1 / 15

Chemical Potential ( m )

Chemical Potential ( m ). How does the chemical potential of a pure gas change with pressure? ( dG / dP ) T = V. G m (P) - Gº m = RT ln (P/Pº) 1 mole of ideal gas. m i (P ) = m i (P˚) + RT ln (P i /P˚). However, since an ideal gas does not interacts with other components in

aria
Télécharger la présentation

Chemical Potential ( m )

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. Chemical Potential (m) How does the chemical potential of a pure gas change with pressure? (dG/dP)T = V 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.

  2. 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.

  3. + - 0 undefined DH DS DG isoT Compression of Ideal gas 0 - + Melting of benzene at its nmp + + 0 Melting of benzene below nmp + + + Heating copper (cst P) + + U CH4 + 2O2→ CO2 + 2H2O(l) 1atm &298K - - - Choices: a = + b = - c = 0 d = undefined

  4. Chemical potential (mi) = (dG/dni)T,P,n´ mi= Gm for a pure substance 0 0 mi(P) = mi(P˚) + RT ln(Pi/P˚) Chemical Equilibrium Starting Point – Gibbs Equation for Free Energy dG = Simidni dG = -S dT + V dP + Simidni Criteria for spontaneity: DGrx < 0 (cst T,P) Goal: DGrx = DGºrx + RT ln Q

  5. Goal: DGrx = DGºrx + RT ln Q dG = Simidni Q = Product of the concentration all reactant reagents raised to the power of their stoichiometric coefficients. Q = Pi [i]ni [i] unitsymbolstd state pressure P 1 atm molarity c 1 M molality m 1 m mole fraction cc = 1 Write Q expression for the reaction … N2(g) + 3H2(g)→ 2NH3(g) gas:DGm = Gm(P) – Gm(Pº) = RT ln (P/Pº) and ….. mi = miº + RT ln (Pi/Pº)

  6. DGrx= DGºrx+ RT ln K and… DGºrx = - RT ln K DGºrx is a constant whereas DGrxchanges as the reaction proceeds. • if DGrx < 0 then the reaction is spontaneous as written • DGrx > 0 then the reverse reaction occurs • DGrx = 0 then the reaction is at equilibrium and Q = K

  7. Ball 5.6 A = 1.0 (yes) B = 1.5 (no) C = 3.0 D = 3.5 dG = Simi dni where mi = (dG/dni)T,P,n′ Example reaction: aA  bB Extent of reaction (ξ) – doesn’t depend on which reagent of the reaction is followed x = (ni,t – ni,0)/ni, or dx = dni/ni dx = dni/ni = dnA/a = dnB/b (don’t forget sign convention) e.g. 2I  I2 dG/dx =DGrx = 0 dni = dx • ni and dG = Simidni so ….. dG = Sini• mi •dx & dG/dx= Sinimi dG/dx =DGrx = bmB– amA G x

  8. Goal: DGrx = DGºrx + RT ln Q from m(P) = mº + RT ln (P/Pº) and mi = dG/dni dG = Si midnidni = nide dG = Sinimide  de dG/de = DGrx = Sinimisub in mi = miº + RT ln (Pi/Pº) DGrx = Sinimiº + SiniRT ln (Pi/Pº)Sinimiº = DG°rx DGrx = DGºrx + Si RT ln (Pi/Pº)niSiln (xi) = lnPi xi DGrx = DGºrx + RT lnPi (Pi/Pº)nisub in Q = Pi (Pi/Pº)ni DGrx = DGºrx + RT ln Q (5.7) at equilibrium Q = K and DGrx= 0 DGºrx = -RT ln KP This can be applied to solution chemistry replacing KP with Kc

  9. Gas Phase Reaction: aA  bB DGrx = bmB – amA = (dG/dξ)P,T mB = mBº + RT ln PB/Pº(repeat for A) DGrx = bmBº - amAº + bRT ln (PB/Pº) – aRT ln (PA/Pº) DGrx = DGºrx + RT ln (PB/Pº)b – RT ln (PA/Pº)a DGrx = DGºrx + RT ln {(PB/Pº)b/(PA/Pº)a} DGrx = DGºrx + RT ln Q

  10. 5.11 SO2(g) + ½O2(g) ↔ SO3(l) K = _____1______ (PSO2) (PO2)½ Write the expression for Q/K. Calculate DG°298 -67.9 kJ/mol 7.98 x 1011 Calculate K Predict direction of reaction if …. 0.200 atm SO2, 0.100 atm O2 & some liquid present in closed flask. DGrx= DGºrx + RT lnQ -67900 + 8.314 • 298 • ln (15.8) = -61060 J forward

  11. 5.16 H2(g) + D2(g) ↔ 2 HD(g) 0.50 0.10 0.0 -x -x +2x 0.50 -x 0.10 -x +2x 0.4167 0.0167 0.1667 K = 4.00 PH2 = 0.50 atm and PD2 = 0.10 atm. in closed system. Calculate the equilibrium partial pressures and e. 4.00 = (2x)2/{(0.50 – x)(0.10 – x)} Write K expression 4.00 = 4x2/(x2 – 0.6x + 0.05) Set up ICE chart -2.4x + 0.2 = 0 & x = 0.0833 atm

  12. Pressure dependence of K (dK/dP)T = 0 for solid/liquid and gas phase reactions where Dn = 0 Gas phase reaction equilibria where Dn  0 are influenced by P. Qualitatively this can be predicted by LeChatelier’s Principle. e.g. N2 + 3H2 2NH3 KP = (PNH3/Pº)2 (PN2/Pº)(PH2/Pº)3 = (PNH3)2 Pº2 (PN2)(PH2)3 = (cNH3)2 Pº2 (cN2)(cH2)3 P2 PNH3 = cNH3 • P KP = Kc (P/P)Dn Kc = P2 • KP

  13. KP vs. Kc e.g. N2 + 3H2 2NH3 PV = nRT and P = nRT/V = cRT KP = (PNH3/Pº)2 (PN2/Pº)(PH2/Pº)3 KP = (cNH3)2 •(RT/Pº)Dn (cN2)(cH2)3 KP = Kc  (RT/P ) Dn If a reaction does not involve any gases then Dn = 0 and Kc is determined from DGºrx just as KP was for a gas phase reaction.

  14. {d(DG/T)/dT}P = -DH/T2 {d(DG/T)/d(1/T)}P = DH DGºrx = - RT ln K Temperature dependence of K the Van’t Hoff equation {d(-RTlnK/T)/d(1/T)}P = DHº {d(lnK)/d(1/T)}P = - DHºrx/R lnK(T2) – lnK(T1) = - DHºrx/R • (1/T2 – 1/T1) Plot lnK vs. 1/T ….. Slope = - DHºrx/R

  15. KPº = 0.245 1.99 4.96 9.35 T (K) = 485 534 556 574 {d(lnK)/d(1/T)}P = - DHºrx/R slope = -11,406 DHo = 94.8 kJ/mol ln KPo PCl5 PCl3 + Cl2 1/T ln KPo vs. 1/T

More Related