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
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
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
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 = 1.381 x 10-23 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.
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 )
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?
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
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
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
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
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.
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?
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.
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
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
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.
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) = 40.79 kj mol -1 What are Δ fus S and Δ vap S for 1 mole H 2 O at normal melting and boiling points?
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
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?
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
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
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( )
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
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)
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)
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.
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 = -2801.3 kj/mol (we found this earlier)
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?
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
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 = -393.5 kj mol -1 (measure with calorimeter) Δ r S = S (CO 2 (g)) - S (C(graphite)) - S (O 2 (g)) = (213.6-5.7-205.0) 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 = -393.5 kj mol -1 - (298 K)(2.9 x 10-3 kj K -1 mol -1 ) = -394.4 kj mol -1 Direct method to find Δ f G (CO 2 (g))
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.
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
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: 0.590 mol ideal gas at 300 K and 1.5 bar is compressed to 6.9 bar. What is ΔG?
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.
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
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
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
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 = 273.15 K in atm K -1 for H 2 O, given: Δ fus H = 6.01 kj/mol V L = 0.0180 L/mol V S = 0.0196 L/mol
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 1 1 1 T 2 T 1 Clausius-Clapeyron Equation
Phase diagram of water: Critical point: L-V no longer distinguishable
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
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?