1. If we define Gi = G° ( Pi = 1 atm ) then Gf = G°(T) + RT ln Pf where Pf must be expressed in atmospheres. In general for 1 mole ideal gas G (T, P) = G°(T) + RT lnP Shows how G depends on P at fixed T if know G°(T). Note G° is a function of T. To distinguish G for 1 mole, call it  (T, P) = ° (T) + RT ln P G f = G i + RT ln Note: P is the partial pressure of a given gas component

2.  is free energy/mole for an ideal gas at T and P. (Called chemical potential) ∆G = ∑Gprod - ∑Greact aA(PA) + bB (PB) = cC (PC) + dD (PD) Consider now the Chemical reaction: For n moles n = n° + nRT ln P G = c[C° + RT ln PC] + d[D° + RT ln PD] - a[A° + RT ln PA] - b[B° + RT ln PB] G = c[C° ] + d[D°] - a[A°] - b[B°]  G° Free Energy for 1 mole of ideal gas D at a partial pressure of PD

3. True for arbitrary values of PC, PD, PA, PB What happens if Pj happen to be those for equilibrium? Since initial and final states are in eq. ∆G is not neg for either direction.  ∆G = 0 G° = - RT ln {(PCeq)c (PDeq )d/(PAeq)a (PBeq)b} G = G° + cRT ln PC + dRT ln PD - aRT ln PA - bRT ln PB G = G° + RT ln {(PC)c(PD )d/(PA)a(PB)b} But Kp = {(PCeq)c (PDeq )d/(PAeq)a (PBeq)b}! Kp

4. Remarkably important formula relates free energy and Kp. ∆G° = -RT ln Kp Since ∆G° is defined for a specific pressure of 1 atm, it is only a fct of T. Kp fct only of T! * ∆G° < 0  exponent > 0 Kp > 1 ∆G° > 0  exponent < 0 Kp < 1 Reactions with a large neg ∆G° tend to proceed to completion. Kp = e -∆G°/RT = 10 -∆G°/2.303RT

5. Write ∆G° = ∆H° - T∆S°  Larger ∆S° larger Kp. ( More chaos/randomness toward products ) ∆H° < 0  larger Kp for more negative (large ∆H°) ∆H° Major T dependence for Kp is in ∆H° term Kp = 10 -∆H°/2.303RT10 ∆S°/2.303R Kp = e -∆H°/RT e ∆S°/R Bonus * Bonus * Bonus * Bonus * Bonus * Bonus Kp = e -∆G°/RT

6. Remember old rule for shift in eq. with T. Equilibrium shifts to left for an exothermic reaction and to right for an endothermic reaction. Shift in Equilibrium with temperature: If ∆S° roughly independent of T, then temperature dependence of Kp is in e -∆H°/RT term. ∆H° < 0 means heat released (exothermic) A  B + heat Kp = e -∆H°/RT e ∆S°/R

7. Think of heat as a reagent that works like common ion effect: A  B + heat ∆H° > 0 heat absorbed (endothermic) heat + A  B When -∆H° /RT < 0 look at e -∆H°/RT = 1/ e+∆H°/RT Increasing T causes ∆ ∆ ∆ ∆H°/RT and hence e+∆H°/RT to get smaller. Kp increases and equilibrium shifts to right. Kp = e -∆H°/RT e ∆S°/R ∆H° < 0 means heat released (exothermic) A  B + heat Therefore, 1 11 1/ e+∆H°/RT gets larger. Exponent -∆H°/RT > 0 and increasing T causes this to shrink so Kp gets smaller. Shift to left.

8. Connecting Kinetics and Equilibria By definition, kinetic processes are not equilibrium processes. In fact, we may think of kinetic processes as the mechanism that nature uses to reach the equlibrium state. has 2 rate constants, we can write: kf[A]e[B]e=kr[C]e[D]e (Equilibrium condition) Where [A]e etc. are the equilibrium concentrations of [A] etc. kf/kr = [C]e[D]e / [A]e[B]e = Kequilibrium If we realize A + B C + D

9. Using the Arrhenius form for the rate constants kf and kr But as we just learned (or you already knew from high school): ln[Keq]= -G0/RTG0= H0 - T S0 Where    H0 is the enthalpy change for the reaction and S0 is the entropy change for the reaction. G0 is the Free Energy k f  A f e EAf/RTk r A r eEAr/RTKeq = kf/kr =(Af/Ar) exp [-(EAf-EAr)/RT] But this gives Keq = e -∆G°/RT = e -∆H°/RT e ∆S°/R

10. Equating these two forms for the equilibrium constant allows us to connect thermodynamics and kinetics! Identify Af / Ar with {e (S/R)} (T “independent” assuming ∆Sº indep of T). EAf - EAr identify with ∆Hº A + B Activated State E Af E Ar C + D +∆Ho = EAf - EAr (∆Ho = Enthalpy change for A+B C+D) } (Af/Ar) exp [-(EAf-EAr)/RT] = {e (S/R)}  { e (-H/RT)}