Idea Transcript
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 .