北京科技大学-冶金物理化学-第三章---作业3

1P52例1-3-7熔渣组成为组元CaOSiO2MnOMgOFeOP2O5w(i)/%42.6819.348.8414.9712.022.15试用分子理论计算FeO,CaO,MnO的活度及活度系数。在1873K测得与此渣平衡的钢液中[O]=0.058%。试确定计算FeO的活度的正确性。解：1)分子理论取100g渣，计算其中各简单氧化物分子的物质的量12.020.16772FeOnmol；14.970.37440MgOnmol；8.840.14171MnOnmol；219.340.37460SiOnmol；42.680.76256CaOnmol；252.150.015142POnmol;设熔渣中存在的复杂氧化物;在分子理论的假设下，熔渣的结构有4种组元：简单氧化物FeO;RO(CaO,MnO,MgO三物之和);复杂氧化物2252,4CaOSiOCaOPO.其中，酸性氧化物全部与碱性氧化物形成复杂氧化物。2220.374ROSiOSiOnnmol;252540.015ROPOPOnnmol;0.167FeOnmol过剩碱22524()240.7620.1250.37420.33240.0150.537CaOMnOMgOCaOSiOCaOPOROnnnnnnmol剩余22524()0.1670.5370.3220.0151.041iFeOCaOSiOCaOPOROnnnnnmol剩余;aaaa0.7620.732,11.041COCOCOCOinaxn;nnnn0.1250.120,11.041MOMOMOMOinaxn;0.1670.160,11.041FeOFeOFeOFeOinaxn2)完全离子理论0.079FeOa,修正后0.166FeOa3)由反应()[][]FeOOFe得[%O]OFeOLa,在1873K时实验得到0.23OL,所以[%O]0.0580.2520.23FeOOaL2P34055已知炼钢炉渣组成如下：组元CaOSiO2MnOMgOFeOP2O5w(i)/%42.6819.348.8414.9712.032.15试用完全离子溶液模型计算1600℃时炉渣中(CaO)、(MnO)、(FeO)的活度及活度系数。在1600℃时测得与此平衡的钢液中w[O]=0.058%,试确定此模型计算(FeO)活度的精确度。假设渣中络阴离子按下列反应形成：44222SiOOSiO;3425223POOOP;已知氧在渣-钢之间分配系数与温度的关系式为734.26320lglg%TOwaFeOLO，离子活度系数修正公式为：17.0)(53.1lg4422SiOxOFe解：（2）据实验所得氧在渣钢种分配比T=1873K时，%63206320lglg2.7342.7340.6401873FeOOaLwOT则：0.640%100.0584.3650.25317FeOawO（3）根据修正公式有：2244lg1.53()0.1710.3220..530.170.3241.09009302FeOxSiO则：220.324102.11FeO22222.110.1170.6770.167FeOFeOFeOaxx3解：（1）据完全离子溶液模型的理论,熔渣的结构为：阳离子：2222,,,CaFeMgMn;阴离子：23444,PO,OSiO取100g炉渣，各物质的摩尔数：42.680.76256CaOnmol;219.340.32260SiOnmol；8.840.12571MnOnmol；14.970.37440MgOnmol；12.030.16772FeOnmol;252.150.0151142POnmol；各阳离子的摩尔数：2+CaOCan=n=0.762mol;2+MgOMgn=n=0.374mol;?2+MnOMnn=n=0.125mol;2+FeOFen=n=0.167mol1.4282+2+2+2+CaMgMnFeinnnn0.7620.3740.1250.167olnm由44222SiOOSiO；3425223POOOP得出复合阴离子的摩尔数为：20.03024-3-22544SiOPOSiOPOn=n=0.322moln=nmol；简单阴离子摩尔数：2225231.42820.32230.01510.7387SiOPOOinnnnmol阴离子总摩尔数：243440.73870.3220.03021.0909jOSiOPOnnnnmol则220.1670.1171.428FeFeinxn;220.7620.5341.428CaCainxn;220.1250.0881.428MnMninxn;220.73870.6771.0909OOjnxn所以:22 0.1170.6770.0792FeOFeOaxx;220.5340.6770.362CaOCaOnxx;220.0880.6770.0596MnOMnOaxx炉渣组成各物质的总的物质的量2251.4280.3220.01511.7651CaOMnOMgOFeOSiOPOnnnnnnnmol各物质的浓度0.7620.4321.7651CaOCaOnxn；0.1250.07081.7651MnOMnOnxn；0.1670.09461.7651FeOFeOnxn；40.3620.8380.432CaOCaOCaOax；0.05960.8420.0708MnOMnOMnOax；0.07920.8370.0946FeOFeOFeOax；559已知炉渣组成如下：组元CaOSiO2MnOMgOFeOAl2O3CaF2W(i)/%47180.2141.8514试计算温度为1873K时炉渣的光学碱度及硫化物容量。解：取100g炉渣，CaOSiO2MnOMgOFeOAl2O3CaF2W%47180.2141.8514M566071407210278n(氧化物i的物质的量）0.8392860.300000.0028170.3500.0250.049020.179487ix(氧化物i在熔渣中的摩尔分数）0.4807980.171860.0016140.2005030.0143220.0280820.102822im(氧化物i中的氧原子数)1211131ixm0.4807980.343720.0016140.2005030.0143220.0842450.102822iiiiixmxxm(氧化物i中阳离子的摩尔当量)0.3915220.2798970.0013140.1632730.0116620.0686020.08373i(氧化物i的光学碱度)10.480.590.780.510.6050.2iix0.3915220.134350.0007750.1273530.0059480.0415040.016746解：熔渣的光学碱度0.7014iix226905464022690546400.7014lg43.625.243.60.701425.22.9661873SCT炉渣硫化物容量31.0810SC660已知高炉渣组成如下：组元CaOSiO2MgOAl2O3W(i)/%43.4635.322.7618.48试碱度法和光学碱度法分别计算温度为1673K时炉渣的硫化物容量。解：一、碱度法已知高炉渣系的SC与碱度有如下关系lg5.571.39SCR取100g高炉渣，其碱度考虑MgO对CaO的碱当量以及Al2O3对SiO2的酸当量。232223(CaO)(MgO)(CaO)(MgO)(AlO)(SiO)(SiO)(AlO)143.4612.76125624021.24884135.32118.48136031023nnnnxxRnnxxnn则5.571.395.571.391.248843.83441010101.46510RSC或者：2231.79(CaO)1.24(MgO)6911lg1.351.6491.66(SiO)0.33(AlO)1.790.43461.240.27669111.351.6491.660.35320.330.184816733.44367SwwCwwT则3.443674103.60010SC参考张家芸《冶金物理化学》P787二、光学碱度取100g炉渣，组元CaOSiO2MgOAl2O3W(i)/%43.4635.322.7618.48M566040102n(氧化物i的物质的量）0.7760710.5886670.0690.1811761.614915ix(氧化物i在熔渣中的摩尔分数）0.4805650.3645190.0427270.11219im(氧化物i中的氧原子数)1213ixm0.4805650.7290380.0427270.3365691.588898iiiiixmxxm(氧化物i中阳离子的摩尔当量)0.3024520.4588320.0268910.211825i(氧化物i的光学碱度)10.480.780.605iix0.3024520.2202390.0209750.1281540.67182解：熔渣的光学碱度0.67182iix226905464022690546400.67182lg43.625.243.60.6718225.24.287751673SCT炉渣硫化物容量55.155210SC861已知炉渣组成如下：组元CaOSiO2MnOMgOFeOAl2O3P2O5W(i)/%4520771632试用光学碱度法计算温度为1873K时炉渣的磷酸盐容量。解：取100g炉渣，CaOSiO2MnOMgOFeOAl2O3P2O5W%4520771632M5660714072102142n(氧化物i的物质的量）0.8035710.333330.0985920.17500.2222220.0294120.014085ix(氧化物i在熔渣中的摩尔分数）0.4793960.1988610.0588180.1044020.1325740.0175470.008403im(氧化物i中的氧原子数)1211135ixm0.4793960.3977220.0588180.1044020.1325740.052640.042013iiiiixmxxm(氧化物i中阳离子的摩尔当量)0.3782030.3137680.0464020.0823640.1045890.0415280.033145i(氧化物i的光学碱度)10.480.590.780.510.6050.4iix0.3782030.1506090.0273770.0642440.0533410.0251250.013258熔渣的光学碱度0.6988iix磷酸盐容量与光学碱度的关系式为34299902999017.5523.7417.550.698823.744.5361873POlgCT炉渣磷酸盐容量3434331POC