1 / 30

Criteria for Spontaneity

Criteria for Spontaneity. isolated dS > 0. Closed dS ≥ dq /T or dq – TdS ≤ 0. Cst V,T: dU – TdS ≤ 0 let A = U - TS dA ≤ 0 . Cst P,T: dH – TdS ≤ 0 let G = H - TS dG ≤ 0 . Free Energy. A = U - TS Helmholtz Free Energy.

salaam
Download Presentation

Criteria for Spontaneity

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Criteria for Spontaneity isolated dS > 0 Closed dS≥ dq/T or dq – TdS≤ 0 Cst V,T: dU – TdS≤ 0 let A = U - TS dA≤ 0 Cst P,T: dH – TdS ≤ 0 let G = H - TS dG ≤ 0

  2. Free Energy A = U - TS Helmholtz Free Energy dA = dU – TdS = dq + dw – TdS if rev then dw = dwmax & dS = dq/T dA = dwmax A called ‘work’ function It is equivalent to the maximum work a system can perform A is minimized for any irrev process in a closed system at cst V & T.

  3. Free Energy G = H - TS Gibbs Free Energy dG = dH – TdS– SdT Because absolute S is never known (3rd Law ignores nuclear spin contributions and isotopic mixing at T = 0 K) the change in G with change is T is undefined (like dividing by 0 in math). Therefore the last term in this equation is simply ignored. dG= dH – TdS = dU + PdV+ V dP – TdScst P & 1st Law dq + dw+ PdV– TdS if rev then dw = dwrev & dS = dq/T dG = dwrev – dwexp = dwnonexp dG is often referred to as the maximum ‘useful’ work G is minimized for any irrev process in a closed system at cst P & T.

  4. Measuring DG • Heating and Cooling: DG is undefined for temperature changes. • DG can only be determined for constant T processes. • However, the change in DG for a process with change in T can be determined. • (more on this later) 2.Isothermal processes: DG = DH – T DS • 3.Phase changes at nmp or nbp: • DG = DH – T DS sub DStr = DHtr/T • DG = DHtr – T DHtr/T = 0 • At the nmp or nbp two phases are in equilibrium and DGtr = 0

  5. 0˚C 0˚C What is DG for conversion of supercooled water at -10˚C to ice at -10˚C? (n = 1) DS = ∫ dqrev/T DS = T1T2CPdT/T DS = DHtr/T -10˚C You can’t measure DG for constant P heating/cooling! -10˚C DH = 1  75.4  10 + 1  -6007 + 1  38.07  -10 = -6312 DS = 75.4  ln(273/263) – 6007/273 + 38.07  ln(263/273) = -20.61 DG = -6312 – 263(-20.61) = -892 J

  6. Measuring DG • 4. Chemical Reactions: • DG298= SiniDGºf,m,i • (DG298 can be determined from thermodynamic table data) or…. • DG298 = DH298 – 298 • DS298 • 5. Chemical Reactions (not at 298): • Find DHT and DST using DCP values inthermodynamic tables • Then…. DGT = DHT – TDST DHºT=DHo298 + DCPºdT DSºT=DSo298 + DCPºdT/T

  7. Free Energy of Mixing IG’sDG = DH – TDS = 0 - TDS = -TDS DS = naRln(V/Va) + nbRln(V/Vb) DS = -naR lnca - nbR lncb DG = -naRTln(V/Va) - nbRTln(V/Vb) DG = naRTlnca + nbRTlncb

  8. H2(g) + ½ O2(g)→ H2O(l) Df DG˚f -285.83 – 298(0.06991) = -306.66 DS˚f = 0.06991 - .130684 – ½ 0.205138 = -0.163343 DG˚f = -285.83 – 298(-0.163343) = -237.15 Spontaneity in chemical systems at cst T & P involves a balance between the natural tendency to minimize enthalpy and to maximize entropy. Increasing T gives entropy a higher stake in that balance. This is why solids melt, proteins unfold, and DNA unwinds at higher values of T.

  9. dU = dq + dw (if closed, rev & wexp) dU = dqrev - PdV dqrev/T = dS & dqrev = TdS dU = TdS - PdV Since U is a state function, this also applies to irreversible processes. Note that in this expression U is a function of S and V or U(S, V). S and V are called the natural variables of U. In theory, if we know how U varies with S and V, then we can determine the other thermodynamic variables of the system.

  10. Hydrophobic Interactions C3H8(aq) C3H8(l) Yes (-) • Is this process spontaneous? A = yes, b = no • What is the sign of DG? For this process? • Does entropy increase during this process? (more disorder) • Does enthalpy decrease during this process? (stronger bonding) Yes No DH298 ~ +8 kJ mol-1 DS298 ~ +80 J mol-1 DG298 = 8 - 298  0.080 = -16 kJ mol-1 Hydrophobic effects are due to the entropy increase of the solvent which must be more ordered when it is in the presence of nonpolar solutes compared to when it is surrounded by other water molecules.

  11. 1. dU = dq + dw(closed sys, exp work only) 2. H = U + PV 3. A = U - TS 4. G = H - TS 5. CV = (dU/dT)V= T(dS/dT)V 6. CP = (dH/dT)P= T(dS/dT)P Gibbs Equations • 1. dU = TdS - PdV(derived from 1st Law) • dH= • dA = • dG = TdS + VdP -SdT - PdV -SdT + VdP

  12. 1. dU = TdS - PdV U (S,V) natural variables 2. dH = TdS + VdP H (S,P) 3. dA = -SdT – PdV A (T,V) 4. dG = -SdT + VdP G (T,P) e.g. dG = (dG/dT)PdT + (dG/dP)TdP (dG/dP)T = V find (dG/dP)T = (dG/dT)P = -S T -P (dU/dS)V= (dU/dV)S = (dH/dS)P = (dH/dP)S = T V (dA/dT)V = (dA/dV)T = -S -P

  13. 1. dU = TdS - PdV (derive from 1st) 2. dH = TdS + VdP 3. dA = -SdT - PdV 4. dG = -SdT + VdP Euler Reciprocity Relations Given z = z (x,y) dz = (dz/dx)Ydx + (dz/dy)xdy let M = (dz/dx)Y & N = (dz/dy)x then..... (dM/dy)x = (dN/dx)y (dT/dV)S = -(dP/dS)V dU = TdS – PdV……. (dT/dP)S = (dV/dS)P dH: dA: dG: Maxwell Relations (dS/dV)T = (dP/dT)V (dS/dP)T = -(dV/dT)P

  14. Starting Points dU = TdS - PdV (derive from 1st) dH = TdS + VdP dA = -SdT – PdV dG = -SdT + VdP CV = (dU/dT)V= T(dS/dT)V(since dU = dqV) CP = (dH/dT)P= T(dS/dT)P (since dH= dqP) a = 1/V (dV/dT)P k = -1/V (dV/dP)T a/k = (dP/dT)V

  15. desired relationships .......... (dS/dT)P (dS/dP)T(dS/dT)V (dS/dV)T (dH/dT)P(dH/dP)T (dG/dT)P (dG/dP)T (dU/dV)T(dU/dT)V (dA/dT)V (dA/dV)T some of the relationships are fairly easy.... (dS/dT)P = CP/T (dH/dT)P = CP (dU/dT)V = CV (dS/dT)V = CV/T Starting Points dU = TdS - PdV dH = TdS + VdP dA = -SdT – PdV dG = -SdT + VdP CV = (dU/dT)V= T(dS/dT)V & CP = (dH/dT)P= T(dS/dT)P a = 1/V (dV/dT)P k = -1/V (dV/dP)T a/k = (dP/dT)V

  16. desired relationships .......... (dS/dT)P(dS/dP)T (dS/dT)V (dS/dV)T (dH/dT)P(dH/dP)T (dG/dT)P (dG/dP)T (dU/dV)T (dU/dT)V (dA/dT)V (dA/dV)T Starting Points dU = TdS - PdV dH = TdS + VdP dA = -SdT – PdV dG = -SdT + VdP CV = (dU/dT)V= T(dS/dT)V & CP = (dH/dT)P= T(dS/dT)P a = 1/V (dV/dT)P k = -1/V (dV/dP)T a/k = (dP/dT)V Check Maxwell relations no result …. ÷ by dP at cst T dH = TdS + VdP from Maxwell (dG)... (dH/dP)T = T(dS/dP)T + V (dS/dP)T = -(dV/dT)P = - Va....... (dH/dP)T = -TVa + V

  17. (dH/dP)T = -TVa + V Water at 303 K n = 1 density = 0.995 g/ml = 995 kg/m3. a = 3.04 x 10-4 K-1. What is DH if the pressure is increased by 100 atm. (1 atm. = 101325 Pa) DH = (-303 • 0.018/995 • 3.04 x 10-4 + 0.018/995) • 100 • 101325 = 166 J Squeezing the liquid does not strengthen bonding What is DH if the pressure is decreased from 1 atm to 0.1 atm? DH = (-303 • 0.018/995 • 3.04 x 10-4 + 0.018/995) • -0.9 • 101325 = -1.5 J

  18. 1. dU = TdS - PdV (derive from 1st) 2. dH = TdS + VdP 3. dA = -SdT - PdV 4. dG = -SdT + VdP (dG/dT)P = -S (dG/dP)T = V (dU/dS)V = T (dU/dV)S = -P (dH/dS)P = T (dH/dP)S = V (dA/dT)V = -S (dA/dV)T = -P (dT/dV)S = -(dP/dS)V (dT/dP)S = (dV/dS)P Maxwell Relations (dS/dV)T = (dP/dT)V CV = (dU/dT)V= T(dS/dT)V & CP = (dH/dT)P= T(dS/dT)P (dS/dP)T = -(dV/dT)P a = 1/V (dV/dT)P k = -1/V (dV/dP)T a/k = (dP/dT)V

  19. Internal P = (dU/dV)T dU = TdS - PdVdVT (dU/dV)T = T(dS/dV)T - P apply Euler/Maxwell from dA = -SdT - PdV (dU/dV)T = T(dP/dT)V – P gas (dU/dV)T = T(a/k) - Psol/liq for an IG show that(dU/dV)T = 0 For water at 298 K Internal P = 298 • 2.07 x 10-4/4.57 x 10-10 – 101325 = 1.35 x 108 Pa or 1330 atm a = 1/V (dV/dT)P k = -1/V (dV/dP)T a/k = (dP/dT)V

  20. The effect of T on Gibbs energy (dG/dT)P = -S or (dDG/dT)P = -DS d(G/T)/dT = G d(1/T)/dT + 1/T dG/dT = -G/T2 – S/T = -(G + TS)/T2 = -(H – TS + TS)/T2 = -H/T2 {d(G/T)/dT}P = -H/T2 or {d(DG/T)/dT}P = -DH/T2 d(G/T) = -H dT/T2 & dT/T2 = d∫ dT/T2 = -d(1/T) d(G/T) = H d(1/T) {d(G/T)/d(1/T)}P = H or {d(DG/T)/d(1/T)}P = DH

  21. {d(DG/T)/d(1/T)}P = DH What is DG400 for the reaction: CO(g) + ½O2(g) → CO2(g) DG298 = -257.2 kJ (table value) -257.2 kJ (DH298 – TDS298) DH400 = DH298+ DCP• DT -284 = -283 - 0.0067 (202) = DS400 = DS298 + DCP• ln (T2/T1) -88.5 = -86.5 - 6.8 (0.2944) = DG400 = DH400 - TDS400 -248.6 = -284 - 400 (-0.0885)

  22. {d(DG/T)/d(1/T)}P = DH What is DG400 for the reaction: CO(g) + ½O2(g) → CO2(g) DG/400 - (-257.2/298) = -283.0 (1/400 – 1/298) DG400 = -248.4 kJ mol-1 DG/400 + 0.8631 = 0.2422 At what T is the reaction at equilibrium? 0 - (-257.2/298) = -283.0 (1/T – 1/298) and T = 3244 K DGT ~ DH - TDS 0 ~ -284 - T (-0.0865) & T = 3283K assumes DH and DS are constant

  23. The effect of P on Gibbs Free Energy (dG/dP)T (dG/dP)T = V dG = -S dT + V dP dG = V dP DG = ∫ V dP DG = GP2 – GP1~V (P2 – P1) solids/liquids (since V is not affected very much by P) DG = GP2 – GP1 = nRT/P dP = nRTln (Pf/Pi) IG Applied to process: e.g. phase change or chemical reaction D(DG) = DGP2 – DGP1~DV (P2 – P1) for a reaction involving gases you can assume DV = is entirely due to gas volumes.

  24. D(DG) = DGP2 – DGP1~DV(P2 – P1) Why is graphite more stable than diamond at 1 atm. P? What does the ° mean in DG°? What is DG° for Cdia→ Cgr? What is DG at P ≠ P° for Cdia→ Cgr? Look at process qualitatively using sign change for ↑ P + + (DGP - DGP°)= DV ( P - P°) + DGP is smaller (-), 0 or + value Process is less likely at higher P Estimate the Pressure at which graphite will change into diamond? dDG = DVm (P – P°) & 0 – 2900 = (3.41 x 10 -6 - 5.29 x 10 -6) (P – 101325) P = 1523 atm

  25. D(DG) = DGP2 – DGP1~DV(P2 – P1) CO(g) + ½O2(g) → CO2(g) Will the reaction become more or less favorable at higher T? Will the reaction become more or less favorable at higher pressures? Increasing T favors reactants …… Increasing P favors products. Putting the same amount of starting material in smaller flask will influence the equilibrium!

  26. dG = -SdT + VdP This Gibb’s equation assumes that there is no change in the amount of substance in the closed system. However, if you open the lid and throw in some substance, G will change. It will also change if there is a chemical reaction or phase changes that how much of a particular substance is present. This concept is handled by expanding the Gibb’s equation above to …… dG = -S dT+ V dP+ Si (dG/dni)T,P,n´ Chemical potential (mi) = (dG/dni)T,P,n´ Gm(P) = Gºm + RT ln (P/Pº) ideal gas or … mi(P) = mi(P˚) + RT ln(Pi/P˚)

  27. Chemical Potential (m) Is Gan extensive or an intensive property? More ‘stuff’ = more free energy Gm (the molar free energy) is an intensive property. G ni Chemical potential (mi) = (dG/dni)T,P,n´ For a pure substance (mi) = Gm.

  28. Chemical potential (mi) = (dG/dni)T,P,n´ However mi depends on other molecules in system. Just as the volume of ethanol changes when mixed with water … the free energy of one mole of a substance has a different impact in a mixture G For a mixture substance (mi) ≠Gm, and mi changes with added moles of i. ni

  29. Chemical Potential (m) How does the chemical potential of a pure gas change with pressure? Gm(P) - Gºm= RT ln (P/Pº) 1 mole of ideal gas mi(P) = mi(P˚) + RT ln(Pi/P˚) However, since an ideal gas does not interacts with other components in a mixture the chemical potential of that gas also applies to partial pressures in an ideal gas mixture.

  30. dG = -SdT + VdP + Si (dG/dni)T,P,n´ dni dG = -SdT + VdP + Simidni What can cause a change in ni? In closed system (no added material). 1) Chemical Reactions … Chapter 5 2) Phase changes … Chapter 6 The natural tendency for systems is to achieve their lowest potential energy in a given force field. In a gravitational field water achieves this by flowing downhill and pooling in valleys. In a chemical system, molecules do this by reacting or changing phases to minimize G (at constant T and P). Thus Free energy is the potential energy of chemical systems, and thus the name chemical potential. Molecules will spontaneously ‘flow’ from higher to lower m.

More Related