The third law: Absolute entropies

T Because there is an absolute zero 2 CpdT ∆S = ò for temperature, the entropy for T T 1 any state can be calculated ...

if we know S(0K). p

T = 0K 1 T C (T) dT S(T) = ò p + S(0K) 0 T

We also defined entropy statistically: W(T) depends on T through the W(T) = number of Boltzmann distribution (fewer accessible states accessible at low T): microstates at T ∝ -∆E/kT S(T) = k lnW(T) Pn e

Nils Walter: Chem 260 This definition gives us a useful S(T = 0K)

Flip three coins P = W = # of probability microstates S = k ln W H H H A

” 1/8 1 0 k

c

i

t

e H H T

g

r 4 possible e H T H B n 3/8 3 1.10 k

e macrostates

o

s

i T H H

“ T T H 8 possible T H T C microstates 3/8 3 1.10 k H T T (distributions) T T T D 1/8 1 0 k

If there is only one accessible microstate, the entropy of the system (macrostate) is given by: This is consistent with our view of entropy SS == kk lnln WW == kk lnln 11 == 00 as a measure of disorder. No disorder (only one possible state) Þ S = 0 @ T = 0 K Nils Walter: Chem 260 The third law of thermodynamics

SS == 00 forfor allall perfectlyperfectly crystallinecrystalline materialsmaterials atat TT == 00 KK If there is only one possible arrangement of atoms (i.e., in a perfect crystal): p SS == kk lnln WW == kk lnln 11 == 00

Since every substance has a finite (nonzero) heat capacity T C (T) dT = 0 and S(T) = ò p + S(0K) 0 T Þ except for a perfect crystal at absolute zero, every substance has a finite positive entropy

Standard absolute or “third law” molar entropies are

tabulated, generally at 1 bar andNils 298Walter: K Chem( ) 260 One can calculate the molar entropy of HCl

98.36 K 158.91 K 188.07 K Solid I Solid II Liquid 12.1 12.6 85.9 Gas 1.3 + 29.5 21.1 9.9 13.5 Sum: S = 185.9 J K-1-1 mol -1-1 Sum: S298.15298.15 = 185.9 J K mol

S = ò C dlnT from 0 K to 16 K using Debye T3-law: C = aT3 Solid I p p ò S = Cp dlnT from 16 K to 98.36 K using experimental Cp S = ∆H/T for each phase transition

ò Solid II S = Cp dlnT from 98.36 K to 158.91 K using experimental Cp ò Liquid S = Cp dlnT from 158.91 K to 188.07 K using experimental Cp ò Gas S = Cp dlnT from 188.07 K to 298.15Nils K using Walter: experimental Chem 260 Cp The exception: An imperfect crystal still has residual entropy

a) Consider CO If there were no preferred O C µ = 0.12 D orientation W = 2N For one mole: W = 2NA S = k ln W = k ln 2NA Actual value: 4.2 J K-1 mol-1 = NAk ln 2 = R ln 2 = 5.8 J K-1 mol-1

b) Consider CH3D No dipole moment, same number of electrons No preferred orientation! W = 4N or for one mole W = 4NA Actual value: 11.7 J K-1 mol-1 S = R ln 4 = 11.5 J K-1 mol-1

Nils Walter: Chem 260 The three laws of thermodynamics Atkins Moore You can’t construct The internal energy of ∆U = q + w a perpetuum mobile 1. an isolated system is Energy Conservation that generates constant energy from nothing ∆S ≥ 0 You can’t construct a everything tends The entropy of the 2. machine that does toward increasing universe tends to nothing but convert disorder increase heat into work The entropy of a S > 0 except for You can never perfectly crystalline 3. perfect crystals at reach absolute substance is zero at T = 0 K zero T = 0 K Summary: The energy of the universe is constant; the entropy of the universe tends always toward a maximum. Rudolf Julius Clausius (1822-1888)Nils Walter: Chem 260