Calculating Complex Equilibrium Concentrations by a Next Guess Factor Method S. Chaston University of Canberra, Belconnen, ACTAustralia 2616 The next guess factor (NGF) method was designed to allow undergraduates to calculate equilibrium concentrations to two or three significant figures with a pocket calculator in two or three iterations without having to solve for t h e roots of nth degree polynomial equations. The method was first developed for solutions of dilute weak acids and weak bases ( I ) , and for calculating the wncentrations of ion pair complexes in saline waters, such a s sea water (2).Spreadsheet programs, however, are ideal for calculations based on this method, particularly for more complex chemical systems and when more than three significant figures are required. This article describes calculations, based on the NGF method, of some complex chemTable 1. Solubility Calculations for Calcium Iodate in Water Thermodynamic Data 1. Ca(lOs)z(s)= ca2+(aq)+ 2103-(aq) Ki = Ksp= 7.079.10-~ 2. ca2+(aa1+ IO3daa1 = ~a~+103'aal Kz = KML = 7.762 Equations for calculations: M = ca2+,L = 103-, ML = ca2+1033. guess pa2+) new [ca21= ( o I ~ [ c ~ ~ ~ ) . N G F 4. Ionic strength,I = 3[ca2+] use the Debye-Hijckel equationb 5-7. h, h, and ~ML: 8. [I031= ( ~ ~ ~ l [ c.afM~ + l 9. [ ~ a ~ + l 0 3Ksp. 1 = KM~l[1037.h. hL 10. 2 4 =Totalca2+= [ ~ a ~ * ] + [ ~ a ~ ' 1 0 3 1 1I . XL = Total 103- = [1037+[ C ~ ~ + I O F I 12. average solubility, Sav = (XM + 0.5XL)iZ 13. NGF= & E M
.
Calc~laton Res~ltsana Condlt ons Temperat~re,25'C. Expermental S o l ~ b~ty", ~ S = 0 00784 mol L 1st 2nd 3rd 4th 5th 6th 3. I O ~ [ M0.7000 ] 0.7145 0.7199 0.7218 0.7225 0.7228 102.1 2.1000 2.1435 2.1597 2.1654 2.1675 2.1684 4. fM 0.5890 0.5865 0.5856 0.5853 0.5852 0.5851 5. h 0.8668 0.8657 0.8653 0.8652 0.8652 0.8651 6. 7. h~ 0.8617 0.8605 0.8601 0.8599 0.8599 0.8599 8. 10'[~7 1.5116 1.5011 1.4974 1.4960 1.4955 1.4954 IO~[ML] 0.0486 0.0491 0.0493 0.0493 0.0493 0.0493 9. 10. 102XM 0.7486 0.7636 0.7692 0.7712 0.7719 0.7722 11. 1o2Zl 1.5602 1.5503 1.5467 1.5454 1.5449 1.5447 12. lo2& 0.7641 0.7692 0.7711 0.7717 0.7720 0.7721 13. NGF 1.0210 1.0075 1.0027 1.0010 1.0003 1.0001 'Ref. 7. qhe Debye-Hiickei equation: logfi= 4.510~[P51(1 + 0.329R.10')): fi= anivity coetticient of an ion; ionic strength. I = 0.5xGL2; CI = concentration of each ionic species;L= charge on the ion; ion size parameters, Ri, are (6): ca2*= 6 and 1X=4; ca2*lC5=3 (as far CC (8)).
622
Journal of Chemical Education
ical systems that have been discussed and treated recently by other methods of calculation in this Journal ( 3 3 ) . The iteration procedure is started by making a n estimate of the approximate value of the equilibrium concentration of one or more of the species in the chemical system that is being calculated. For example, in the calculation of the pH of a solution of a weak acid whose analytical concentration CHBis known ( I ) , a n approximate estimate of the equilibrium concentration of HB, W ] , , may be used to start the calculation of the concentrations of all of the other species with which it is in equilibrium, such a s in Table 2. Now the sum of all of the calculated values of the concentrations of the species containing B, XB, will equal CHB when the true value of [HB] is chosen. But if a chosen value, [HBI,, is too high, it follows that XB also will be high will be and greater than CHB,and the ratio, NGF = C& less than one. The principle of the method is that the product, NGFGIB],, will give a truer value for [HBI than the chosen value [HB], and provides the starting value of [HB] in the next iteration. Next guess factors, NGF, are derived from equations that define the system, such a s conservation equations and equations based on equilibrium constants. In the above example, the NGF is a ratio derived from the conservation equation, CHB= XB, that is rearranged to produce a ratio (i) that equals one when the true value of [HB] is chosen, and (ii) whose denominator reflects the error in the chosen value oT[HBl. The second requirement liil for an NGF Tor HR here is satisfied bv outtinr XB in the denominator. It is sometimes found t h a t the square root or even the cube root of the ratio in a n NGF produces iterations that converge more efficiently, a s in eqs 52 and 53 in Table 4. Example: Solubility of Ca(lO3)z Russo and Hanania (3) solved the calcium iodate solubility problem by a n iteration method that involved fmding the appropriate root to a cubic equation with the help of a comp&e; algorithm a t the heg&ning of each iteration. Their method also used the published analytical solubility of the salt. The NGF method described here does not involve a cubic equation nor does it use the published solubility of the salt. The first round of calculations in Table 1 is begun by guessing a n approximate value of the equilibrium concentration of the calcium ion, [Ca2+l.Equation 3 gives the new value for [Ca2+la t the beginning of each subsequent round of calculation after the first and is obtained from the product of the [Ca2+land the NGF values from the previous round. The NGF value in Table 1 is obtained from eq 13 and is based on the relationships of the calcium and iodate concentrations to the solubility. The average solubility, S, equals XM, as defined in eqs 10-12, only when the true equilibrium value of [Ca2+lis used. Because XM is a function of the guessed value for [Ca2+l,the ratio in eq 1 3 that has XM in the denominator qualifies a s a suitable NGF. Another possible NGF is the square root of 0.5XLZM that gives almost identical results, round by round, a s are obtained in Table 1. The published analytical solubility of Ca(I031~ of S = 0.00784 moll. (7) is slightly greater than
Table 2. Equations for Equilibrium Concentrations in TINOdHN02 Mixtures Thermodynamic Data 14. Hf(aq)+ NOz-(aq) = HNOz(aq) 15. HN@(aq)+H+(aq)= NOt(aq)+ H z 0 16. 2HNOz(aq)= NzOs(aq)+ H z 0 17. Tlt(aq)+ NOz-(aq)= TINOz(aq)
K14 = 103.15moll^ Kis = lo4." mol/L 6 6=
~OIIL
~ i =71ooe5~
O Y L
Equations for the Calculations 18. guess [TIt]:new guess [n'] = (old guess [TI+]),NGF(Tlt) 19. guess [HNOz]:new guess [HNOz]= (old guess [HNOz]).NGqHNOz) 20.1 =[TIT for first guess; otherwise I = O.S(ZGZ')~ 21-25. h,, h+,fo~.,hm-, ho+: use the Debye-Huckel equationb 26. [H']~= {fi4[HNOzJ (I+ fi7[Tl+])I(fH+. hm-)I+ {KwlfH+.f o ~ - l 27. [OH] = Kw/[Htl. &I+. .foK fH+. h0228. [NO21= Ki~HN02lI[H*l~ 29. [TIN@]=K17(Tl+][N021. h+.ho& 30. [N203] = KIS[HNOZ]~ 31. [NOT = Ki5[H+l[HN@l'fHJf~o+ 32. total N =ZN = [HN@]+[N027+[TIN021+ 2[N2031 +[NOi] 33. NGF(HNOz)=CHNOZEN 34. total TI = ZTI = [Tlf]+[TINOd 35. NGF(Tlf)= Cn+/ZTI
.
*Ref. 9. %ee Table 1:ion size parameters (6). R,,are H' = 9, NO' = 3, TI* = 2.5, OH= 3, N02- = 3.
the calculated solubility of S, = 0.00772 morn, that was obtained from the thermodynamic data in Table 1.
Example: TINOdHNOz Mixtures Cobranchi and Eyring ( 4 ) recently discussed the calculation of equilibrium concentrations in T 1 N 0 & N 0 2 mixtures by the continuation method. This method involved the development of two simultaneous quintic polynomial expressions from the conservation eauations for these mixtures. They were solved for 25 mots by a computer program, CONSOL8T. I n order to get the NGF calculation in Table 3 o n t h e T1NO3RWO2 s y s t e m started, guesses must be made about the equilibrium wncentrations of the two species. HNO. and Tlt. in the mixture. For this'reasoi there are two NGF equations, 33 and 35, in 2, which produce an NGF value for HNOz and another value for ~ 1 +resnectivelv, , ~~~h NGF are based , on conservation equations, where CHNOZ = zNs and = When the guessed value of [HNOZ] i n the first calculat~on round is larger than the true value (as in Table 31, the calculated TN . from eq 32 will be greater than C ~ ~ and ~ z the , NGF (HN02)value from eq 33 will be less than one. A new and more accurate value for IHN021 is then obtained from eq 19, which becomes the basis for the iteration in the second round. The same procedure is used &
to obtain the progressive improvement of guessed values of [Tl']. The procedure that is set up in Table 2, also may be used for calculating the pH of solutions of HNOz on its own, by putting CnNoaequal to a vanishingly small number, such a s lo4 mol/L. Under these conditions, the pH obtained for 0.001 moVL HNOz was 3.257, and the value of mil was 5.680 x 104moVL. As expected, the additionof 0.001 m o m of Tlf increases the value of [Hi] to 5.749 x 104(see Table 31, but the pH, nevertheless, increases very slightly to 3.258, because the other effect of Tlt was to decrease the activity coefficient f-, by changing the ionic strength of the solution.
Example: Si-C-0 System Wai and Hutchison (5) recently discussed the calculation of complex chemical equilibria by the free energy minimization (FEM)techniaue and took the examnle of the calculation of the equ~hbriumcomposition in mole fractions of all sneclesobta~nedfrom the reduction of slhcon dlox~debv cardon. The FEM computer program was set u p in the ladoratory for students, who were asked to run the program to determine equilibrium compositions of the Si-C-0 system a t different initial compositions; i.e., for different values of R in Table 4. The NGF method described here is simple enough for students to set up the entire calculation for themselves without computer programming. Apart from the usual basic equations of chemical equilibrium in Table 4 (eqs 4750) that are involved, the equations for calculating the three NGF values (eqs 5 1 5 3 ) are ratios that are derived from the conservation equations that define the system. When all of the reactants in the system are in the gas phase the mole fractions of three of the major gaseous species (CO. Si. SiO) have to be messed i n order to eet the calculati'on btarted, and three-NGF values are neided to adjust these values. I n Table 4 the mole fractions of CO
Table 3. Equilibrium Concentrations in TINOdHNOz Mixtures Calculation Results and Conditions: 0.001 moWL Temperature 25%; Cn~o3= 0.001 moliL; CHNOZ= 1st
2nd
3rd
4th
51h
6th
1.0000
0.9960
0.9962
0.9963
0.9963
0.9963
IO~[
[email protected] 1O . OOO
0.4472 1.1614
0.4313 1.5858
0.4269 1.5752
0.4257 1.5722
0.4254 1.5714
0.9645
0.9553
0.9557
0.9559
0.9559
0.9559
0.9666 0.9965
0.9587 0.9556
0.9590 0.9560
0.9592 0.9956
0.9592 0.9562
0.9592 0.9562
24. 25. 26. 27. 28. 29.
hm-
0.9646 ho+ 0.9646 104[~t] 6.1824 10"[0~1 1.6175 1 0 ~ [ ~ 0 2 16.1405 ~ O ~ [ T I N 4.0443 O~]
0.9556 0.9556 5.898 1.6955 5.8586 3.7712
0.9560 0.9560 5.7901 1.7271 5.7513 3.7061
0.9561 0.9561 5.7601 1.7361 5.7215 3.6883
0.9562 0.9562 5.7518 1.7386 5.7133 3.6833
0.9562 0.9562 5.7494 1.7393 5.7109 3.6819
30. 31. 32.
1 0 ~ [ ~ ~ 0 33.9622 ] 1015[~0'] 2.1683 IO~(ZN) 1.1182
3.1690 1.8521 1.0369
2.9477 1.7535 1.0102
2.8887 1.7268 1.0028
2.8725 1.7195 1.0007
2.8679 1.7174 1.0002
33. 34.
NGF(HN@) 0.8943 IO~(ZTI) 1.0040 NGF(Tl+) 0.9960 DH 3.2235
0.9645 0.9997
0.9899 I .oooo
0.9971 1.0000
0.9992 1.0000
0.9998 1.oooo
1.OOO3 3.2476
1.0001 3.2555
1.0000 3.2577
1.0000 3.2583
1O . OOO 3.2585
18.
IO~[TI*]
19. 20.
fo31
". 22. 23,
35,
36.
'+ fH+ 6,
Volume 70 Number 8 August 1993
623
when the pressure is greater than 1.3513 atm, Q i s less thanK4%,solid S i c forms and NFG(Q) becomes redundant as well, leaving NGF(C0) the only factor i n the calculation. Table 4 is set up for calculations within the range from YIR = 1to YIR = 2, which is when neither elemental carbon nor liquid silicon dioxide are present a t equilibrium. When YIR is less than one there i s not enough oxygen to convert all carbon into carbon monoxide, and when YIR i s greater than 2 there i s a surplus of silicon dioxide over available carbon.
Table 4. Calculation of Equilibrium Mole Fractions of Gaseous Products in the SIC-0 system Thermodynamic data(3000K)' 37.3Si(g)+ CO(g)= SiO(g) + SinC(g) 38. 2SizC(g)= SiCn(g)+ 3Si(g) 39. 2Si(g)= Siz(g) 40. 3Si(g)= Sis(g) 41. Si(l)= Si(g) 42. Si(l)+ CO(g) =Sic($ + SiO(g)
107 = 4.2580 108 = 0.02070 K3g = 0.491 0 ffio = 0.7012 ffii = 0.1040 &2= 0.1093
Conclusion The set ofthree NGF eqs 51,52,53,inTable 4 is only one set of several wssible sets that can be used to solve this problem. It is a useful exercise to give senior students the oroblem of'settinrr un additional sets for calculatina the Sib-0 system, before KU, 45. guess Q = XsdXco; new guess Q = (old Q'NGF(Q; when O < K22, SiC(s)forms,use 0 =ffi2. 46. calculate moles SiO; Xsio = QXco 47. calculate moles SizC; Xsec = 107. p2. ( ~ s i ) ~ l Q 48. calculate moles SiCz; Xsicn = K38 .( x s ~ ~ c ) ~(Xd3 /P~. 49. calculate moles Sin = Xsiz = P . (xs~)' 50. calculate moles Sia = Xsi3 = ffio . (XS~ 51. calculate NGF(C0)= (G+Xsicn- XSO- Xsi- Xsiz- XsidEC 52. calculate NGF(Si) = (ZCl(R. Si)]0333; Si=total Si moles in gas 53. calculate NGF(Q = ((Y. ZC - R.Xco)lR. ~ s i o ) ~ ' 54. calculate R = (ZCESi);ZC =total moles of C in gas phase 55. calculate Y = (ZOESi);ZO = total 0 moles in the gas phase. 56. calculate G =ZX; y(; =calculated total moles of gas phase 'All thermodynamic data were taken from the JANAF Tables (10). ?he sum of all of ths mole fractionsof the gaseous reactants.
+.
and Si are guessed directly, but the mole fraction of SiO is linked to the mole fraction of CO by guessing the ratio Q = XsidXco. Equation 51 gives the N G F value for CO that is based on the requirement that the sum of all of the mole fractions of the gaseous reactants equals one; i.e., G = LY,,and i s a rearrangement of this conservation equation. The sum X! is chosen for the denominator i n eq 51,because when the guessed value for CO is high ZC also will be high. The N G F value for Si from eq 52,is the cube mot of XCIR.ZSi, which i s a rearrangement of the conservation equation, XCESi = R.The sum m i i s i n the denominator because its value depends on the guessed value ofXsi. The cube root is involved i n eq 52,because i t dampens down the effect of the NGF(Si) factor on the convergence of the iteration process, and because Xs; i s cubed in eq 47. The value ofNGF(Q) is obtained from eq 53 which i s based on the ratio ZOEC = YIR that is rearranged to become sensitive to the guessed values of Q a n d Xco. This factor produces a better iteration a s a square root, because i t i s sensitive to two guesses, Q and xco. The calculation in Table 5 is based on the conditions of 3000 K. ZCnSi = 2, IORSi = 2,which are obtained when two moles of carbon are reacted with one mole of silicon dioxide i n a n electric furnace. The system pressure of 0.702 atmospheres i s t h e maximum pressure under these conditions that allows all ofthe species to be in the gas phase only, a t equilibrium. Above this pressure liquid silicon forms, t h e NGF(Si) factor becomes redundant, and the value of Xsi becomes equal to K d P . But
624
Journal of Chemical Education
Literature Cited 1. Chsston, S. InAustmlion Chemisfry fisourcs Bwk:C. L. Fogkani, Ed.;RoyalAushalian Chmieai Institute,1991; Vol. 10, p54. 2. Chaston, S. In AusVolian Chemistry RosoumBook; C. LFogliani, Ed.; Royal Aushaiian Chemical Institute, 1992; Vol. 11,p145. 3. Rurso, S. 0.: Hanania, G. I. H.J. C k m . Edve 1989,66,148. 4. Cobranchi, D. P ; Eyrhg,E. M. J Chem.Educ 1991,68,40. 5. Wai, C. M.; Hutchison. S. G. J. Chem Educ 1989.66, 546. 6. KoiUloC I. M.; Einng, P. J., Eds. h 6 e on Andyliml Chemistry, 2nd d.Wiley: NewYork, 1wB;Part1, Vol.1,539. 7. Smith.R.M.; Msrtel1.A. E.C~itlml Stabilie Constants, Vol.4, InorgonieCompkxas: Flenum:New York, 1976. 8. Diek8on.A. 0.; Whitfield. M. Marine Ckem. 1981. 10.321. 9. Siuen, L. G.; Martell, A. E. Stobaity Constants; Supplement No. 1,Special Publica60" N o 25; The Chemical Society: London, 1971. 10 JANAF Thrmoeh~mleolTkbles;U.S. Government Doe. No. NSRDS-NBS37,Znd 4.
Table 5. Results of Calculations of Equilibrium Mole Fractions of Gaseous Products in the Si-C-0 System Calculation Results and Conditions temperature, T = 3000 K molar ratio of ZCWi. R = 2 total gas molecules, G = ZX = 1
43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56.
guess Xco guess Xsi guess 0 calc Xsio calcxsnc calc Xsoa calcXsi2 calcXsi3 NGF(C0) NGF(Si) NGF(Q) calcR calcY calc G
pressure, P = 0.7020 atm molar ratio ZOESi, Y=2 condensed phases: nil
I sl
2nd
3rd
4th
5th
6th
0.677 0.1460 0.1430 0.0968 0.0457 0.0281 0.0073 0.0011 0.9974 1.0081 1.0263 2.0488 2.0352 1.0020
0.6752 0.1472 0.1468 0.0991 0.0456 0.0274 0.0075 0.0011 0.9961 1.0041 1.0062 2.0246 2.0214 1.0030
0.6726 0.1478 0.1477 0.0993 0.0459 0.0274 0.0075 O.OOH 0.9980 1.0073 1.0064 2.0104 2.0072 1.0016
0.6712 0.1480 0.1486 0.0998 0.0458 0.0272 0.0076 0.0011 0.9991 1.0006 1.0019 2.0034 2.0024 1.0007
0.6706 0.1481 0.1489 0.0998 0.0458 0.0271 0.0076 0.0011 0.9997 1.0001 1.0008 2.0008 2.0004 1.0002
0.6705 0.1481 0.1490 0.0999 0.0458 0.0271 0.0076 0.0011 0.9999 1.0000 1.0001 2.0000 2.0000 1.0001
