Chemical Potential (  ) 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  i (P) =  i (P˚) + RT ln(P i /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.

What can cause a change in n i ? In closed system (no added material). dG = -SdT + VdP +  i (dG/dn i ) T,P,n´ dn i 1) Chemical Reactions … Chapter 5 2) Phase changes … Chapter 6 dG = -SdT + VdP +  i  i dn i 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 .

HSGHSG undefined isoT Compression of Ideal gas Melting of benzene at its nmp Melting of benzene below nmp Heating copper (cst P) CH 4 + 2O 2 → CO 2 + 2H 2 O (l) 1atm & 298K U Choices: a = +b = -c = 0d = undefined

Chemical Equilibrium Starting Point – Gibbs Equation for Free Energy dG = -S dT + V dP +  i  i dn i Criteria for spontaneity:  G rx < 0 (cst T,P) 0 dG =  i  i dn i Goal:  G rx =  Gº rx + RT ln Q Chemical potential (  i ) = (dG/dn i ) T,P,n´  i = G m for a pure substance  i (P) =  i (P˚) + RT ln(P i /P˚)

Goal:  G rx =  Gº rx + RT ln Q Q = Product of the concentration all reactant reagents raised to the power of their stoichiometric coefficients. Q =  i [i] i [i] unit symbol std state pressure P 1 atm molarity c 1 M molality m 1 m mole fraction   = 1 gas:  G m = G m (P) – G m (Pº) = RT ln (P/Pº) and …..  i =  i º + RT ln (P i /Pº) dG =  i  i dn i Write Q expression for the reaction … N 2(g) + 3H 2(g) → 2NH 3(g)

 Gº rx is a constant whereas  G rx changes as the reaction proceeds. if  G rx < 0 then the reaction is spontaneous as written  G rx > 0 then the reverse reaction occurs  G rx = 0then the reaction is at equilibrium and Q = K  G rx =  Gº rx + RT ln K and…  Gº rx = - RT ln K

Example reaction: aA  bB Extent of reaction (ξ) – doesn’t depend on which reagent of the reaction is followed  = (n i,t – n i,0 )/ i, or d  = dn i / i d  = dn i / i = dn A /a = dn B /b (don’t forget sign convention) e.g. 2I  I 2 dG =  i  i dn i where  i = (dG/dn i ) T,P,n′ dn i = d  i and dG =  i  i dn i so ….. dG =  i i  i d  & dG/d  =  i i  i dG/d  =  G rx = b  B – a  A G  dG/d  =  G rx = 0 Ball 5.6 A = 1.0 (yes) B = 1.5 (no) C = 3.0 D = 3.5

 Gº rx = -RT ln K P This can be applied to solution chemistry replacing K P with K c Goal:  G rx =  Gº rx + RT ln Q from  (P) =  º + RT ln (P/Pº) and  i = dG/dn i dG =  i  i dn i dn i = i d  dG =  i  i  i d  d  dG/d  =  G rx =  i i  i sub in  i =  i º + RT ln (P i /Pº)  G rx =  i i  i º +  i i RT ln (P i /Pº)  i i  i º =  G° rx  G rx =  Gº rx +  i RT ln (P i /Pº) i  i ln (x i ) = ln  i x i  G rx =  Gº rx + RT ln  i (P i /Pº) i sub in Q =  i (P i /Pº) i  G rx =  Gº rx + RT ln Q (5.7) at equilibrium Q = K and  G rx = 0

Gas Phase Reaction: aA  bB  G rx = b  B – a  A = (dG/dξ) P,T  B =  B º + RT ln P B /Pº (repeat for A)  G rx = b  B º - a  A º + bRT ln (P B /Pº) – aRT ln (P A /Pº)  G rx =  Gº rx + RT ln {(P B /Pº) b /(P A /Pº) a }  G rx =  Gº rx + RT ln Q  G rx =  Gº rx + RT ln (P B /Pº) b – RT ln (P A /Pº) a

5.11 SO 2(g) + ½O 2(g) ↔ SO 3(l) compound  H° f kJ/mol  G° f kJ/mol S° J/mol SO 3(l) SO 2(g) O 2(g) Write the expression for Q/K. Calculate  G° 298 Calculate K Predict direction of reaction if … atm SO 2, atm O 2 & some liquid present in closed flask kJ/mol 7.98 x  G rx =  Gº rx + RT ln Q ln (15.8) = J forward K = _____1______ (P SO2 ) (P O2 ) ½

5.16 H 2(g) + D 2(g) ↔ 2 HD (g) K = 4.00 Calculate the equilibrium partial pressures and . Write K expression P H2 = 0.50 atm and P D2 = 0.10 atm. in closed system. Set up ICE chart x -x +2x x x +2x = (2x) 2 /{(0.50 – x)(0.10 – x)} 4.00 = 4x 2 /(x 2 – 0.6x ) -2.4x = 0 & x = atm

Pressure dependence of K (dK/dP) T = 0 for solid/liquid and gas phase reactions where  n = 0 Gas phase reaction equilibria where  n  0 are influenced by P. Qualitatively this can be predicted by LeChatelier’s Principle. e.g. N 2 + 3H 2  2NH 3 K P = (P NH3 /Pº) 2 (P N2 /Pº)(P H2 /Pº) 3 = (P NH3 ) 2 Pº 2 (P N2 )(P H2 ) 3 = (  NH3 ) 2 Pº 2 (  N2 )(  H2 ) 3 P 2 P NH3 =  NH3 P K  = P 2 K P K P  = K    P/P  )  n

K P vs. K c K P = (P NH3 /Pº) 2 (P N2 /Pº)(P H2 /Pº) 3 PV = nRT and P = nRT/V = cRT K P = (c NH3 ) 2  RT/Pº)  n (c N2 )(c H2 ) 3 e.g. N 2 + 3H 2  2NH 3 K P  = K c  (RT/P  )  n If a reaction does not involve any gases then  n = 0 and K c is determined from  Gº rx just as K P was for a gas phase reaction.

Temperature dependence of K the Van’t Hoff equation {d(lnK)/d(1/T)} P = -  Hº rx /R {d(  G/T)/dT} P = -  H/T 2 {d(  G/T)/d(1/T)} P =  H  Gº rx = - RT ln K {d(-RTlnK/T)/d(1/T)} P =  Hº Plot lnK vs. 1/T ….. Slope = -  Hº rx /R lnK(T 2 ) – lnK(T 1 ) = -  Hº rx /R (1/T 2 – 1/T 1 )

ln K P o vs. 1/T ln K P o 1/T slope = -11,406  H o = 94.8 kJ/mol PCl 5  PCl 3 + Cl 2 {d(lnK)/d(1/T)} P = -  Hº rx /R K P º = T (K) =