What is Thermo, Stat Mech, etc.?

Transcription

1 What is hermo, Stat Mech, etc.? Rando om Kinetics & Mechanics Macroscopic hermo Coher rent Statistical Mechanics ic Atom Classical Microscopici Quantum Molec cular hermodynamics is a funny subject. he first time you go through it, you don't understand it at all. he second time you go through it, you think you understand it, except for one or two small points. he third time you go through h it, you know you don't understand it, but by then you are so used to it, it doesn't bother you any more. -- Arnold Sommerfield Ludwig Boltzmann, who spent much of his life studying statistical mechanics, died in 1906, by his own hand. aul Ehrenfest, carrying on the work, died similarly in Now it is our turn to study statistical mechanics. erhaps it will be wise to approach the subject cautiously. -- David L. Goodstein Our approach will be to focus on the macroscopic, thermodynamic picture with occasional insight from the microscopic picture via statistical mechanics. aul ercival Modified by Jed Macosko 9/10/008

2 What is hermodynamics? Knowledge Humanities Science Social (sciences) hysics Chemistry Biology Analytical hysical Inorganic, Organic Macroscopic hermodynamics Molecular Quantum Chemistry Spectroscopy Statistical hermodynamics Kinetics Equilibrium Change Structure aul ercival 1/10/004

3 Energy, Work and Heat Energy is the capacity to do work. Its classification into: kinetic potential (by motion) (by position) e.g. thermal chemical, electrical Is purely arbitrary! Heat and work are not types of energy, but are processes involving transfer of energy. hey appear and disappear at the system boundary. hey are path variables. Heat is the transfer of energy from one body to another of lower temperature. Convention: if heat flows into the system, q > 0. Work is the transfer of energy by some mechanism other than temperature difference. Convention: if work is done on the system, w > 0. Heat stimulates random motion. Work stimulates organized motion. Work degrades into heat.! qualitative observations by Count Rumford (Ben hompson)! quantitative measurements by James Joule aul ercival 1/10/004

4 erminology 1 A system is a particular sample of matter or region of space. An isolated system does not interact with its surroundings. system + surroundings = universe A closed system does not allow passage of mass over its boundaries, in contrast to An open system. An adiabatic system has boundaries which permit no flow of heat. It is insulated. A system is in a definite state when all its properties have definite values. A system at equilibrium is time independent; it is not affected by the history of the system. Extensive properties depend on the amount of substance in the system, e.g. n,. otal roperty = property(part) Intensive properties are independent of amount, e.g.,. otal roperty = property of part aul ercival 1/10/004

5 matter open system heat surroundings matter closed system heat surroundings matter isolated system heat surroundings aul ercival 1/10/004

6 erminology State variables (state functions) uniquely determine the state of a system at equilibrium. wo samples of a substance with the same state variables are in the same state. he change in a state variable depends only on the initial and final states, independent of path. ath functions depend on the process and therefore vary with path. A cyclic process is one in which the initial and final states are the same, i.e. no change in the state variables. In contrast, path functions generally have non-zero values for cyclic processes, dependent on the path. A reversible process is one that can be reversed by an infinitesimal modification of a variable. he system is in equilibrium with the surroundings at all times. his is an idealized situation, useful as a theoretical limit, but All real processes are irreversible. It is possible to restore the system or the surroundings to their original states but not both. An equation of state is the functional relationship between the properties of a system, e.g, the ideal gas law. aul ercival 1/10/004

7 Ideal Gases -- Review Ideal gases obey the ideal gas law: = nr ressure a 1 N m - 1 J m -3 1 kg m -1 s - olume m 3 emperature K number of moles gas constant 10 3 dm litres n = w M N = L = L -1-1 R = J K mol 3 no. of atoms in 1 g 1 C but if is in atm ( a) and in litres, R = dm atm K mol isotherms isobars isochors aul ercival 1/10/004

8 Mixtures of Ideal Gases If the ideal gas law applies to each component, i i = n R i partial pressure i nr n = i i tot i = n =χ tot tot Dalton s Law of artial ressures i tot i = ni = tot ntot i e.g. for two components: pressure = + A B A B =χ =χ A B χ A χ B 0 composition Real gases are ideal only at the low density limit. Why? aul ercival 1/10/004

9 Real Gases have non-zero volume at low and high have repulsive and attractive forces between molecules short range, important at high longer range, important at moderate At low pressure, molecular volume and intermolecular forces can often be neglected, i.e. properties ideal. irial Equations R 1 B C = R B C = B is the second virial coefficient. C is the third virial coefficients. hey are temperature dependent. an der Waals Equation a + ( b) = R = m = n aul ercival 1/18/005

10 Compressibility Factor also known as compression factor Z = = R ideal.0 C H 4 CH 4 Z 1.0 H NH 3 0 he curve for each gas becomes more ideal as Z aul ercival 1/18/005

11 he van der Waals Equation 1 a + ( b) = R Intermolecular attraction = internal pressure R a = b Z a = = R b R molecular volume excluded volume ( ) 3 3 r / π = πσ a a a R R R = 1+ b + b + 3 ( R ) (boring algebra) Z 1 a = b +... R R he initial slope depends on a, b and : positive for b> a/ R molecular size dominant negative for b< a/ R forces dominant zero at = a/ Rb Boyle emperature ~ ideal behaviour over wide range of aul ercival 1/18/005

12 Condensation of Gases Real gases condense don t they? supercritical fluid c 1 liquid c c 1 gas c, c and c are the critical constants of the gas. Above the critical temperature the gas and liquid phases are continuous, i.e. there is no interface. aul ercival 1/18/005

13 he van der Waals Equation he van der Waals Equation is not exact, only a model. a and b are empirical constant. R a ab + + = 0 3 b he cubic form of the equation predicts 3 solutions 0 b here is a point of inflection at the critical point, so slope: curvature: R a = = ( b) R 6a = = 0 ( b) 3 4 a c = c = 3b c = 7b 8a 7Rb 3 a 7 Z = = = = c c c B c Rc 8 Rb 8 aul ercival 1/18/005

14 he rinciple of Corresponding States Reduced variables are dimensionless variables expressed as fractions of the critical constants: = = = r r r c c c Real gases in the same state of reduced volume and reduced temperature exert approximately the same reduced pressure. hey are in corresponding states. If the van der Waals Equation is written in reduced variables, 3 + ( 3 1) = 8 r r r r Since this is independent of a and b, all gases follow the same curve (approximately). Z 1.0 r = 1.5 r = 1. r = 1.0 r aul ercival 1/18/005

15 roperties of Real Gases as 0 Real gases have interactions between molecules. hese change when the gas is compressed, but they need not go to zero as 0. e.g. consider Z For an ideal gas: For a real gas: and in the limit: Z = = () 1 = 0 R Z B C R = = Z = B + C + Z lim = B 0 0 Not all properties of real gases tend to ideal values as 0. aul ercival 1/18/005

16 van der Waals Johannes Diderik van der Waals, Nobel rize in hysics aul ercival CHEM 360 Spring 005 1/18/005

17 Energy, Work and Heat Energy is the capacity to do work. Its classification into: kinetic potential (by motion) (by position) e.g. thermal chemical, electrical Is purely arbitrary! Heat and work are not types of energy, but are processes involving transfer of energy. hey appear and disappear at the system boundary. hey are path variables. Heat is the transfer of energy from one body to another of lower temperature. Convention: if heat flows into the system, q > 0. Work is the transfer of energy by some mechanism other than temperature difference. Convention: if work is done on the system, w > 0. Heat stimulates random motion. Work stimulates organized motion. Work degrades into heat.! qualitative observations by Count Rumford (Ben hompson)! quantitative measurements by James Joule aul ercival /0/006

18 he First Law of hermodynamics " For finite changes of state: " For infinitesimal changes: U is the internal energy of the system. " When a system changes from one state to another along an adiabatic path, the amount of work done is the same, whatever the means employed. For For " he energy of an isolated system is constant. For q > 0 for heat flow into the system w > 0 for work done on the system ad final initial U = q+ w du = δ q +δw q = 0, w = U U = U q 0, q = U w= w w q = 0, w= 0 U = 0 No perpetual motion machines! " In any cyclic transformation the work done by a system on its surroundings is equal to the heat withdrawn from the surroundings. δ w= δq du = 0 " he energy of the universe is constant. ad!!! U system = U surroundings aul ercival /0/006

19 emperature wo systems in thermal equilibrium are at the same temperature. If system A is in thermal equilibrium with system B, and A is in thermal equilibrium with C, then B must be in thermal equilibrium with C. his is a statement of the zeroth law of thermodynamics. Ideal gas temperature: Unit of temperature: 1 Kelvin = lim 0 nr = ( triple point of water) he freezing point of water at 1 atm is K. he boiling point of water at 1 atm is K. he Celsius scale is defined as t/ C = / K It is possible to define an absolute temperature scale (Kelvin scale) by considering the work done in an isothermal reversible expansion/compression. 1 w= nrln aul ercival /0/006

20 ressure olume Work 1 Expansion 1 > / mg A = ex = mg/ A m m h,, 1, 1, 1 A = area of piston w = = ( mg) h ( ex A) h ( ) = ex equation of state (need not be isotherm) 1 Expansion into a vacuum ex = 0 w= 0 Compression < 1 ex = ex ( ) w= ex aul ercival /0/006

21 ressure olume Work Multi-stage Expansion 1 3 w= 4 5 ( ) + 3( 3 ) + 4( 4 3) + ( ) Reversible Expansion Make steps so small that hen d 0, d 0 δ w= d ex ( ) = + d d int d int w= dw= d path 1 For ideal gases int / = int 1 nr wrev = nrln = nrln and at fixed temperature 1 1 aul ercival /0/006

22 ressure olume Work compression 3 expansion Consider the cyclic path ( 3 1) 0 1( 1 3) ( )( ) w= = Consider the cyclic reversible path = w d d = d d = Even for a cyclic process w depends on path!!! du = 0 δ w = δq aul ercival /0/006

23 Energy vs. Enthalpy For a change in state at constant volume, no expansion work is done, so U = q, du =δq However, for a change in state at constant pressure, U = q + w, du =δq d 1 1 = = δ U du q d 1 ( ) U U = q 1 1 ( ) ( 1 1) ( ) ( ) U + U + = q U + U + = q = = constant Enthalpy H = U + H = q, dh =δq H, being a function of state variables only, is also a state variable. For a general change of state ( and may both change), ( ) H = U + = U dh = du + d + d aul ercival /0/006

24 Heat Capacity ransfer of heat to a system may result in a rise in. δ q= Cd path function, so C depends on conditions. Define: δ q at constant volume, no work = Cd δ q = C d at constant pressure, only work From 1 st Law, du =δ q +δ w =δq d du =δ q f o r d = 0 C U = ex assume no other work Similarly dh = du + d d = ( δq d) + d d =δ q for d= 0 C H = dh = du + d = du + nrd For ideal gases ( ) C d = C d + nrd C = C + R aul ercival /0/006

25 he Relation Between C and C C But since H U C = U U = + U U du = d + d U U U = + H = U + C C U = + work needed to overcome U = intermolecular forces internal pressure expansion per degree For ideal gases U nr = 0, = C C = nr For liquids and solids is so small that C C aul ercival /0/006

26 adiabatic = insulated: Adiabatic Expansion 1 1 w du C d = = + 1 (0 for ideal gases) = C if C is independent of q = 0, δ q = 0 U = w du =δw U U du = d + d For adiabatic expansion w! 0, U! 0! 1 Free expansion: ex = 0 w= 0, = 0 Fixed pressure: Reversible expansion: w= ex w= U = C ex = C = ex w= 1 (, ) d Substitute appropriate equation of state. Not useful if changes. aul ercival /0/006

27 Adiabatic Expansion Reversible adiabatic expansion of ideal gases: du =δw δ q = 0 nr C d = d = d ideal gases only 1 1 C d = R d 1 1 C ln / = Rln / ( 1) ( 1) = ( C C ) ln ( / 1) ( ) = ( γ ) ( ) ln / 1 ln / 1 1 = 1 1 γ 1 C γ= C Also, since 1 1 γ = = or = γ γ 1 1 = 1 1 ( γ 1)/ γ 1 1 γ 1 aul ercival /0/006

28 ypes of Expansion Work Summary For all types of expansion δ w= d ex For reversible changes In an irreversible expansion <, ( w) < ( w ) ex ( ) ex = =, in general rev For isothermal expansions is held constant by leakage of heat into the system from the surroundings. For an ideal gas In an adiabatic expansion U = 0, U = 0, q = w q = 0 so the internal energy must provide for the work: w = U For reversible expansion of an ideal gas through a given volume change from the same initial state, ( w ) > ( w ) isothermal adiabatic since U is continuously replenished by heat intake to keep the temperature constant. aul ercival /0/006

29 he Joule Expansion Experiment For a closed system, the state function U is determined by and alone: U U du = d + d Cd U = + d Joule tried to measure this partial derivative. 1. Gas in A, vacuum in B. A B. Open valve. 3. Any change in? He found no change in temperature when the gas expanded from A to A + B, i.e. q = 0. Also, no work was done (free expansion), so w = 0. Conclusion: U = 0 and hence U = 0 or U = U( ) only Strictly only true for ideal gases. ( ) Not true for liquids and solids, but since U U / and is very small, the effect of on U is usually ignored. aul ercival 1/5/004

30 ressure Dependence of Enthalpy For a closed system, the state function H is determined by and alone: H H dh = d + d H = Cd + d For ideal gases H = U + = U + nr H U = = 0 Not true for real gases, liquids and solids! dh = du + d + d H U C d + d = C d + d + d + d At fixed temperature, d = 0 H U = + + small for liquids and solids Investigate real gases at constant enthalpy, i.e. dh = 0 H = C H aul ercival 1/5/004

31 he Joule-hompson Experiment Joule-hompson Coefficient µ J = H ump gas through throttle (hole or porous plug) 1 > 1 Work done by system w= Since q = 0 Keep pressures constant by moving pistons. 1 1 U = U U1 = w= 1 1 U + = U H = H constant enthalpy 1 Measure change in of gas as it moves from side 1 to side. µ J = 0 A modern, more direct experiment uses similar apparatus but with a H q lim heater to offset the temperature drop. = 0 aul ercival 1/5/004

32 he Linde Refrigerator For most gases at room temperature µ J > 0 so sudden (adiabatic) expansion results in a drop in. his is the basis of operation of the Linde refrigerator and gas liquefaction. cooling fins cold gas compressor expansion nozzle In general, µ J depends on and can even change sign. inversion temperature H H > < 0 0 lines of constant enthalpy aul ercival 1/30/005

33 he Molecular Interpretation of U What is U? U can be related to thermochemical observables: U C =, U = C d for ideal gases 1 ( ) ( 0) U U C d = Can U be calculated from molecular properties? U = u = N < u > ignoring intermolecular forces molecules Classically, <u> is given by the Equipartition Law: he average energy of each different mode of motion of a molecule is ½k. k = Boltzmann constant = R/L = J K -1 Implicit in this law is the concept that there is no coupling between molecular modes of motion. otal: ranslation: Rotation: ibration: no. of vib. modes u =ε tr +ε rot + εv ib( +εe l) ε = mv = mx! + my! + mz! tr rot Ix x Iy y Iz z ε = ω + ω + ω ε = + = vib mx! kx per mode 0 3n 6 for non-linear molecules = 3n 5 for linear molecu les aul ercival 1/5/004

34 he Molecular Interpretation of U (cont.) When quantum effects can be ignored, the average energy of every quadratic term in the energy expression has the same value, ½k. Monoatomic gases Diatomic gases Linear polyatomics Non-linear molecules U < > =, = = u k C R 3 ( ) 3 ( 1 3n 5) < u >= k, C = 3.5R < u >= + + e.g. for 3 3 ( 3n 6) < u >= + + e.g. for k n= 3, C = 6.5R k n = 3, C = 6. 0R Experimental observations do not agree with these predictions, except for monoatomic gases. In particular:! he predicted values are too high.! Experimental values are temperature-dependent. Accurate calculations are provided by statistical mechanics, where it can be shown that the equipartition principle only holds in the limit of high temperature, specifically for the condition k >> E, where E is the appropriate spacing of energy levels. Equipartition free transfer of energy between modes aul ercival 1/5/004

35 emperature Dependence of C U C =, where U U0 = N <ε tr >+<ε rot >+ = C + C + C + C C C tr rot tr rot vib el = for all 1.5R { R linear = f or k # ε 1. 5R non-linear ( ) rot C = 0 at very low, where < ε rot > 0 C vib = R for each vibrational mode, at high el C = 0 almost always, since electronic excitation takes great energy rot e.g. for a diatomic molecule: C R 7 5 t + r + v t + r two atoms t 3 t 0 0 aul ercival 1/5/004

36 emperature Dependence of H Interactions between molecules also contribute to the heat capacity of real systems. In particular, first-order phase changes often involve large energy changes. hese are usually measured at constant pressure and expressed as enthalpies. gas H solid solid 1 liquid H s-s H f H v 0 0 In general, H a b c = H C = = b+ c + for each phase aul ercival 1/31/004

37 hermochemistry he study of energy changes that occur during chemical reactions:! at constant volume U = q no work! at constant pressure H = q only work For practical reasons most measurements are made at constant, so thermochemistry mostly deals with H. reaction products H = H H reactants If H > 0 the reaction is endothermic. If H < 0 the reaction is exothermic. For comparison purposes we need to refer H to the same and. o define a standard reaction enthalpy each component of the reaction must be in its standard state the most stable form at 1 bar pressure and (usually) 5 C. 1 bar = 10 5 a 1 atm = bar aul ercival /6/005

38 Reaction Enthalpy 1 Hess s Law he standard enthalpy change in any reaction can be expressed as the sum of the standard enthalpy changes, at the same temperature, of a series of reactions into which the overall reaction can be formally divided. Combine chemical equations as if mathematical equations, e.g. A + B C H1 C + D E + F H F B + G H3 A + D E + G H H = H1+ H + H3 Standard Reaction Enthalpy H H H o o 98 o 500 reaction enthalpy at 1 bar and at standard or some other aul ercival /6/005

39 Reaction Enthalpy Standard (molar) enthalpy of formation H o f Heat of formation of a substance from its elements, all substances being in their standard state. By definition, for all elements o H f = 0 f H o Enthalpy of combustion H for total oxidation of a substance H o c c H o e.g. C 6 H 1 O 6 + 6O 6CO + 6 H O c H = -808 kj mol -1 Enthalpy of hydrogenation H when an unsaturated organic compound becomes fully saturated e.g. C 6 H 6 + 3H C 6 H 1 H = -46 kj mol -1 Enthalpy of atomization Bond dissociation enthalpy H for the dissociation of a molecule into its constituent gaseous atoms e.g. C H 6 (g) C(g) + 6H(g) H = 883 kj mol -1 Bond strength single bond enthalpy An average value taken from a series of compounds and often combined for a rough estimate e.g. H (C H 6 ) = H (C-C) + 6 H (C-H) aul ercival /6/005

40 emperature Dependence of H he temperature dependence of reaction enthalpies can be expressed in terms of the dependence of the enthalpies of the reaction components: wher e H ( ) = H ( ) + Cd 1 = + H ( ) H ( ) Cd 1 C = C C p p p products reactants 1 1 p p his is known as Kirchoff s Law. e.g. A + B C + D A + B C + D H H ( p p )( ) H ( ) = C (A) + C (B) 1 + H ( ) 1 ( ) 1 ( ) ( Cp(C) Cp(D) )( 1) ( ) p p = H ( ) + C C products reactants assuming that the C p values are independent. aul ercival /6/005

41 Reactions at Constant olume ( ) ( ) H = U + r r products reactants For solids and liquids For ideal gases ( ) 0, so ( ) gas H U = n R, so H U + n R gas e.g. C H (g) + O (g) 3CO (g) + 3H O(l) (- ) H = U + R r r he relationship between H and U is particularly important when relating thermochemical enthalpies to molecular properties, e.g. for a single bond energy U = H R as seen in the case of O (g) O(g). In practice, R is usually so much smaller than H that it is often ignored. aul ercival /6/005

42 Enthalpies of Ions in Solution Enthalpy of solution H for solution of a substance in a stated amount of solvent Enthalpy of dilution H for dilution of a solution to a lower concentration Enthalpy of solution to infinite dilution infinite amount of solvent o H soln for an he enthalpy of formation for a species in solution can be o o found by combining H soln with the H f of the gaseous species: 1 1 o -1 H (g) + Cl (g) HCl(g) H = 9.31 kj mol f HCl(g) HCl(aq) H = kj mol o -1 soln H (g) + Cl (g) HCl(aq) 1 1 H (ion) = H + H o o o f f soln = kj mol -1 o H f for individual ions in solution can only be found if one is arbitrarily fixed. By convention this is H + (aq). + o + H(g) H(aq) + e Hf ( Ha ) = 0 + ( Cl ) ( HCl ) ( H ) ( HCl ) 1 q H = H H = H o o o o f aq f aq f aq f aq he standard state for a substance in solution (not just ions) is a concentration of 1 mole solute in 1 kg solution (1 molal). aul ercival /6/005

43 Enthalpy of Formation of an Ionic Solid Consider individual steps in the formation of NaCl. o 1. Na(s) Na(g) H subl ( Na) H = I + R. Na(g) Na + (g) + e o ( Na) 1 o 3. ½Cl (g) Cl(g) H ( ) Cl-Cl H = E R 4. Cl(g) + e Cl o (g) ( Cl ) 5. Na + (g) + Cl (g) Na + (aq) + Cl (aq) o o NaCl(aq) Hsol ( Na + ) + Hsol ( Cl ) o Na(s) + ½Cl (g) NaCl(aq) Hf ( NaClaq) o o 1 o Hf ( NaClaq ) = Hsubl ( Na ) + I(Na) + H ( Cl-Cl) o o EA(Cl) + Hsol( Na + ) + Hsol( Cl ) Step 5 could be creation of solid NaCl instead of solution 5'. Na + (g) + Cl o (g) NaCl (s) H lattice ( NaCl) leading us to the enthalpy of formation of solid NaCl: Na(s) + ½Cl (g) NaCl(s) 1 ( NaCl ) ( Na ) (Na) ( Cl-Cl) o E (Cl) + H ( NaCl) H = H + I + H o o o f s subl A lattice A aul ercival /6/005

44 he Born-Haber Cycle Enthalpy changes can also be expressed in a diagram, e.g. I ( Na) + R Na + (g) + e + Cl(g) Na+(g) + Cl (g) E A ( Cl ) + R ( ) o H subl Na ( ) 1 o H Cl-Cl H H o f o f ( NaCl ) Na(g) + Cl(g) Na(s) + Cl(g) Na(s) + ½Cl (g) ( NaClaq) s NaCl(aq) NaCl(s) o H sol o H lattice Since H is a state variable, the sum of enthalpy changes around the cycle must be zero. Consequently, if all but one of the enthalpy changes is known, it can be readily calculated. his is equivalent to using Hess s Law to sum reaction steps. aul ercival /6/005

45 Spontaneous Change Who needs entropy, anyway? direction of spontaneous change he direction of spontaneous change is that which! leads to chaotic dispersal of the total energy! moves from a state of low intrinsic probability towards one of greater probability. Work is needed to reverse a spontaneous process. We need a quantity entropy to describe energy dispersal, i.e. the probability of a state. Spontaneous processes are irreversible. hey generate entropy Reversible processes do not generate entropy but they may transfer it from one part of the universe to another. aul ercival /13/005

46 Entropy 1 " Entropy is a state variable (property) which determines if a state is accessible from another by a spontaneous change. " Entropy is a measure of chaotic dispersal of energy. " he natural tendency of spontaneous change is towards states of higher entropy. " here are both thermodynamic (how much heat is produced?) and statistical definitions (how probable is a state?). hey both become equivalent when statistics is applied to a large number of molecules. Consider a falling weight which drives a generator and thus results in heat q being added to the reservoir (the surroundings). reservoir Define a system variable S ds(surr) = δq/ generator Use stored energy to restore the weight to its original height. he reservoir gives up δq rev to the system, and there is no overall change in the universe. δq ds(sys) = ds(surr) = rev aul ercival /13/005

47 In general, Entropy ds(sys) + ds(surr)! 0 ds(sys)! ds(surr) Equality for reversible processes only or, for the system, δq ds! Clausius inequality For an isolated system, q = 0 hence S! 0 Isothermal rocesses S= q / rev H H fus e.g. S( fusion) " = S( vap) " = routon s Rule: S m ( vap) 85 J K mol -1-1 Can be used to estimate H vap if b is known. Not good for associated liquids. emperature ariation C ( S) = d 1 C and ( S) d = 1 Absolute Entropy δ q = Cd rev C = S ( ) S(0) d S solid b vap liquid gas H b v aul ercival /13/005

48 Entropy 3 Entropy depends on robability. Consider the number of ways Ω of arranging n molecules between two sides (A and B) of a container. he probability A that all molecules are on side A depends on the ratio of Ω Α to the total number of arrangements. A B Ω = 1 Ω = = 1 A tot A Ω = 1 Ω = 4 = 1 A tot A 4 Ω = 1 Ω = 16 = 1 A tot A 16 State A becomes less and less probable as n increases. Conversely, the probability of the less ordered, roughly evenly distributed states, increases. Since entropy is a measure of disorder, it follows that S depends on Ω. Boltzmann equation Ω = 1 Ω = = n A tot A S = kln Ω n ( AND ) = xi y x+ y = x + y Since x y, ln ln ln aul ercival /13/005

49 he Second Law of hermodynamics # An isothermal cyclic process in which there is a net conversion of heat into work is impossible. # No process is possible in which the sole result is the absorption of heat from a reservoir and its conversion into work. It is possible to convert all work into heat! # It is impossible for heat to be transformed from a body at a lower temperature to one at a higher temperature unless work is done. # he entropy of an isolated system increases during any natural process. he universe is an isolated system. S(sys) < 0 is allowed provided S(sys) + S(surr) > 0 # All reversible Carnot cycles operating between the same two temperatures have the same thermodynamic efficiency. # here is a state function called entropy S that can be calculated from ds = δq rev /. he change in entropy in any process is given by ds δq/, where the inequality refers to a spontaneous (irreversible) process. he 1 st Law uses U to identify permissible changes of state. he nd Law uses S to identify natural changes among the permissible ones. aul ercival /13/005

50 he hird Law of hermodynamics # If the entropy of every element in its stable state at = 0 is taken as zero, every substance has a positive entropy which at = 0 may become zero, and does become zero for all perfect crystalline substances, including compounds. Nernst Heat heorem he entropy change accompanying transformation between condensed phases in equilibrium, including chemical reactions, approaches zero as 0. lim 0 S = 0 ractical consequence: Set S(0) = 0 for elements by convention. Apply Nernst to determine S(0) for all else. # It is impossible to reach absolute zero in a finite number of steps. he 1 st Law says U cannot be created or destroyed. he nd Law says S cannot decrease. he 3 rd Law says zero Kelvin cannot be reached. aul ercival /13/005

51 Given he Clausius Inequality Substitute into the 1 st Law: All exact differentials, so path independent. But Even more generally, Conditions for: δ ds =" q rev du = ds d ( ) d! 0 ex surr thermal equilibrium and du =δ q +δ w =δq exd ds" d =δq d δq ds = + ex ( ) = + δ w = d ( ) surr dsuniv dssurr d ex ex d If >, d > 0; if <, d < 0 Clausius Inequality δq ds! ex rev Fundamental Equation of hermodynamics mechanical equilibrium In general, i.e. any path Equal for reversible change aul ercival /13/005

52 he Fundamental Equation of hermodynamics Combine or du =δq d with du = ds " d 1 ds = du + d δ ds =" q rev Reversible change but true for all paths since d exact his fundamental equation generates many more relationships. U U Example 1: Comparison with du = ds + d S S U U = and = S Example : Consider that du is exact and cross differentiate. = S S S a Maxwell relation Example 3: S S = S S = = S cyclic rule and again! = = =α κ aul ercival /13/005

53 How Entropy Depends on and Compare with 1 ds = du + d U U U du = d + d = C d + d C 1 U ds = d + d + S S ds = d + d S = C S 1 U = + S = C d 0 for ideal gases 1 S = d = nr d = nrln / ( ) 1 For any substance, For ideal gases, = C α + κ ds d d S = C ln + nrln 1 1 assuming C is independent aul ercival /13/005

54 How Entropy Depends on and 1 ds = du + d First problem: replace du ; second problem: replace d. Use But du = dh d d 1 ds = dh d and both are solved! H H H dh = d + d = C d + d Compare with C 1 H ds = d + d S S ds = d + d S = C C S = d 1 = = S d nr d ( ) ( ) S = nrln / = nrln / S 1 H = 0 for ideal gases 1 1 aul ercival /13/005

55 Entropy Changes in Spontaneous rocesses Entropy is a state function, so S(sys) = S S 1 independent of path his can be used to calculate S for an irreversible process. " Consider isothermal expansion of a gas from 1 to : For the reversible case For the irreversible case ( ) S(sys) = nrln / reversible and irreversible cases 1 S(surr) = S(sys) S(surr) > S(sys) q w S(surr) = = e.g. for free expansion, w = 0 S(surr) = 0, S(univ) = S(sys) > 0 " Consider freezing of supercooled water at < 73 K reversible water, 0 C ice, 0 C rev. water, irreversible rev. ice, o 73 ( Hfus) S(sys) = C(l) ln + + C(s) ln Hfus( ) 1 S(surr) = = Hfus(73) + [ C(l) C(s) ]( 73) { } aul ercival /13/005

56 Entropy of Mixing " Consider the mixing of two ideal gases :,, 1 n 1,, n,, 1 + n 1 +n n S = nrln = nrln = nrln χ n1+ n n S = n Rln = n Rln = n Rln χ 1+ n1+ n ( ) ( ) S = nrln χ n Rln χ = n + n R χ ln χ +χ ln χ mix In general S = χ χ mix ntot R iln i his expression applies to the arrangement of objects (molecules) just as well as fluids (gases and liquids). For example, arrange N identical atoms in N sites in a crystal: Ω= N!/ N! = 1 S = kln Ω= 0 Compare with the arrangement of two types of atom, A and B. Ω= N! S k N NA N N! N! = A B i ( ln! ln! ln!) Application of Stirling s approximation ( ) leads to S = kn( χ ln χ +χ ln χ ) config A A B B B ln z! = zln z z aul ercival /13/005

57 Using Entropy to Achieve Low C S = d Cln / ( ) 1 o achieve lower temperatures, S must be reduced. if C is constant Choose some property X that varies with S, i.e. S = f(x,). his could be the pressure of a gas or, for example, the magnetic moment of a paramagnetic salt (whose energy varies with magnetic field). 1. Alter X isothermally. Entropy changes.. Restore X by a reversible adiabatic process. 3. Repeat cycle. S S S X, q = S X q = 0, S = 0 X 1 curves coincide at 0 K, a consequence of the Nernst Heat heorem X aul ercival /13/005

58 Spontaneous Change For a system in thermal equilibrium with its surroundings, δq ds! Clausius inequality At constant volume: At constant pressure: dq ( ds ) ( du ) For convenience, define: = du ds du!" 0 "!" 0 dq ds U, S, # 0 = dh!" dh 0 ( dh ), # S 0 no work isolated system work only A = U S G = H S da = du ds Sd dg = dh ds Sd hen the conditions for spontaneous change become: ( da ), # 0 ( dg ), # 0 aul ercival //004

59 Helmholtz Energy A Helmholtz energy; Helmholtz free energy; Helmholtz function; Maximum work function A for Arbeit For spontaneous change at constant and da = du ds # Note that it is the total function A that tends to a minimum; this is not the same as minimizing U and maximizing S. 0 Maximum Work Combine du =δ q +δw and ds! δq Word done by the system du # ds + δw ( δw) # ds du equality for reversible change ( δ w) max = ds δq rev δ w rev = δw rev A system does maximum work when it is operating reversibly. But ( da) = du ds = du δ qrev =δwrev herefore, for macroscopic changes wmax = A= S U constant (-w) can be more or less than U according to the sign of S. For S > 0, heat flows into the system to fuel the extra work. aul ercival //004

60 ( ), 0 Gibbs Energy G Gibbs free energy, Gibbs function! ery important in chemistry since it tells whether a particular reaction can proceed at a given and. For spontaneous change, G, dg # G = Gproducts Greactants # 0 for reactions can be calculated from tabulated data!!! G ( ) = H ( ) S ( ) If H is and S is then G the reaction proceeds -ve +ve < 0 at all temperatures +ve -ve > 0 at no temperatures -ve -ve if < H / S +ve +e if > H / S Maximum Work: ( dg) = dh ds ( dg) = du + d ds, =δ q +δ w + d ds rev rev max rev =δ w + d =δw (non- ) w (non- ) G max = constant, aul ercival //004

61 Basic hermodynamic Relations 1 Laws du =δq d (1) ds =δq Definitions H = U + A = U S G = H S (3) (4) (5) Fundamental Equations du = ds d dh = ds + d da = Sd d dg = Sd + d (6) (7) (8) (9) artial Differentials U U = = S H H = = S A A = S = G G = S = rev / S S () aul ercival //004

62 Basic hermodynamic Relations Maxwell Relations From (6) From (7) From (8) From (9) = S S = S S S = S = (10) (11) (1) (13) hermodynamic Equation of State du = ds d U S = (6) Substituting (1) U = U α = κ aul ercival //004

63 How Free Energy Depends on G = H S G = H S dg = d Sd (1) () (3) definition constant fundamental eqn. But G G H = S = G G H = ( G/ ) 1 G G = from (1) (4) (5) Gibbs-Helmholtz Equation ( G/ ) H = substitute (4) in (5) alternative form: ( G/ ) (1/ ) = H By applying the Gibbs-Helmholtz equation to both reactants and products of a chemical reaction, ( G/ ) H = aul ercival //004

64 How Free Energy Depends on G = G = G G1 = d 1 For solids and liquids does not change much with, so G ( ) G ( ) + ( ) 1 1 nr For a perfect gas = G = nrln( / ) where G is the standard free energy defined at =1 bar 1!! G = G ( ) + nrln( / ) he chemical potential for a pure substance is the molar Gibbs energy: For a perfect gas µ= G = G n µ=µ! +R ln( /! ) aul ercival //004

65 he Chemical otential So far, our thermodynamic relations apply to closed systems. However, G, U, H, etc. are extensive properties. artial molar quantities are used to describe quantities which depend on composition. In general Chemical otential J J i J = n i = nj i i i n,, j i G G dg = d + d n,, n... n,, n... J = U, H, S, A, G, G G + dn + dn + n 1 1 n n,,... n,,, n... G µ i = n i,, nj i du = ds d + µ dn Fundamental Equations i i dh = ds + d + µ dn etc. U H A µ i = n = i n = i S, i n S,, i i aul ercival //004

66 Free Energy of Mixing 1 Consider the mixing of two ideal gases A and B at constant and. Before: where After mixing: G = n µ (pure) + n µ 1 A A B B! A A R 0! { ln( / )}! { ln( / ) ln } G = n µ + R p A A A A 0 = n µ + R + R χ A A 0 A = n µ (pure) + n Rlnχ G = G + G < G A A A A A B 1 (pure) µ (pure) =µ + ln( / ) negative Change: G = G G mix 1 = n Rln χ + n Rln χ A A B B ( ln ln ) = nr χ χ +χ χ A A B B ( ln ln ) G = nr χ χ +χ χ mix A A B B G mix < 0 Mixing is spontaneous. S = G / = nr χ lnχ Entropy of Mixing ( ) mix mix n, Enthalpy of Mixing Hmix = Gmix + Smix = 0 Similarly, Umix mix 0 i = = for ideal gases and solutions i i aul ercival //004

67 Free Energy of Mixing For a binary system, G mix is at its minimum and S mix at its maximum when χ A =χ B = S mix G mix Equilibrium and Mixing: Consider an equilibrium between two ideal gases G = G (pure) + G (pure) + G 0 χ Α χ Β 0 A B mix ( ) nr( ln ln ) = n χ µ +χ µ + χ χ +χ χ A A B B A A B B 1 χ Α χ Β 0 A# B G G pure Gtotal G mix he equilibrium composition is determined by the minimum in G total. A composition B aul ercival //004

68 Reaction Equilibrium 1 A # B Consider Suppose an amount dx of A turns into B. dg =µ dn +µ dn = µ dx +µ dx, G = µ B µ A x hen ( ) A A B B A B, µ A and µ B depend on composition, and therefore change during the reaction. If µ A >µ B the reaction proceeds from A B. If µ A <µ B the reaction proceeds from B A. At equilibrium µ =µ A B G G slope = x 0 eq x 1 A B eq eq ( ) ( ) eq eq!!! ( A B ) =µ B µ A = rxn eq B = exp ( G! rxn / R) µ + Rln p / =µ + Rln p /!!!! A A B B R ln p / p G p p eq A aul ercival //004

69 Reaction Equilibrium A reaction such as A + 3B C + D can be written as 0 = A 3B +C +D or completely generally as 0= ν R = 0 hen G ξ, i i = νµ and at equilibrium, ν iµ i = 0 i i eq! pi νµ i i +ν ir ln = 0! i p! i Grxn + Rln = 0! i eq i i i ν G = Rln K! rxn i K p ν i i = i! eq aul ercival //004

70 Consider K! i Equilibrium Constants ν p = = K aa + bb " yy + zz ν= y+ z a b hermodynamic Equilibrium Constant! y! ( py / ) ( pz/ ) eq! a! ( A / ) ( B/ ) i i! eq p eq p Some books still refer to the pressure equilibrium constant y ( py) ( p ) eq Z ( p ) ( p ) eq p = a b A eq B eq! y! ( χc tot / ) ( χd tot / )! a! ( χatot / ) ( χbtot / ) eq z! eq eq tot K = = K b x! z eq z eq b eq which in general has pressure units he equilibrium constant can also be expressed in mole fractions: and in concentrations: K! i ν p = n R / = c R i i i ν ν ν p cr cr = = = i i! i i!! eq i! eq K c where K! y! ([ C]/ c ) ([ D]/ c ) eq!! ([ A]/ c ) ([ B]/ c ) eq c = a b eq z eq for this example aul ercival 3/6/005

71 Equilibrium Calculations stoichiometry initial moles during reaction mole fraction partial pressure NO + Cl NOCl ( ) 1 a 1 a a ( ) 1 a 1 a a 3 a 3 a 3 a ( ) 1 a 1 a a 3 a 3 a 3 a!!! K x a ( 3 a) a 3 a = = 3 1 ( a) ( 1 a) ( 3 a) 3 a =!! K Kx ( ) ( 1 a) he value of a at equilibrium (and thus the equilibrium composition of the reaction mixture) depends on pressure. aul ercival 3/6/005

72 emperature Dependence of K Gibbs-Helmholtz ( G/ ) H = substitute G = Rln K! rxn van t Hoff Equation (ln K) = rh R! or (ln K) rh (1/ ) = R! ln K slope H = R! 1/ Exothermic reactions: H < 0 so K falls with increasing. Endothermic reactions: H > 0 so K rises with increasing. Integrating,! or from rxn = ln ( ) 1 1 ln K! H = K ( ) R G R K 1 1 ln K ( ) H = + R S R!! aul ercival 3/6/005

73 ressure Dependence of Equilibrium But, x ( / ) K = K ( ) ln K = ln K νln / x ( ) Since K! = exp G! / R, ln K ln!! ν!! = ln K ν ln + νln x!! = ν ln K x ν = = R K! = 0 ideal gas Although the thermodynamic equilibrium constant does not depend on pressure, the K for mole fraction does if ν 0 he equilibrium composition depends on pressure if ν 0 Le Chatelier s rinciple A system at equilibrium, when subjected to a perturbation, responds in a way that tends to minimize the effect. aul ercival 3/6/005

74 Equilibria Involving Condensed Matter e.g. CaCO (s) " CaO(s) + CO (g) 3 where but G ξ, At equilibrium, ( CaCO ) ( CaO) ( CO ) = µ +µ +µ 3!! ( CO ) ( CO ) ln ( / ) µ =µ +R p CO! ( CaCO3) ( CaCO3)! µ ( CaO) µ ( CaO) µ µ G ξ, = 0 ( eq ) ( eq ) G! + Rln p /! = 0 CO CO K = p!! K depends only on the partial pressures of the gaseous reaction components. A special case is the evaporation of a liquid: L(l) " G(g) eq ( G ) K = p!!! ln K ln H = = R! vap aul ercival 3/6/005

75 hase Equilibria Consider a closed system of a single component. he chemical potential determines which phase is stable at a particular and. µ tends to a minimum. At the melting point m, µ(s) = µ(l) At the boiling point b. µ(l) = µ(g) hese points depend on temperature and pressure. dg = d Sd µ = S µ = µ solid liquid gas µ higher pressure m b phase 1 phase x x ' phase transition at higher aul ercival 3/6/005

76 he Clapeyron Equation Consider two phases α and β in equilibrium: (,, ) (,, ) µα =µβ If small changes in and are made such that α and β are still in equilibrium: dµα,, = dµβ,, ( ) ( ) S( α ) d + ( α ) d = S( β ) d + ( β) d [ ( α) ( β )] = [ ( α) ( β) ] d S S d d S H = = d Melting Integrating, d H m = d m H ( ) m m 1 = ln m m( 1) H m m m m H m > 0 and usually m > 0 m increases with pressure not for water! aul ercival 3/6/005

77 he Clausius-Clapeyron Equation d Hvap Hvap aporization = d vap ( g) Assuming the vapour is an ideal gas, Integrating, dln = d H R vap H vap 1 1 ln = R 1 1 g ( ) = R/ he normal boiling point is the temperature at which the vapour pressure becomes standard, i.e. 1 bar. Sublimation solid gas he liquid is not stable at any temperature. riple oint: solid, liquid and gas are all in equilibrium his happens at the pressure where the sublimation temperature and the boiling temperature coincide. At the triple point, vapour pressure of liquid = vapour pressure of solid triple and triple are fixed. aul ercival 3/6/005

78 he hase Rule How many intensive variables are needed to describe fully a system of C components and phases?! wo for temperature and pressure.! How many for the composition of each phase? ake mole fractions of each component in each phase ( C 1) C-1 because for each phase i 1 but since the phases are in equilibrium, µ ( phase 1) =µ ( phase ) = ( 1)C variables are redundant χ = Number of independent concentration variables otal number of variables (degrees of freedom) ( 1) ( 1) = C C= C F = C + hase: A state of matter that is uniform throughout, in both chemical composition and physical state. Component: he number of components is the minimum number of independent species necessary to define the composition of all phases in the system. Reactions and phase equilibria must be taken into account. aul ercival 3/6/005

79 hase Diagrams of ure Materials F = C + with C = 1 F = 3 For single phase regions there are degrees of freedom. For phase boundaries there is 1 degree of freedom. At the triple point there is no freedom. ressure e.g. CO c 3 solid liquid riple point gas supercritical fluid Critical point 3 emperature c aul ercival 3/6/005

80 he hase Diagram of Water ressure / atm FLUID ICE riple point AOUR LIQUID SEAM Critical point emperature / C here are other solid phases at much higher pressures. aul ercival 3/6/005