Lecture 6 Free Energy

Similar documents
OCN 623: Thermodynamic Laws & Gibbs Free Energy. or how to predict chemical reactions without doing experiments

Lecture 20. Chemical Potential


The Second Law of Thermodynamics (Chapter 4)

Introduction Statistical Thermodynamics. Monday, January 6, 14

Some properties of the Helmholtz free energy

Lecture 6. NONELECTROLYTE SOLUTONS

Effect of adding an ideal inert gas, M

UNIVERSITY OF SOUTHAMPTON

Practice Examinations Chem 393 Fall 2005 Time 1 hr 15 min for each set.

CHAPTER 6 CHEMICAL EQUILIBRIUM

Chapter 5. Simple Mixtures Fall Semester Physical Chemistry 1 (CHM2201)

Chapter 3. Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc.

Thermodynamic Laws, Gibbs Free Energy & pe/ph

where R = universal gas constant R = PV/nT R = atm L mol R = atm dm 3 mol 1 K 1 R = J mol 1 K 1 (SI unit)

Thermodynamics. Chem 36 Spring The study of energy changes which accompany physical and chemical processes

Module 5 : Electrochemistry Lecture 21 : Review Of Thermodynamics

MS212 Thermodynamics of Materials ( 소재열역학의이해 ) Lecture Note: Chapter 7

Chapter 19 Chemical Thermodynamics

Identify the intensive quantities from the following: (a) enthalpy (b) volume (c) refractive index (d) none of these

Lecture Notes 2014March 13 on Thermodynamics A. First Law: based upon conservation of energy

Thermodynamics of Reactive Systems The Equilibrium Constant

Rate of Heating and Cooling

Adiabatic Expansion/Compression

= 16! = 16! W A = 3 = 3 N = = W B 3!3!10! = ΔS = nrln V. = ln ( 3 ) V 1 = 27.4 J.

University of Washington Department of Chemistry Chemistry 452/456 Summer Quarter 2014

Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry S H 2 = S H 2 R ln P H2 P NH

Lecture 3: Thermodynamics

1 mol ideal gas, PV=RT, show the entropy can be written as! S = C v. lnt + RlnV + cons tant

Chemistry 163B Absolute Entropies and Entropy of Mixing

ADIABATIC PROCESS Q = 0

Thermodynamic Variables and Relations

Chapter 19 Chemical Thermodynamics Entropy and free energy

Chapter 19 Chemical Thermodynamics

4/19/2016. Chapter 17 Free Energy and Thermodynamics. First Law of Thermodynamics. First Law of Thermodynamics. The Energy Tax.

Thermodynamics: Lecture 6

Thermodynamic Processes and Thermochemistry

Chapter 19 Chemical Thermodynamics

Lecture 14. Entropy relationship to heat

ESCI 341 Atmospheric Thermodynamics Lesson 12 The Energy Minimum Principle

Thermodynamics of solids 5. Unary systems. Kwangheon Park Kyung Hee University Department of Nuclear Engineering

Chapter 17. Free Energy and Thermodynamics. Chapter 17 Lecture Lecture Presentation. Sherril Soman Grand Valley State University

Chemistry. Lecture 10 Maxwell Relations. NC State University

Equations: q trans = 2 mkt h 2. , Q = q N, Q = qn N! , < P > = kt P = , C v = < E > V 2. e 1 e h /kt vib = h k = h k, rot = h2.

Affinity, Work, and Heat

Problem: Calculate the entropy change that results from mixing 54.0 g of water at 280 K with 27.0 g of water at 360 K in a vessel whose walls are

Ch. 19 Entropy and Free Energy: Spontaneous Change

Solutions to Problem Set 6

Chpt 19: Chemical. Thermodynamics. Thermodynamics

UNIVESITY OF SWAZILAND FACl.JLTY OF SCIENCE AND ENGINEERING DEPARTMENT OF PHYSICS

General Physical Chemistry I

8.21 The Physics of Energy Fall 2009

Physical Biochemistry. Kwan Hee Lee, Ph.D. Handong Global University

Chapter 19. Chemical Thermodynamics

MME 2010 METALLURGICAL THERMODYNAMICS II. Fundamentals of Thermodynamics for Systems of Constant Composition

Chemistry 2000 Lecture 9: Entropy and the second law of thermodynamics

du = δq + δw = δq rev + δw rev = δq rev + 0

TODAY 0. Why H = q (if p ext =p=constant and no useful work) 1. Constant Pressure Heat Capacity (what we usually use)

CHEMICAL ENGINEERING THERMODYNAMICS. Andrew S. Rosen

Physics is time symmetric Nature is not

Entropy in Macroscopic Systems

THE SECOND LAW OF THERMODYNAMICS. Professor Benjamin G. Levine CEM 182H Lecture 5


SCORING. The exam consists of 5 questions totaling 100 points as broken down in this table:

Concentrating on the system

THERMODINAMICS. Tóth Mónika

Express the transition state equilibrium constant in terms of the partition functions of the transition state and the

NENG 301 Week 8 Unary Heterogeneous Systems (DeHoff, Chap. 7, Chap )

Thermodynamic Third class Dr. Arkan J. Hadi

Chapter Eighteen. Thermodynamics

Ideal Gas Law. Deduced from Combination of Gas Relationships: PV = nrt. where R = universal gas constant

Last Name or Student ID

Mixtures. Partial Molar Quantities

5.60 Thermodynamics & Kinetics Spring 2008

Chapter 11 Spontaneous Change and Equilibrium

Chemistry 1A, Spring 2007 Midterm Exam 3 April 9, 2007 (90 min, closed book)

Chapter 17.3 Entropy and Spontaneity Objectives Define entropy and examine its statistical nature Predict the sign of entropy changes for phase

THERMODYNAMICS. Dr. Sapna Gupta

Lecture 7: Kinetic Theory of Gases, Part 2. ! = mn v x

PY Introduction to thermodynamics and statistical physics Summer Examination suggested solutions Dr. Asaf Pe er

rate of reaction forward conc. reverse time P time Chemical Equilibrium Introduction Dynamic Equilibrium Dynamic Equilibrium + RT ln f p

Chapter 19 Chemical Thermodynamics Entropy and free energy

Statistical thermodynamics (mechanics)

Name: Discussion Section:

PHYSICS 210A : STATISTICAL PHYSICS HW ASSIGNMENT #4 SOLUTIONS. ( p)( V) = 23kJ,

MASSACHUSETTS INSTITUTE OF TECHNOLOGY SPRING 2007

Thermodynamics 2013/2014, lecturer: Martin Zápotocký

Entropy Changes & Processes

Atkins / Paula Physical Chemistry, 8th Edition. Chapter 3. The Second Law

4) It is a state function because enthalpy(h), entropy(s) and temperature (T) are state functions.

Chapter 19: The Kinetic Theory of Gases Questions and Example Problems

Entropy and Free Energy in Biology

General Chemistry I. Dr. PHAN TẠI HUÂN Faculty of Food Science and Technology Nong Lam University. Module 4: Chemical Thermodynamics

MP203 Statistical and Thermal Physics. Jon-Ivar Skullerud and James Smith

Challa Vijaya Kumar University of Connecticut Module 4. Physical Chemistry 1 (Thermodynamics) Module 4. Open Source Textbook. Challa Vijaya Kumar

Statistical mechanics of biological processes

Lecture. Polymer Thermodynamics 0331 L First and Second Law of Thermodynamics

Classical Thermodynamics. Dr. Massimo Mella School of Chemistry Cardiff University

Introduction. Statistical physics: microscopic foundation of thermodynamics degrees of freedom 2 3 state variables!

Chapter 2: Equilibrium Thermodynamics and Kinetics

Transcription:

Lecture 6 Free Energy James Chou BCMP21 Spring 28

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 = k B 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 Equilibrium E1 E2 Picture from hyperphysics.phy-astr.gsu.edu

III. The Boltzmann & Gibbs Distribution. ( ) = A! exp "E a % p E a # & ( $ k B 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 " A F! dr A B 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 = 1 7 erg =.239 cal. At the single molecule level, k B T is commonly used because 1 k B T! 4 pn nm. In thermodynamics, work typically involves volume change. B W =!# F " dr =! # P " A dr A B A dr F op V 2 W =! # P " dv Ideal Gas V1 F int

Example: A mole of ideal gas initially has a pressure of 2 atm, and then expands at a constant temperature of 3 K against the atmosphere pressure (1 atm) until equilibrium is reached. How much work is done by the system? dr F op F int Ideal Gas Heat Block 3 K Irreversible P op =1 atm P = 1 atm dr F int Ideal Gas Heat Block 3 K W = P op ( V 2! V 1 ) =1247 J

Reversible P int = P op dr A reversible process has welldefined path. F int Ideal Gas Heat Block 3 K A W = V 2 " P op! dv = nrt dv V = nrt ln V 2 " V1 V 2 V1 V 1 = (8.314 J mol -1 K -1 ) (3 K) ln 2 =1728 J P V B

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?! T = de $ # & " ds % N,V T is defined at constant V. So T is not related to work. In this definition, replace de with Q. Q = T!S 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. dr F op F int Ideal Gas Heat Block 3 K Energy of one mole of ideal gas: E = 3 2 RT. Since T = const,!e =. V 2 But W =! # P " dv op =!1728 J. V 1

Introducing the definition of free energy G dr F op Now how about!g =!( E + PV ) " T!S? F int Ideal Gas Heat Block 3 K N ( ) " k B ln( V 1 ) = Rln V 2 N!S = k B ln V 2 = Rln P 1 V 1 P 2 = ( 8.314 J mol -1 K -1 )ln2 = 5.76 J K -1!G =!E +! PV ( ) " T!S = "( 3 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 = Q B + W B = T!S B " P!V B.!E =!E B +!E a = "!E B = #!E a, T!S B = #!E a + P!V B Principle of maximum entropy: T!S tot = T!S B + T!S a = "!E a + P!V B + T!S a #.!V B = "!V a!g a!e a + P!V a " T!S a #

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 (G a ) to be G a = ( E + PV! TS) a. a will move to a new equilibrium to minimize G a and during the process, useful work can be obtained. Normally, for biological systems, T and P of the surrounding are constant, and hence!g a =!E a + P!V a " T!S a # represents the maximum useful work can be extracted from a. a Surrounding, B Not equilibrium A Useful Work 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 + N 1! 1 + N 2! 2 1 2 G is a function of both E and N i, at constant T and P. G( E,N 1,N 2 ) = E + PV! TS, E = E kin + N 1 " 1 + N 2 " 2

Define chemical potential µ of specie i to be µ i = dg dn i T,P,N j, i! j Remember what is temperature? 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 µ 1 in the 2-component ideal gas? G( E,N 1,N 2 ) = E + PV! TS, E = E kin + N 1 " 1 + N 2 " 2? µ 1 = dg dn 1 = de! T ds = " 1! T ds dn 1 dn 1 dn 1 *# 2! 3N / 2 & S = k B ln, % ( 2m) $ ( kin +, 3N /2 "1)! ' 3 2 k BT ( ) 3N / 2 V N 1 ( ) "3N 1 2 N! 2!! - /./ Don t worry about this T ds = 3 dn 1 2 k T ln " mk T % " B B $ ' ( k # 2!! 2 & B T ln N % 1 $ ' # V & Concetration, C 1

µ 1 =! 1 + k B T lnc 1 " 3 2 k B T ln mk B T 2#! 2 Define the standard chemical potential µ 1 =! 1 " 3 2 k T ln mk T B B C 2#! 2 1 ( ) 3 / 2 We obtain! µ 1 = k B T ln C $ 1 # & + µ 1 " C 1 % Also true for aqueous solution For gas, C = 1 mole / 22 L. For solutes in aqueous solution, C = 1 mole / L = 1 M. For solvent, C = the molarity of the pure solvent e.g. C of H 2 O is 55.5 M.

The Concept of Chemical Equilibrium 1 2 "!G = dg % " $ ' dn 1 + dg % $ ' dn 2, dn 1 = (dn 2 # dn 1 & # dn 2 & " =! dg % " $ ' dn 2 + dg % $ ' dn 2 # dn 1 & # dn 2 &!G dn 2 = µ 2 " µ 1 The difference in chemical potential is the available chemical energy to do work per unit of molecule. At equilibrium,!g = " µ 2 = µ 1 " chemical equilibrium

When does a chemical reaction stop? A + B! AB At chemical equilibrium, µ AB! ( µ A + µ B ) = µ i = Nk B T ln( C i C ) i + µ i = RT ln( C i C ) i + µ i µ AB! µ A! µ B RT " C + ln AB / C $ AB # $ ( C A / C A ) C B / C B ( ) % ' &' =

!G The equilibrium constant µ AB! µ A! µ B RT " C + ln AB / C $ AB # $ ( C A / C A ) C B / C B ( ) % ' &' = Define equilibrium constant K eq 1 M # K eq = exp!"g $ % RT & ' ( = C AB / C AB C A / C A ( )( C B / C B ) # K eq = exp!"g $ % RT & ' ( = C AB C A C B pk! "log 1 K eq

The general equation for!g at any point of the reaction 1 2!G =!G + RT ln C 2 C 1!G RT ln C 2 C 1!G ( fixed)!g = "RT ln K eq

An example of cellular chemical potential ATP + H2O! ADP + P i 1 mm 55.5 M.1 mm 1 mm typical cellular concentration!g =!G + RT ln ( C ADP C ADP )( C P C P ) ( C ATP C ADP )( C H2O C H2O )!G can be determined if equilibrium C i 's are known.!g can also be calculated from standard free energy of formation!g f.!g =!G f, ADP +!G f, Pi "!G f, ATP "!G f, H2O = "3.3 kj/mol

ATP + H2O! ADP + P i "G = #3.3 kj/mol 1 mm 55.5 M.1 mm 1 mm typical cellular concentration!g =!G + RT ln ( C ADP C ADP )( C P C P ) ( C ATP C ADP )( C H2O C H2O ) ( 1. "1!3 )( 1. "1!4 ) =!3.3 + 2.748 ln =!53.1 kj / mol 1. "1!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 A B B " F! dl = # " du = # U B #U A A F =! du dl A ( ) Entropic Force Similarly, in the case of isothermal compression of ideal gas from V A to V B, the total kinetic energy stays the same, but the free energy G changes. dl F Ideal Gas T Reservoir B V " B # F! dl = # " dg!g = "T!S = "k B NT V & B % ( A V A $ ' F =! dg dl V A

What is the entropic force exerted on the wall by an ideal gas at constant T? dl T Reservoir L F F =! dg dl ( ) F =! de dl! d PV dl G = E + PV! TS + T ds dl 3N / S = k B ln! = k B ln E 2 kin " V N 3N / 2 ( ) + const = k B ln( E kin ) + k B ln( V N ) + const ds = k B N V dv = k B N L dl F = T ds dl = k B NT L PV = Nk B T

Osmotic force is similar to the expansion force of an ideal gas movable semi-permeable wall solute H2O F 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. F = k B NT L, F A = N V k B T P osmotic = Ck B T 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. G a = E a + PV a! TS a 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 "G a. The chemical potential of molecular specie i in a mixture is defined as µ i = 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, µ A = µ B, or G =.