If we define G i = G° ( P i = 1 atm ) then

1. Gf = Gi + RT ln 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 Note: P is the partial pressure of a given gas component 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

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

3. 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} 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  Kp G° = - RT ln {(PCeq)c(PDeq )d/(PAeq)a(PBeq)b} But Kp = {(PCeq)c(PDeq )d/(PAeq)a(PBeq)b}!

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

5. Bonus * Bonus * Bonus * Bonus * Bonus * Bonus Write ∆G° = ∆H° - T∆S°  Kp= e -∆H°/RT e ∆S°/R Kp= e -∆G°/RT Kp= 10-∆H°/2.303RT10∆S°/2.303R 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

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: Kp= e -∆H°/RT e ∆S°/R 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

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

8. If we realize A + B C + D 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

9. - E / RT - E / RT = k = A e k A e Ar Af r r f f Using the Arrhenius form for the rate constants kf and kr Keq = kf/kr = (Af/Ar) exp [-(EAf-EAr)/RT] 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 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! {e (S/R)}  { e (-H/RT)} (Af/Ar) exp [-(EAf-EAr)/RT] = Identify Af / Ar with {e (S/R)} (T “independent” assuming ∆Sº indep of T). EAf - EAr identify with ∆Hº Activated State EAf EAr } A + B +∆Ho = EAf - EAr (∆Ho = Enthalpy change for A+B C+D) C + D