The Second Law of Thermodynamics (Chapter 4)

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1 The Second Law of Thermodynamics (Chapter 4) First Law: Energy of universe is constant: ΔE system = - ΔE surroundings Second Law: New variable, S, entropy. Changes in S, ΔS, tell us which processes made possible by first law actually occur spontaneously. spontaneous not spontaneous

2 Entropy: Statistical Definition What is the entropy change for gas to go from V 1 to V 2? Start with 1 atom: After removing barrier: Probability of finding atom in V 2 = 1 Probability of finding atom in V 1 = 1/2

3 Entropy: Statistical Definition N atoms: Probability of finding all N atoms in V 1 = (1/2) N Call this probability W: W = (1/2) N What is probability of finding N = 100 atoms all in half the container? W = (1/2) 100 = 8 x 10-31

4 Entropy: Statistical Definition S, entropy is a measure of probability, W. Want S to be extensive, like U and H. Is W extensive? No: In our example, W = (1/2) N, so W is not proportional to N (amount of material). Define entropy: S = k B ln W Boltzmann s constant: k B = x J K -1 Entropy extensive: S = k B ln W = k B ln (1/2) N = N k B ln (1/2) With this definition S is proportional to N.

5 Entropy: Statistical Definition Example: ΔS = S 2 - S 1 for isothermal expansion from V 1 to V 2. S 1 = k B ln W 1 S 2 = k B ln W 2 ΔS = S 2 - S 1 = k B (ln W 2 - ln W 1 ) = k B ln W 2 /W 1 We know: W 1 = (c V 1 ) N c = constant W 2 = (c V 2 ) N ΔS = k B ln W 2 /W 1 = k B ln (V 2 /V 1 ) N = k B N ln (V 2 /V 1 ) ΔS = N k B ln (V 2 /V 1 ) = n R ln (V 2 /V 1 )

6 Entropy: Statistical Definition ΔS = N k B ln (V 2 /V 1 ) = n R ln (V 2 /V 1 ) Example: Calculate ΔS when 2.0 moles of ideal gas expand isothermally from 1.5 L to 2.4 L. What is the probability that it spontaneously contracts back to 1.5 L?

7 Entropy: Thermodynamic Definition Consider isothermal, reversible expansion of ideal gas. First law: ΔU = q + w. For ideal gas, if T is constant ΔU = 0. w rev = - n R T ln (V 2 /V 1 ) = -q rev (ΔU = 0, so w = -q) Then: q rev = n R T ln (V 2 /V 1 ) = TΔS ΔS = q rev / T Entropy change = heat absorbed in reversible process temperature at which process occurs

8 Entropy: Thermodynamic Definition ΔS = q rev / T Entropy change = heat absorbed in reversible process temperature at which process occurs We have focused on system. What about surroundings? Surroundings (rest of universe) are VERY large; any change to surroundings is reversible: q surroundings (q surroundings ) rev ΔS surr = change of entropy of surroundings = q surr / T

9 Entropy Change of Universe ΔS univ = ΔS sys + ΔS surr ΔS surr = change of entropy of surroundings = q surr / T ΔS univ = ΔS sys + q surr / T Reversible, isothermal expansion: ΔS sys = q sys / T Also: heat emitted by system = heat absorbed by surroundings q sys = - q surr ΔS univ = q sys / T + q surr / T = - q surr / T + q surr / T = 0

10 Entropy Change of Universe Reversible, isothermal expansion: ΔS univ = 0 Irreversible, isothermal expansion: ΔS sys > q sys / T Because q sys = - w sys, and system does less work if process is irreversible. ΔS univ = ΔS sys + ΔS surr = ΔS sys + q surr / T = ΔS sys - q sys / T > 0

11 Entropy Change of Universe Reversible, isothermal expansion: ΔS univ = 0 Irreversible, isothermal expansion: ΔS univ > 0 Second law of thermodynamics: Entropy of universe (or isolated system) increases in an irreversible, spontaneous process and remains unchanged in a reversible process.

12 Example: 0.50 moles of ideal gas at 20 C expands isothermally against a constant pressure of 2.0 atm from 1.0 L to 5.0 L. Calculate the change in entropy of the system, surroundings and universe. Is the process spontaneous?

13 Entropy changes We saw that for isothermal expansion of ideal gas: ΔS = n R ln (V 2 /V 1 ) What is the entropy change for: 1. Mixing gases? 2. Phase transition? 3. Heating? 4. Chemical reaction? 1. Entropy change due to mixing of ideal gases.

14 1. Entropy change due to mixing of ideal gases. For gas A: For gas B: ΔS A = n A R ln ((V A +V B )/V A ) ΔS B = n B R ln ((V A +V B )/V B ) Entropy of mixing: Δ mix S = ΔS A + ΔS B Recall: V n (constant T, p) Avogadro s Law For gas A: ΔS A = n A R ln ((n A +n B )/n A ) = -n A R ln (n A /(n A +n B )) ΔS A = -n A R ln x A x A = mole fraction A

15 1. Entropy change due to mixing of ideal gases. Entropy of mixing: Δ mix S = ΔS A + ΔS B ΔS A = -n A R ln x A x A = mole fraction A ΔS B = -n B R ln x B x B = mole fraction B Δ mix S = -n A R ln x A -n B R ln x B Δ mix S > 0 (x A, x B < 1) Spontaneous process

16 2. Entropy change due to phase transition. At 0 C and 1 atm, ice and water are in equilibrium. If we start with ice, heat is absorbed to form water. Enthalpy change: Δ H = q p Heat of fusion (melting): Δ fus H = q p Entropy of fusion: Δ fus S = q p /T f = Δ fus H/T f (melting at equilibrium is reversible, so ΔS = q p /T ) T f = fusion or melting point = 273 K for water at 1 atm. Similarly, for boiling: Δ vap S = Δ vap H/T b T b = boiling point = 373 K for water at 1 atm.

17 2. Entropy change due to phase transition. Example: Given Δ fus H (H 2 O) = 6.01 kj mol -1 Δ vap H (H 2 O) = kj mol -1 What are Δ fus S and Δ vap S for 1 mole H 2 O at normal melting and boiling points?

18 3. Entropy change due to heating. Consider heating system at constant p, T 1 --> T 2. Entropy change: ds = dh T = C pdt T Integrate from T 1 to T 2 : S 2 ds = S 1 T 2 T C p dt dt = C 2 p T T T 1 T 1 (We are assuming C p does not vary with T.) Then: ΔS = S 2 S 1 = C p ln T 2 = nc p ln T 2 T 1 T 1

19 3. Entropy change due to heating. Example: 200 g H 2 O is heated from 10 C to 20 C at constant p. Given C p (H 2 O) = 75.3 J K -1 mol -1 at these temperatures, what is ΔS?

20 Third law of thermodynamics: Every substance has a finite, positive entropy, but at absolute zero, entropy may be zero, as for a perfect crystalline substance. At T = 0 K, crystal usually in only 1 state ---> W(crystal) = 1. (Recall: W = probability for system to be in a state.) S = k B ln W = k B ln (1) = 0 for perfect crystal at T = 0 K. Absolute U, H cannot be determined, only ΔU and ΔH. Absolute S can be determined, since S = 0 at T = 0 K. T dt At other temperatures we calculate: S = C p T 0

21 Third law of thermodynamics: Every substance has a finite, positive entropy, but at absolute zero, entropy may be zero, as for a perfect crystalline substance. At temperature, T, we calculate: S = C p dt T Entropy as function of T for a particular substance: Measure C p as function of T, then find S using: S = T 0 C p dt T S T 0 T

22 4. Entropy change due to chemical reaction. Δ r S = ν S (products) - ν S (reactants) S = standard molar entropy (J mol -1 K -1 ) ν = stoichiometric coefficient (no units) Example: Standard molar entropy change at 298 K for H 2 (g) + (1/2)O 2 (g) --> H 2 O( )

23 Gibbs and Helmholtz Energy We would like to decide if a process occurs spontaneously by focusing on the system. Spontaneous: ΔS univ = ΔS sys + ΔS surr > 0 = ΔS sys + q surr / T > 0 = ΔS sys - q sys / T > 0 At constant P: q sys = ΔH sys At constant V: q sys = ΔU sys (Now only system variables!) At constant P, T: ΔS sys - ΔH sys / T > 0 Spontaneous process

24 Gibbs Energy At constant P, T: ΔS sys - ΔH sys / T > 0 Spontaneous process All variables are system variables: we can drop sys subscript. Gibbs energy, G: G = H - TS Isothermal process: ΔG = ΔH - TΔS Therefore: ΔG < 0 spontaneous dg = 0 equilibrium G is a state function (like H and S)

25 Helmholtz Energy ΔS univ = ΔS sys - q sys / T > 0 At constant V: q sys = ΔU sys At constant V, T: ΔS sys - ΔU sys / T > 0 Spontaneous process Helmholtz energy, A: Isothermal process: ΔA A = U - TS = ΔU - TΔS Therefore: ΔA < 0 spontaneous da = 0 equilibrium A is a state function (like U and S)

26 Helmholtz Energy Interesting relation between ΔA and maximum possible work Isothermal process: ΔA = ΔU - TΔS Maximum work attained during reversible process, where TΔS = q rev Then: ΔA = ΔU - q rev = (q rev + w rev ) - q rev (Note: ΔU = q rev + w rev ) Therefore: ΔA = w rev, which is the maximum possible work the system can do.

27 Example: C 6 H 12 O 6 (s) + 6O 2 (g) --> 6CO 2 (g) + 6 H 2 O(l) Assume: T = 25 C, what is maximum possible work (Δ r A)? Δ r U = Δ r H = kj/mol (we found this earlier)

28 Chemical Reactions: If reaction is studied at constant p (usual case), then we compute Gibbs energy of reaction, Δ r G, to decide if the reaction is spontaneous. Δ r G = ν Δ f G (products) - ν Δ f G (reactants) Δ f G = standard molar Gibbs energy of formation (kj mol -1 ) ν = stoichiometric coefficient (no units) Δ f G listed for many compounds in Appendix 2. How are values of Δ f G measured?

29 Chemical Reactions: Δ f G listed for many compounds in Appendix 2. How are values of Δ f G measured? Example: C(graphite) + O 2 (g) --> CO 2 (g) Δ r G = ν Δ f G (products) - ν Δ f G (reactants) Δ r G = Δ f G (CO 2 (g)) - Δ f G (C(graphite)) - Δ f G (O 2 (g)) Convention: Δ f G (C(graphite)) = Δ f G (O 2 (g)) = Δ f G (most stable form) = 0 Δ r G = Δ f G (CO 2 (g)) How do we measure Δ r G? Δ r G = Δ r H - T Δ r S

30 Chemical Reactions: Δ f G listed for many compounds in Appendix 2. How are values of Δ f G measured? Example: C(graphite) + O 2 (g) --> CO 2 (g) How do we measure Δ r G? Δ r G = Δ r H - T Δ r S Δ r H = kj mol -1 (measure with calorimeter) Δ r S = S (CO 2 (g)) - S (C(graphite)) - S (O 2 (g)) = ( ) J K -1 mol -1 = 2.9 J K -1 mol -1 Δ f G = Δ r G = Δ r H - T Δ r S = kj mol -1 - (298 K)(2.9 x 10-3 kj K -1 mol -1 ) = kj mol -1 Direct method to find Δ f G (CO 2 (g))

31 Dependence of Gibbs energy on T and p: Some thermodynamic relations: G = H - TS --> dg = dh - TdS - SdT H = U + pv --> dh = du + pdv + Vdp du = dq + dw, dq = TdS, dw = -pdv Putting together: dg = Vdp - SdT Constant p, variable T: dp = 0 --> dg = -SdT G T p dh = TdS - pdv + pdv + Vdp = S S > 0, so G decreases with increasing T at constant p.

32 Dependence of Gibbs energy on T and p: dg = Vdp - SdT Constant p, variable T: dp = 0 --> dg = -SdT G T p = S S > 0, so G decreases with increasing T at constant p. Constant T, variable p: dt = 0 --> dg = Vdp G p T = V V > 0, so G increases with increasing p at constant T. Consider ideal gas at constant T: G 2 dg = Vdp = nrt G 1 p 2 p 1 dg = Vdp dp --> ΔG = nrt ln p 2 p p 2 p 1 p 1

33 Dependence of Gibbs energy on T and p: Consider ideal gas at constant T: G 2 p 2 dg = Vdp = nrt G 1 p 1 p 1 dg = Vdp dp --> ΔG = nrt ln p 2 p p 2 p 1 Example: mol ideal gas at 300 K and 1.5 bar is compressed to 6.9 bar. What is ΔG?

34 Gibbs Energy and Phase Equilibria At some T and p it s possible that for a substance 2 phases can coexist. The condition for this is: G α = G β α = phase (such as solid), β = different phase If T or p changes so G solid > G liquid some solid melts to liquid, lowering G of the substance. This helps us understand phase diagrams. At given p, T: G α is lowest for phase indicated.

35 How does molar Gibbs energy of solid, liquid or vapor depend on p and T? Consider T: G solid T p G T G liquid = S solid T p = S < 0 for any phase. p G = S vapor liquid T p = S vapor G decreases with increasing T at constant p, for any phase. In general: S vap >> S liquid > S solid

36 How does molar Gibbs energy of solid, liquid or vapor depend on p and T? Consider p: G p T = V > 0 for any phase. G increases with increasing p at constant T, for any phase. In general: V vap >> V liquid, V solid Usually: V liquid > V solid (not H 2 O!) p 2 > p 1 T f > T f T b > T b

37 More quantitatively: Clausius-Clapeyron Equation Consider 2 phases, α and β. At equilibrium, G α = G β, dg α = dg β Recall: Then: dg α = V α dp S α dt = dg β = V β dp S β dt ( S β S α )dt = ( V β V α )dp Or: dp dt = ΔS ΔV = At equilibrium: ΔS = ΔH T Change in molar entropy for α --> β Change in molar volume for α --> β --> dp dt = ΔH TΔV T = phase transition temperature, such as T f or T b. Clapeyron Equation gives slope of phase equilibria. Clapeyron Equation

38 More quantitatively: Clausius-Clapeyron Equation dp dt = ΔH TΔV Clapeyron Equation T = phase transition temperature, such as T f or T b. Clapeyron Equation gives slope of phase equilibria. Example: Calculate slope of solid-liquid equilibrium curve, dp/dt, at T = K in atm K -1 for H 2 O, given: Δ fus H = 6.01 kj/mol V L = L/mol V S = L/mol

39 More quantitatively: Clausius-Clapeyron Equation Calculation of dp/dt easier for liquid --> vapor, solid --> vapor transition: Δ vap V = V vapor V condensed V vapor = RT / p (ideal gas) dp dt = Δ vaph TΔ vap V Δ vaph Δ vaph TV vap T(RT / p) = p Δ vaph RT 2 Note: Everything on the right is > 0, so dp/dt > 0. Rearrange and integrate: p 2 dp = p p 1 T 2 T 1 Δ vap H dt --> RT 2 ln p 2 = Δ H vap R p T 2 T 1 Clausius-Clapeyron Equation

40 Phase diagram of water: Critical point: L-V no longer distinguishable

41 Phase diagram of CO 2 : Critical point: L-V no longer distinguishable Liquid unstable for p < 5 atm. At p = 1 atm, only solid, vapor

42 Phase Rule: Gibbs derived phase rule: f = c - p + 2 c = number of components p = number of phases present f = degrees of freedom (number of intensive variables, such as T, p, composition, that can be changed independently without disturbing number of phases in equilibrium) Example: What is f for pure water (a) in vapor phase? (b) along phase boundary? (c) at triple point?

Homework Problem Set 8 Solutions

Homework Problem Set 8 Solutions Chemistry 360 Dr. Jean M. Standard Homework roblem Set 8 Solutions. Starting from G = H S, derive the fundamental equation for G. o begin, we take the differential of G, dg = dh d( S) = dh ds Sd. Next,