What is herm, Stat Mech, etc.? Rand m Kinetics & Mechanics Macrscpic herm Cher rent Statistical Mechanics ic Atm Classical Micrscpici Quantum Mlec cular hermdynamics is a funny subject. he first time yu g thrugh it, yu dn't understand it at all. he secnd time yu g thrugh it, yu think yu understand it, except fr ne r tw small pints. he third time yu g thrugh h it, yu knw yu dn't understand it, but by then yu are s used t it, it desn't bther yu any mre. -- Arnld Smmerfield Ludwig Bltzmann, wh spent much f his life studying statistical mechanics, died in 1906, by his wn hand. aul Ehrenfest, carrying n the wrk, died similarly in 1933. Nw it is ur turn t study statistical mechanics. erhaps it will be wise t apprach the subject cautiusly. -- David L. Gdstein Our apprach will be t fcus n the macrscpic, thermdynamic picture with ccasinal insight frm the micrscpic picture via statistical mechanics.
Energy, Wrk and Heat Energy is the capacity t d wrk. Its classificatin int: kinetic ptential (by mtin) (by psitin) e.g. thermal chemical, electrical Is purely arbitrary!??????? Heat and wrk are nt types f energy, but are prcesses invlving transfer f energy. hey appear and disappear at the system bundary. hey are path variables. Heat is the transfer f energy frm ne bdy t anther f lwer temperature. Cnventin: if heat flws int the system, q > 0. Wrk is the transfer f energy by sme mechanism ther than temperature difference. Cnventin: if wrk is dne n the system, w > 0. Heat stimulates randm mtin. Wrk stimulates rganized mtin. Wrk degrades int heat. qualitative bservatins by Cunt Rumfrd (Ben hmpsn) quantitative measurements by James Jule
erminlgy 1 A system is a particular sample f matter r regin f space. An islated system des nt interact with its surrundings. system + surrundings = universe A clsed system des nt allw passage f mass ver its bundaries, in cntrast t An pen system. An adiabatic system has bundaries which permit n flw f heat. It is insulated. A system is in a definite state when all its prperties p have definite values. A system at equilibrium is time independent; it is nt affected by the histry f the system. Extensive prperties depend n the amunt f substance in the system, e.g. n,. tal rperty = prperty(part) Intensive prperties are independent f amunt, e.g.,. tal rperty = prperty f part
matter pen system heat surrundings matter clsed system heat surrundings matter islated system heat surrundings
erminlgy State variables (state functins) uniquely determine the state f a system at equilibrium. w samples f a substance with the same state variables are in the same state. he change in a state variable depends nly n the initial and final states, independent f path. ath functins depend n the prcess and therefre vary with path. A cyclic prcess is ne in which the initial and final states are the same, i.e. n change in the state variables. In cntrast, path functins generally have nn-zer values fr cyclic prcesses, dependent n the path. A reversible prcess is ne that can be reversed by an infinitesimal mdificatin f a variable. he system is in equilibrium with the surrundings at all times. his is an idealized situatin, useful as a theretical limit, but All real prcesses are irreversible. It is pssible t restre the system r the surrundings t their riginal states but nt bth. An equatin f state is the functinal relatinship between the prperties f a system, e.g, the ideal gas law.
Ideal Gases -- Review Ideal gases bey the ideal gas law: = nr ressure a 1 N m - 1 J m -3 1 kg m -1 s - lume m 3 emperature K number f mles gas cnstant 10 3 dm 3 10 3 litres n = w M N = L = 6.0 10 L -1-1 R = 8.3145 J K ml 3 n. f atms in 1 g 1 C but if is in atm ( 1.0135 10 5 a) and in litres, 3-1 -1 R = 0.0806 dm atm K ml 1 1 3 3 3 1 istherms isbars ischrs
Mixtures f Ideal Gases If the ideal gas law applies t each cmpnent, i i = n R i partial pressure i = nr i n i tt i = n =χ tt tt Daltn s Law f artial ressures i tt i = ni = n tt i tt e.g. fr tw cmpnents: pressure = + A B A B =χ =χ A B χ A 0 1 1 χ B 0 cmpsitin Real gases are ideal nly at the lw density limit. Why?
Real Gases have nn-zer vlume at lw and high have repulsive and attractive frces between mlecules shrt range, imprtant at high lnger range, imprtant at mderate At lw pressure, mlecular vlume and intermlecular frces can ften be neglected, i.e. prperties ideal. irial Equatins R 1 B C = + + + K R B C = 1+ + + B is the secnd virial cefficient. C is the third virial cefficients. hey are temperature dependent. K = m = n an der Waals Equatin a + ( b) = R
Cmpressibility Factr als knwn as cmpressin factr Z = = R ideal.0 C H 4 CH 4 Z 1.0 H NH 3 0 he curve fr each gas becmes mre ideal as Z 1.0 1 3
he van der Waals Equatin 1 a + ( b) = R Intermlecular attractin = internal pressure R a = b a Z = = R b R mlecular vlume excluded vlume ( ) 3 3 r / π = πσ 4 3 3 1 a a a R R R = 1+ b + b + 3 ( R ) K (bring algebra) Z Z 1 a = b +... R R he initial slpe depends n a, b and : psitive fr b > a / R mlecular l size dminant negative fr zer at b< a/ R frces dminant = a/ Rb Byle emperature ~ ideal behaviur ver wide range f
Cndensatin f Gases Real gases cndense dn t they? supercritical fluid c 1 liquid c 1 c gas c, c and c are the critical cnstants f the gas. Abve the critical temperature the gas and liquid phases are cntinuus, i.e. there is n interface.
he van der Waals Equatin he van der Waals Equatin is nt exact, nly a mdel. a and b are empirical cnstant. R a ab + + = 0 3 b he cubic frm f the equatin predicts 3 slutins??????? Hw t slve? 0 b here is a pint f inflectin at the critical pint, s slpe: curvature: R a = + = 0 ( b ) 3 R 6a = = 0 ( b) 3 4 a c = c = 3b c = 7b 8a 7Rb 3 a 7 Rc 8 Rb 8 c c Z c = = B = = c
he rinciple f Crrespnding States Reduced variables are dimensinless variables expressed as fractins f the critical cnstants: r = r = r = c c c Real gases in the same state f reduced vlume and reduced temperature exert apprximately the same reduced pressure. hey are in crrespnding states. If the van der Waals Equatin is written in reduced variables, 3 + ( 3 1 ) = 8 r r r r Since this is independent f a and b, all gases fllw the same curve (apprximately). Z 1.0 r = 1.5 r = 1 1. r = 1.0 r
rperties f Real Gases as 0 Real gases have interactins between mlecules. hese change when the gas is cmpressed, but they need nt g t zer as 0. e.g. cnsider Z Fr an ideal gas: Z = = () = R 1 0 Fr a real gas: Z = = 1+ B + C + K R Z = B + C + K and in the limit: Z lim = B 0 0 Nt all prperties f real gases tend t ideal values as 0.
van der Waals Jhannes Diderik van der Waals, 1837-193 Nbel rize in hysics 1910 http://www.s-he.cm/stamp.html h / t
Carbn Dixide Image cpied frm: http:// science.nasa.gv/headlines/y003/0aug_supercriticalc.htm Original surce: http:// www.chem.leeds.ac.uk/eple/cmr/criticalpics.html
artial Differentiatin fr functins f mre than ne variable: f=f(x, y, ) ake area as an example A = xy Fr an increase Fr an increase Δx in x, Δ A = yδx Δy in y, Δ A = xδy 1 y cnstant x cnstant Fr a simultaneus increase ( )( ) Δ A = x+δ x y+δ y xy = yδ x+ xδ y+δxδy x ΔA Δx Δy ΔA Δx ΔA x y x y Δ y 1 = Δ + Δ +Δ Δ A = xy ΔA 1 y In the limits Δx 0, Δy 0 A A ΔA da = dx + dy x y y x ttal differential partial differential fr a real single-value functin f f tw independent variables, ( + δ, ) (, ) f lim f x x y f x y = x δ x δ x 0 y
Cnsider artial Derivative Relatins f ( xyz,, ) = 0, s z= zxy (, ) z z dz = dx + dy x y y x artial derivatives can be taken in any rder. z z z z = x = xy yx y y x y aking the inverse: 1 z x = xy z find the third partial derivative: z y x y x z z dz = 0 dy = dx y xy ( z/ y) ( / ) x z x y z x y z x = = y x y x Chain Rule: x y z 1 y = zx xy z and y x z x z y z y x = 1
artial Derivatives in hermdynamics Frm the generalized equatin f state fr a clsed system, f (,, ) = 0 six partial derivatives can be written: but given the three inverses, e.g and the chain rule 1 = = 1 there are nly tw independent basic prperties f matter. By cnventin these are chsen t be: the cefficient f thermal expansin (isbaric), and the cefficient f isthermal cmpressibility. he third derivative is simply 1 α= 1 κ = ( ) ( / ) / α = = κ
he Euler Relatin Suppse δ = (, ) + (, ) z A x y dx B x y dy Is δz an exact differential, i.e. dz? dz is exact prvided A B = y x x y crssdifferentiatin because then A B z A z = x = y y y x z B z = = y x y x y x x he crllary als hlds. State functins have exact differentials. ath functins d nt. New thermdynamic relatins may be derived frm the Euler relatin. e.g. given that it fllws that du = ds d = S S
Energy, Wrk and Heat Energy is the capacity t d wrk. Its classificatin int: kinetic ptential (by mtin) (by psitin) e.g. thermal chemical, electrical Is purely arbitrary! Heat and wrk are nt types f energy, but are prcesses invlving transfer f energy. hey appear and disappear at the system bundary. hey are path variables. Heat is the transfer f energy frm ne bdy t anther f lwer temperature. Cnventin: if heat flws int the system, q > 0. Wrk is the transfer f energy by sme mechanism ther than temperature difference. Cnventin: if wrk is dne n the system, w > 0. Heat stimulates randm mtin. Wrk stimulates rganized mtin. Wrk degrades int heat. qualitative bservatins by Cunt Rumfrd (Ben hmpsn) quantitative measurements by James Jule
he First Law f hermdynamics q > 0 fr heat flw int the system w > 0 fr wrk dne n the system Fr finite changes f state: Fr infinitesimal it i changes: Δ U = q+ w U is the internal energy f the system. du =δ q+ δw When a system changes frm ne state t anther alng an adiabatic path, the amunt f wrk dne is the same, whatever the means emplyed. Fr Fr q= 0, w = U U =ΔU ad final initial q 0, q =Δ U w = w w he energy f an islated system is cnstant. Fr q= 0, w= 0 Δ U = 0 N perpetual mtin machines! In any cyclic transfrmatin the wrk dne by a system n its surrundings is equal t the heat withdrawn frm the surrundings. δ w= δq du = he energy f the universe is cnstant. Δ U = Δ U system ad surrundings 0
emperature w 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 f the zerth law f thermdynamics. Ideal gas temperature: Unit f temperature: 1 Kelvin = lim 0 nr = ( triple pint f water) 73.16 he freezing pint f water at 1 atm is 73.15 K. he biling gpint f water at 1 atm is 373.1 K. he Celsius scale is defined as t/ C = / K 73.15 It is pssible t define an abslute temperature scale (Kelvin scale) by cnsidering the wrk dne in an isthermal reversible expansin/cmpressin. w= nrln 1
ressure lume Wrk 1 Expansin 1 > / mg A = ex = mg/ A m m h,, 1, 1, 1 A = area f pistn w = = ( mg) h ( ex A) h ( ) = ex 1 1 equatin f state 1 (need nt be istherm) 1 Expansin int a vacuum ex = 0 w = 0 Cmpressin < 1 ex = ex ( ) w= ex 1 1 1 1
ressure lume Wrk Multi-stage Expansin 1 3 w = 4 5 ( ) + 3( 3 ) + 4( 4 3) + ( ) 1 5 5 4 Reversible Expansin Make steps s small that d 0, d 0 hen δ w = ex d = + d d ( ) int d int path 1 int w= dw= d Fr ideal gases int = nr / and at fixed temperature wrev = nr ln nr ln = 1 1 1
ressure lume Wrk 3 1 4 cmpressin 3 expansin Cnsider the cyclic path 1 3 4 1 0+ 3 ( 3 1 ) + 0+ 1 ( 1 3 ) ( )( ) w= + + + = 3 1 3 1 Cnsider the cyclic reversible path 1 3 1 = + 3 1 w d d 1 3 3 3 = d d = 1 1 0 Even fr a cyclic prcess w depends n path du = 0 δ w = δ q
Energy vs. Enthalpy Fr a change in state at cnstant vlume, n expansin wrk is dne, s Δ U = q, du =δq Hwever, fr a change in state at cnstant pressure, Δ U = q + w, du =δq d 1 1 Δ = = δ U du q d 1 ( ) U U = q 1 1 ( ) ( 1 1) ( ) ( ) U + U + = q U + U + = q = = 1 1 1 1 cnstant Enthalpy H = U + Δ H = q, dh =δq H, being a functin f state variables nly, is als a state variable. Fr a general change f state ( and may bth change), ( ) Δ H =Δ U +Δ =Δ U + Δ + Δ +ΔΔ dh = du + d + d
Define: δ q = Cd δ q = C d Heat Capacity ransfer f heat t a system may result in a rise in. δ q = Cd path functin, s C depends n cnditins. Frm 1 st Law, at cnstant vlume, n wrk at cnstant pressure, nly wrk du =δ q +δ w =δq exd assume n du =δ q f r d = 0 ther wrk C U = Similarly dh = du + d d = ( δq d) + d d = δqq fr d= 0 C H = dh = du + d = du + nrd Fr ideal gases ( ) C d = C d + nrd C = C + R
he Relatin Between C and C C But since H U C = U U = + U U du = d + d U U U = + H = U + U C C = + wrk needed t vercme U = intermlecular frces internal pressure expansin per degree Fr ideal gases U nr = 0, = C C = nr Fr liquids and slids is s small that C C
adiabatic = insulated: Adiabatic Expansin 1 q= 0, δ q= 0 Δ U = w du =δw U U du = d + d 1 K w du C d = = + 1 (0 fr ideal gases) = C Δ if C is independent f Fr adiabatic expansin w 0, ΔU 0 1 Free expansin: ex = 0 w = = 0, Δ = 0 Fixed pressure: Reversible expansin: w= Δ ex w= Δ U = C Δ exδ Δ = C = ex w= 1 (, ) d Substitute apprpriate equatin f state. Nt useful if changes.
Adiabatic Expansin Reversible adiabatic expansin f ideal gases: du =δw δ q = 0 nr C d = d = d ideal gases nly 1 1 C d = R d 1 1 C ln ( / 1 ) = R ln ( / 1 ) = ( C C ) ln ( / 1) C ln ( / 1) = ( γ 1) ln ( / 1) γ= C γ 1 1 = 1 Als, since 1 r γ = 1 1 1 γ = r = 1 γ 1 1 = 1 1 ( γ 1)/ γ 1 1 γ 1
ypes f Expansin Wrk Summary Fr all types f expansin δ w= d ex Fr reversible changes In an irreversible expansin ( ) ex = =, ex in general <, ( w) < ( w ) Fr isthermal expansins is held cnstant by leakage f heat int the system frm the surrundings. U Fr an ideal gas = 0, Δ U = 0, q = w In an adiabatic expansin q = 0 s the internal energy must prvide fr the wrk: rev w =ΔU Fr reversible expansin f an ideal gas thrugh a given vlume change frm the same initial state, ( w ) > ( w ) isthermal adiabatic since U is cntinuusly replenished by heat intake t keep the temperature t cnstant. t
he Jule Expansin Experiment Fr a clsed system, the state functin U is determined by and alne: U U du = d + d U = Cd + d Jule tried t measure this partial derivative. 1. Gas in A, vacuum in B. A B. Open valve. 3. Any change in? He fund n change in temperature when the gas expanded frm A t A + B, i.e. q = 0. Als, n wrk was dne (free expansin), s w = 0. Cnclusin: ΔU = 0 and hence U = 0 r U = U ( ) nly Strictly nly true fr ideal gases. ( ) Nt true fr liquids and slids, but since ΔU U / Δ and Δ is very small, the effect f Δ n U is usually ignred.
ressure Dependence f Enthalpy Fr a clsed system, the state functin H is determined by and alne: H H dh = d + d H = Cd + d Fr ideal gases H = U + = U + nr H U = = 0 Nt true fr real gases, liquids and slids! dh = du + d + d H U C d + d = C d + d + d + d At fixed temperature, d = 0 H U = + + small fr liquids and slids Investigate real gases at cnstant enthalpy, i.e. dh = 0 H = C H
he Jule-hmpsn Effect (nt in text) Jule-hmpsn Cefficient μ J = H ump gas thrugh thrttle (hle r prus plug) 1 > Keep pressures cnstant by mving pistns. Wrk dne by system w= 1 1 Since q = 0 Δ U = U U1 = w= 1 1 U + = U + 1 1 1 H = H cnstant t enthalpy 1 Measure change in f gas as it mves frm side 1 t side. Δ μ J = Δ Δ 0 A mdern, mre direct experiment uses similar apparatus but with a H q = lim heater t ffset the temperature drp. Δ 0 Δ 1
he Linde Refrigeratr Fr mst gases at rm temperature μ J > 0 s sudden (adiabatic) expansin results in a drp in. his is the basis f peratin f the Linde refrigeratr and gas liquefactin. cling fins cld gas cmpressr expansin nzzle In general, µ J depends n and can even change sign. inversin i temperaturet H H > < 0 0 lines f cnstant enthalpy
he Mlecular Interpretatin f U What is U? ΔU can be related t thermchemical bservables: U C =, Δ U = C d fr ideal gases 1 ( ) ( ) 0 = 0 U U C d Can U be calculated frm mlecular prperties? U = u = N < u > mlecules l ignring intermlecular frces Classically, <u> is given by the Equipartitin Law: he average energy f each different mde f mtin f a mlecule is ½k. k = Bltzmann cnstant = R/L =1.381 10-3 J K -1 Implicit in this law is the cncept that there is n cupling between mlecular mdes f mtin. tal: ranslatin: Rtatin: ibratin: n. f vib. mdes u =ε tr +ε rt + εv ib( +εe l) ε = mv = mx& + my& + mz& 1 1 1 1 tr 1 1 1 rt Ix x I y y Iz z 1 1 ε = ω + ω + ω ε vib = + = mx& + kx per mde 3n 6 fr nn-linear mlecules = 3 n 5 fr linear mlecules les
he Mlecular Interpretatin f U (cnt.) When quantum effects can be ignred, the average energy f every quadratic term in the energy expressin has the same value, ½k. 3 U Mnatmic gases < u > = k, C 15 1.5 = = R 3 Diatmic gases < u >= ( + 1+ 1 ) k, C = 3.5R Linear plyatmics Nn-linear mlecules 3 ( n ) < u >= + 1+ 3 5 k e.g. fr n= 3, C = 6.5R 3 3 ( 3n 6) < u >= + + k e.g. fr n = 3, C = 6.0 R Experimental bservatins d nt agree with these predictins, except fr mnatmic gases. In particular: he predicted values are t high. Experimental values are temperature-dependent. Accurate calculatins are prvided by statistical mechanics, where it can be shwn that the equipartitin principle nly hlds in the limit f high temperature, specifically fr the cnditin k >> ΔE, where ΔE is the apprpriate spacing f energy levels. Equipartitin free transfer f energy between mdes
emperature Dependence f C U C =, where U U0 = N <ε tr >+<ε rt >+ = C + C + C + C C C tr rt tr rt vib el = fr all 1.5R = { R linear f r k 1. 5R nn-linear Δε ( K) rt C = 0 at very lw, where < ε rt > 0 C vib = R fr each vibratinal mde, at high el C = 0 almst always, since electrnic excitatin ti takes great energy rt e.g. fr a diatmic mlecule: C R 7 t + r + v 5 t+r 5 tw atms t 3 t 0 0
emperature Dependence f H Interactins between mlecules als cntribute t the heat capacity f real systems. In particular, first-rder phase changes ften invlve large energy changes. hese are usually measured at cnstant pressure and expressed as enthalpies. gas H slid slid 1 liquid ΔH s-s ΔH f ΔH v 0 0 In general, H a b c = + + + K H C = = b+ c + K fr each phase
hermchemistry he study f energy changes that ccur during chemical reactins: at cnstant vlume ΔU = q n wrk at cnstant pressure ΔH = q nly wrk Fr practical reasns mst measurements are made at cnstant, s thermchemistry mstly deals with ΔH. Δ H = H H reactin prducts reactants If ΔH > 0 the reactin is endthermic. If ΔH < 0 the reactin is exthermic. Fr cmparisn purpses we need t refer ΔH t the same and. define a standard reactin enthalpy each cmpnent f the reactin must be in its standard state the mst stable frm at 1 bar pressure and (usually) 5 C. 1 bar = 10 5 a 1 atm = 1.0135 bar
Reactin Enthalpy 1 Hess s Law he standard enthalpy change in any reactin can be expressed as the sum f the standard enthalpy changes, at the same temperature, f a series f reactins int which the verall reactin can be frmally divided. Cmbine chemical equatins as if mathematical equatins, e.g. A + B C ΔH1 C + D E + F ΔH F B + G Δ H 3 A + D E + G ΔH Δ H =Δ H1+Δ H +ΔH3 Standard Reactin Enthalpy ΔH ΔH ΔH 98 500 reactin enthalpy at 1 bar and at standard r sme ther
Reactin Enthalpy Standard (mlar) enthalpy f frmatin ΔHf ΔfH Heat f frmatin f a substance frm its elements, all substances being in their standard state. By definitin, iti fr all elements Δ = H f 0 Enthalpy f cmbustin ΔH fr ttal xidatin f a substance ΔH c Δ c H e.g. C 6 H 1 O 6 + 6O 6CO + 6 H O Δ c H = -808 kj ml -1 Enthalpy f hydrgenatin ΔH when an unsaturated rganic cmpund becmes fully saturated e.g. C 6 H 6 + 3H C 6 H 1 ΔH = -46 kj ml -1 Enthalpy f atmizatin Bnd dissciatin enthalpy ΔH fr the dissciatin f a mlecule int its cnstituent gaseus atms e.g. C H 6 (g) C(g) + 6H(g) ΔH = 883 kj ml -1 6 Bnd strength single bnd enthalpy An average value taken frm a series f cmpunds and ften cmbined fr a rugh estimate e.g. ΔH (C H 6 ) = ΔH (C-C) + 6 ΔH (C-H)
emperature Dependence f ΔH he temperature dependence f reactin enthalpies can be expressed in terms f the dependence f the enthalpies f the reactin cmpnents: H ( ) = H ( 1 ) + Cd p 1 H ( ) H ( ) Cd Δ =Δ + Δ 1 wher e Δ Cp = Cp Cp prducts 1 reactants p his general phenmenn is knwn as Kirchff s Law. e.g. A + B C + D A + B C + D Δ H ΔH ( p p )( ) Δ H ( ) = C (A) + C (B) 1 +Δ H ( ) 1 ( ) 1 ( ) ( Cp (C) Cp (D) )( 1 ) + + 1 ( ) p p =Δ H ( ) + C C Δ prducts reactants assuming that the C p values are independent.
Reactins at Cnstant lume ( ) ( ) Δ H =Δ U + r r prducts reactants Fr slids and liquids ( ) 0, s Δ ΔH ΔU Δ =Δn R, Fr ideal gases ( ) gas s ΔH Δ U +Δn R e.g. CH(g) + O (g) 3CO (g) + 3H O(l) 9 3 6 5 (- ) Δ H =Δ U + R r r he relatinship between ΔH and ΔU is particularly imprtant when relating thermchemical enthalpies t mlecular prperties, e.g. fr a single bnd energy ΔU = ΔH R as seen in the case f O (g) O(g). In practice, R is usually s much smaller than ΔH that it is ften ignred. gas
Enthalpies f Ins in Slutin Enthalpy f slutin ΔH fr slutin f a substance in a stated amunt f slvent Enthalpy f dilutin ΔH fr dilutin f a slutin t a lwer cncentratin Enthalpy f slutin t infinite dilutin infinite amunt f slvent ΔH sln fr an he enthalpy f frmatin fr a species in slutin can be fund by cmbining ΔH sln with the ΔH f f the gaseus species: 1 1-1 H (g) + Cl (g) HCl(g) Δ H = 9.31 kj ml f HCl(g) H Cl(aq) ΔHH = 75. 14 kj ml -1 slnl H (g) + Cl (g) HCl(aq) 1 1 Δ H (in) =Δ H +ΔH f f sln = 167.45 kj ml -1 ΔH f fr individual ins in slutin can nly be fund if ne is arbitrarily fixed. By cnventin this is H + (aq). + + H(g) H(aq) + e Δ Hf ( Ha ) = 0 + ( Cl ) ( HCl ) ( H ) ( HCl ) 1 q Δ H =ΔH Δ H =ΔH f aq f aq f aq f aq he standard state fr a substance in slutin (nt just ins) is a cncentratin f 1 mle slute in 1 kg slutin (1 mlal).
Enthalpy f Frmatin f an Inic Slid Cnsider individual steps in the frmatin f NaCl. 1. Na(s) Na(g) ΔH subl ( Na). Na(g) Na + (g) + e ( Na ) Δ H = I + R 1 3. ½Cl (g) Cl(g) ΔH ( ) Cl-Cl 4. Cl(g) + e Cl (g) ( Cl ) Δ H = E R 5. Na + (g) + Cl (g) Na + (aq) + Cl (aq) Δ H Na +ΔH NaCl(aq) ( + sl ) sl ( Cl ) Na(s) + ½Cl (g) NaCl(aq) ΔHf ( NaClaq) 1 Δ Hf ( NaClaq ) = Δ Hsubl ( Na ) + I(Na) + ΔH ( Cl-Cl) EA(Cl) +Δ Hsl( Na + ) +ΔHsl( Cl ) Step 5 culd be creatin f slid NaCl instead f slutin 5'. Na + (g) + Cl (g) NaCl (s) ΔH lattice ( NaCl) leading us t the enthalpy f frmatin f slid NaCl: Na(s) + ½Cl (g) NaCl(s) 1 ( NaCl ) ( Na ) (Na) ( Cl-Cl) E (Cl) +Δ H ( NaCl ) Δ H = Δ H + I + ΔH f s subl A lattice A
A Brn-Haber Cycle fr NaCl Enthalpy changes can als be expressed in a diagram, e.g. I ( Na) + R Na + (g) + e + Cl(g) Na+(g) + Cl (g) E A ( Cl ) + R ( Na ) ΔH subl Na ( ) 1 ΔH Cl-Cl ΔH ΔH f f Na(g) + Cl(g) Na(s) + Cl(g) Na(s) + ½Cl (g) ( NaClaq) ( NaCl ) s NaCl(aq) NaCl(s) ΔH sl ΔH lattice Since H is a state variable, the sum f enthalpy changes arund the cycle must be zer. Cnsequently, if all but ne f the enthalpy changes is knwn, it can be readily calculated. his is equivalent t using Hess s Law t sum reactin steps.
Spntaneus Change (S, why d we need entrpy, anyway?) directin f spntaneus change he directin f spntaneus change is that which leads t chatic dispersal f the ttal energy mves frm a state f lw intrinsic prbability twards ne f greater prbability. Wrk is needed t reverse a spntaneus prcess. We need a quantity entrpy t describe energy dispersal, i.e. the prbability f a state. Spntaneus prcesses are irreversible. hey generate entrpy Reversible prcesses d nt generate entrpy but they may transfer it frm ne part f the universe t anther.
Entrpy 1 Entrpy is a state variable (prperty) which determines if a state is accessible frm anther by a spntaneus change. Entrpy is a measure f chatic dispersal f energy. he natural tendency f spntaneus change is twards states f higher entrpy. here are bth thermdynamic (hw much heat is prduced?) and statistical definitins (hw prbable is a state?). hey bth becme equivalent when statistics is applied t a large number f mlecules. Cnsider a falling weight which drives a generatr and thus results in heat q being added t the reservir (the surrundings). reservir Define a system variable S ds(surr) = δq/ generatr Use stred energy t restre the weight t its riginal height. he reservir gives up δq rev t the system, and there is n verall change in the universe. ds(sys) (y) = ds (surr) = δq rev
In general, Entrpy ds(sys) + ds(surr) 0 ds(sys) ds(surr) Equality fr reversible prcesses nly r, fr the system, δq ds Clausius inequality Fr an islated system, q = 0 hence Δ S 0 Isthermal rcesses Δ = / e.g. S q / rev ΔHfus Δ S( fusin) = Δ S( vap) = rutn s Rule: m ( vap ) ΔSS vap = 85 J K ml -1-1 ΔH Can be used t estimate ΔH vap if b is knwn. Nt gd fr assciated liquids. emperature ariatin Δ = ( S) and 1 ( S) C d Δ = C 1 Abslute Entrpy d δ = Cd q rev S b vap liquid gas = + C 0 0 0 S ( ) S(0) d slid Δ H v b
Entrpy 3 Entrpy depends n rbability. Cnsider the number f ways Ω f arranging n mlecules between tw sides (A and B) f a cntainer. he prbability A that all mlecules are n side A depends n the rati f Ω Α t the ttal number f arrangements. A B Ω = 1 Ω = = 1 A tt A Ω = 1 Ω = 4 = 1 A tt A 4 Ω = 1 Ω = 16 = 1 A tt A 16 Ω = 1 Ω = = n A tt A State A becmes less and less prbable as n increases. Cnversely, the prbability f the less rdered, rughly evenly distributed states, increases. Since entrpy is a measure f disrder, it fllws that S depends n Ω. Bltzmann equatin S = kln Ω n ( AND ) = x y x y = x + y Since x y, ln ln ln +
he Secnd Law f hermdynamics An isthermal cyclic prcess in which there is a net cnversin f heat int wrk is impssible. N prcess is pssible in which the sle result is the absrptin f heat frm a reservir and its cnversin int wrk. It is pssible t cnvert all wrk int heat! It is impssible fr heat t be transfrmed frm a bdy at a lwer temperature t ne at a higher temperature unless wrk is dne. he entrpy f an islated system increases during any natural prcess. he universe is an islated system. ΔS(sys) ) < 0 is allwed prvided d ΔS(sys) ) + ΔS(surr) ) > 0 All reversible Carnt cycles perating between the same tw temperatures have the same thermdynamic efficiency. here is a state functin called entrpy S that can be calculated frm ds = δq rev /. he change in entrpy in any prcess is given by ds δq/, where the inequality refers t a spntaneus (irreversible) prcess. he 1 st Law uses U t identify permissible changes f state. he nd Law uses S t identify natural changes amng the permissible nes.
he hird Law f hermdynamics If the entrpy f every element in its stable state at = 0 is taken as zer, every substance has a psitive entrpy which at = 0 may becme zer, and des becme zer fr all perfect crystalline substances, including cmpunds. Nernst Heat herem he entrpy change accmpanying transfrmatin between cndensed phases in equilibrium, including chemical reactins, appraches zer as 0. lim 0 Δ S = 0 ractical cnsequence: Set S(0) = 0 fr elements by cnventin. Apply Nernst t determine S(0) fr all else. It is impssible t reach abslute zer in a finite number f steps. he 1 st Law says U cannt be created r destryed. he nd Law says S cannt decrease. he 3 rd Law says zer Kelvin cannt be reached.
Given he Clausius Inequality ds = δ q rev Substitute int the 1 st Law: and du = ds = d All exact differentials, s path independent. But du =δ q+δ w =δq d ex ds= d =δq d δq ds = + ex ( ) ex ex ex d ( ) d 0 ex δ w = d If >, d > 0; if <, d < 0 Clausius Inequality Even mre generally, Cnditins fr: δq ds = + thermal equilibrium rev Fundamental Equatin f hermdynamics ( ) mechanical equilibrium In general, i.e. any path Equal fr reversible change surr surr dsuniv dssurr d
he Fundamental Equatin f hermdynamics Cmbine du =δq d with du = ds = d ds= 1 r ds = du + d δq q rev Reversible change but true fr all paths since d exact his fundamental equatin generates many mre relatinships. U U du = ds + d S S U U = and = S Example 1: Cmparisn with Example : Cnsider that du is exact and crss differentiate. = S S S a Maxwell relatin Example 3: S S = S cyclic rule S = and again! = S = = =α κ
Hw Entrpy Depends n and Cmpare 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 = + Δ S = C d 0 fr ideal gases 1 Δ S = d = nr d = nrln / ( ) 1 Fr any substance, Fr ideal gases, (eqs 3.7.4, 6.1.6) C α = + κ ds d d Δ S = C ln + nrln 1 1 assuming C is independent
Entrpy Depends n and (nt in text) 1 ds = du + d First prblem: replace du ; secnd prblem: replace d. du = dh d d Use and dbth are slved! But 1 ds = dh d H H H dh = d + d = C d + d Cmpare with C 1 H ds = d + d S S ds = d + d unnatural variables S = C C S 1 H = Δ S = d 1 Δ = = 0 fr ideal gases S d nr d ( ) ( ) ΔS S = nr ln / = nr ln / 1 1
Entrpy Changes in Spntaneus rcesses Entrpy is a state t functin, s ΔS(sys) = S S 1 independent f path his can be used t calculate ΔS fr an irreversible prcess. Cnsider isthermal expansin f a gas frm 1 t : Δ S(sys) = nrln / reversible and irreversible cases Fr the reversible case Fr the irreversible case ( ) 1 Δ S(surr) = ΔS(sys) Δ S (surr) > Δ S (sys) q w Δ S(surr) = = e.g. fr free expansin, w = 0 Δ S(surr) = 0, Δ S(univ) =Δ S(sys) > 0 Cnsider freezing f supercled water at < 73 K reversible water, 0 C ice, 0 C rev. water, irreversible rev. ice, 73 ( ΔHfus) Δ S(sys) = C(l) ln + + C(s) ln 73 73 ΔH 1 Δ fus( ) S (surr) = Hfus (73) [ C (l) C (s) ]( 73) = Δ + { }
Entrpy f Mixing Cnsider the mixing f tw ideal gases :,, 1 n 1,, n,, 1 + n 1 +n n Δ S = nrln = nrln = nrln χ 1 1 1 1 1 1 1 1+ n1+ n Δ S = nr ln = nr ln = nr ln χ 1+ n1+ n n ( ) ( ) Δ S = nrln χ n Rln χ = n + n R χ ln χ +χ ln χ mix 1 1 1 1 1 Δ S = n R χ ln χ In general mix tt ln his expressin applies t the arrangement f bjects (mlecules) just as well as fluids (gases and liquids). Fr example, arrange N identical atms in N sites in a crystal: Ω= N!/ N! = 1 S = kln Ω= 0 Cmpare with the arrangement f tw types f atm, A and B. Ω= N! S k ln N! ln NA! ln NB! N! N! Δ = A B i i i ( ) ln z! = zln z z Applicatin f Stirling s apprximatin ( ) leads t Δ S = kn ( χ ln χ +χ ln χ ) cnfig A A B B
Using Entrpy t Achieve Lw C Δ S = d Cln / ( ) 1 achieve lwer temperatures, S must be reduced. if C is cnstant Chse sme prperty X that varies with S, i.e. S = f(x,). his culd be the pressure f a gas r, fr example, the magnetic mment f a paramagnetic salt (whse energy varies with magnetic field). 1. Alter X isthermally. Entrpy changes.. Restre X by a reversible adiabatic prcess. 3. Repeat cycle. S S ΔS Δ X, q = ΔS X q= 0, Δ S = 0 X 1 curves cincide at 0 K, a cnsequence f the Nernst Heat herem X 0 0 1
Heat Engines A heat engine is a system capable f transfrming heat int wrk by sme cyclic prcess. he nd Law states that an isthermal cyclic prcess can nt prduce net wrk. he efficiency f a heat engine is defined as the rati f the wrk prduced t the heat input: w q q q ε= = = 1 q q q H L L H H H high temperature reservir @ H q H engine q L lw temperature reservir @ L w ut = q H - q L A heat pump is a heat engine in reverse. Wrk is needed t transfer heat frm a lwer t a higher temperature reservir. Carnt Cycle ideal gas all steps reversible istherm @ H adiabatic expansin adiabatic cmpressin istherm @ L
he Carnt Cycle A H B D L C step w q ΔU A B B C C D D A nr l( ln( / ) w 0 H B A AB ΔU 0 C ( ) BC H L nr ln( / ) w 0 L D C CD Δ U 0 C ( ) DA H L tal nr( ) ln( / ) w H L A B cyc 0 ( )ln( / ) ε= w = nr = 1 q nr ln( / ) ut H L A B L in H A B H ( H L) ε= H fr best efficiency, maximize H minimize i i L
Entrpy Changes in the Carnt Cycle 1 H w = cycle = d area f plt 4 L 3 = q cycle Isthermal expansin (1 ): q Δ U = 0 w = q > 0 Δ S = > 0 Isthermal cmpressin (4 3): Δ S < 0 Adiabatic steps ( 3 and 1 4): q = 0 Δ S = 0 S 3 4 L 1 H ( ) L( S4 S3) ( )( S S ) q = S S tt H 1 + = H L 1 = area f S plt = w cycle
Spntaneus Change Fr a system in thermal equilibrium with its surrundings, δq ds Clausius inequality At cnstant vlume: At cnstant pressure: dq = du ds du 00 d 0 0 ( ds ) ( du ) dq ds Fr cnvenience, define: U, S, = 0 dh 00 ( dh ), S dh A = U S da = du ds Sd G = H S dg = dh ds Sd 0 n wrk islated system wrk nly hen the cnditins fr spntaneus change becme: ( da ), 0 ( dg ), 0
Helmhltz Energy F (r A) Helmhltz lt energy; Helmhltz lt free energy; Helmhltz lt functin; Maximum wrk functin (F fr Functin r A fr Arbeit) Fr spntaneus change at cnstant and da = du ds 0 Nte that it is the ttal functin A that tends t a minimum; this is nt the same as minimizing U and maximizing S. Maximum Wrk Cmbine du =δ q +δw and ds δq Wrd dne by the system du ds + δw ( δw) ds du ( δ w) max = ds δq rev δ w rev = δw rev equality fr reversible change A system des maximum wrk when it is perating reversibly. But ( da) = du ds = du δ qrev =δw herefre, fr macrscpic changes wmax = Δ A= ΔS ΔU cnstant (-w) can be mre r less than ΔU accrding t the sign f ΔS. Fr ΔS S > 0, heat flws int the system t fuel the extra wrk. rev
Gibbs Energy G Gibbs free energy, Gibbs functin ery imprtant in chemistry since it tells whether a particular reactin can prceed at a given and. Fr spntaneus change, ΔG, ( ), 0 dg Δ G = Gprducts Greactants 0 fr reactins can be calculated frm tabulated data Δ G ( ) =ΔH ( ) ΔS ( ) If ΔH is and ΔS is then ΔG the reactin prceeds -ve +ve < 0 at all temperatures +ve -ve > 0 at n temperatures -ve -ve if < ΔH / ΔS +ve +e if > ΔH / ΔS Maximum Wrk: ( dg) = dh ds ( dg) = du + d ds, = δ q + δ w + d ds rev rev rev =δ w + d =δw (nn- ) max Δ w (nn- ) G max =Δ cnstant,
Basic hermdynamic Relatins 1 Laws du = δq d K (1) ds =δq K() rev / Definitins H = U + K(3) A = U S K(4) G = H S K(5) Fundamental Equatins du = ds d dh = ds + d da = Sd d K(6) K(7) K(8) dg = Sd + d K(9) artial Differentials U U = S S = H H = S S = A A = S = G G = S =
Basic hermdynamic Relatins Maxwell Relatins Frm (6) Frm (7) Frm (8) Frm (9) = S S = S S K(10) K(11) S = K (1) S = K(13) hermdynamic Equatin f State du = ds d K(6) U S = Substituting (1) U = U α = κ
Hw Free Energy Depends n G = H S Δ G =ΔH ΔS dg = d Sd K(1) K() K(3) definitin cnstant fundamental eqn. But G G H = S = G G H = ( G/ ) 1 G G = frm (1) K(4) K(5) Gibbs-Helmhltz Equatin ( G/ ) H = substitute (4) in (5) alternative frm: ( G/ ) (1/ ) = H By applying the Gibbs-Helmhltz equatin t bth reactants and prducts f a chemical reactin, Δ ( G/ ) ΔH =
Hw Free Energy Depends n G G = Δ = 1 = 1 G G G d Fr slids and liquids des nt change much with, s G ( ) G ( ) + ( ) 1 1 nr Fr a perfect gas = Δ G = nrln( / ) 1 G = G ( ) + nrln( / ) where G is the standard free energy defined at =1 bar he chemical ptential fr a pure substance is the mlar Gibbs energy: G μ= G = n Fr a perfect gas μ=μ +R ln( / ) like eq 5.3.5
he Chemical tential S far, ur thermdynamic relatins apply t clsed systems. Hwever, G, U, H, etc. are extensive prperties. artial mlar quantities are used t describe quantities which depend n cmpsitin. J Ji = n J = U, H, S, A, G, i n,, j i J = nj i i In general Chemical tential i G G dg = d + d n,, n... n,, n... 1 1 G G + dn + dn + n K 1 1 n n,,... n,,, n... G μ i = n i 1 3,, nj i du = ds d + μ dn Fundamental Equatins i i dh = ds + d + μ dn etc. U H A μ = i n = n = n i i i S, S,, i i
Free Energy f Mixing 1 Cnsider the mixing f tw ideal gases A and B at cnstant and. Befre: G = n μ (pure) + n μ 1 A A B B (pure) μ (pure) =μ +R ln( / ) where A A 0 After mixing: { ln( / )} { μ + ln( / ) + ln χ } G = n μ + R p A A A A 0 = n μ + R + R χ A A 0 A = naμ A(pure) + narln χa G = G + G < G negative A B 1 Change: Δ 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 ) Δ G = nr χ χ +χ χ mix A A B B Δ G mix < 0 Mixing is spntaneus. Δ S = ΔG / = nr χ lnχ Entrpy f Mixing ( ) mix mix n, Enthalpy f Mixing Δ Hmix =Δ Gmix + Δ Smix = 0 Similarly, il l Umix mix 0 i Δ = Δ = fr ideal gases and slutins i i
Free Energy f Mixing Fr a binary system, ΔG mix is at its minimum i and ΔS mix at its maximum when χ =χ = 0 A B 0.5 ΔS mix ΔG mix 0 0 0. 0.4 0.6 0.8 1 1 χ Α χ Β 0 Equilibrium and Mixing: 0 0. 0.4 0.6 χ Α 0.8 1 1 χ Β 0 A G = GA(pure) + GB(pure) +ΔGmix = n( χ μ +χ μ ) + nr( χ ln χ +χ ln χ ) Cnsider an equilibrium between tw ideal gases 0 A A B B A A B B B G G pure Gttal G mix he equilibrium cmpsitin is determined by the minimum in G ttal. A cmpsitin B
Cnsider A B Reactin Equilibrium 1 Suppse an amunt dx f A turns int B. dg =μ dn +μ dn = μ dx +μ dx, G G lik h t i = μb μa x hen ( ) A A B B A B, like what is dne in.5.10 µ A and µ B depend n cmpsitin, and therefre change during the reactin. If µ A >µ B the reactin prceeds frm A B. If µ A <µ B the reactin prceeds frm B A. At equilibrium μ =μ A B G G slpe = x 0 eq x 1 A B eq eq ( ) ( ) eq eq ( A B ) =μb μ A =Δ rxn eq B exp ( Grxn / R ) μ + Rln p / =μ + Rln p / A A B B R ln p / p G p B = exp Δ / see536 5.3.6 eq p A
Reactin Equilibrium A reactin such as A + 3B C + D can be written as 0 = A 3B +C +D r cmpletely generally as 0= ν R = 0 hen p y g y i i i G ξ, = νμ and at equilibrium, 0 i νiμ i = i i eq pi νμ i i +νi R ln = 0 i p i Δ Grxn + Rln = 0 i eq rxn i i ν Δ G = Rln K K ν i pi = i eq
Cnsider K Equilibrium Cnstants aa + bb yy+ zz Δν= y + z a b hermdynamic Equilibrium Cnstant ν ( Y / ) y p ( pz/ ) eq a ( A / ) ( B/ ) p p p i i = = i eq Sme bks still refer t the pressure equilibrium cnstant K y ( p Y ) ( p ) eq Z ( p ) ( p ) eq p = a b A eq B eq z eq z eq b eq which in general has pressure units he equilibrium cnstant can als be expressed in mle fractins: and in cncentratins: K y ( χc tt/ ) ( χd tt / ) a ( χatt / ) ( χbtt / ) eq eq tt K = = K b x eq z eq Δν p = n R / = c R i i i ν ν ν p cr cr i i i i = = = i eq i eq???? cnfusing K ntatin c where K y ([ C]/ c ) ([ D]/ c ) eq ( [ A ]/ c ) ( [ B ]/ c ) eq c = a b eq z eq fr this example
Equilibrium Calculatins stichimetry initial mles during reactin mle fractin partial pressure NO + Cl NOCl 1 + 1 0 ( ) 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 = = 1 1 K = Kx ( 3 a ) a ( 3 a ) 3 ( a) ( a) ( 1 a) ( 3 a) 3 a he value f a at equilibrium (and thus the equilibrium cmpsitin f the reactin mixture) depends n pressure.
emperature Dependence f K Gibbs-Helmhltz ( ΔG/ ) ΔH = substitute Δ G = Rln K rxn van t Hff Equatin (ln K) = ΔrH R r (ln K) ΔrH (1/ ) = R ln K slpe ΔH = R 1/ Exthermic reactins: ΔH < 0 s K falls with increasing. Endthermic reactins: ΔH > 0 s K rises with increasing. Integrating, r frm Δ Grxn = R ln K K ( ) ΔH 1 1 ln = K ( ) R 1 1 Δ H ln K ( ) = + R Δ S R
But, ressure Dependence f Equilibrium K Since K = exp ( ΔG / R), = 0??????? ( / ) K = K Δν x ( ) ln K = ln K Δνln / x ln K ln = ln K Δν ln +Δνln x = Δν??????? ln K x Δν Δ = = ideal gas R Althugh the thermdynamic equilibrium cnstant des nt depend n pressure, the K fr mle fractin des if Δν 0 he equilibrium cmpsitin depends n pressure if Δν 0 Le Chatelier s rinciple A system at equilibrium, when subjected t a perturbatin, respnds in a way that tends t minimize the effect.
Equilibria Invlving Cndensed Matter e.g. CaCO 3(s) CaO(s) + CO (g) where but G ξ, μ At equilibrium, ( CaCO ) ( CaO) ( CO ) = μ +μ +μ 3 ( CO ) ( CO ) ln ( / ) μ =μ +R p CO ( CaCO3) μ ( CaCO3) μ( CaO) μ ( CaO) G ξ, = 0 ( eq ) ( eq ) Δ G + Rln p / = 0 CO CO K = p K depends nly n the partial pressures f the gaseus reactin cmpnents. A special case is the evapratin f a liquid: L(l) G(g) eq ( G ) K = p ln K ln ΔH = = R vap
hase Equilibria Cnsider a clsed system f a single cmpnent. he chemical ptential determines which phase is stable at a particular and. µ tends t a minimum. At the melting pint m, µ(s) = µ(l) At the biling pint b. µ(l) = µ(g) hese pints depend n temperature and pressure. dg = d Sd μ = S μ = µ slid liquid gas µ higher pressure m b phase 1 phase x x ' phase transitin at higher
he Clapeyrn Equatin Cnsider tw phases α and β in equilibrium: (,, ) =μβ (,, ) μα If small changes in and are made such that α and β are still in equilibrium: dμα = dμβ Melting (,, ) (,, ) S( α ) d + ( α ) d= S( β ) d + ( β) d [ ( α ) ( β ) ] = [ ( α ) ( β ) ] d S S d d Δ S ΔH = = d Δ Δ d ΔH m = d Δ m Δ H ( ) ln = Δm m( 1) ΔH m Δ Δ Δ m m Integrating, 1 Δ H m > 0 and usually Δ m > 0 m m m m increases with pressure nt fr water!
aprizatin he Clausius-Clapeyrn Equatin d ΔHvap ΔHvap = d Δvap ( g) Assuming the vapur is an ideal gas, g ( ) = R/ Integrating, dln = d ΔH R vap ΔH vap 1 1 ln = 1 R 1 he nrmal biling pint is the temperature at which the vapur pressure becmes standard, i.e. 1 bar. Sublimatin slid gas he liquid is nt stable at any temperature. riple int: slid, liquid and gas are all in equilibrium his happens at the pressure where the sublimatin temperature t and the biling temperature t cincide. id At the triple pint, vapur pressure f liquid = vapur pressure f slid triple and triple are fixed.
he hase Rule Hw many intensive i variables are needed d t describe fully a system f C cmpnents and phases? w fr temperature and pressure. Hw many fr the cmpsitin f each phase? ake mle fractins f each cmpnent in each phase C 1 C-1 because fr each phase χ i = 1 ( ) but since the phases are in equilibrium, μ ( phase 1) =μ ( phase ) =K ( 1)C variables are redundant Number f independent cncentratin variables tal number f variables (degrees f freedm) ( ) ( ) = C 1 1 C= C F = C + hase: A state f matter that is unifrm thrughut, in bth chemical cmpsitin and physical state. Cmpnent: he number f cmpnents is the minimum number f independent species necessary t define the cmpsitin f all phases in the system. Reactins and phase equilibria must be taken int accunt.
hase Diagrams f ure Materials F = C + with C = 1 F = 3 Fr single phase regins there are degrees f freedm. Fr phase bundaries there is 1 degree f freedm. At the triple pint there is n freedm. ressure e.g. CO c 3 slid liquid riple pint gas supercritical fluid Critical pint 3 emperature c
he hase Diagram f Water ressure / atm FLUID 18 LIQUID ICE 10 1.0 riple pint SEAM AOUR Critical pint 0 100 374 emperature / C here are ther slid phases at much higher pressures.
Kinetics and Mechanism t determine / describe / predict the curse f reactins qualitatively: identify reactants, prducts, intermediates identify elementary reactin steps quantitatively: measure reactin rates rate cnstants explre effects f,, [H + ], relate rate cnstants t mlecular prperties develp theries make predictins Rate data (grwth f prducts, decay f reactants) are the basic experimental input t reactin kinetics. he shapes f kinetic data plts depend n rate cnstants and cncentratins. heir functinal frm reactin rder cnc. A + B C time
Empirical Chemical Kinetics time dependence f reactant and prduct cncentratins e. g. fr A + B 3Y + Z d[a] 1 d[b] 1 d[y] d[z] rat e = dt = dt = 3 dt = dt In general, a chemical equatin is written 0= ν X he extent t f a reactin (the advancement) ξ= i ( t ) n ( 0 ) n t Fr an infinitesimal advancement dξ the cncentratin f h t t/ d t h b d d f each reactant/prduct changes by [ ] By definitin, rate = dξ dt = ν ν [ X ] 1 i i d dt i i X i i i = ν ξ Reactin rates usually depend n reactant cncentratins, rate rate cnstant = k x [ A] [ B] y rder in B ttal t rder = x + y In elementary reactin steps the rders are always integral, but they may be fractinal in multi-step reactins. he mlecularity is the number f mlecules in a reactin step. i i
Measurement f Reactin Rates Sampling f reactin mixture, fllwed by chemical analysis, chrmatgraphy, spectrscpy Quenching f whle reactin r sample, fllwed by Matrix islatin is a particular quenching technique used t study therwise shrt-lived species. Flw A B he time between mixing and measurement is varied by changing the distance r the flw rate. mixing chamber detectr Real time analysis invlves measurement f a quantity that varies thrughut the reactin. his culd be: a general prperty f the system pressure, vlume, cnductance, ptical rtatin, a mlecular prperty: usually by spectrscpy Accurate initiatin is necessary fr real time experiments.
d[a] Zer rder: = k0 dt Simple Rate Laws [A] -[A] t = kt 0 0 t [A] k 1 0 0 [A] = 0 t d[a]!st rder: = k dt half-life [A] t t = = 1 1 k1 [A] 1 [A] e kt 0 ln [A] ln[a] 0 t 0 t nd rder: d[a] = k[a] dt [A] t [A] 0 t [A] 0 = + 1+ k [A] t 0 [A] = [A] + kt t -1-1 t 0 1 = k [A] 1 0 [A] -1 0 t
Secnd-rder Kinetics: [A] [B] k A + B prducts rat e = da kab a [ A ], b [B] dt = = = ( )( ) = k a x b x 0 0 dx da But sinc e a= a 0 x, = dt dt dx = k a0 x b0 x dt ( )( ) x dx kt = 0 ( a0 x)( b0 x) x 1 1 1 = a0 b 0 a 0 0 x b0 x ( ) ( ) ( ) 1 ab ln 0 = ( a0 b0) a0 b dx r mre usefully, a a ln 0 = ln + 0 0 b b 0 ( ) a b kt
Determinatin f Order 1 Integral Methds est the data against an apprpriate integral rate law. e.g. ln[a] vs t r 1/[A] vs t. Differential e Methds lt the rate directly: x y ρ= ka b lgρ= lg k + xlg a+ ylgb Determine initial slpes r tangents frm a single curve fr different initial cncs. a(t) a(t) t t and plt versus cncentratin: lg ρ slpe = x 0 lg a
Determinatin f Order Half-life lif Methd Zer-rder: the reactant is used up in tw half-lives. 1 st -rder: the half-life is cnstant in time: nd rder: the half-life increases with time In general, t 1 = ( n 1) ka n 1 1 n 1 0 t = ln / k 1 1 1st-rder nd rder a(t) a(t) Islatin Methd t ½ t ½ t ½ t ½ tgether with ne f the ther methds All except ne reactant is added in large excess, s that their cncentratins d nt vary significantly. hen the ther reactant has pseud-first rder kinetics: ρ = k [A][B] ] =λ [B ] fr [A] [B]
Data Analysis Classical methds f data analysis are ften useful t explre the rder f reactins, r t display the results (e.g. a semi-lg plt t demnstrate expnential decay). Hwever, these methds shuld be avided fr quantitative data analysis, since errrs (and thus weighting) can be distrted. e.g. [A] [A] -1 t 0 t 0 Mdern data analysis uses cmputer methds fr direct curve fitting, e.g. by chi-square minimizatin. æexpt - theryö χ = å ç è exptl. errr ø data pints A gd fit has χ /(n. f degs. freedm) = 1.
emperature Dependence he Arrhenius law is an empirical descriptin f the dependence f the rate cnstant: k = Ae E a / R he pre-expnential factr is ften interpreted as a cllisin rate. Cllisin thery predicts ½ dependence fr A. k A E R a ln = ln -, ln k 0 1/ n(e) he expnential factr describes the fractin f cllisins with sufficient energy fr reactin, as predicted by the Bltzmann distributin: E a he activatin energy is defined by E E a = R dln k d Curvature in the Arrhenius plt is ften attributed t tunneling, but ther reasns include cmplex reactins, and dependence f the pre-expnential factr, which arises naturally in mst theries.
Oppsing Reactins Relaxatin A At equilibrium da dz = k a k z = = dt dt ka k 1 k 1 1 1 0 = k z 1 eq 1 eq Z K [Z] = = [A] k eq 1 eq k 1 If equilibrium is disturbed by an amunt x, s that dz dx da at () = aeq xt (), zt () = zeq + xt (), = = dt dt dt da = k1 aeq x k 1 zeq + x dt dx = ( k 1+ k 1 ) x= dt x= x exp k + k t ( ) ( ) { ( ) } 0 1 1 ( ) a a a a 1 + ( q) = ( q) 1 ( ) ( z z) ( z z ) 1 + 1 eq 0 eq e k k t eq eq 0 e k k t cnc z eq a eq = time his expnential relaxatin f cncentratins is the basis fr several jump methds f studying fast reactin kinetics.
arallel Reactins Cmpetitin Cnsider a mlecule that can react by tw different rutes: A k b k c B C Define a = [A], b = [B], c =[C] [C]. he verall decay f A depends n bth reactins: da = ka+ ka= ( k + k) a a= a dt ( ) e k b + k c t b c b c 0 he rate f frmatin f each prduct depends n bth rate cnstants: db ( b c) k + k t = ka b = ka b 0e dt b kb a dt kb db/ dt = = = dc ( b c) c c / c c 0e k + k t c k adt k dc dt = ka= ka dt k b [ B ] yieldfb k = [ C] = yield f C c his is the basis fr cmpetitin kinetics, whereby an unknwn rate cnstant is determined frm a knwn rate cnstant and the rati f cmpetitive prducts. he abve treatment assumes kinetic cntrl. In cntrast, at equilibrium, [B] eq [C] eq [B] eq K b k b k-c = Kb, = Kc, = = [A] [A] [C] K k k eq eq eq c c -b
da ka Cnsecutive Reactins Simplest case - tw first-rder steps k1 k A B C a= a 1 = 1 0 e kt dt db k1 kt 1 kt = ka 1 kb b= a0 e e dt k k 1 dc k k kb dt = kt 1 1 kt c = a0 1 e e k k + 1 k k 1 da db dc + + = 0 a+ b+ c= a0 dt dt dt Fr k 1 >> k the kinetics can be cnsidered as tw steps: 1. At shrt times b increases as a falls.. At lnger times (k 1 t >> 0), c increases as b falls. cnc. b ( ) c a 0 1 e kt b a0 e kt a c t
he Steady-State Apprximatin k1 k A B C f r k k 1 db k k = = + dt k k k k ( ) kt 1 1 k k1 t ka 1 kb kae 1 0 e 1 1 cnc. a c After the inductin perid, i.e. fr k t >> 0, b da kt dc 1 kt db 1 kt k 1 1 = kae 1 0, kae 1 0, kae 1 0 dt dt dt k k 1 db da dc, dt dt dt Althugh b is nt cnstant, it changes at a much smaller rate than a r c. his is the essence f the steady-state state apprximatin. t
An Example f a Cmplex Mechanism Cnsider the verall reactin NO + O NO It is fund experimentally t be third rder verall, secnd rder in NO, first rder in O. It is much t fast t be a termlecular prcess est the mechanism: NO k k 1-1 N O k N O + O NO Apply the steady-state apprximatin t [N O ] d dt [ ] k [ ] k [ ] k [ ][ ] N O = NO N O N O O = 0 1 1 [ NO ] = k k 1 [ NO] k [ O ] 1 1 d kk rate = [ NO] = k[ NO][ O] = dt k k + kk [ NO] [ O ] + [ O ] 1 1 1 In the limit f k -1 >> k [O ], rt a e = [ NO] [ O ] k 1 his is an example f a pre-equilibrium equilibrium mechanism.
Atm/Radical Cmbinatin Reactins In lw pressure gases, atms seem t react slwer than expected because the cmbinatin prduct falls apart in the perid f a mlecular vibratin (~ 10-14 s). A + B (A-B)* A + B U D e r(a-b) If A and B are plyatmic radicals, (A-B)* may live lnger (e.g. 10-9 s), by distributin f D e amng different vibratinal mdes. Fr efficient reactin a third bdy is needed: 1.. A+ B AB* AB* + M AB+ M k eff d[ab] ρ= = k[m][ AB* ] dt k1[a][b] = k [M] k 1+ k[m] ressure kk 1 [M] = [A][B] k 1+ k[m] he effective rate cnstant depends n pressure ([M]).
Enzyme Kinetics 1 Anther example f a pre-equilibrium mechanism is ne used t mdel the kinetics f enzyme actin: E + S k1 k ES k 1 E + enzyme substrate bund enzyme prduct enzyme regenerated Applying the steady-state apprximatin t the bund state, d[es] dt Rearranging: Rate: = k [E][S] k [ES] k [ES] = 0 1 1 k 1 1 [ES] = [E][S] = [E] 0 [ES] [S] k 1+ k k 1+ k [ES] = k k1[e] 0[S] k + k 1 + k 1 [S] d[] kk 1 [E] 0[S] ρ= = k[es] = dt k + k + k [S] 1 1 ( ) ρ= k[e] 0[S] K + [S] M with K M = k + 1 k k 1 Michaelis-Menten Menten Michaelis cnstant
Enzyme Kinetics substrate enzyme bund enzyme prduct Write the Michaelis-Menten equatin in reciprcal frm: 1 1 KM 1 KM 1 ρ = k [E] + k [E] [S] = ρ + ρ [S] 0 0 max max ρ ρ max ρ -1 slpe K = ρ M max -1 ρ max [S] [S] -1 Lineweaver-Burke plt
Diffusin-limited Kinetics Fr fast reactins in liquids, id the rate-determining t i step can be diffusin f the reactants t frm the encunter pair: k { } D R A+ B AB prducts k -D Apply the steady-state apprximatin t {AB}: d AB = k A B k-d + k AB = 0 { } D[ ][ ] ( R) { } dt kdkr rate = kr { AB} = [ A][ B] k + k k -D R he effective rate cnstant has tw limits: Slw diffusin : Fast diffusin : k k k = k if k k D R eff D R -D k-d + kr kd kr k = K k = k if k k k eff {AB} R act -D R -D Intermediate situatins can be described by: k eff k k k ( k k / k ) k k = = = k + k k + ( k k / k ) k + k D R D D R -D D act -D R D D R -D D act ake the inverse: 1 1 1 = + k k k eff D act