5.60 Thermodynamics & Kinetics Spring 2008

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1 MIT OpenCourseWare Thermodynamcs & Knetcs Sprng 2008 For nformaton about ctng these materals or our Terms of Use, vst:

2 5.60 Sprng 2008 Lecture #29 page 1 PPLICTIOS: CHEMICL D PHSE EQUILIRI pply statstcal mechancs to develop mcroscopc models for problems you ve treated so far wth macroscopc thermodynamcs 0 Products Reactants Separated atoms E Product & reactant energy levels ε r = -D 0,r ε p = -D 0,p Chemcal equlbra Gas phase: Calculate Kp from mcroscopc propertes a + b F cc + dd ΔG = RT lnk p = cg C + dg D ag + bg eed G 0 for each speces G = + pv = kt lnq + pv = kt lnq + kt Q = q (! q trans ) nt lnq = ln(q trans q nt ) ln! = ln (q trans q nt ) ln + = ln (q trans q nt ) + G = kt lnq + kt = kt ln(q trans q nt ) = kt ln( q ) For molecules n the gas phase, nternal degrees of freedom are rotatons and vbratons and electronc states.

3 5.60 Sprng 2008 Lecture #29 page 2 q nt = qrot qvb q elec D kt Dssocaton energy D 0 from ground electronc level q elec = e 0 Usually no other electronc levels matter q = e nε 0 kt = 1 + e ε 0 kt 2ε 0 kt 3ε 0 kt 1 vb + e + e + " = 1 e ε 0 kt n ote we ve set the zero of vbratonal energy as the lowest vbratonal level. The zero-pont vbratonal energy has been ncluded n q elec by usng the dssocaton energy rather than the bottom of the electronc potental energy. kt (300 K) 200 cm -1 Most molecular vbratonal frequences > 500 cm -1 q vb 1 - eed to calculate t, but t s not large We have not treated q rot. Levels are OT evenly spaced: ε rot = J(J + 1)ε 0,rot where J = 0, 1, 2,, and ε 0,rot 1 cm -1. Hgh-T lmt: q rot = kt/ε 0,rot We ve seen that q trans 10 30, q trans / For multple speces n chemcal equlbrum, need to use for each speces and use partal pressure values n pv term. G = kt ln(q ) = n RT ln(q ) G = RT ln(q ) = G + RT ln( p p ) q kt qkt G = RT ln = RT ln pv pv Example: smple reacton at 300 K: - + C-D -C + -D ssume all molecules have hgh vbratonal frequences q vb 1 for all speces Translatonal & rotaton: hgh-t lmts ssume smlar rotatonal energes (depends on moments of nerta) q trans & q rot contrbutons are all equal

4 5.60 Sprng 2008 Lecture #29 page 3 G q q C D Δ = RT ln = RT lnk p q qcd K = q q C D = e D 0,C kt p q q e D 0, kt e CD e D 0,D kt D 0, CD kt (D 0, C +D 0,D D 0, D 0,CD ) kt = e Only dssocaton energes matter snce the other degrees of freedom contrbute equally. Usually energy effects domnate n chemcal equlbra. Relatvely smple case: Sold-state chemcal reacton a F b e.g. somerzaton n an organc molecular crystal molecules of, molecules of, + = o translaton or rotaton, no change n pv all the change s n the electronc and vbratonal energy If vbratonal frequences are the same, then only the dfference n dssocaton energes s needed. Ω( )e Q = e E kt E kt = E E mcrostates energes Recall the entropy due to mxng of two speces! ( D!! E kt Q = Ω( E ) e = e + D ) kt! (D kt ) (D kt ) = e e!! E =0 =0 energes! q = e ( ) D D kt ΔD 0 kt = q q = (q + q ) = q (1 + s) where s = q = e 0!! = The bnomal theorem gves a smple closed form for the sum. It s convenent here to redefne the zero of energy as D so q = 1 and ( ) = + 2 Q = 1 + s 1 as + a s + a s " + a s The ground state of the system has all molecules

5 5.60 Sprng 2008 Lecture #29 page 4 System energy s E( ) = ΔD 0. Probablty that the system has ths energy s ( ) a s p = Q The average number of molecules s p ( ) = = 2 3 as 1 + 2a 2 s + 3a 3 s + " + a s s Q lnq = = Q Q s lns and E = U = ΔD 0 If vbratonal energes dffer then q vb also needs to be ncluded. In general, low vbratonal frequency favors a speces snce the levels are closely spaced, so there are many low-energy states of that speces. For chemcal reactons nvolvng covalent bonds, the bond energes domnate over entropy. ut the treatment we ve gven could also be used for much more subtle changes lke rotaton of CO molecules n a CO crystal. ote that ths treatment assumes no nteractons between dfferent unt cells. Ths gves rse to a farly gentle T-dependence of the equlbrum.

6 5.60 Sprng 2008 Lecture #29 page 5 Sold-sold phase equlbra Phase transtons between dfferent crystallne phases may be easer to treat then chemcal reactons, because there s just one state at any temperature: the crystal n phase α or the crystal n phase β. Problem set problem Ensten model for crystallne phases α and β, frequences ν α, ν β ndng energy per atom: ε α, ε β 0 Phase α Phase β E ε β ε α ndng energy s lke dssocaton energy. It ncludes zero-pont vbratonal energy, so no need to account for ths separately. You can predct the phase transton temperature based on a smple frst prncples model. Phase transton only occurs f the crystal wth stronger bndng energy, e.g. α phase, also has hgher vbratonal frequency. ote that ths treatment assumes complete cooperatvty: the crystal s ether all one or all the other phase. Ths gves rse to an abrupt T-dependence of the equlbrum. Many ntermedate cases occur.

7 5.60 Sprng 2008 Lecture #29 page 6 Example: helx-col transton n bopolymers. Interactons make few long segments of helx or col more stable than many short segments. (Smlar for magnetc and nonmagnetc domans n a ferromagnetc crystal, and n all sorts of nteractng systems). In ths case the T-dependence s not completely abrupt, but not nearly as gradual as n the non-nteractng lmt. Extra example (usng QM result for q trans ) Can also calculate sold-gas phase equlbra,.e. vapor pressure over the crystal. ssume monatomc deal gas only translatonal partton functon. 32 g 2πmkT Q = q trans! = 2 V! g = ktlnq g h 2πmkT 32 g lnq = ln 2 V ln + h lnq g 2 πmkt 32 2 πmkt V = ln 2 V ln = ln 2 T,V h h g μ = 2 πmkt 32 kt ktln 2 (deal gas) h p 32 μ g = g T,V Condton for phase equlbrum: μ s =μ g 32 (kt) 52 s kt g 2π m μ = ε 3kTln = μ = ktln 2 hν E h p 2 m hν ε=ktln = ktln 2 m 2 12 h p kt (kt) p (2πm) 32 3 ν E p = 12 ε kt (kt) e π (kt) E ( ) ν E 3 π 32 3 gves vapor pressure p(t) over the crystal! Strong bndng energy or low T gve low pressure as expected. Low vbratonal frequency also gves low p, snce ths allows hgher entropy n the crystal. So you can calculate the p-t phase dagram that you descrbed before n macroscopc terms only.