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Mme 3514 Materials Thermodynamics

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Mme 3514 Materials Thermodynamics MME 3514 MATERIALS THERMODYNAMICS Chemical Reaction Equilibria Continued Partial oxygen pressure grid lines 표 ∆퐺 PO2 0 1 atm XO2, O2 X, O 2 T 700 °C -11 10 atm Ellingham diagram offers a simple and useful way to estimate equilibrium oxygen pressures as a function of temperature 표 For constant 푃푂2 values, ∆퐺 vs T is represented by straight lines with R ln푃푂2slope and ∆퐺표 = 0 intercept The intersections of the constant oxygen partial pressure lines and the X-XO2 equilibirum line give the equilibrium oxygen partial pressures for this reaction at various temperatures Carbon is extensively used in materials processing The two oxides of carbon, CO and CO2 are both gaseous species and it is important to know how they are distributed in an environment containing these gases Ellingham for the oxides yield another line that represent the equilibirum between CO and CO2: 표 2C 푠 + 2O2 푔 = 2CO2 푔 , ∆퐺 = −394321 − 0.84푇 퐽 표 2C s + O2 g = 2CO g , ∆퐺 = −223532 − 175.4푇 퐽 표 2CO s + O2 g = 2CO2 g , ∆퐺 = −565110 − 173.72푇 퐽 표 ∆퐺 CO/CO2 CO2 CO 978.4 K, 1 atm T Reduction of metal oxides in industrial practice involves both carbon and CO as the reducing agent since carbon in solid state does not promote high reduction rates due to small contact area with the metal oxide For example hematite is in contact with carbon in oxygen blast furnace and reacts as: Fe2O3 s + 3C s = 2Fe s + 3CO g This reaction proceeds slowly and the main reduction reaction occurs by CO: 3Fe2O3 s + CO g = 2Fe3O4 s + 3CO2 g Fe3O4 s + CO g = 3F푒푂 s + CO2 g F푒푂 s + CO g = F푒 s + CO2 g Consider the oxidation of iron, as represented in the Ellingham diagram 표 2F푒 s + O2 g = 2F푒푂 s , ∆퐺 = 푅푇 ln 푃푂2 2 표 푃 퐶푂2 2CO s + O2 g = 2CO2 g , ∆퐺 = −푅푇 ln 2 푃 퐶푂푃푂2 At equilibrium temperature both reactions are in equilibrium with the same partial oxygen pressure 2 표 표 푃 퐶푂2 The net reaction is 2Fe s + 2CO2 g = 2FeO s + 2CO s , ∆퐺 = ∆퐺 퐹푒−퐶푂= −푅푇 ln 2 푃 퐶푂 표 표 푃퐶푂2 ∆퐺 퐹푒= ∆퐺 퐶푂 − 2푅푇 ln 푃퐶푂 CO/CO2 grid lines can be related to the oxygen partial pressure in a system using the CO-CO2-O2 equilibrium 2 표 푃 퐶푂2 ∆퐺 = −RT ln 2 푃 퐶푂푃푂2 표 푃퐶푂2 ∆퐺 = RT ln 푃푂2 − 2RT ln 푃퐶푂 Partial pressure of oxygen is converted to the ratio of CO and CO2 gases as 표 푃퐶푂 RT ln 푃푂2 = ∆퐺 − 2RT ln 푃퐶푂2 ∆퐺표 = −565110 + 173.72T J 푃퐶푂 RT ln 푃푂2 = −565110 + (173.72 − 2R ln )푇 푃퐶푂2 Constant CO/CO2 ratios fall on straight lines on Ellingham diagram since RT lnPO2 is the ordinate: Intercept at -565110 Joules and slopes of 173.72-2R ln(PCO/PCO2) P CO/CO2 표 ∆퐺 PO2 0 CO/CO 2 -565 kJ P eqm T Teqm PCO/ PCO2 is constant along a straight line that passes through -565 kJ at 0 K and can be read from the ratio scale At the intersection of the CO/CO2 lines with a metal oxidation line, all metal X, XO2, O2, CO and CO2 coexist at equilibrium The metal is reduced by CO at higher PCO/CO2 than the equilibrium ratio At lower ratios CO2 is reduced while metal oxide forms Example – What is the equilibrium PCO/CO2 for CO to reduce MnO at 1000 C? 표 2푀푛 s + O2 g = 2푀푛푂 s , ∆퐺 1273 = −140 푘푐푎푙 표 2CO g + O2 g = 2CO2 g , ∆퐺 1273 = −82 푘푐푎푙 The oxygen partial pressure for the Mn oxidation reaction at equilibrium is calculated as 표 ∆퐺 = 푅푇 ln 푃푂2 −140000 = 1.987154 ∗ 1273 ∗ ln 푃푂2 −24 푃푂2 = 10 푎푡푚 The free energy change for the carbon oxidation reaction is used to calculate PCO/CO2 2 표 푃 퐶푂2 ∆퐺 = −푅푇 ln 2 푃 퐶푂푃푂2 2 푃 퐶푂2 −82000 = −1.987154 ∗ 1273 ∗ ln 2 −24 푃 퐶푂 ∗ 10 푃 퐶푂 = 105 푃퐶푂2 H2/H2O grid lines In industrial processes involving H2 as the reducing agent, the equilibrium between hydrogen and water should be taken into account as well as the equilibrium between metal and metal oxide 표 2푋 s + O2 g = 2푋푂 s , ∆퐺 푋 표 2H2 g + O2 g = 2퐻2푂 푔 , ∆퐺 퐻 표 표 2푋푂 s + 2H2 g = 2푋 s + 2퐻2푂 푔 , ∆퐺 퐻 − ∆퐺 푋 At equilibrium temperature both reactions are in equilibrium with the same partial oxygen pressure 표 ∆퐺 푋 = 푅푇 ln 푃푂2 2 표 푃 퐻2푂 푃퐻2 표 푃퐻2 ∆퐺 퐻 = −푅푇 ln 2 = 푅푇 ln 푃푂2 + 2푅푇 ln = ∆퐺 푋 + 2푅푇 ln 푃 퐻2푃푂2 푃퐻2푂 푃퐻2푂 The net reaction is 표 표 푃퐻2 ∆퐺 푋= ∆퐺 퐻 − 2푅푇 ln 푃퐻2푂 H2/H2O grid lines can be related to the oxygen partial pressure in a system using the H2-H2O-O2 equilibrium 표 2H2 g + O2 g = 2퐻2푂 푔 , ∆퐺 = −492887 + 109.62푇 퐽 2 표 푃 퐻2푂 ∆퐺 = −RT ln 2 푃 퐻2푃푂2 표 푃퐻2푂 ∆퐺 = RT ln 푃푂2 − 2RT ln 푃퐻2 Partial pressure of oxygen is converted to the ratio of H2 and H2O gases as 표 푃퐻2 RT ln 푃푂2 = ∆퐺 − 2RT ln 푃퐻2푂 ∆퐺표 = −492887 + 109.62T J 푃퐻2 RT ln 푃푂2 = −492887 + (109.62 − 2R ln )푇 푃퐻2푂 Constant H2/H2O ratios fall on straight lines on Ellingham diagram since RT lnPO2 is the ordinate: Intercept at -492887 Joules and slopes of 109.62-2R ln(PH2/PH2O) 푃퐻2 is constant along a straight line that passes through -492387 J at 0 K and can be 푃퐻2푂 read from the scale on the far right of the Ellingham diagram At the intersection of the H2/H2O lines with a metal oxidation line, all metal X, XO2, O2, H2 and H2O coexist at equilibrium The metal is reduced by H2 at higher PH2/H2O than the equilibrium ratio At lower ratios H2O is reduced while metal oxide forms PH2/H2O PCO/CO2 ∆퐺표 PO2 0 H2/H2O Peqm T Teqm Example – What is the equilibrium PH2/H2O for H2 to reduce TiO2 at 1600 C? 표 푇푖 s + O2 g = TiO2 s , ∆퐺 1873 = −136 푘푐푎푙 표 2H2 g + O2 g = 2H2O g , ∆퐺 1873 = −68 푘푐푎푙 The oxygen partial pressure for the Ti oxidation reaction at equilibrium is calculated as 표 ∆퐺 = 푅푇 ln 푃푂2 −136000 = 1.987154 ∗ 1873 ∗ ln 푃푂2 −16 푃푂2 = 10 푎푡푚 The free energy change for the hydrogen oxidation reaction is used to calculate PH2/H2O 2 표 푃 퐻2 ∆퐺 = −푅푇 ln 2 푃 퐻2푂푃푂2 푃 −68000 = −136000 + 2 ∗ 1.987154 ∗ 1873 ln 퐻2 푃퐻2푂 푃 퐻2 = 104 푃퐻2푂 Sulfidation of metals Sulphide ion has affinity for metal cations and many valuable metal ores are found in nature in sulphide forms Economically important sulfides are Acanthite Ag2S, Chalcocite Cu2S, Bornite Cu5FeS4, Galena PbS, Sphalerite ZnS, Chalcopyrite CuFeS2 Sulfidation reactions are in many ways thermodynamically similar to oxidation reactions Sulphur in gaseous form can be considered the same with oxygen and the activity of metal sulfides can be taken as unity 2Mn s + S2 g = 2Mn푆 s 표 ∆퐺 = 푅푇 ln 푃푆2 The methods for calculation of the equilibrium partial sulphur pressure at a constant temperature are the same as finding the equilibrium partial oxygen pressure Ellingham diagram showing the standard formation free energies of sulfides The equilibrium partial pressure of S2 and the ratio of PH2S/H2 can be determined graphically from the figure in the same way as described previosly for oxygen, carbon and hydrogen Example – Determine the sulphur partial pressure in equilibrium with MnS at 700 C 표 2Mn s + S2 g = 2MnS s , ∆퐺 973 = −410 푘퐽 표 ∆퐺 = 푅푇 ln 푃푆2 = −410000 = 8.314 ∗ 973 ∗ ln 푃푆2 −22 푃푆2 = 10 Reduction of metal sulphurs by H2 Just as the metal oxides are reduced by hydrogen to form metal and H2O, metal sulphides are reduced by hydrogen to form H2S and metal 표 2푀푛 s + S2 g = 2푀푛푆 s , ∆퐺 푀 표 2H2 g + S2 g = 2퐻2푆 푔 , ∆퐺 퐻푆 표 표 2푀푛푆 s + 2H2 g = 2푀푛 s + 2퐻2푆 푔 , ∆퐺 퐻푆 − ∆퐺 푀 At equilibrium temperature both reactions are in equilibrium with the same partial sulphur pressure 표 ∆퐺 푀 = 푅푇 ln 푃푆2 2 표 푃 퐻2푆 푃퐻2 표 푃퐻2 ∆퐺 퐻푆 = −푅푇 ln 2 = 푅푇 ln 푃푆2 + 2푅푇 ln = ∆퐺 푀 + 2푅푇 ln 푃 퐻2푃푆2 푃퐻2푆 푃퐻2푆 The net reaction is 표 표 푃퐻2 ∆퐺 푀= ∆퐺 퐻푆 − 2푅푇 ln 푃퐻2푆 H2/H2S grid lines can be related to the sulphur partial pressure in a system using the H2-H2S-S2 equilibrium 표 2H2 g + S2 g = 2퐻2푆 푔 , ∆퐺 = −152900 + 99.37푇 퐽 2 표 푃 퐻2푆 ∆퐺 = −RT ln 2 푃 퐻2푃푆2 표 푃퐻2푆 ∆퐺 = RT ln 푃푆2 − 2RT ln 푃퐻2 Partial pressure of sulphur is converted to the ratio of H2 and H2S gases as 표 푃퐻2 RT ln 푃푆2 = ∆퐺 − 2RT ln 푃퐻2푆 푃퐻2 RT ln 푃푆2 = −152900 + (99.37 − 2R ln )푇 푃퐻2푆 Constant H2/H2S ratios fall on straight lines on Ellingham diagram since RT lnPS2 is the ordinate: Intercept at -152900 Joules and slopes of 99.37-2R ln(PH2/PH2S) Example – What is the equilibrium PH2/H2S for H2 to reduce MnS at 700 C? 2푀푛 s + S2 g = 2푀푛푆 s 표 ∆퐺 푀푛 = −410000 퐽 = 푅푇 ln 푃푆2 −22 푃푆2 = 10 푎푡푚 푃퐻2 RT ln 푃푆2 = −152900 + (99.37 − 2R ln )973 푃퐻2푆 푃 퐻2 = 10−9 푃퐻2푆 푃 MnS is reduced to Mn at 퐻2 > 10−9 푃퐻2푆 Determining the composition of reaction system under equilibrium 푎퐴(푔) + 푏퐵(푔) + 푒퐸(푔) 푐퐶(푔) + 푑퐷(푔) + 푒퐸(푔) Consider the reacting A, B and inert E to produce C and D and inert E 푐 푑 푃퐶 푃퐷 퐾 = 푎 푏 푃퐴 푃퐵 The partial pressures of the components are expressed as a function of the total P: 푛퐴.
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