Thermodynamics: the Second Law 자연과학대학 화학과 박영동 교수 The Direction of Nature and spontaneity

Thermodynamics: the Second Law 4.1 Entropy 4.1.1 The direction of spontaneous change 4.1.2 Entropy and the Second Law 4.1.3 The entropy change accompanying expansion 4.1.4 The entropy change accompanying heating 4.1.5 The entropy change accompanying a phase transition 4.1.6 Entropy changes in the surroundings 4.1.7 Absolute entropies and the Third Law of thermodynamics 4.1.8 The statistical entropy 4.1.9 Residual entropy 4.1.10 The standard reaction entropy 4.1.11 The spontaneity of chemical reactions 4.2 The Gibbs energy 4.2.12 Focusing on the system 4.2.13 Properties of the Gibbs energy

The direction of spontaneous change the chaotic dispersal of energy the dispersal of matter One of spontaneous process is the chaotic dispersal of matter. This tendency accounts for the spontaneous tendency of a gas to spread into and fill the container it occupies. It is extremely unlikely that all the particles will collect into one small region of the container. (In practice, the number of particles is of the order of 1023.) Another spontaneous process is the chaotic dispersal of energy (represented by the small arrows). In these diagrams, the orange spheres represent the system and the purple spheres represent the surroundings. The double-headed arrows represent the thermal motion of the atoms.

Can we convert heat to work completely? ch04f03 The Second Law denies the possibility of the process illustrated here, in which heat is changed completely into work, there being no other change. The process is not in conflict with the First Law, because the energy is conserved.

The Second Law 1. Reversible vs Irreversible processes Reversible: A process for which a system can be restored to its initial state, without leaving a net influence on the system or its environment. * idealized, frictionless; * proceeds slowly enough for the system to remain in thermodynamic equilibrium. Irreversible: Not reversible * natural; * proceeds freely, drives the system out of thermodynamic equilibrium; * interacts with environment, can not be exactly reversed

Example: Gas-piston system under a constant temperature * Slow expansion and compression * Rapid expansion and compression

The Carnot cycle and entropy The cycle operates with one mole ideal gas. 1) 1 →2: Isothermal expansion 2) 2 → 3: Adiabatic expansion

3) 3 → 4: Isothermal compression 4) 4 → 1: Adiabatic compression

4. Work for adiabatic expansion from chap. 2 4. Work for adiabatic expansion For an adiabatic process for an ideal gas,

The net heat transfer and the net work over the Carnot cycle are: 2→3, 4→1, Then, So the system absorbs heat and performs net work in the Carnot cycle, which behaves as a heat engine.

The Carnot cycle -summary q w ∆U ∆H ∆S 1→2 RThln(V2/V1) -RThln(V2/V1) Rln(V2/V1) 2→3 cV(Tc-Th) cP(Tc-Th) 3→4 -RTcln(V2/V1) RTcln(V2/V1) -Rln(V2/V1) 4→1 cV(Th-Tc) cP(Th-Tc) cycle R(Th-Tc)ln(V2/V1) -R(Th-Tc)ln(V2/V1)

For the Carnot cycle, we can also have: This relationship also hold for the reversed Carnot cycle. This is called the Carnot’s Theorem: (4.1) The change of S is independent of path under a reversible process. S is a state function, means a system property. It is called entropy.

3. The Second Law and its Various Forms To get the second law, we use the Clausius Inequality, i.e., for a cyclic process, which indicates during the cycle, heat must be rejected to the environment; heat exchange is larger at high temperature than at low temperature under reversible conditions; the net heat absorbed is smaller under the irreversible condition than under the reversible condition.

Now, consider two cycles as shown in the plot. For the cycle which contains one reversible process and one irreversible process, (4.2) For the cycle which have two reversible processes, (4.3)

The difference between (4.2) and (4.3) gives Because states 1 and 2 are arbitrary, we have the second law, (4.4) Combine (4.1) and (4.4), we have or It indicates that the heat absorbed by the system during a process has an upper limit, which is the heat absorbed during a reversible process. The first law relates the state of a system to work done on it and heat it absorbs. The second law controls how the systems move to the thermodynamic equilibriums, i.e., the direction of processes.

The Second Law of Thermodynamics The Second Law of Thermodynamics establishes that all spontaneous or natural processes increase the entropy of the universe DStotal = DSuniv = DSsys + DSsurr In a process, if entropy increases in both the system and the surroundings, the process is surely spontaneous

Several simplified forms of the second law: 1) For an adiabatic process, (4.4) becomes (4.5) If the adiabatic process is reversible, then It is also isentropic (S is constant). 2) For an isochoric process, (4.1) becomes (4.6) Because only state variables are involved, it holds for either reversible or irreversible processes. (4.5) and (4.6) show that : Irreversible work can only increase entropy; heat transfer can either increase or decrease entropy.

5. Thermodynamic Equilibrium * Consider an adiabatic process, the second law becomes For an irreversible condition, or is the entropy at the initial state. When reaches the maximum, the state is in thermodynamic equilibrium because the entropy can not increase anymore.

Calculate the changes in entropy as a result of the transfer of 100 kJ of energy as heat to a large mass of water (a) at 0°C (273 K) and (b) at 100°C (373 K). (a) ΔS at 0°C (273 K) (b) ΔS at 100°C (373 K)

Heat engines The engine will not operate spontaneously if this change in entropy is negative, and just becomes spontaneous as ΔStotal becomes positive. This change of sign occurs ΔStotal = 0, which is achieved when the efficiency, η, of the engine, the ratio of the work produced to the heat absorbed, is

Refrigerators, and heat pumps

Refrigerator power No thermal insulation is perfect, so there is always a flow of energy as heat into the sample at a rate proportional to the temperature difference. The rate at which heat leaks in can be written as A(Th – Tc ), where A is a constant. Calculate the minimum power, P, required to maintain the original temperature difference? Assume the refrigerator is operating at 100% of its theoretical efficiency. Express P in terms of A, Th, Tc.

The entropy change with isothermal expansion ch04f04

The entropy change with heating ch04f05 In case heat capacity is constant,

The entropy change with heating In case heat capacity is temperature dependent, ch04f06

The entropy change with a phase transition When a solid (a), melts, the molecules form a more chaotic liquid, the disorderly array of spheres (b). As a result, the entropy of the sample increases. At phase transition, the temperature stays constant, T = Ttr.

Absolute entropies and the Third Law of thermodynamics S (0) = 0 for all perfectly ordered crystalline materials. ch04f08

nonmetallic solids, Debye T 3 -law: At temperatures T << TD, ch04f09 nonmetallic solids, Debye T 3 -law: At temperatures T << TD, Cv,m = aT 3 , and Sm (T) = 1 3 Cv,m  

Standard Molar Entropies In general, the more atoms in its molecules, the greater is the entropy of a substance ch04f10

The statistical entropy The 19 arrangements(W ) of four molecules in a system with three energy levels and a total energy of 4ε. U = 4ε

A simple 4-particle system (in equal-spaced energy levels) U = nε U = 0 U = ε U = 2ε U = 3ε U W 1 4 2 10 3 20 35 5 56 6 84 7 120 8 165 9 220 286 W = 1 W = 20 W = 4 W = 10 W = C(n+3, n) = (n+3)(n+2)(n+1)/6 S = kB ln(W )

A simple 4-particle system W = C(n+3, n) = (n+3)(n+2)(n+1)/6 U = nε S = kB ln(W ) From S = kB ln W, U = nε, T = dU/dS Cv = dU/dT

Two–Level system a very simplified version In case when ε >> kT, almost no particle can reach upper state. According to Boltzmann, the probability of a particle at level 1 at T is p1 = exp(-ε/kT ). Energy U = 0 U = ε p1 = exp(-ε/kT )  A two-level system

Two–Level system a very simplified version ε 1.66E-19 "=100 kJ/mol" kB 1.36E-23 　J/K T, K U(T), J/mol Cv(T), J/mol 0.001 1 10 100 9.00E-49 1.10E-48 1000 4.96E-01 6.06E-03 U(T) = Nε exp(-ε/kT ) Cv,m = (∂U/ ∂T) = Nε exp(-ε/kT ) (-ε/k) (-1/T 2) = N (ε 2/kT 2) exp(-ε/kT ) ≈ 0 1. This is why we do not consider heat capacity contributions from subatomic particles such as electrons, etc. 2. This is why heat capacity approaches 0 when T approaches 0 for every material.

W and Energy Level Spacing At a given temperature, the number of arrangements corresponding to the same total energy is greater when the energy levels are closely spaced than when they are far apart.

Residual entropy for CO molecules, there are 2N possible arrangements at T= 0. Sm = NAk ln 2 = R ln 2 = 5.8 J K−1 mol−1

Ice at 0 K W = ( 3 2 ) N, S = k ln ( 3 2 ) N = Nk ln ( 3 2 ) , Sm = R ln ( 3 2 ) = 3.4 J K−1 mol−1.

The spontaneity of chemical reactions at constant V, qV = ΔU. ∆𝑆 𝑠𝑢𝑟𝑟 =− ∆𝑈 𝑇 q ∆𝑆 𝑢𝑛𝑖𝑣 = ∆𝑆+ ∆𝑆 𝑠𝑢𝑟𝑟 =∆𝑆− ∆𝑈 𝑇 ≥0 ∆𝑈 −𝑇∆𝑆 ≤0 if we define A ≡ U – TS, Helmholtz Free Energy, ΔA = ΔU –TΔS at constant V. So, at constant T and V, ΔA = ΔU –TΔS ≤ 0.

The spontaneity of chemical reactions at constant p, qp = ΔH. q if we define G ≡ H – TS, Gibbs Free Energy, ΔG = ΔH –TΔS at constant T. So, at constant T and p, ΔG = ΔH –TΔS ≤ 0.

Gibbs Energy, G Spontaneity criterion ΔSuniv > 0. at constant pressure and temperature, ΔGsystem < 0.

Standard molar Gibbs Energies Δr 𝐺 ⊖ = Δr 𝐻 ⊖  TΔr 𝑆 ⊖ The Born equation ∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 r 𝜀 r = relative permittivity For water, 𝜀 r = 78.54 at 25 ℃ ∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑟 i /pm ×6.86× 10 4 kJ mol −1

Gibbs Energies of solvation The Born equation ∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 r 𝜀 r = 𝜀 𝜀 0 = relative permittivity For water, 𝜀 r = 78.54 at 25 ℃ ∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑟 i /pm ×6.86× 10 4 kJ mol −1

Gibbs Energies of solvation The Born equation 𝑉(𝑟)= 𝑄 1 𝑄 2 4π𝜀𝑟 𝜀= permittivity of the medium For water, 𝜀 r = 78.54 at 25 ℃ ∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑟 i /pm ×6.86× 10 4 kJ mol −1 ∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 r

Gibbs Energies of solvation The Born equation 𝑉(𝑟)= 𝑄 1 𝑄 2 4π𝜀𝑟 𝜀= permittivity of the medium 𝜀 𝜀 𝑉 𝑟 = 𝑄 1 𝜙 𝑟 𝜙(𝑟)= 𝑄 2 4π𝜀𝑟 𝑟 i 𝑤= 0 𝑧 i 𝑒 𝜙 𝑟 i 𝑑𝑄 = 1 4𝜋𝜀 𝑟 i 0 𝑧 i 𝑒 𝑄𝑑𝑄 = 𝑧 i 2 𝑒 2 𝑒 2 8π𝜀 𝑟 i ∆ solv 𝐺 ⊖ = 𝑧 i 2 𝑒 2 𝑁 A 8πε 𝑟 i − 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 r =− 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 𝜀 0 =− 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 r

Combining the First and Second Laws 𝑑𝑈=𝑇𝑑𝑆 −𝑝𝑑𝑉 𝑈(𝑆, 𝑉) 𝑑𝑈= 𝜕𝑈 𝜕𝑆 𝑉 𝑑𝑆 − 𝜕𝑈 𝜕𝑉 𝑆 𝑑𝑉 𝜕 2 𝑈 𝜕𝑉𝜕𝑆 = 𝜕 2 𝑈 𝜕𝑆𝜕𝑉 𝜕𝑈 𝜕𝑆 𝑉 =𝑇 𝜕𝑈 𝜕𝑉 𝑆 =−𝑝 𝜕𝑇 𝜕𝑉 𝑆 =− 𝜕𝑝 𝜕𝑆 𝑉

The Maxwell relations 𝑈 𝑆, 𝑉 ;𝑑𝑈=𝑇𝑑𝑆 −𝑝𝑑𝑉;𝑇= 𝜕𝑈 𝜕𝑆 𝑉 ;𝑝=− 𝜕𝑈 𝜕𝑉 𝑆 ; 𝜕𝑇 𝜕𝑉 𝑆 =− 𝜕𝑝 𝜕𝑆 𝑉 𝐻 𝑉, 𝑇 ;𝑑𝐻=𝑇𝑑𝑆+𝑉𝑑𝑝; 𝑇= 𝜕𝐻 𝜕𝑆 𝑝 ; 𝑉= 𝜕𝐻 𝜕𝑝 𝑆 ; 𝜕𝑇 𝜕𝑝 𝑆 = 𝜕𝑉 𝜕𝑆 𝑝 𝐺 𝑉, 𝑇 ;𝑑𝐺=𝑉𝑑𝑝 −𝑆𝑑𝑇; 𝑉= 𝜕𝐺 𝜕𝑝 𝑇 ; 𝑆=− 𝜕𝐺 𝜕𝑇 𝑝 ; 𝜕𝑉 𝜕𝑇 𝑝 = − 𝜕𝑆 𝜕𝑝 𝑇 𝐴 𝑉, 𝑇 ;𝑑𝐴=−𝑝𝑑𝑉 −𝑆𝑑𝑇;𝑝=− 𝜕𝐴 𝜕𝑉 𝑇 ;𝑆=− 𝜕𝐴 𝜕𝑇 𝑉 ; 𝜕𝑝 𝜕𝑇 𝑉 = 𝜕𝑆 𝜕𝑉 𝑇

Properties of the Gibbs energy 𝑑𝐺=𝑉𝑑𝑝 −𝑆𝑑𝑇 𝐺(𝑝, 𝑇) 𝜕𝐺 𝜕𝑇 𝑝 =−𝑆 𝜕𝐺 𝜕𝑝 𝑇 =𝑉 𝜕𝐺 𝜕𝑇 𝑝 = 𝐺−𝐻 𝑇 𝜕(𝐺/𝑇) 𝜕𝑇 𝑝 =− 𝐻 𝑇 2 𝜕(∆𝐺/𝑇) 𝜕𝑇 𝑝 =− ∆𝐻 𝑇 2

the Gibbs energy vs p 𝜕𝐺 𝜕𝑝 𝑇 =𝑉 𝜕 𝐺 𝑚 𝜕𝑝 𝑇 = 𝑉 𝑚 𝑑𝐺=𝑉𝑑𝑝 −𝑆𝑑𝑇 𝐺(𝑝, 𝑇) 𝜕𝐺 𝜕𝑝 𝑇 =𝑉 𝜕 𝐺 𝑚 𝜕𝑝 𝑇 = 𝑉 𝑚 𝑑𝐺=𝑉𝑑𝑝 −𝑆𝑑𝑇 𝐺(𝑝, 𝑇) 𝐺 𝑚 𝑝 f = 𝐺 𝑚 𝑝 i + 𝑝 i 𝑝 f 𝑉 𝑚 𝑑𝑝 𝐺 𝑚 𝑝 f = 𝐺 𝑚 𝑝 i + 𝑉 𝑚 ( 𝑝 f − 𝑝 i ) ⊖ 𝐺 𝑚 𝑝 f = 𝐺 𝑚 𝑝 i +𝑅𝑇 ln 𝑝 f 𝑝 i 𝐺 𝑚 𝑝 f = 𝐺 𝑚 𝑝 i +𝑅𝑇 ln(𝑓 𝑝 f 𝑝 i ) 𝐺 𝑚 𝑝 = 𝐺 𝑚Θ +𝑅𝑇 ln(𝑓𝑝⊖)

The Gibbs energy and the fugacity 𝐺 𝑚 𝑝 f = 𝐺 𝑚 𝑝 i +𝑅𝑇 ln 𝑝 f 𝑝 i 𝐺 𝑚 𝑝 f = 𝐺 𝑚 𝑝 i +𝑅𝑇 ln(𝑓 𝑝 f 𝑝 i ) 𝐺 𝑚 𝑝 = 𝐺 𝑚 Θ+𝑅𝑇 ln(𝑓𝑝Θ) 𝑓:fugacity 𝑓=𝜙𝑝 𝜙 :fugacity coefficient ⊖ ln 𝜙 = 0 𝑝 𝑍−1 𝑝 d𝑝

dSU,V  0 (b) dUS,V  0 dSH,p  0 (b) dHS,p  0 A = U – TS G = H – TS Property Equation Comment Criteria of spontaneity dSU,V  0 (b) dUS,V  0 Constant volume (etc)*   dSH,p  0 (b) dHS,p  0 Constant pressure (etc) Helmholtz energy A = U – TS Definition Gibbs energy G = H – TS

Property Equation Comment (a) dSU,V  0 (b) dUS,V  0 Criteria of spontaneity (a) dSU,V  0 (b) dUS,V  0 Constant volume (etc)* (a) dSH,p  0 (b) dHS,p  0 Constant pressure (etc) Helmholtz energy A = U – TS Definition Gibbs energy G = H – TS   (a) dAT,V  0 (b) dGT,p  0 Constant temperature (etc) Equilibrium dAT,V = 0 Constant volume (etc) Maximum work dwmax = dA, wmax = A Constant temperature dGT,p = 0 Maximum non-expansion work dwadd,max = dG, wadd,max = G Constant temperature and pressure Standard Gibbs energy of reaction Δr 𝐺 ⊖ = Δr 𝐻 ⊖  TΔr 𝑆 ⊖ ∆ r 𝐺 ⊖ = J 𝜈 J ∆ f 𝐺 m ⊖ (J) Practical implementation Ions in solution Δf 𝐺 ⊖ (H+, aq) = 0 Convention Born equation ∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 r Solvent a continuum

Property Equation Comment Fundamental equation dU = TdS  pdV No additional work Fundamental equation of chemical thermodynamics dG = Vdp  SdT Variation of G (G/p)T = V and (G/T)p = S Composition constant Gibbs–Helmholtz equation ((G/T)/T)p = H/T 2 Pressure dependence of G Gm(pf) = Gm(pi) + VmΔp Incompressible substance G(pf) = G(pi) + nRT ln(pf/pi) Perfect gas 𝐺 m = 𝐺 m ⊖ +𝑅𝑇 ln(𝑝/ 𝑝 ⊖ ) Fugacity 𝐺 m = 𝐺 m ⊖ +𝑅𝑇 ln(𝑓/ 𝑝 ⊖ ) Definition Fugacity coefficient   f = p ln 𝜙 = 0 𝑝 𝑍−1 𝑝 d𝑝 Determination

Calculate the change in molar entropy when one mole of argon gas is compressed from 2.0 dm3 to 500 cm3 and simultaneously heated from 300 K to 400 K. Take CV,m = (3/2)R p V T = 300 K T = 400 K

Calculate the change in entropy when 100 g of water at 80°C is poured into 100 g of water at 10°C in an insulated vessel given that Cp,m = 75.5 J K−1 mol−1.

The enthalpy of vaporization of chloroform (trichloro-methane), CHCl3, is 29.4 kJ mol−1 at its normal boiling point of 334.88 K. (a) Calculate the entropy of vaporization of chloroform at this temperature. (b) What is the entropy change in the surroundings?

Suppose that the weight of a configuration of N molecules in a gas of volume V is proportional to VN. Use Boltzmann’s formula to deduce the change in entropy when the gas expands isothermally.

Without performing a calculation, estimate whether the standard entropies of the following reactions are positive or negative: