusses diatomic and polyatomics.
TL-9Literature Values and Trends
Phase: Gas, liquid, solidMass
# of atoms
Table 21.2
Table 21.3
What are the trends for:
Tables are often a combination of statistical
thermodynamics and calorimetric values.
TL-10Entropy and Molecular Structure
CH3 CH3
OAcetone Trimethylene oxide
Function of Mass
From what E term?
O
C3 H6 O
298 J·K-1·mol-1 274 J·K-1·mol-1Why?
TL-11Sometimes there is not agreement
3.160=calcS
6.155exp =SexpSScalc >
expSSS calcresidual −=
NW 2=
2ln2ln BN
Bresidual NkkS ==
7.52ln == RS
OC ≡ At 81.6 K:
J·K-1·mol-1
J·K-1·mol-1
CO has a very small dipole moment so the molecules do not have a strong tendency to line up in an energetically favorable way. As a result, in the crystal (i.e., low T form) gets “locked” into its own orientation and cannot find the state of lowest energy (i.e., where W = 1).
J·K-1·mol-1
WkS B ln=
3.161exp =+= SSS rescalc J·K-1·mol-1
TL-12Entropy changes of chemical reactions
As used in Homework #3…
zZyYbBaA +→+
][][][][ BbSAaSZzSYySSrooooo −−+=Δ
