other examples Ch 13 - 13.7(also available as a Mathcad...

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Unformatted text preview: 13.7 (also available as a Mathcad worksheet) “C32 : ,Vng “casz {Vs} Species balance table: . Since "C : 1 (solid), and P : 1 bar : standard state pressure. Species Initial Final yi K3 : X f (1 7 X) C i i i 52 1 17 X (1 — X) :> or (:52 0 X X X=Ka/(1+Ka) 1 Using CHEMEQ I find K3050D C) : 8.478 and Ka(1000° C) : 6.607 . Therefore X(750D C) : 0.894 and X(1000° C) : 0.869 . X : yc52 is the percentage equilibrium conversion of sulfur. 13.4 A G0 7 Kg : exp — "m : exp —2866 : 0.3147 RT 8.314X298.15 . [P V, : “ma ; where ai : L : exp{ ’1 (P71bar)} gimme )2 (1:13 :1bar) RT TPoynfing correction terms assumed inmmpressible solid adiamoud : expiZdJ-m (13,1 bar)/RT} : 6Xp{(zdjm 7 [£02, 1 bar)} 57mph“,a exp{Zy(P—1bar)/RT} RT where 76m :lzg—m:3.4188 cc/mol:3.4188x104 ms/mol 3.51 g/cc 7 a h :ug—ll'm’l: 5.3333 cc/mol: 5.3333x10’6 m3/mol grp 2.25 g/cc 34188—53333 P—l cc—bar 7A 6° 4 m0.3147:( X ) : "m— : 2866 J/mol RT RT R 2P7] bar: 2866 “”101 :1497.0 J/cc:14970 bar 1.9145 cc/mol or P: 14971 bar. Thus for P < 14971 bar ; 0‘11"“ > K“ and graphite is stable phase for P > 14971 bar ; 0‘11”“ < Ka and diamond is stable phase "mph :> Need a hydraulic press capable of exerting 14971 bar to convert pencil leads to diamonds. (Also, should consider a higher temperature!) 13.20 Using the data in the problem statement, Tables 3.4 and Appendix AIV, I find a - a P Kama K) — “5‘02 C02 — 148.1 — am: — yco, , (1) IQCalooflsro, 1 bar since the activity of all the solids are unity. 3 a ' a a Kagoso K) — 5‘302 F3304 0° — 0.0277 — ”CO — yco , (2) arage-51025111302 aco2 J’co2 and a a3- 1,12 K0 (750192M203973x1014 2;: lb” ,3 3 _ 1,12 ' 1,12 P a1105102130, “0, y 02 P :> yo! = 1.242 ><1()_28 (3) 1 bar From eqn. (2) we have yco ~ 0.0277 , while from spectroscopic observations yoo E 10—4 yoo, J’co2 Also, fi‘om eqn. (1), P002 ~148 bar , while fi‘om the probe, the total atmospheric pressure is only between '15 and 105 bar. Finally, fi‘om Eqn. (3), we conclude there is no 02 in the atmosphere, compared to a trace from spectroscopic observations. Conclusions? Somewhat ambiguous! Calculations and data are not in quantitative agreement, but are certainly in qualitative agreement. Consider the uncertainty in all the measurements, the atmospheric model is undoubtedly a reasonable one, and can not be rejected. ...
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