Geological Sciences 4550: Geochemistry

Transcription

1 Geochemstry Problem Set Four Soluton Due Sept. 28, Assumng deal soluton behavor for the followng: a.) Show that the bolng pont of a substance s ncreased when another substance s dssolved n t. The chemcal potental of each component n a lqud soluton and n a or n equlbrum wth t must be equal. Denotng the component of nterest as, we have: µ lq = µ a Each of the chemcal potentals may be wrtten as: µ α = µ α + RT ln X b where α denotes the phase. The standard state chemcal potental, µ, s the chemcal potental of n the pure phase, α, of (for example, the chemcal potental of water n pure water or). Substtutng b n to a: µ lq + RT ln X lq = µ + RT ln X c Rearrangng: µ µ lq = RT ln X lq X d The chemcal potental of n the pure substance s therefore just the Gbbs Free Energy of that substance: µ α = G α = H α TS α c Therefore, the dfference between the chemcal potentals s the Gbbs Free Energy change of orzaton, ΔG v : µ µ lq = ΔG v e The Gbbs Free Energy change of bolng, may also be wrtten as: G v = G - G lq = H v T S V f Substtutng equaton f nto e, we have: ΔH v TΔS v = RT ln X lq g X where ΔH v and ΔS v are the heat and entropy of orzaton of pure. At the orzaton temperature of the pure substance, the Gbbs Free Energy change must be 0 (snce the two phases are n equlbrum, thus: H v = T v S V h where T v s the bolng pont of pure. Substtutng g nto f, we have: rearrangng: (T v T)ΔS v = RT ln X (T v T) T lq X = R ln X lq ΔS v X j 1

2 Geochemstry Problem Set Four Soluton Due Sept. 28, 2015 The entropy of orzaton wll always be postve. So long as the concentraton of n the or s greater than n the lqud (.e., X lq /X < 1), the rght hand sde of the equaton wll be negatve. If that s the case, then T must be greater than T v,.e., the bolng pont of the soluton wll be greater than that of the pure substance. b.) By how much wll the bolng pont of water be elevated when 10% salt s dssolved n t? When salt water bols, essentally none of the salt goes n the or, so that X 1. Thus equaton h above smplfes to: (T v T ) T Solvng for T, we have: T = = R lq ln X ΔS v T v R ln X lq ΔS + 1 v From Table 2.2., we fnd the entropy of orzaton of water s = J/K-mol (at 25 C; we assume t s ndependent of temperature, whch wll not be strctly true). Usng R=8.314 and X as 0.9, we calculate the bolng pont of an deal aqueous soluton wth 10% salt to be 375.9K or C. 2. Fnd G for the reacton: Pb 2+ + Mn Pb + Mn 2+ Whch sde of the reacton s favored? (Hnt: use the data n Table 3.3) The E for the two reactons s lsted n Table 4.3 as and V respectvely. Subtractng one from the other, the E for ths reacton s 1.05 V. Accordng to equ. 4.41a, G = -nfe. n wll be two snce two electrons are exchanged. F s 96,485 J/V. so G for ths reacton s kj/mol. The rght sde of the reacton s favored. 3. Construct a pε ph dagram for dssolve speces of uranum: UO and U(OH) 5, and the two sold phases UO2 and U3O8 at 25 C and 0.1 MPa. Assume the actvty of dssolved uranum s fxed at The followng free energes of formaton should provde suffcent nformaton to complete ths task. speces G o f U(OH) 5 (aq) UO (aq) UO2 (s) U3O8 (s) H 2 O Values are n kj/mole, standard state s 25 C and 0.1 MPa. R = J/mole- K. 2

3 Geochemstry Problem Set Four Soluton Due Sept. 28, 2015 We start by wrtng the possble reactons between these speces. Some general rules for wrtng these reactons are: 1. Frst balance the element of prncpal nterest (U n ths case). 2. Next balance any oxygen defct usng H 2 O. 3. Balance any hydrogen defct usng H Balance any remanng charge defct usng e. Our reactons are: ΔG (kj/mol) log K 1) U(OH) H + UO 2(s) + 3H 2 O x = ) U(OH) H + UO e + 3H 2 O 3) 3U(OH) x = U 3 O 8(s) + H + + 7H 2 O + 4e x x = ) UO e UO 2(s) = ) 3UO e + 2H 2 O U 3 O 8(s) + 4H x x = ) 3UO 2(s) + 2H 2 O U 3 O 8(s) + 4H + + 4e x x = Equlbrum constant expressons for these reactons (wth actvtes of solds and H 2 O as 1) are: 1) log K = log ([UO 2 ]/[ U(OH) 5 - ]) + ph log K = -log[u(oh) 5 - ] + ph 2) log K = log ([UO 2 2+ ]/[U(OH) 5 - ] + ph -2pe log K = ph - 2pe 3) log K = log[u 3 O 8 ]-3log[U(OH) 5 - ]-ph-4pe log K = -3log[U(OH) 5 - ] - ph -4pe 4) log K = log[uo 2 ] - log[uo 2 2+ ] + 2pe log K = - log[uo 2 2+ ] + 2pe 5) log K = log[u 3 O 8 ] -3log[UO 2 2+ ]-4pH+2pe log K = -3log[UO 2 2+ ] - 4pH + 2pe 6) log K = log(u 3 O 8 ]-3log[UO 2 ]) - 4pH -4pe log K = - 4pH - 4pe Rearrangng the above equatons, so that pe s a functon of ph, and settng the actvtes of aqueous speces to 10-6 we have: 1) ph = log K-6 ph = ) pe = -(log K)/2 + ph/2 pe = ph/2 3) pe = -log K/4 +18/4 - ph/4 pe = ph/4 4) pe = log K/2-3 pe =

4 Geochemstry Problem Set Four Soluton Due Sept. 28, ) pe = log K/2-18/2 + 2pH pe = pH 6) pe = -log K/4 - ph pe = 8.78 ph 4. Show that: G ex = B + C 2 + D 3 reduces to: G ex = (W G1 + W G2 X 1 )X 1 wth approprate substtuton for a bnary system n whch X1 = 1- X2. Substtutng X 1 = (1 ) G ex = { W G1 +W G2 (1 )}(1 ) G ex = W G1 2 { 2 +W G2 W G2 }(1 ) Rearrangng: G ex = W G2 + (W G1 2W G2 ) (W G2 W G1 ) 4

5 Geochemstry Problem Set Four Soluton Due Sept. 28, 2015 Lettng B = W G2, C = W G1 2W G2 and D = W G2 W G1, then G ex = A+ B + C 2 +D 3 5. Interacton parameters for the enstatte dopsde sold soluton have been determned as follows: WH- En = 34.0 kj/mol, WH- D = kj/mol (assume WV and WS are 0). (a) Use the asymmetrc soluton model to calculate Greal as a functon of X2 (let dopsde be component 2) curves for ths system at 100 K temperature ntervals from 1000 K to 1500 K. Label your curves. The key equaton s: Gex = (W G1 + W G2 X 1 )X To express ths as a functon of, we need only note that X 1 = 1-. Makng ths substtuton, we have: G ex = (W G1 +W G2 (1 ))(1 ) Expandng ths: Gex = (W G2 W G1 ) (W G1 2W G2 ) + W G 2 Snce we may assume W S s 0, then W G = W H. We also note that ΔG real = ΔG deal + ΔG ex where ΔG deal = RT (X 1 ln X 1 + ln ) = RT ((1- )ln (1- ) + ln )). As an example, the calculaton for 1000K s shown n the spreadsheet below: The G-bar-X dagram s shown below. 5

6 Geochemstry Problem Set Four Soluton Due Sept. 28, ! 1500! 1000! 1000 K! 1100 K! 1200 K! 1300 K! 1400 K! 1500 K! Greal (J)! 500! 0! -500! -1000! -1500! -2000! 0.0! 0.1! 0.2! 0.3! 0.4! 0.5! 0.6! 0.7! 0.8! 0.9! 1.0! X2 (D)! (b) What s the maxmum mole fracton of dopsde that can dssolve n enstatte n ths temperature range? The maxmum would be about 20% at 1500K. At hgher mole fractons of dopsde, two solds would be stable, each of whch would be a lmted sold soluton. (c) Sketch the correspondng T X phase dagram. 6

7 Geochemstry Problem Set Four Soluton Due Sept. 28,