Module - 7 Extraction of metals from halides Learning objectives Importance of halide metallurgy Naturally occurring halides and halides produced from oxidic ores Extraction of nuclear reactor metals U, Pu, Th, Zr and Be Extraction of other reactive metals Mg, alkali and alkalli earth metals R.E metals, Titanium Special importance of nuclear reactor metals and titanium in Indian context. Extraction of Uranium Ores are very low grade and complex Initially physical or chemical methods of beneficiation produce a concentrate Concentrate is treated for production of pure intermediate and for recovery of valuable by products Pure intermediate is reduced to produce metal Then there is final refining and consolidation of metal. Chemicals methods Acid leaching with various concentrations 2U 3 O 8 + 6H 2 SO 4 + (O 2 ) = 6UO 2 SO 4 +6H 2 O Also amenable to bioleaching Th 3 (PO 4 ) 4 +6H 2 SO 4 = 3Th(SO 4 ) 2 + 4H 3 PO 4 3BeO.Al 2 O 3.6SiO 2 ( After heating and quenching from C) + 6H 2 SO 4 = 3BeSO 4 + Al 2 (SO 4 ) 3 + 6SiO 2 + 6H 2 O Alkati leaching Th 3 (PO 4 ) 4 +12NaOH = 3ThO 2 + 4Na 3 PO 4 + 6H 2 O ZrSiO 4 + 4NaOH = Na 2 ZrO 3 + Na 3 SiO 3 + 2H 2 O 2U 3 O 8 + O Na 2 CO 3 + 6H 2 O = 6Na 4 UO 2 (CO 3 ) + 12NaOH Acids are stronger leaching agents Pure oxides are precipitated from leach liquors Chlorination breakdown MO 2 + C + 2Cl 2 = MCl 4 + CO 2 MO 2 + 2C + 2Cl 2 = MCl 4 + 2CO 2 If sufficient carbon is present then CO/CO 2 ratio is governed by temperature Relatively less stable oxides can be chlorinated without use of carbon. Reactions with Cl 2 and F 2 can be used to break down complex minerals to produce halides of different metals at different temperatures. Reduction of metal halides Metallothermic reduction of halides allows oxygen free operation and, therefore, superior metallic product. Choice of reduction method will depend on Thermodynamic feasibility and kinetics The heat balance Melting and boiling points of constituents Densities of metal and slag Uranium isotopes U 238 (99.28%) - Not fissionable U 235 (0.7%) Fissionable U 234 (0.005%) - Not important In nature: U 233 fissionable, potentially most important. U 238 is a Fertile material because neutron irradiation converts it to Pu 239 which is fissionable and source of far greater energy than that obtainable from fission of U 235 Nuclear reactors exploit fissionable atoms. Produced by neutron radiation of Th 232 Separate U 235 Fission Energy (Products) Neutron Separate U 238 Separate Pu 239 Energy Thermal neutronsTh 232 U 233 Energy Uranium U n U 239 Np 239 Pu min (Half life) 2.33 d. Th n Th 233 Pa 233 U min 27.4 d Np and Pu are transuranic elements. One gram of U can release energy nearly 4 x 10 7 times greater than that released by explosion of one molecule of TNT. Number of neutrons emitted by a fissile nucleus per neutron absorbed (eta value) U 233 = 2.30 0.02 U 235 = 2.06 0.02 Pu 239 = 2.03 0.02 U 233 Th fuel cycle is thus highly promising. Fission reaction: U 235 +n Fission products + neutrons + energy (Atomic products are rejected in opposite directions at extremely high velocities carrying enormous energies) (24,360 yr) Extraction of plutonium On irradiation only small amounts of U 238 are converted to Pu 239 This is extracted through extensive chemical engineering techniques. The aim will be to recover PuO 2 then convert it to PuF 4 or PuCl 3 for calciothermic reduction, using a booster reaction, e.g. Ca + I 2 = CaI 2 + heat The reaction provides additional heat. Also, CaI2 dissolves CaF 2 /CaCl 2 to form a low melting slag. Ca-reduction is done in a bomb reactor using inert atmosphere. PuF 4 +2Ca = Pu +2CaF 2 2PuCl 3 + 3Ca= 2Pu + 3CaCl 2 Extraction of thorium Ores Simple oxides ( Th,U) O 2, ThSiO 4 (Thorite) or complex oxides containing one or more of Y, Er, Ca, Nb, Ta,Fe.Ti, Ce, Zr, Pb, Sn etc in complex phosphates and silicates. Common in beach sands of India Monazite - ( Ce, La, Y, Th) PO 4 Indian monazite resources are the richest and most extensive. Separation Process for Monazite Sun dried beach sand contains 60-75% Ilmenite, 5-7% garnet, 5-6% Zircon, 2-4% Rutile, 0.5-5% Monazite, 8-28 % silica and others. After screening to remove lime shells and trashes, low intensity magnesite separator removes Ilmenite ( highly paramagnetic). Then from tailings high intensity magnetic separator recovers monazite ( weakly magnetic). Other constituents are removed by electrostatic separators or air tables. Monazite then goes for chemical treatment. In India an alkali process is followed. Zirconium Most important property is low value absorption cross-section of thermal neutrons, good corrosion resistance and high temperature mechanical strength. Zirconium alloys are used as cladding material in reactors. In a nuclear reactor, when a given mass of fuel material is undergoing fission, fast moving neutrons generated face the following possibilities 1.They may encounter additional fissile mass, producing more neutrons. 2.They may encounter a fertile atom and produce another fissile atom ( e.g. U Pu 239 ) 3.They may encounter some other atom without any useful result 4.They may escape altogether If (1) predominates then the fission is accelerated ( heat generated is removed by a coolant) The cladding element must not absorb neutrons. (4) is minimized by using moderator rods that slow down neutrons. For a given mixture of fissile and nonfissile atoms, there is a certain critical size beyond which the proportion of neutrons that escape is so reduced that the condition for a nuclear reaction to take place is attained. Zr alloys, which have low capacity for absorbing neutrons, allow the pile to be kept as small as possible. Titanium Extraction Very important metal today Very high strength to weight ratio, nearly double that of steel, corrosion resistance better than that of 18-8 stainless steel. Ti alloys retain strength even at high temperatures and show less creep. Applications : Jet engine components (45%), Air frames (25%), Missiles and spacecraft(20%) Chlorination of TiO 2 TiO 2 (s) + 2Cl 2 (g) = TiCl 4 (g) + O 2 (g) At C G 0 = -132 K Cal / mole of TiCl 2 G = 30,000 + R p p a. TiO 2 p Cl 2 p p TiCl 4 Assume that reaction proceeds i.e. G < 0 This will be possible if = 1.25 If total pressure is 1 atm then 0.3 atm p TiCl 4 Thus, without carbon, there is not much conversion. o2o2 TiCl 4 TiO 2 (s) + 2C (s) + 2Cl 2 (g) = TiCl 4 (g) + 2CO (g) G 0 = -76 K.Cal. G 0 = -76,000 + R p TiCl 4 p 2 co.a.a TiO 2 p 2 cl 2. a c 2 For G 0 = 0 1.2 x p3p3 TiCl 4 p 2 cl 2 Since + + p co = 1 atm p Cl 2 p TiCl 4 P co = 2p TiCl 4 p Cl 2 We get 1 - p TiCl 4 3 (1 - ) 3 / 3 x 14 p Cl 2 p Cl 2 2 For E q m. 1 = 3 x p Cl 2 p2p2 p2 Cl 2 p3p3 p p3p3 Since