Physical Chemistry I for Biochemists. Chem340. Lecture 16 (2/18/11)

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1 hyscal Chemstry I or Bochemsts Chem34 Lecture 16 (/18/11) Yoshtaka Ish Ch4.6, Ch & HW5 4.6 Derental Scannng Calormetry (Derental hermal Analyss) sample = C p, s d s + dh uson = ( s )Kdt, [1] where K =(1/) s a heat conductance re = C p, re d re = ( re )Kdt [] By calculatng [1]/dt []/dt, ( s - re )K = sample /dt re /dt =(1/K){ sample /dt re /dt} = (1/K)( p /dt) By ntegratng p /dt over tme q p = (C p, s s C p, re re ) +H uson 1

2 Chemcal and Structural Change Detected by DSC Sold /dt H uson Sold Lqud p dt dt Lqud me q p = (C p, s s C p, re re ) +H uson HW5 4.31) he gure below shows a DSC scan o a soluton o a 4 lysozyme mutant. From the DSC data, determne m, the excess heat capacty C and the ntrnsc and transton excess heat capactes C nt and C trs at = 38 K. In your calculatons, use the extrapolated ed curves, shown as dashed lnes n the DSC scan.

3 DSC Curve Upon roten Denaturaton (/d) p = C p () C p () Denatured (Unolded) Intrnsc excess heat capacty: C nt ransent excess heat capacty: C trs Natve (Folded) Meltng Heat Cap. or denaturaton C p den = C D () - C N () Excess heat capacty C p = C nt () + C trs () A Ch 5.1 he Unverse has a natural drecton o change =1L = 1 bar =3 K =1L = bar C 5K 3K B Adabatc expanson = 1/ bar = 3K D Q. Why cannot we expect B A or D C n a spontaneous way? 4K 3

4 Ch 5.3 Introducng Entropy reversble Entropy: S ds [5.1] reversble S s a new state uncton ds For a natural process S n an solated system Second Law o hermodynamcs: So our world s entropy s always ncreasng! 5.4 Calculatng Changes n Entropy or Ideal Gas ds reversble =1L = 1 bar =3 K =1L = bar Case 1. Adabatc reversble process ds = reversble / = S = Case. Isothermal reversble process (, ) (, ) q reversble = -w reversble = S reversble q d reversble nr nr ln( d nr ln( / ) / ) Q. What s the sgn o S or sothermal reversble compresson? 4

5 Enthalpy Change or Irreversble rocess Case For sothermal rreversble process (, ) (, ) S must be calculated or an equvalent reversble process. Ex. Isothermal adabatc rreversble expanson (, ) (, ) Isothermal reversble expanson (, ) (, ) reversble qreversble S nr ln( / ) Q. What s the sgn o S or sothermal rreversble expanson? Q. What s the sgn o S or sothermal rreversble compresson? Q. Does sothermal rreversble compresson occur naturally? =1L = 1 bar =3 K Irreversble adabatc & sothermal expanson =1L = bar (1L, 1 bar, 3K) (L,.5 bar, 3K) = 1/ bar = 3K Calculated Change n Enthalpy (contnued) Case3 (3 ). Reversble change n or a xed reversble ncvd S nc ln( / ) Case4 (4 ) ). Reversble change n or a xed S reversble ncd nc ln( / ) Case 5 (reversble) & 5 (rreversble) process: (, ) (, ) By calculatng S or (, ) (, ) (, ) S nr ln( / ) nc ln( / ) Case 6 & 6 (, ) (, ) By calculatng S or (, ) (, ) (, ) (/ = / ) S nr ln( / ) nc ln( / ) nr ln( / ) nc ln( / ) 5

6 Sample Queston (p9) For a reversble adabatc expanson (, ) (, ), show that S = usng ( / ) = ( / ) -1 and S nr ln( / ) nc ln( / ) S nr ln( nr ln( [Q] / ) nc / ) ( 1 ) nc { nc ( 1) nr)ln( ([Q1] nc v p nc v nr)ln( ln( ln( / ) / ) / ) / ) Ex. 5.3 (p89) One mole o a monatomc deal gas ntally at =5K and =1 atm expands adabatcally aganst a constant external pressure o 1. atm untl the gas pressure also equals 1. atm. Assume that C m,m = 3R/. (a) Calculate and o the gas and. (b) Calculate S or the gas. Q. How to calculate S once we obtan,, and? (b) S nr ln( / ) nc ln( / ) (a) s obtaned by [Q1] = nr /.. [1] Use = nr / ext. [] & q = nc,m [Q ( - ] ) = - ext ( - ) [3]. By substtutng n [3] wth [], nc,m ( ) = - ext (nr / ext ) = -nr +nr ext / n(c v,m +R) = n(c v,m +R ext / ) = (C v,m +R ext / ) /(C v,m +R) =(3/+1/1) /(3/+1) =3K 6

7 S n vaporzaton and uson (p9) Durng uson and vaporzaton, s const. LqudGas dg under a constant reversble q Svavorzaton Sold Lqud under a constant reversble vaporzaton H vaporzaton vaporzaton S uson reversble q reversble uson H uson uson How can we prove that S s a state uncton? HW6 1. reversble = du -Dw (U/) d C d d reveresble 1 ds d C d For deal gas, 1 C nr d d ds / For lqud/sold, ds 1 d C d Cv d d 7

8 ds S or lqud & sold (p91) 1 Cv d C d d d S d Cv d ( ) Cv ln( / ) [5.] S d C Correct or Sold & Lqud d ( ) C ln( / ) Q Is the equaton vald or an deal gas? 1 1 & 1 1 nr [5.3] 5.5 Usng Enthalpy to Calculate the Natural Drecton o a rocess n an Isolated System ds L = -/ L =C v d L / L ds R = / R =C v d R / 5K 3K R ds = (1/ R -1/ L ) > ds> or ds<? 4K 4K C C S dl dr 5K L 3K R C ln( 4K / 5K ) C ln( 4K / 3K ) B 4 K C v ln K K K K 4K 16 Cv ln Cv ln 4K 1K 16 1 A 4K Q.How much s S or BA? Is S postve? 8

9 5. Heat Engnes and the Second Law o hermodyamcs Isotherm: -d = -(nr/)d a b: q ab = -w ab =nr hot ln( b / a )> c d: q cd = -w cd = nr cold ln( d / c )< Adabatc: q= da & bc: q cd = q bc = U = w cycle + q ab +q cd = w cycle = -(q ab +q cd ) < ( q ab > q cd ) Ecency = w / q ab = q ab +q cd / q ab < 1 Carnot Cycle 9

10 S= or Carnot Cycle? S = q ab / hot + q cd / cold = nr hot ln( b / a )/ hot + nr cold ln( d / c )/ cold = nrln( b / a ) + nrln( d / c ) = nrln( b d / a c ) d a & bc are adabatc processes: cold -1 d = hot -1 a & hot -1 b = cold -1 c ( 1 1 d b ) -1 = ( a c ) -1 b d / a c =1 S = q ab / hot + q cd / cold = w cycle = -(q ab +q cd ) = nr hot ln( b / a ) + nr cold ln( d / c ) = nr hot ln( b / a ) - nr cold ln( b / a ) = nr( hot - cold )ln( b / a ) Q1. Explan the meanng o (H/), n and d(h/), n? (δh/δ),n = -C p μ J- and (δh/δ) p,n = C p (H/), n 1 st Dervatve o H wth respect to or xed and n. (H/), n s the pressure dependence o enthalpy at constant temperature and constant number o partcles. 1

11 4.6) From the ollowng data at 5 C, calculate the standard enthalpy o ormaton o FeO(s) and o Fe O 3 (s): H recton Fe O 3 (s) + 3C(graphte) Fe(s) + 3CO(g) 49.6 A 3 B FeO(s) + C(graphte) Fe(s) + CO(g) C C(graphte) + O (g) CO (g) D CO(g) + 1/ O (g) CO (g) 8.98 H (FeO(s)) s calculated or the ollowng reacton: (1) Fe(s) + 1/O (g) FeO(s) (-A B) (C+1/D) -B (Fe O 3 : A = ; CO : C+D =) =1 =1/ =1 -B = 1 B = -1 & A = C+1/D =1/ & C+D = C + ½(-C) = ½ C = 1 D= ) From the ollowng data at 5 C, calculate the standard enthalpy o ormaton o FeO(s) and o Fe O 3 (s) H recton A Fe O 3 (s) + 3C(graphte) Fe(s) + 3CO(g) 49.6 B FeO(s) + C(graphte) Fe(s) + CO(g) C C(graphte) + O (g) CO (g) D CO(g) + 1/ O (g) CO (g) 8.98 (1) Fe(s) + 3/O (g) Fe O 3 (s) (-A B) (C+1/D) -A (FeO: B = ; CO : C+D =) [Q1 ] [Q ] [Q3 ] B = & -A = 1 C+1/D = 3/ 11

12 4.8) Calculate H reacton, at 65 K or the reacton 4NH 3 (g) +6NO(g) 5N (g) + 6H O(g) usng the temperature dependence o the heat capactes rom the data tables. (-1811 kj) H reacton, H reacton, K C ( ' ) d ' K (See able.4 n Appendx B or H ) C = {-4A NH3 (1) - 6A NO (1) + 5 A N (1) + 6A HO (1) } 4 n + { X A X () }(/K) + { X A ' X (3) }(/K) + { X A () }(/K) C 3 ( ') d ' B( n) ; B( n) { X AX ( n)} K n n X 98 K H reacton,98.15 K 4H ( NH 3) 6H ( NO) 6H ( H O) C = B(1) + B()/K + B(3)(/K) + B(4)(/K) 3 C ( ') d ' B(1)( ) + B()( - )/K +C(3)( NH3 NO N HO Coecent A E+4 A E E+4 / 7.36E+ A E E E-5.88E E E+6 3/3-1.66E+4 A E E E E E E+146E E+9 4/4 6.3E E+3 J Reacton enthalpy at K E+3 kj Reacton enthalpy at 65 K E+3 kj 1

13 4.18 Calculate the sngle bond enthalpes and energes or S F, S Cl, C F, N F, O F, H F: Substance SF 4(g) SCl 4(g) CF 4(g) NF 3(g) OF (g) HF(g) H kj mol a) SF 4 he average S-F sngle bond enthalpy s calculated rom the ormaton enthalpes and the ollowng assocated reactons: H [Q1] S s F kjmol -1 g SF4 g 1/ F g[q] 79.4 kjmol -1 F g S s [Q3] S g 45. kjmol -1 H g S g 4 F g reacton SF 4 Q3 (Mod). Now, we would lke to derve H reacton () at a pressure slghtly hgher than 1 bar at a standard temperature. Derve H reacton( () usng g J-, C pa p,a,, H reacton( (1 bar). Use dh A = C pa d C p, A J- d + (H/n), dn A (A = B, X, or Y) assumng that J- and C p,a are constant. Specy the path o (,, n) you would lke to use. Assume a reacton 3B X + Y. Step 1: 1 bar Step : 3B(1 bar) X(1 bar) + Y(1bar) Step 3: 1 bar 13