Transcript Lecture 19
If we define Gi = G° ( Pi = 1 atm ) then
Gf = Gi + RT ln
pf
pi
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
Note: P is the partial pressure
of a given gas component
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
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:
aA(PA) + bB (PB) = cC (PC) + dD (PD)
∆G = ∑Gprod - ∑Greact
Free Energy for 1 mole
of ideal gas D at a partial
pressure of PD
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°
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}!
∆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.
Bonus * Bonus * Bonus *
Bonus * Bonus * Bonus
Write ∆G° = ∆H° - T∆S°
Kp = e
-∆G°/RT
Kp = e -∆H°/RT e ∆S°/R
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
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
∆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
When -∆H° /RT < 0
Increasing T causes
Kp = e -∆H°/RT e ∆S°/R
look at e -∆H°/RT = 1/ e+∆H°/RT
∆H°/RT and hence e+∆H°/RT to get smaller.
Therefore, 1/ e+∆H°/RT gets larger.
Kp increases and equilibrium shifts to right.
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.
If we realize A + B
C+D
kr
kf
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
Using the Arrhenius form for the rate constants kf and kr
k f = Af e
- EAf / RT
k r = Ar e
- EAr / RT
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
Equating these two forms for the equilibrium constant allows us
to connect thermodynamics and kinetics!
(Af/Ar) exp [-(EAf-EAr)/RT] =
{e (S/R)} { e (-H/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
}
C+D
+∆Ho = EAf - EAr
(∆Ho = Enthalpy
change for
A+B C+D)