Download Lecture 6 Free Energy

Lecture 6 Free Energy
James Chou
BCMP201 Spring 2008
A quick review of the last lecture
I. Principle of Maximum Entropy
Equilibrium = A system reaching a state of
maximum entropy.
Equilibrium = All microstates are equally
probable.
S = kB ln!
II. Temperature
T = A measure of the tendency of an object to spontaneously give up energy to its
surroundings.
T = ( dE dS ) N,V
Not in Equilibrium
E1
E2
Picture from hyperphysics.phy-astr.gsu.edu
Equilibrium
III. The Boltzmann & Gibbs Distribution.
# "E &
p( E a ) = A ! exp% a (
$ kB T '
When a small system is in thermal equilibrium with the large surrounding, the
probability distribution of the energy state of the small system follows the Boltzmann
distribution and is dependent only on the T of the surrounding.
Where does the energy for doing work come from?
Today’s lecture outline
Free Energy, !G
The concept of chemical potential
Chemical equilibrium, how !G governs the direction
of a chemical reaction
Work
Work can be harnessed from a system in the process of going from non-equilibrium to
equilibrium. At equilibrium, the system can no longer do work.
W =
B
" F ! dr
A
B
A
Sign convention: Work done on the system is positive
Work done by the system is negative.
Some commonly used units: 1 N m = 1 Joule = 107 erg = 0.239 cal.
At the single molecule level, kBT is commonly used because 1 kBT ! 4 pN nm.
In thermodynamics, work typically
involves volume change.
B
B
A
A
W = ! # F " dr = ! # P " A dr
V2
W = ! # P " dV
V1
Fop
dr
Fint
Ideal Gas
Example:
Fop
A mole of ideal gas initially has a pressure of 2
atm, and then expands at a constant temperature
of 300 K against the atmosphere pressure (1 atm)
until equilibrium is reached. How much work is
done by the system?
Irreversible Pop = 1 atm
P = 1 atm
dr
Fint
Ideal Gas
Heat Block 300 K
W = Pop (V2 ! V1 )
= 1247 J
dr
Fint
Ideal Gas
Heat Block 300 K
Reversible Pint = Pop
A reversible process has welldefined path.
dr
Fint
A
Ideal Gas
Heat Block 300 K
P
B
W =
V2
"P
op
V1
V2
dV
V
= nRT ln 2
V1
V1 V
! dV = nRT "
= (8.314 J mol-1 K-1) (300 K) ln 2
=1728 J
V
What is heat?
368 K
298 K
T1
T2
Heat (Q) is amount of energy transferred between different objects. An isolated body does
not have heat.
What is the relation between heat and temperature?
! dE $
T =# &
" dS % N ,V
Q = T !S
T is defined at constant V. So T is not related
to work. In this definition, replace dE with Q.
Heat
The thermodynamic relation for internal energy E
!E = Q + W = T !S " P!V
Problem: For biological systems at constant T, !E is a poor indicator of the work done by
or done to the systems.
3
Energy of one mole of ideal gas: E = RT .
2
Since T = const, !E = 0 .
Fop
dr
Fint
Ideal Gas
Heat Block 300 K
V2
But W = ! # Pop " dV = !1728 J .
V1
Introducing the definition of free energy G
Now how about !G = ! ( E + PV ) " T!S ?
Fop
dr
Fint
Ideal Gas
Heat Block 300 K
!S = k B ln(V2N ) " k B ln(V1N ) = Rln
V2
P
= Rln 1
V1
P2
-1
-1
-1
= (8.314 J mol K ) ln2 = 5.76 J K
0
0
!G = !E + ! ( PV ) " T !S
(
)
= " ( 300 K ) 5.76 J K -1 = "1728 J
G of the system, a, in equilibrium with the universe decreases as
S of the universe increases
a
Surrounding, B
For the surrounding B, !E B = QB + W B = T!SB " P!VB .
!E = !E B + !E a = 0
"
!E B = #!E a ,
T!SB = #!E a + P!VB
Principle of maximum entropy: T!Stot = T!SB + T!Sa = "!E a + P!VB + T!Sa # 0 .
!VB = "!Va
!Ga
!E a + P!Va " T!Sa # 0
A small system a is in thermal contact with a large surrounding B. B will stay in its original
equilibrium because a is too small to affect it. Define the Gibbs Free Energy of a (Ga) to be
Ga = ( E + PV ! TS) a .
a will move to a new equilibrium to minimize Ga and during the process, useful work can be
obtained.
Normally, for biological systems, T and P of the surrounding are constant, and hence
!Ga = !E a + P!Va " T!Sa # 0
represents the maximum useful work can be extracted from a.
a
Not equilibrium
A
Useful Work
Surrounding, B
B
Equilibrium
The four thermodynamic quantities and their relations to useful work
Internal Energy: E
!E = Q + W exp ansion + W useful = TdS " PdV + W useful
At constant S and V, W useful = !E .
Enthalpy: H = E + PV
!H = !E + P!V + V!P = T!S + V!P + W useful
At constant S and P, W useful = !H .
Helmholtz free energy: F = E ! TS
"F = "E ! T"S ! S"T = !P"V ! S"T + W useful
At constant T and V, W useful = "F .
Gibbs free energy: G = E + PV ! TS
"G = "E + P"V + V"P ! T"S ! S"T = V"P ! S"T + W useful
At constant T and P, W useful = "G .
The concept of chemical potential
G = E + PV ! TS
E kin + N1!1 + N 2!2
1
2
G is a function of both E and Ni , at constant T and P.
G( E,N1,N 2 ) = E + PV ! TS, E = E kin + N1"1 + N 2"2
Define chemical potential µ of specie i to be
µi =
dG
dN i T ,P,N
Remember what is temperature?
j
, i! j
T=
dE
dS
N ,V
T describes the tendency of a system to give up energy. T depends
on kinetic energy of molecules.
µ describes the tendency of a molecular specie to chemically react.
µ depends on both concentration and internal energy of the
molecule.
What is the chemical potential µ1 in the 2-component ideal gas?
G( E,N1,N 2 ) = E + PV ! TS, E = E kin + N1"1 + N 2"2
µ1 =
dG dE
dS
dS
=
!T
= "1 ! T
dN1 dN1
dN1
dN1
3
kB T
2
*# 2! 3N / 2 &
3N / 2 N 1
"3N
S = kB ln,%
2m
)
V
2
!
!
((
)
(
)
kin
N!
,+$ ( 3N /2 "1)!'
" mk T %
"N %
dS 3
T
= kB T ln$ B 2 ' ( k B T ln$ 1 '
# 2!! &
#V &
dN1 2
?
1 - Don’t worry about this
/
2 /.
Concetration, C1
3
mk T
µ1 = !1 + kB T lnC1 " k B T ln B 2
2
2#!
Define the standard chemical potential
3
mk BT 0 3 / 2
µ = !1 " kB T ln
C
2 ( 1)
2
2#!
0
1
We obtain
! C1 $
µ1 = kB T ln# 0 & + µ10
" C1 %
Also true for aqueous solution
For gas, C0 = 1 mole / 22 L.
For solutes in aqueous solution, C0 = 1 mole / L = 1 M.
For solvent, C0 = the molarity of the pure solvent
e.g. C0 of H2O is 55.5 M.
The Concept of Chemical Equilibrium
1
2
" dG %
" dG %
!G = $
' dN1 + $
'dN 2 , dN1 = (dN 2
# dN1 &
# dN 2 &
" dG %
" dG %
= !$
' dN 2 + $
'dN 2
# dN1 &
# dN 2 &
!G
= µ2 " µ1
dN 2
The difference in chemical potential is the available chemical
energy to do work per unit of molecule.
At equilibrium, !G = 0 " µ2 = µ1 " chemical equilibrium
When does a chemical reaction stop?
A + B ! AB
At chemical equilibrium,
µ AB ! ( µ A + µ B ) = 0
µi = NkBT ln ( Ci Ci0 ) + µi0 = RT ln ( Ci Ci0 ) + µi0
0
0
"
%
µ AB
! µ A0 ! µ B0
C AB / C AB
+ ln $
'=0
0
0
RT
$# ( C A / C A ) ( C B / C B ) '&
The equilibrium constant
!G 0
0
0
"
%
µ AB
! µ A0 ! µ B0
C AB / C AB
+ ln $
'=0
0
0
RT
$# ( C A / C A ) ( C B / C B ) '&
Define equilibrium constant Keq
1M
0
# !"G 0 &
C AB / C AB
K eq = exp %
=
(
$ RT '
C A / C A0 C B / C B0
(
# !"G 0 &
C AB
K eq = exp %
=
$ RT (' C AC B
)(
)
pK ! "log10 K eq
The general equation for !G at any point of the reaction
1
C2
!G = !G + RT ln
C1
2
0
RT ln
C2
C1
!G
0
!G 0 ( fixed)
!G 0 = "RT ln K eq
An example of cellular chemical potential
ATP + H2O ! ADP + Pi
1 mM
55.5 M
0.1 mM
!G = !G + RT ln
0
(C
(C
1 mM
ADP
ATP
typical cellular concentration
)(
0
CADP
CP CP0
)(
)
0
0
CADP
CH2O CH2O
)
!G 0 can be determined if equilibrium Ci 's are known.
!G 0 can also be calculated from standard free energy
of formation !G 0f .
!G 0 = !G 0f , ADP + !G 0f , Pi " !G 0f , ATP " !G 0f , H2O = "30.3 kJ/mol
ATP + H2O ! ADP + Pi
1 mM
55.5 M
0.1 mM
!G = !G + RT ln
0
(C
(C
1 mM
ADP
ATP
"G 0 = #30.3 kJ/mol
typical cellular concentration
)(
0
CADP
CP CP0
)(
)
0
0
CADP
CH2O CH2O
)
1.0 "10 )(1.0 "10 )
(
= !30.3 + 2.748 ln
= !53.1 kJ / mol
!3
!4
1.0 "10!3
Living organisms constantly perturb G away from its minimum
so that there is free energy to do work.
Entropic Force
Force in Mechanics
In classical mechanics, potential energy U of an object can be defined in terms of the work
required to move the object from A to B with no net change in kinetic energy.
"
B
B
A
F ! dL = # " A dU = #(U B # U A )
F =!
A
B
dU
dL
Entropic Force
Similarly, in the case of isothermal compression of ideal gas from VA to VB, the total kinetic
energy stays the same, but the free energy G changes.
F
dL
Ideal Gas
T Reservoir
"
B
A
F ! dL = # "V dG
VB
A
F =!
dG
dL
# VB &
!G = "T!S = "k B NT% (
$ VA '
What is the entropic force exerted on the wall by an ideal gas at constant T ?
dL
F
T Reservoir
L
F =!
dG
dL
F =!
dE d ( PV )
dS
!
+T
dL
dL
dL
G = E + PV ! TS
0
3N / 2
3N / 2
S = kB ln! = k B ln( E kin
" V N ) + const = kB ln( E kin
) + kB ln(V N ) + const
dS = kB
F =T
N
N
dV = kB dL
V
L
dS
NT
= kB
dL
L
PV = Nk B T
Osmotic force is similar to the expansion force of an ideal gas
movable semi-permeable wall
Since the volume of water roughly stays the
same, the entropy of water does not change.
However, the solutes still want to maximize
entropy by expansion, just like an ideal gas.
solute
H2O
F
F = kB
NT
,
L
Posmotic = CkBT
F N
= kB T
A V
van’t Hoff equation
concentration
Take home messages
For a small system a in thermal contact with a large surrounding B, maximizing
entropy is equivalent to minimizing the Gibbs free energy .
Ga = Ea + PVa ! TSa
During the process of a going from non-equilibrium to equilibrium at constant T
and P, the maximum useful work that can be transduced is "Ga .
The chemical potential of molecular specie i in a mixture is defined as
µi =
dG
dN i T ,P,N
j
, i! j
It is the tendency of that molecular specie to chemically react, which
depends on both its concentration and internal energy of the molecule.
In a chemical reaction A ! B , the difference between A and B is amount of
chemical energy available to do work per unit of molecule. At chemical
equilibrium, µA = µB , or G = 0 .