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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

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Chemical Potential ( m )

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  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

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