Why are Group 15 elements polyatomic
The elements of group 15
1 The elements of group 15
2 Non-metals: occurrence and representation of C & NN occurrences (examples) elementary in the air (78 vol%) lithosphere: nitrates M (N 3) (M = Na, K) biosphere: red representation (examples) a) distillation of air b) Binding of atmospheric oxygen: 4 N Cu 4 N Cu c) oxidation of ammonia NH 3 + HN 2 NH 2 d) decomposition of azides 2 NaN Na + 3 N 2 The mad scientist There has been no lack of attempts to produce other nitrogen modifications. N 6 is suspected to be in space, but like N 10, it is extremely explosive. According to calculations, however, the compound FeN 10 is a stable molecule. N N + N N N N N N N N + N N N N Fe N N N N N N N N N + N N N
3 Non-metals: Occurrence and presentation of: 1669: The Hamburg Henning Brand, discovered by reducing evaporated urine residues. 1779: Discovery of Ca 5 (4) 3 (H, F, Cl) apatites in minerals 1811: Discovery of hospholipids (lecithin) in brain fat Historical RR Glycerin 1820: First synthesis of alkyl phosphites 3 RH + H 3 4 = (R) H : Concept of rtho-, meta- and yrophosphates 1843: patenting of the fertilizer superphosphate 1929: discovery of adenosine trihosphate 1951: first 31 NMR spectra 1960: concept of seudorotation in F: Nobel Prize for the Wittig reaction h 3 = CR 2 + R 2 C = h 3 = + R 2 C = CR 2 HX phosphate The chemiluminescence of is due to the reaction of the 4 vapor with 2 / H 2. Emitting species () 2, H, etc.
4 Representation of white: The electrothermal process for the production of non-metals Apatites: Ca 5 (4) 3 (H, F, Cl) The strongly endothermic reduction of phosphorus oxides (C kj mol C) is carried out in an electric arc furnace at approx energy-intensive synthesis: approx kj per kg). 2 Ca 3 (4) Si C 6 CaSi C + 4 feed: apatite; the 4 10 part is reduced by C. F is bound as [SiF 6] 2. roduct gases: 4 is precipitated in H 2 slag: spray towers Ca is bound by Si 2
5 Occurrences (examples) As Mixed metal arsenides / sulfides: Fe [SAs] Semi-metals: Occurrence and representation of As, Sb, Bi Arsenic sulfides and oxides: As 4 S 4 (Realgar) representation (examples) a) Fe [SAs] + FeS + As b) 2 As C 4 As + 3 C 2 c) AsH 3 + As H 2 (high purity) Sb sulfides: Sb 2 S 3 (gray pearlescent), thioantimonates: M m Sb n S k (M = b, Hg, Cu, Ag) xide: Sb 2 3 (white spike luster) Metal antimonides: Ag 3 Sb (antimony silver) Solid as Sb Bi relatively rare chalcogenides: Bi 2 3, Bi 2 S 3, Bi 2 Se 3 double sulfides: bbi 2 S 4 (bs Bi 2 S 3) dignified (rarely) As S As S a) Sb 2 S 3 + Fe 2 Sb + 3 FeS b) Roasting reduction process: Sb 2 S Sb S 2 Sb C 2 Sb + 4 C a) Bi 2 S Fe 2 Bi + 3 FeS (same principle as metallothermal processes) b) Roasting reduction process Bi 2 S Bi S 2 Bi C 4 Bi + 3 C 2 Realgar S As S As gray cast gloss (stibnit)
6 Dinitrogen: N 2 N 2 only occurs in a molecular modification. According to the 8-N rule, it is three-binding. The bond in N 2 can be described by a σ bond and two orthogonal π bonds. Lewis Formula: N, N. M.p .: 210 C; Bp .: 196 C, therefore use as a coolant e.g. for vacuum apparatus and NMR magnets. N N N N N N N N N N N N N N N N N N The crystal structure of N 2 .. N N N N N N N N N N can be derived from the most cubic packing of spheres.
7 phosphorus: olymorphic and allotropic Like all other elements of the 15th group, phosphorus obeys the 8-N rule and is three-bonded in all modifications. In contrast to N 2, there is no stable modification with multiple bonds. 2 molecules ,, are only observed in the dilute gas phase at elevated temperatures. In general, the following applies: All multiple bond systems with non-metal elements from the higher periods (n> 2) are less stable than their homologues from the 2nd period. They do not appear in the element modifications under standard conditions. Molecules with E = E or E E multiple bonds can only be isolated if ligomerization or polymerization is kinetically prevented by large substituents on the atoms E (double bond rule).
8 Structures of phosphorus modifications white phosphorus Mp C Sdp C 40 C = bar white 4 red (structure unknown) stability violet phosphorus black phosphorus Top view of a layer Linking pattern in the layer: cis-decalin
9 Stwald's step rule G # 1 Modification 1 (e.g. 4) G # 2 Modification 2 (e.g. red) G # 3 Modification 3 (e.g. purple) When an element is separated from the gas phase, the modifications gradually change from the most unstable to the most stable run through. Modification 4 (e.g. black)
10 The structures of arsenic and antimony The atoms of the elements of group 15 (VEK: ns 2, np 3) also obey the 8-N rule and are without exception linked to their neighbors via three 2z-2e bonds. In addition, a lone pair of electrons is isolated on each atom. Replacing the 2z-2e bonds between the layers in Si, Ge by a pair of electrons on each element atom and offsetting the layers creates the structure of gray, metallic α-arsenic and antimony. weak contacts (As-As: 312 pm; Sb-Sb: 336 pm) between the layers produce a distorted octahedral [3 + 3] coordination for each atom (vdw-contact As-As: 370 pm) localized 2z-2e -Bondings within the As-As layers: 252 pm; Sb-Sb: 291 pm
11 Reactivity of the elements: polyanionic compounds of the nictogenic The formal step-by-step reductive degradation of an element structure can be studied particularly well with the elements of the 15 group. This is achieved experimentally by reacting the elements of the 15th group (nictogenic, n) with a reducing metal, M. Common reactions: a) x M z + y n M x (n y xz); directly from the elements (solid-state reaction). b) x [M (NH 3) n] z + + z [e (nh 3) m] + y n M x (n y xz) + (m + n) NH 3; in liquid ammonia General: Assuming that the valence electrons, z, of the metal are completely transferred to the more electronegative element, X, and that X in the olyanion is normal-valent, the number of covalent 2Z-2e bonds on each atom X in the Anion can be calculated according to: b XX = 8 VEK X. The valence electron concentration, VEK X, is calculated from the number of valence electrons of the element (e.g. nictogenic: 5) plus the (formal) charge per atom X.
12 Reactivity of the elements: polyanionic compounds of the nictogenic With the simple rule, b anion / atom = 8 VEK anion / atom, one can often correctly deduce the structure of the olyanion (nota bene: there are also cases in which the rule fails). Example 1: Ni Ni = Ni VEK = = 7 b = 1 Each atom in Ni is bound to another, i.e. Ni can be formulated as 2Ni 2+ (2) 4, which contains the diatomic (2) 4 dianion. 4 (2) 4 is iso (valence) electronic (VEK = 14) and isostructural to the two-atom molecule Cl 2.General: The structure of an olyanion in compounds of the type M x (ny xz) corresponds to that of a modification of the isovalence-electronic element of group 14 17, if the number b xx of the covalently bound neighboring atoms matches (Zintl, Klemm & Busmann). More generally: Isoelectronic atom groupings (often) take on similar structures
13 Example 2: In 3 In 3 = In 3+ + (3) 3 b = 8 (5 + 1) = 2 -Valence electrons Reactivity of the elements: polyanionic compounds of nictogenic A cyclic (3) 3 ion would formally correspond to this rule: Formal charge / atom However, a cyclic S 3 is not a stable modification of sulfur. In 3 is actually to be understood as a dimeric formula unit In 2 6 that the (6) 6 ion, which is isostructural to S 6, contains: In general, the charges in an olyanion can be distributed as follows: Every element with N valence electrons that forms 8-N bonds the formal charge 0. For each missing bond it receives a negative formal charge.
14 Reactivity of the elements: polyanionic compounds of the nictogenic The structural diversity of the olyanions (much larger than that of the elements) is due to the a) gradual reductive breakdown of the XX bonds in the element X anions with X atoms of different formal charges 0, 1, 2, etc. (Nota bene: b XX can be odd). Formally successive reductive degradation of the six-ring layer structure of the nictogens (a more reactive modification of the element is usually used for synthesis, e.g. red and not black phosphorus): black phosphorus [(6) 6] anion in In 3 Sb 0 Sb Sb 0 0 [(6) 4] Anion in Ba 3 (Ba 2 6) b = 2 for 1; b = 3 for 0 Sb Sb Sb [(Sb 6) 8] anion in Sr 2 Sb 3 (Sr 4 Sb 6) b SbSb = 2 for Sb 1; b SbSb = 1 for Sb 2 As 0 As As As [(As 4) 6] anion in Sr 3 As 4 b AsAs = 2 for As 1; b AsAs = 1 for As 2
15 Reactivity of the elements: polyanionic compounds of the nictogenic and b) due to the influence of the lattice of the countercations M z + on the structure, so that olyanions with the same formal charge at the X centers can have completely different structures (nota bene: b XX remains the same !) [(6) 4] Anion in M 4 6 M = K, Rb, Cs b = 8 (5 + 4/6) = 2.33 delocalized 10π electronic structure with partial double bond character Isolated anions with identical atoms [(6 ) 4] Anion in (Ba 2 6) b = 8 (5 + 4/6) = 2.33 four with b = 2, two with b = 3 localized σ-bonds polymeric chain structure 8 - unit of violet polymeric [(15)] Tube anion in K 15, which is reminiscent of the structure of violet phosphorus.
16 Excursion: Zintl hasen compounds with an electropositive cationic component (alkali / alkaline earth / lanthanoid) and an anionic component from a main group element of moderately high electronegativity. The anionic substructure fulfills the concept of the 8-N rule, but the compounds are not built up like salt but have a (semi) metallic character (gloss, electrical conductivity) (cf. the structures of Si, Ge, α-sn, As , Sb, Bi, in which the elements also behave normally with regard to their structures, but have (semi) metallic properties). Zintl-Line Cationic components in Zintl-hasen Anionic components in Zintl-hasen Li Be B C N Na Mg Al Si S K Ca Ga Ge As Se Rb Sr In Sn Sb Te Cs Ba Tl b Bi o element combinations intermetallic salt-like, ionic increasingly metallic
17 Zintl rabbits Classic Zintl rabbits: Na + Tl: The Tl anion lattice isoelectronic with C takes on a diamond structure. Sr 2+ (Ga) 2: The Ga anion lattice, which is isoelectronic with C, adopts a graphite structure
18 Zintl rabbits Silicides are other typical Zintl rabbits in which the topographical linkage in the anion lattice is derived from the Zintl concept, but the detailed structure depends heavily on the cation lattice. Silicides with trigonal-planar Si centers (α-thsi 2) are particularly unusual [cf. also planar (6) 4 with trigonal-planar centers]. (Si 4) 4 analogous to 4 in NaSi (Si 4) 6 in Ba 3 Si 4 example VEK X b SiSi NaSi 5 3 CaSi Ba 3 Si (Si) analogous to gray As in CaSi 2 (Si) network with trigonal-planar Si Atoms in α-thsi 2 CaSi 6 2 complicated (Si) network in α-srsi 2
19 Zintl-hasen Note: The highly charged olyanions are not stable in isolated form, but only in contact (in an electric field) with the countercation lattice. In Zintl rabbits in particular, the specification of charges in the anion lattice is a pure formalism since the bond between anions and cations is not purely ionic. The highly charged anions in the mostly insoluble Zintl rabbits are strongly reducing and cannot be dissolved out of the solid structure in intact molecular form. Exceptions: Some binary (composed of two elements) M m X n cage compounds (cluster = grouping of three identical atoms with direct XX bonds) can be brought into solution by complexing the cation M with a crown ether as a salt and / or by reaction with a Electrophilically converted into a neutral molecular compound. soluble in ethereal solvents 3 N N Na + H + (7) 3 7 H 3 molecular olyphosphane
20 element halides: group 15 selected nitrogen-halogen compounds: NX 3 (X = F, Cl), N 2 F 4, N 2 F 2 NF 3: f H 0 (298K) = kj mol -1 stable compound NCl 3: f H 0 (298K) = kj mol -1 explosive! N X X X Cu, 670K 2 NF 3 N 2 F 4 + CuF 2 F F F F F F F F
21 Element halides: Group 15 phosphorus-halogen compounds: X 3 (X = F, Cl, Br, I) X 5 (X = F, Cl, Br), I 5 is not known. F 158 pm 152 pm F F Axial F F Equatorial X d α X X X d (pm) α () F Cl Br I In solution, F 5 is a fluctuating molecule: Berry seudorotation
22 Element halides: Group 15 F 5 is a strong Lewis acid and forms stable complexes with many donors such as amines or ethers. The hexafluorophosphate ion, [F 6] -, is produced by the reaction of H 3 4 with concentrated HF. [F 6] - is isoelectronic to [SiF 6] 2-: [F 6] - and [SiF 6] 2- F 5 and [SiF 5] -
23 Group 15: Nitrogen-Oxygen Compounds General Chemistry - Part Inorganic Chemistry II: Nitrogen-Oxygen Compounds xide N 2 NN 2 N 2 2 N 2 3 N 2 4 N 2 5 Acids H 2 N 2 2 HN 2 HN 3 Hypo-nitrous Acid Nitrous Acid Nitric Acid
24 Group 15: Nitrogen-Oxygen Compounds [NH 4] N 3 H = -124 kj / mol N H 2 Careful heating to 200 C, at T> 300 C explosive decomposition occurs! 2 [NH 4] N 3 = 2 N H 2 Explosion of Toulouse Explosion of ppau September 21st
25 Group 15: Nitrogen-oxygen compounds 0 +1 Isoelectronic and isostatic to C 2 and N 3 -:: NN:: 1:::: NN: 1 +1:: NN Can also be understood as an N 2 complex . The decomposition> 600 C into the elements is exothermic (H = -82 kj / mol) but kinetically inhibited. N 2 acts as an oxygen atom transfer (oxygenation) reagent although it does not maintain breathing. It is used as a loosening agent (E 942) in whipped cream and ice cream (easily soluble in fats under pressure). In low concentrations, N 2 provokes convulsive laughter and has a weakly numbing effect: Due to the poor solubility, a safe, sufficiently deep anesthesia can only be achieved with very high concentrations and after a relatively long time. 20% oxygen is added to the laughing gas.
26 nitrogen oxides C t 4 NH H = -906 kj / mol stwald method + 3-1 reduction of nitrites: +2 reduction of nitrates: 4 N + 6H 2 KN 2 + KI + H 2 S 4 N + K 2 S 4 + HI HN Cu 3 Cu (N 3) N + 12H 2 0
27 nitrogen oxides N H = + 180 kj / mol 2 N thunderstorms
28 Nitrogen oxides Today's nitric acid production is based on catalytic ammonia combustion ("stwald process") Wilhelm stwald (I) 4NH N 2 + 6H 2 (g) ΔH 0 = kj mol -1 (II) 4NH N + 6H 2 (g) ΔH 0 = kj mol -1 reaction (I) thermodynamically more favorable T = C, t-rh nets, residence time in the ms range Nobel reis in Chemie 1909 reactor diameter: 1-3 m woven or knitted nets (meshes / cm 2) wire diameter up to 0.1 mm
29 Nitrogen oxides π * The highest occupied M (SM) is a π * -rbital σ 1 N-lone pair π σ * σ 2 N - π-bonds bond order B .. number of electrons in binding rbital 1 -lone pair minus number of electrons in antibonding rbitals divided by the number of binding partners (8 3): 2 = 2.5 N 1 N - σ-bond
30 Nitrogen oxides N forms a dimer with a π * π * bond: NNN is easily oxidized (I = 9.25 ev; compare N 2: 15.6 ev) and forms many end-on bound N complexes in which N is 1e - or 3e - Ligand works: [Fe (H 2) 6] 2+ + N [Fe (H 2) 5 (N)] 2+ ring sample as analytical evidence for Nx H 2 N is considered a 3e donor ligand. N forms stable nitrosylium salts: 2010 + X - (X = HS 4, BF 4, Cl 4)
31 N reacts with 2 spontaneously to form N 2: 2 N N 2> 150 C N 2 decomposes again from 150 C to N + 2; at 600 C the decomposition is> 99% metal nitrates decompose to xides, N 2 and 2 (oxidizing agents in solid rockets): b (n 3) C b + 2 NNNN::: 17 e::: -11 C: colorless crystals 27 C: 80% N 2 4/20% NC: 10% N 2 4/90% NNNNN Blue liquid
32 Nitrogen oxides A mixture of N 2 4 and N 2 H 4 ("Aerozin-50") was used as rocket fuel for the lunar lander in the American "Apollo" program: When hydrazine is mixed with N 2 4, a strongly exothermic reaction occurs which causes spontaneous combustion and combustion with red flame on: NN 2 H 4 = 3 NH 2 H 0 = kj Also with newer rockets, e.g. those of the type Titan 23G (last start), N 2 4 / Aerozin-50 is used as fuel.
33 Nitrogen oxides 2 - N nitrites + 2e N 2 is the anhydride of nitric acid & nitric acid HN 2 is unstable and disproportionate N +3 H 2 HN 3 + HN / 3 HN 3 + 2/3 N + 1/3 H 2 N 2 -2e 2 2/3 2 2 N 2 + H HN 3 + N nitrylium ion: (cf. C 2, N 3-) [N 2] + [HS 4] is the active agent of the nitrating acid HN 3 / H 2 S 4
34 Nitrogen oxides N 2 3 H 2 2 NH N Fe / Fe Haber-Bosch t stwald 2 2 N 2 H 2, HN 3 The annual production of HN 3 is approx. 30x10 6 tons [NH 4] N 3 as fertilizer (80%) Plastics (cyclohexanone) (5-10%) Nitrated aromatics and org. Nitro compounds (5-10%) explosives: KN 3 / S / C = black powder Al / NaN 3 / methyl methacrylate / benzene = incendiary agents Mg / NaN 3 = hotoblitz Mg / Sr (N 3) 2 / chlorinated rubber = colored torches
35 Nitrogen oxides The main air pollutants are: C, N and hydrocarbons from combustion processes; approx. 70% from KW / truck traffic C m H n + (m + n / 4) 2 m C 2 + n / 2 H 2 CC 2 N + C 0.5 N 2 + C 2 Lambda probe λ = catalyst t / rh / d Supplied 2 2 Requirement with complete combustion ptimal λ value 1 Zr 2 is a -conductor Zr Zr 2
36 phosphorus-oxygen compounds oxidation of phosphorus Both white and red phosphorus react strongly exothermically (sometimes explosively) with oxidants (oxygenating agents) H 0 = KJ mol -1 phosphorus must be stored under water.Fires with white phosphorus are difficult to extinguish because they will ignite again after the water has evaporated. In the dark, white phosphorus glows by itself. This phenomenon is based on chemiluminescence that occurs when the 4 6 formed on the surface is converted into the more stable 4 10. During this oxidation, energy is given off in the form of heat and light.
37 phosphorus chemistry Phosphorus forms numerous xo acids with the following structural principles: 1. All atoms are tetrahedral and have at least one = group. 2. All xo-acids contain at least one (acidic) -H group 3. Some compounds contain -H groups, which cannot be ionized 4. Chain formation takes place either via - bridges or directly via - bonds. H H + H H H H
38 H H H H H H H H H H H H H H H H H H H H phosphinic acid phosphonic acid (rtho) phosphoric acid diphosphoric acid (yrophosphoric acid) diphosphonic acid hypophosphorous acid phosphorus chemistry
39 hosphorchemie phosphoric acid: A) H 2 4 H 3 4 furnace process Concentrated pure acid for medical purposes and food B) Ca 5 (4) 3 F + H 2 S 4 3 H CaS 4 2H 2 + HF fluorapatite 10 H 2 wet process 30 70 % Acid for technical purposes (fertilizer, metal processing) Anhydrous phosphoric acid forms colorless hygroscopic crystals. The commercially available concentrated phosphoric acid is an 85% solution. With the exception of its caustic effect, it is not harmful to health: the human body itself contains phosphoric acid or its derivatives (AD, AT, ...). Phosphoric acid is used as a preservative in food (E338). 1 liter of Coca Cola contains around 0.5 g of pure phosphoric acid.
40 hosphorchemie 9.5 ph 4.5 NaH 2 4 Na 2 H 4 Na 3 4 pk 1 s = 2.15 pk 2 s = 7.20 pk 3 s = 12.4 Base [ml] M Na 2 H 4 / M KH 2 4 is used for human blood serum (ph 25 C = 7.413) Dihydrogen phosphates are generally readily soluble in water. The concentration of [H 2 4] is maximum at pH = 4.7. Hydrogen phosphates [H 4] 2 (maximum concentration at pH = 9.75) and phosphates  3 (noticeable concentrations at pH> 12) are generally sparingly soluble (with the exception of alkali metal salts). Many hydrogen phosphates and phosphates dissolve in dilute mineral acids because they are protonated to dihydrogen phosphates.
41 hosphorchemie hosphato - molybdate hosphato-metallate [Mo] tetrahedron 2 [Mo 6 21] H + + H NH 4 + [NH 4] 3 [(Mo)] 6 H 2 Qualitative detection of phosphate Canary yellow precipitate of the complex compound triammonium dodecamolybdate phosphate-6hydrate.
42 phosphor chemistry apatite Ca 5 (4) 3 (H, F) framework mineral (bones, teeth)
43 hosphorchemie condensation to olyphosphoric acids H H H H H H H 2 + H H H H H H H 2 H H H H H corner-linked tetrahedra H 2 3 [H 3] n H 2 4 H n + 2 n 3n + 4
44 hosphorchemie 2 MH 2 4 M 2 HH 2 2 M 2 H 4 MH 2 NaH Na 2 H 4 Na n M 2 + n / n M 3 diphosphates (yrophosphates) (foods, gels) sodium triphosphate (water softeners in detergents) chain polyphosphates linear metaphosphates HELICES
45 phosphor chemistry fertilizers (> 80% of the degraded natural calcium phosphates): 2 Ca 5 (4) 3 F + 7 H 2 S 4 + H 2 7 CaS Ca (H 2 4) 2 H HF Ca 5 (4) 3 F + 7 HH 2 superphosphate 5 Ca (H 2 4) 2 H 2 + HF triple superphosphate (contains 3% by weight of soluble 2 5) (NH 4) 2 (H 4) is used as a readily soluble, nitrogenous plant fertilizer in the stage Water treatment Flocculation of colloids Detergents Softeners Lubricant additives Use of phosphates Mineral flotation Solution extraction for metal extraction Food additives Biological olarity as acids Abrasives in toothpaste Fertilizers Pet food Flame retardants for cellulose Propellants in baking powders Alcohol-free beverages Metal coatings Catalysts Water softening Detergents and oils for metals
46 Redox chemistry using nitrogen compounds as an example Nitrogen compounds occur in 10 different oxidation levels: 3, 2, 1, 1/3, 0, +1, +2, +3, +4, +5. Their relative stabilities to each other can be estimated with the help of the standard reduction potentials. Nernst equation for the reduction reaction: x + e Red E = E n lg [Red] [x] Nernst equation for the xidation reaction: Red x + e E = E n lg [x] [Red] with [A] a ( a)
47 General Chemistry - Part Inorganic Chemistry II: Redox chemistry of nitrogen compounds: Thermodynamics G = nf E Redox couple x + e Red Red.potentia l E red [V] Partial reaction G Comment Li + / Li 3.04 Li + + e Li + Red.mittel N 2 / HN ½N 2 + H + + e HN 3 + reducing agent N 2 / N 2 HNH + + 4e N 2 H + 5 N 3 / HN NH + + 2e HN 2 + H 2 HN 2 / N HN 2 + H + + e N + H 2 NH 3 H + / NH NH 3 HH + + 2e NH H 2 + reducing agent x. Agent x. Agent x. Agent HN 3 / NH HN H + + 2e x. Agent NH N 2 F 2 / F ½F 2 + e F x.average
48 Hydroxylamine as a redoxamphoter compound Many nitrogen compounds are redoxamphoter. Example: Hydroxylamine preparation by hydrogenation of N 2 N + 3 H 2 t, d 2 NH 2 HH 2 S 4 [NH 3 H] 2 S 4 Hydroxylamine: pure (melting point 32 C, colorless) explosive NH 3, N 2, H 2 more basic than ammonia: pk b (8.2) 49 Example 1: Reduction: Disproportionation NH 2 H 2 NH 3 + N H 2 2 NH 2 H + 4H + + 4e 2 NH H 2; E red = V xidation: 2 NH 2 H N 2 + 4H e + H 2; E ox = V G red = () = kj mol 1 G ox = () = 19.3 kj mol 1 G R = G red + G ox = kj / mol 50 Disproportionation Example 2: Reduction: 3 HN 2 2 N + HN 3 + H HN 2 + 2H + + 2e 2 N + 2H 2; E red = V xidation: HN 2 + H 2 HN 3 + 2H e; E ox = 0.94 VG red = () = kj mol 1 G ox = () = kj mol 1 GR = G red + G ox = 8.1 kj mol 1 The reaction is slightly exothermic, but N escapes and the reaction is completely dependent on the right side moved. 51 Disproportionation General: Disproportionation occurs exothermically if the (tabulated) standard (reduction) potential of the reduction reaction is more positive than the (tabulated) standard (reduction) potential of the oxidation reaction. 2 (A) X (A +) X n + (A) X m + e (A) X m E red (A / A) 2 (A) X (A +) X ne E ox (A + / A) = E ox (A + / A) 52 Reduction: xidation: comproportionation: / 3 N 2 H HN 2 HN 3 + H H 2 3 HN e + 10 H + HN H 2; E red = V G red = 10F E = 961 kj / mol 3 N 2 H HN e + 13 H +; E ox = VG ox = 10F E = +330 kj / mol GR = G red + G ox = 630 kj / mol General: Comproportioning is possible if the standard potential from the highest to the middle oxidation level is more positive than the standard potential of the lowest to the middle xidation level: (A +) X n + (A) X m 2 (A) X + e (A +) X n E red (A + / A) 2 (A) X (A) Xm e E ox (A / A) 53 Calculation of redox potentials through thermodynamic cycle processes: roblem: Often the potentials of the redox pairs involved in a reaction are not tabulated, e.g. the potentials of the redox pairs N 2 H 5 + / HN 3 and HN 2 / HN 3 in the comproportionation reaction, N 2 H HN 2 HN 3 + H H 2, are not available. However. they can be calculated from known redox potentials using cyclic processes: Example 1: The potentials E (N 2 / N 2) and E (N 2 / [NH 3 H] +) are known, we are looking for E (N 2 / [NH 3 H ] +): E red V +1 N 2 0 N 2 + 2e - + 2e - + 4e - E red 3? VE red V -1 2NH 3 H + G 3 = G 1 + G 2 with G = nf E follows: n 3 FE 3 = n 1 FE 1 + (n 2 FE 2) n 3 FE 3 = F (n 1 E 1 + n 2 E 2) n 3 E 3 = n 1 E 1 + n 2 E 2 4 E 3 = (2 1.77) + (2 1.87) = = 0.2 VE 3 = 0.05 V 54 Thermodynamic cycle processes: Example 2: We are now looking for E (HN 3 / N 2 H 5+) E ox V 1/3 3/2 N 2 e 0 +6 e +5 e 2 HN 3 3/2 N 2 H 5 + E red 3? V E red V 5 E 3 = E E 2 = = 1.71 V E 3 = V 55 Thermodynamic cycle processes: finally one obtains the potential E (HN 2 / HN 3) according to: E red V + 6e E red V / 2 N 2 +3 e 3/2 N 2 + ee 1/3 3 HN 2 HN 3 E red 4? VE red V 10 E 4 = 6 EE 2 + E 3 = = 9.96 VE 4 = V Thus the comproportionation N 2 H HN 2 HN 3 + HH 2 can be estimated thermodynamically (GR = 630 kj / mol, vide supra p. 12 ), even if the potentials of the partial reactions were initially unknown.
49 Example 1: Reduction: Disproportionation NH 2 H 2 NH 3 + N H 2 2 NH 2 H + 4H + + 4e 2 NH H 2; E red = V xidation: 2 NH 2 H N 2 + 4H e + H 2; E ox = V G red = () = kj mol 1 G ox = () = 19.3 kj mol 1 G R = G red + G ox = kj / mol
50 Disproportionation Example 2: Reduction: 3 HN 2 2 N + HN 3 + H HN 2 + 2H + + 2e 2 N + 2H 2; E red = V xidation: HN 2 + H 2 HN 3 + 2H e; E ox = 0.94 VG red = () = kj mol 1 G ox = () = kj mol 1 GR = G red + G ox = 8.1 kj mol 1 The reaction is slightly exothermic, but N escapes and the reaction is completely dependent on the right side moved.
51 Disproportionation General: Disproportionation occurs exothermically if the (tabulated) standard (reduction) potential of the reduction reaction is more positive than the (tabulated) standard (reduction) potential of the oxidation reaction. 2 (A) X (A +) X n + (A) X m + e (A) X m E red (A / A) 2 (A) X (A +) X ne E ox (A + / A) = E ox (A + / A)
52 Reduction: xidation: comproportionation: / 3 N 2 H HN 2 HN 3 + H H 2 3 HN e + 10 H + HN H 2; E red = V G red = 10F E = 961 kj / mol 3 N 2 H HN e + 13 H +; E ox = VG ox = 10F E = +330 kj / mol GR = G red + G ox = 630 kj / mol General: Comproportioning is possible if the standard potential from the highest to the middle oxidation level is more positive than the standard potential of the lowest to the middle xidation level: (A +) X n + (A) X m 2 (A) X + e (A +) X n E red (A + / A) 2 (A) X (A) Xm e E ox (A / A)
53 Calculation of redox potentials through thermodynamic cycle processes: roblem: Often the potentials of the redox pairs involved in a reaction are not tabulated, e.g. the potentials of the redox pairs N 2 H 5 + / HN 3 and HN 2 / HN 3 in the comproportionation reaction, N 2 H HN 2 HN 3 + H H 2, are not available. However. they can be calculated from known redox potentials using cyclic processes: Example 1: The potentials E (N 2 / N 2) and E (N 2 / [NH 3 H] +) are known, we are looking for E (N 2 / [NH 3 H ] +): E red V +1 N 2 0 N 2 + 2e - + 2e - + 4e - E red 3? VE red V -1 2NH 3 H + G 3 = G 1 + G 2 with G = nf E follows: n 3 FE 3 = n 1 FE 1 + (n 2 FE 2) n 3 FE 3 = F (n 1 E 1 + n 2 E 2) n 3 E 3 = n 1 E 1 + n 2 E 2 4 E 3 = (2 1.77) + (2 1.87) = = 0.2 VE 3 = 0.05 V
54 Thermodynamic cycle processes: Example 2: We are now looking for E (HN 3 / N 2 H 5+) E ox V 1/3 3/2 N 2 e 0 +6 e +5 e 2 HN 3 3/2 N 2 H 5 + E red 3? V E red V 5 E 3 = E E 2 = = 1.71 V E 3 = V
55 Thermodynamic cycle processes: finally one obtains the potential E (HN 2 / HN 3) according to: E red V + 6e E red V / 2 N 2 +3 e 3/2 N 2 + ee 1/3 3 HN 2 HN 3 E red 4? VE red V 10 E 4 = 6 EE 2 + E 3 = = 9.96 VE 4 = V Thus the comproportionation N 2 H HN 2 HN 3 + HH 2 can be estimated thermodynamically (GR = 630 kj / mol, vide supra p. 12 ), even if the potentials of the partial reactions were initially unknown.
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