The Third Law of Thermodynamics TL-1 Nernst suggested that the change in entropy for chemical reactions approached 0 as the temperature approached 0 K. as Δr S → 0 T → 0 I think that the entropy of a pure substance approaches 0 at 0 K! Walther Nernst 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. Statistical Mechanics and the 3rd Law TL-2 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 S= kln W will be in its lowest energy state B and therefore W = 1, S = 0. p = 1 and all other p ’s = 0. S = 0. S= k −B∑ p jln j p0 j j 1st and 2nd 3Law vs rd Law TL-3 The 1st and 2nd Law of Thermodynamics introduced new state functions. The 3rd Law of Thermodynamics simply provides an absolute scale for entropy. Law 2 Law 1 qδ rev = TdS dU= δ rev w+δ rev q − PdV dU= TdS− PdV1st and 2nd Law dH d U= () PV+ = dU+ + PdV VdP dH= TdS+ 1st VdP and 2nd Law EX-TL1 Results of EX-TL1 TL-4 ⎛ ∂S ⎞ CV ⎛ ∂S ⎞ C ⎜ ⎟ = ⎜ ⎟ = P ⎝ ∂T ⎠V T ⎝ ∂T ⎠P T ⎛ ∂S ⎞ 1 ⎡ ⎛ ∂U ⎞ ⎤ ⎛ ∂S ⎞ 1 ⎡⎛ ∂H ⎞ ⎤ ⎜ ⎟ =⎢P +⎜ ⎟ ⎥ ⎜ ⎟ = ⎜ ⎟ −V VT∂ ∂V ⎢ ⎥ ⎝ ⎠T ⎣ ⎝ ⎠T ⎦ ⎝PT∂ ⎠T ⎣⎝ ∂P ⎠T ⎦ We work at constant pressure most of the time… ⎛ ∂S ⎞ C ⎜ ⎟ = P Integrate with respect to T at constant P ⎝ ∂T ⎠P T T CT2 P() ΔSSTST =()()2 −1 = dT ∫T1 T T CT2 P() If T1 = 0 K ΔSST =()2 = dT ∫0 T Phase Transitions!! TL-5 TCT2 () P If we know Cp(T) we can find ΔS ΔSST =()2 = dT ∫0 T What happens at phase transitions? q rev Δtrs H Δtrs S = At constant P: ΔS = T trs trs T trs Practical Absolute Entropies TL-6 Table 21.1 Figure 21.1 N2 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). TheoryLow T and Debye TL-7 s 3 CTTP()→ as T → 0 Less than 15 K 3 12π 4 ⎛ T ⎞ Peter Debye ⎜ ⎟ 0 <TT≤ CP = R⎜ ⎟ low 5 ⎝ ΘD ⎠ 4 12π R T′ CT() ST()= T2 dT= P 3 ∫0 5ΘD 3 Partition Functions and 3rd Law TL-8 Remember: ⎛ ∂ lnQ ⎞ S= kB ln Q + B k⎜ T ⎟ ⎝ ∂T ⎠NV, E− j/ k B T ∑Ej e E− / k T 1 S= kln ej B + j B ∑ E− j/ k B T j T ∑e j 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. Literature Values and Trends TL-9 Table 21.2 What are the trends for: Phase: Gas, liquid, solid Mass # of atoms Tables are often a combination of statistical thermodynamics and Table 21.3 calorimetric values. Entropy and Molecular Structure TL-10 Function of Mass From what E term? Acetone Trimethylene oxide O O C3H 6 O HC3 CH3 298 J·K-1·mol-1 274 J·K-1·mol-1 Why? Sometimes there is not agreement TL-11 CO≡ At 81.6 K: -1 -1 Scalc 160= J·K . 3·mol SScalc > exp -1 -1 Sexp 155= J·K . 6·mol CO has a very small dipole moment SSSresidual=calc − exp so the molecules do not have a strong tendency to line up in an N W = 2 S= kB ln W energetically favorable way. As a N result, in the crystal (i.e., low T form) Sresidual=lnB k 2 =NkB ln 2 gets “locked” into its own orientation and cannot find the state of lowest SR=ln 2 = 5J·K . 7-1·mol-1 energy (i.e., where W = 1). -1 -1 SSScalc=res +exp 161 = J·K . 3 ·mol Entropy changes of chemical reactions TL-12 As used in Homework #3… aA bB+ → yY + zZ o o o o o SΔ ySr = Y[][][][] zS + Z − aS A − bS B.