Pourbaix基础理论及计算

1、1THE GEOCHEMISTRY OF NATURAL WATERSREDOX REACTIONS AND PROCESSES - IICHAPTER 5 - Kehew (2001)N-O2-H2O SYSTEM2N-O2-H2O SYSTEMSpeciesGr (kJ mol-1)SpeciesGr (kJ mol-1)NH4+-79.4NO3-110.8NH30-26.5NO2-37.2N20H2O(l)-237.13pHNH4+NH30N2NO2-NO3-Preliminary mapping of species in pe-pH space for the N-O2-H2O sy

2、stem. More oxidized species are at high pe and more reduced species at low pe. Nitrite (NO2-) is a metastable species.4NH30/NH4+ BOUNDARYAs before, we write a reaction between the speciesNH3 + H+ NH4+By definition, the boundary between these two species is wheresooNHHNHaaaK34oNHNHaa34KpHlog5Gr = GfN

3、H4+ - GfNH3oGr = (-79.4) - (-26.5) = -52.9 kJ mol-1So this is a vertical line at pH = 9.27.27. 9)K)(298.15molKJ314. 8(303. 2molJ900,52log111K27. 9pH6pH02468101214pe-12-8-4048121620T = 25oCpH2 = 1 atmpO2 = 1 atmO2H2OH2H2ONH4+NH30We plot the first boundary from upper to lower water stability limit, be

4、cause at this point, we do not know where the other boundaries will intersect it.7N2(g)/NH4+ BOUNDARYWe write the reactionN2 + 4H+ + 3e- NH4+To plot this boundary, we have to fix both Naq m NH4+ and pN2. For Naq we choose 10-3 mol L-1, which is near the drinking water standard for nitrate nitrogen.

5、For pN2 we choose the atmospheric value of 0.77 atm.212443NHeNHpaaaK24log34loglog21NNHppepHaK8Gr = GfNH4+ - GfN2Gr = (-79.4) - (0) = -79.4 kJ mol-191.13)K)(298.15molKJ314. 8(303. 2molJ400,79log111K)11. 0(34391.1321pepHpHpe486.163pHpe3462. 59pH02468101214pe-12-8-4048121620T = 25oCpH2 = 1 atmpO2 = 1 a

6、tmO2H2OH2H2ONH4+NH30N2(g)Naq = 10-3 mol L-1pN2 = 0.77 atmThe N2/NH4+ boundary intersects the NH4+/NH3 boundary at pe = -6.78 and pH = 9.27. The field of NH4+ is now totally enclosed by boundaries intersecting at angles 180. The next logical boundary to calculate is the N2/NH3 boundary. 10N2(g)/NH3o

7、BOUNDARYWe now write the reaction N2 + 3H+ + 3e- NH3We choose Naq m NH3 = 10-3 mol L-1 and pN2 = 0.77 atm as before.212333NHeNHpaaaKo23log33loglog21NNHppepHaKo11Gr = GfNH3 - GfN2Gr = (-26.5) - (0) = -26.5 kJ mol-164. 4)K)(298.15molKJ314. 8(303. 2molJ500,26log111K)11. 0(33364. 421pepHpHpe356. 73pHpe5

8、3. 212pH02468101214pe-12-8-4048121620T = 25oCpH2 = 1 atmpO2 = 1 atmO2H2OH2H2ONH4+NH30N2(g)Naq = 10-3 mol L-1pN2 = 0.77 atmNow the NH3 field is totally enclosed. We would suspect that the next boundary to calculate is the N2(g)/NO3- boundary.13N2(g)/NO3- BOUNDARYStarting with the reactionNO3- + 6H+ +

9、 5e- N2 + 3H2OTo be consistent, we choose Naq m NO3- = 10-3 mol L-1 and pN2 = 0.77 atm as before.321265NOHeNaaapK23log56loglog21NNOppepHaK14Gr = 3GfH3O + GfN2 - GfNO3- Gr = 3(-237.1) + (0) - (-110.8) = -600.5 kJ mol-12 .105)K)(298.15molKJ314. 8(303. 2molJ500,600log111KpHpe62 .1025pHpe5645.20)11. 0(5

10、632 .10521pepH15pH02468101214pe-12-8-4048121620T = 25oCpH2 = 1 atmpO2 = 1 atmO2H2OH2H2ONH4+NH30N2(g)Naq = 10-3 mol L-1pN2 = 0.77 atmNO3-Final Eh-pH diagram for the N-O2-H2O system. Note that NO3- should be present in significant quantities only in waters containing free oxygen. Ammonium ion and ammo

11、nia will be present only in very reducing waters.16ANY WATER CONTAINING SIGNIFICANT Fe2+ SHOULD REDUCE NO3-!z We can see this from the pe-pH diagrams.z We can also see this from a simple calculation. We first write the following reaction5Fe2+ + NO3- + 6H+ 5Fe3+ + N2 + 3H2OGr = 3GfH3O + 5GfFe3+ - GfN

12、O3- - 5GfFe2+ Gr = 3(-237.1) + 5(-16.7) - (-110.8) - 5(-90.0) = -234.0 kJ mol-1565233212FeNOHFeNaaaapK17Let us assume that for a given water, the pH was measured to be 6, pN2 = 0.77 atm, and99.40)K)(298.15molKJ314. 8(303. 2molJ000,234log111K21023FeFeaapHaaapKFeFeNON6log5loglog23321215)6(610log599.40

13、log23212NONap-116 Lmol 1077. 83NOa18NO2-/NO3- BOUNDARYDenitrification of NO3- to N2 proceeds via several intermediate steps, nitrite (NO2-) being the first intermediate. We next calculate the NO2-/NO3- boundary.NO3- + 2H+ + 2e- NO2- + H2OWe assume a NO3- = a NO2-. 3222NOHeNOaaaaKpepHaaKNONO22loglog3

14、219Gr = GfH2O + GfNO2- - GfNO3- Gr = (-237.1) + (-37.2) - (-110.8) = -163.5 kJ mol-164.28)K)(298.15molKJ314. 8(303. 2molJ500,163log111KpHpe264.282pHpe32.14pepH2264.2820pH02468101214pe-12-8-4048121620T = 25oCpH2 = 1 atmpO2 = 1 atmO2H2OH2H2ONH4+NH30N2(g)Naq = 10-3 mol L-1pN2 = 0.77 atmNO3-NO3-NO2-The dotted line is the NO3-/NO2- boundary. It is drawn as a dotted line because it is metastable. We can see this because NO3- is reduced at the NO3-/N2(g)