If we define GIf we define Gii = G° ( P = G° ( Pii = 1 atm ) then = 1 atm ) then

GGff = G°(T) + RT ln P = G°(T) + RT ln Pff

where Pwhere Pff must be expressed in atmospheres. must be expressed in atmospheres.

In general for 1 mole ideal gas In general for 1 mole ideal gas

G (T, P) = G°(T) + RT lnPG (T, P) = G°(T) + RT lnP

Shows how G depends on P at fixed T if know G°(T).Shows how G depends on P at fixed T if know G°(T).

Note G° is a function of T.Note G° is a function of T.

To distinguish G for 1 mole, call it To distinguish G for 1 mole, call it (T, P) = (T, P) = ° (T) + RT ln P° (T) + RT ln P

GGff = G = Gii + RT ln + RT ln pf

pi

Note: P is the Note: P is the partial pressurepartial pressureof a given gas componentof a given gas component

is free energy/mole for an ideal gas at T and P. (Called chemicalis free energy/mole for an ideal gas at T and P. (Called chemicalpotential) potential)

∆∆G = ∑GG = ∑Gprodprod - ∑G - ∑Greactreact

aaA(PA(PAA) + ) + bbB (PB (PBB) = ) = ccC (PC (PCC) + ) + ddD (PD (PDD))

Consider now the Chemical reaction:Consider now the Chemical reaction:

For n moles nFor n moles n = n = n° + nRT ln P ° + nRT ln P

G = G = cc[[CC° + RT ln P° + RT ln PCC] + ] + dd[[DD° + RT ln P° + RT ln PDD] - ] - aa[[AA° + RT ln P° + RT ln PAA] - ] - bb[[BB° + RT ln P° + RT ln PBB]]

GG = c = c[[CC° ] + d° ] + d[[DD°] - a°] - a[[AA°] - b°] - b[[BB°] °] G° G°

Free Energy for 1 moleFree Energy for 1 moleof ideal gas D at a partialof ideal gas D at a partial

pressure of Ppressure of PDD

True for True for arbitraryarbitrary values of P values of PCC, P, PDD, P, PAA, P, PBB

What happens if PWhat happens if Pj j happen to be those for happen to be those for equilibriumequilibrium??

Since initial and final states are in eq. ∆G is not neg for either Since initial and final states are in eq. ∆G is not neg for either direction. direction. ∆G = 0 ∆G = 0

G° G° = - RT ln {(P= - RT ln {(PCeqCeq))cc (P(PDeqDeq ) )dd/(P/(PAeqAeq))aa (P(PBeqBeq))bb}}

G = G = G° G° + + ccRT ln PRT ln PCC + + ddRT ln PRT ln PDD - - aaRT ln PRT ln PAA - - bbRT ln PRT ln PBB

G = G = G° G° + RT ln {(P+ RT ln {(PCC))cc(P(PDD ) )dd/(P/(PAA))aa(P(PBB))bb}}

But But KKpp = = {(P{(PCeqCeq))cc (P(PDeqDeq ) )dd/(P/(PAeqAeq))aa (P(PBeqBeq))bb}!}!

KKpp

Remarkably important formula relates free energy and KRemarkably important formula relates free energy and Kpp. .

∆∆G° = -RT ln KG° = -RT ln Kpp

Since ∆G° is defined for a specific pressure of 1 atm, it is only a Since ∆G° is defined for a specific pressure of 1 atm, it is only a

fct of T. fct of T. KKpp fct only of T! fct only of T! **

∆∆G° < 0 G° < 0 exponent > 0 K exponent > 0 Kpp > 1 > 1

∆∆G° > 0 G° > 0 exponent < 0 K exponent < 0 Kpp < 1 < 1

Reactions with a large neg ∆G° tend to proceed to completion. Reactions with a large neg ∆G° tend to proceed to completion.

KKpp = e = e -∆G°/RT-∆G°/RT = = 1010 -∆G°/2.303RT-∆G°/2.303RT

Write ∆G° = ∆H° - T∆S° Write ∆G° = ∆H° - T∆S°

Larger ∆S° larger KLarger ∆S° larger Kpp. ( More chaos/randomness toward products ). ( More chaos/randomness toward products )

∆∆H° < 0 H° < 0 larger K larger Kpp for more negative (large for more negative (large ∆H°∆H°) ∆H°) ∆H°

Major T dependence for KMajor T dependence for Kpp is in ∆H° term is in ∆H° term

KKpp = = 1010 -∆H°/2.303RT-∆H°/2.303RT1010 ∆S°/2.303R∆S°/2.303R

KKpp = e = e -∆H°/RT -∆H°/RT e e ∆S°/R∆S°/R

Bonus * Bonus * Bonus * Bonus * Bonus * Bonus * Bonus * Bonus * BonusBonus * Bonus * Bonus

KKpp = e = e -∆G°/RT-∆G°/RT

Remember old rule for shift in eq. with T. Equilibrium shiftsRemember old rule for shift in eq. with T. Equilibrium shiftsto left for an exothermic reaction and to right for an to left for an exothermic reaction and to right for an endothermic reaction.endothermic reaction.

Shift in Equilibrium with temperature:Shift in Equilibrium with temperature:

If ∆S° roughly independent of T, then temperature dependence If ∆S° roughly independent of T, then temperature dependence

of Kof Kpp is in is in e e -∆H°/RT -∆H°/RT term. term.

∆∆H° < 0 means heat released (exothermic)H° < 0 means heat released (exothermic)A A B + heat B + heat

KKpp = e = e -∆H°/RT -∆H°/RT e e ∆S°/R∆S°/R

Think of heat as a reagent that works like common ion effect:Think of heat as a reagent that works like common ion effect:A A B + heat B + heat

∆∆H° > 0 heat absorbed (endothermic)H° > 0 heat absorbed (endothermic)heat + A heat + A B B

WhenWhen -∆H° /RT < 0 look at -∆H° /RT < 0 look at e e -∆H°/RT-∆H°/RT == 1/ 1/ ee+∆H°/RT+∆H°/RT

Increasing T causesIncreasing T causes ∆H°/RT∆H°/RT and henceand hence ee+∆H°/RT+∆H°/RT to get smaller. to get smaller.

KKp p increases and equilibrium shifts to right. increases and equilibrium shifts to right.

KKpp = e = e -∆H°/RT-∆H°/RT e e ∆S°/R∆S°/R

∆∆H° < 0 means heat released (exothermic)H° < 0 means heat released (exothermic)A A B + heat B + heat

Therefore,Therefore, 1/ 1/ ee+∆H°/RT+∆H°/RT gets larger. gets larger.

Exponent Exponent -∆H°/RT-∆H°/RT > 0 and increasing T causes > 0 and increasing T causes this to shrink so Kthis to shrink so Kpp gets smaller. Shift to left. gets smaller. Shift to left.

Connecting Kinetics and EquilibriaConnecting Kinetics and EquilibriaBy definition, By definition, kinetic processeskinetic processes are not are not equilibrium processesequilibrium processes..In fact, we may think of kinetic processes as the mechanismIn fact, we may think of kinetic processes as the mechanismthat nature uses to reach the equlibrium state.that nature uses to reach the equlibrium state.

has 2 rate constants, we can write:has 2 rate constants, we can write:

kkff[A][A]ee[B][B]ee=k=krr[C][C]ee[D][D]ee (Equilibrium condition) (Equilibrium condition)Where [A]Where [A]e e etc. are the equilibrium concentrationsetc. are the equilibrium concentrationsof [A] etc. of [A] etc. kkff/k/krr = = [C][C]ee[D][D]ee / / [A][A]ee[B][B]e e = K= Kequilibriumequilibrium

If we realize A + B C + DIf we realize A + B C + Dk f⏐ → ⏐ ⏐ k r

← ⏐ ⏐ ⏐

Using the Arrhenius form for the rate constants kUsing the Arrhenius form for the rate constants kff and k and krr

But as we just learned (or you already knew from high school):But as we just learned (or you already knew from high school):

ln[Kln[Keqeq]= -]= -GG00/RT/RT GG00== HH0 0 - T - T SS0 0

Where Where HH0 0 is the enthalpy change for the reaction and is the enthalpy change for the reaction and SS00 is the is theentropy change for the reaction. entropy change for the reaction. GG0 0 is the Free Energy is the Free Energy

kkff AAff ee EEAfAf // RTRT kkrr AArr ee EEArAr // RTRT

KKeqeq = k = kff/k/krr = = (A(Aff/A/Arr) exp [-(E) exp [-(EAfAf-E-EArAr)/RT])/RT]

But this gives KBut this gives Keqeq = e = e -∆G°/RT-∆G°/RT = e = e -∆H°/RT -∆H°/RT e e ∆S°/R∆S°/R

Equating these two forms for the equilibrium constant allows usEquating these two forms for the equilibrium constant allows us to connect thermodynamics and kinetics!to connect thermodynamics and kinetics!

Identify AIdentify Aff / A / Arr with {e with {e ((SS/R)/R)} (T “independent” assuming ∆Sº} (T “independent” assuming ∆Sº indep of T).indep of T).EEAfAf - E - EArAr identify with ∆Hº identify with ∆Hº

A + BA + B

Activated StateActivated State

EEAfAf EEArAr

C + DC + D

+∆H+∆Hoo = E = EAfAf - E - EArAr

(∆H(∆Hoo = Enthalpy = Enthalpy change forchange forA+B A+B C+D)C+D)

}}

(A(Aff/A/Arr) exp [-(E) exp [-(EAfAf-E-EArAr)/RT] =)/RT] = {e {e ((SS/R)/R)} } { e { e (-(-HH/RT)/RT)}}