NAME and Section No. it is found that 0.6 mol of O

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1 NAME and Secton No. Chemstry 391 Fall 7 Exam III KEY 1. (3 Ponts) ***Do 5 out of 6***(If 6 are done only the frst 5 wll be graded)*** a). In the reacton 3O O3 t s found that.6 mol of O are consumed. Fnd the extent of reacton, ξ n n = ν ξ and n n =.6. ( n n ) ν ξ ( ) ( ) O O O O O = =.6 3 =.. b) It s found that K P s ndependent of T for a partcular chemcal reacton. What does ths tell you about the reacton? Ths s true f ΔH reacton =.. c). Defne the term trple pont. What condtons are satsfed? At the trple pont three phases are n equlbrum. Ths means that p = p = p a b c T = T = T a b c μ = μ = μ a b c 1

2 NAME and Secton No. d). Show that U A μ = = n n S, V, nj n T, V, nj n The Gbbsan dfferentals for U at fxed S and V and A at fxed T and V wth the ncluson of the chemcal potental terms are: du = TdS pdv + μ dn da = SdT pdv + μ dn The above defntons of the chemcal potental for these ndependent varables show they are equal. e). In settng up thermodynamc tables for the entropy, how s the thrd law of thermodynamcs used, and what s the conventon for the entropy? The thrd law states that the entropy change approaches zero as T as long as the substance s n nternal equlbrum. Ths feature along wth the conventon that the entropy at T= of the elements s taken as then sets the entropy to be zero for compounds at T= (and 1 atm pressure). f). Defne an deal soluton n terms of the chemcal potental of ts speces. Explan the meanng of all symbols used and ndcate what they depend on. In an deal soluton the chemcal potental s expressed as ( ) μ = μ T, p + RTlnx * where the standard state * μ corresponds to pure speces and depends on T and p. The expresson holds for all values of the mole fractons, x.

3 NAME and Secton No.. (3 ponts). A) Whch of the followng reactons are spontaneous at 5 C n the drecton wrtten? Whch are exothermc or endothermc? HCl(g) + NH 3 (g) = NH 4 Cl(s) Al O 3 (s) + 3S(s) = 3SO (s) + 4Al(s) (at 5 C) ΔH f (kj/mole) ΔG f (kj/mole) HCl(g) NH 3 (g) NH 4 Cl(s) Al O 3 (s) SO (s) Δ G = ΔG ΔG rxn f, prods f, reacts For frst reacton: ( ) ( ) ( ) Δ G rxn = kj/mol = 9. kj/mol Spontaneous snce Δ G rxn <. ( ) ( ) ( ) Δ H rxn = kj/mol = kj/mol Exothermc snce Δ H rxn <. For second reacton: ( )( ) ( ) ( )( ) ( )( ) Δ G rxn = kj/mol = kj/mol Not spontaneous snce Δ G rxn > ( )( ) ( ) ( )( ) ( )( ) Δ H rxn = kj/mol = kj/mol Endothermc snce Δ H rxn > 3

4 NAME and Secton No. B) In whch drecton wll the equlbrum shft f the temperature s ncreased? To answer ths queston, obtan the van t Hoff equaton by evaluatng ( G/ T) T) the equlbrum constant. and relate to P G 1 G H TS S H = + P = = T T T T T T ( G/ T) T) Snce the equlbrum constant s related to Δ G = RTln K t then follows that rxn P dln K P ΔH rxn dt p Δ G rxn by =, whch s the van t Hoff equaton. RT For Δ < the equlbrum constant decreases as T ncreases and for H rxn Δ > the H rxn equlbrum constant ncreases as T ncreases. 4

5 NAME and Secton No. 3. ( ponts) Usng the phase rule, fnd the number of degrees of freedom, f = c + p, for: nd A) H 3 PO 4 n water n equlbrum wth water vapor; dsregard salt onzaton. Two phases p= Two components c nd = No reacton or other restrctons so f=+-= B) The same as A) but take nto account H 3 PO 4 onzaton nto all possble ons (up to 3- PO 4 ) for H 3 PO 4 : HPO = H + HPO = HPO H HPO HPO = H + PO. - Two phases p= Sx components c=6 Three onzaton equlbra r=3 lsted above One charge balance, to balance H + wth all the catons, a=1. cnd = c r a= 6 3 1= f=+-=. Stll. 5

6 NAME and Secton No. 4. ( ponts) A bartender who clearly has not studed thermodynamcs attempts to prepare 1 cm 3 of some potable by mxng 3 cm 3 of ethanol wth 7 cm 3 of water. Does she succeed? The fgure wll be of help. The densty of ethanol s.785 g/cm 3 and that of water s 1. g/cm 3. The molecular weght of ethanol s 46.1 g/mol and that of water s 18 g/mol. Need to frst fgure out the mols for the amounts of water and ethanol: 3 3 ( )( ) ( ) n HO= 7 cm 1g/cm / 18g/mol = 3.89 mol 3 3 ( )( ) ( ) n ETOH = 3 cm.785g/cm / 46.1g/mol =.51mol The mol fractons are: HO HO ( ) x = n / n tot = 3.89 / =.88 xetoh netoh n tot ( ) = / =.51/ =.1 6

7 NAME and Secton No. Estmate from the fgure that at ths composton V = HO V = ETOH 3 18.cm /mol 3 5.5cm /mol Construct total volume: V n V n V ( )( ) ( )( ) = ETOH ETOH + HO HO = cm cm = 97.6 cm. She doesn t succeed. 7

8 NAME and Secton No. Mathematcal expressons: n ( n+ 1) xdx x n n = /( + 1); =,1,,... 1 dx = ln x x ( n 1) ( ) n x dx= x / n 1 ; n=,3,... x y z Euler: = 1 y z x x y z x y Inverter: = 1 y x z z 8

9 NAME and Secton No. n dμ + SdT Vdp = (Gbbs-Duhem) νμ = (reacton equlbrum). μ = μ + RTln a ; a = γ x; 9