材料化学(双语)第四章重点

1、11Chapter 2 Structure and property of materials材料化学热力学材料化学热力学24.24.2 Ellingham Diagrams and its Ellingham Diagrams and its applicationapplication（埃灵罕姆图的应用埃灵罕姆图的应用）0GABT000GHT S 4.2.1 Ellingham Diagrams 4.2.1 Ellingham Diagrams G0 -T lineLinear relationshipB=B=S S0 03stableunstable4Slope（斜率）（斜率）of G0

2、-T line: If the number of gas decrease during the oxidation process, S00, slop is positive.（金属氧化物形成过程中，反应中有气体，而生成物为固体，反应结果是熵减少，斜率为正。即在高温下） If the number of gas increase during the oxidation process, S00, (- S0)0, slop is negative If the number of gas unchange during the oxidation process, S0=0, (- S

3、0)=0，slop is zero, that is to say, G0 is unrelated to temperature.22C(s) +O (g)=CO (g)22C(s) +O (g)=2CO(g)5 we can study the we can study the thermodynamics and thermodynamics and oxidative propertiesoxidative properties of various materials of various materials in in a large temperature rangea larg

4、e temperature range according to according to Ellingham diagrams, and then provide Ellingham diagrams, and then provide providence and data for the development providence and data for the development of new materials.of new materials.4.2 ApplicationApplication of Ellingham Ellingham DiagramsDiagrams

5、（埃灵罕姆图的应用埃灵罕姆图的应用）61) Equilibrium and Control of oxide formation（氧化物生成平衡及控制）（氧化物生成平衡及控制）201OlnGRTP2O ,eqTP Under a certain Under a certain temperature, we can temperature, we can control the direction of the control the direction of the interactioninteraction（相互作用）（相互作用） by modify the Oby modify the

6、 O2 2 pressure.pressure. 72) Comparison of stability of oxide（氧化物稳定性比较）（氧化物稳定性比较） Metal oxide with Metal oxide with G G0 0- -T T line in the below or the -line in the below or the - G G0 0 has biggerhas bigger negative value, the oxide is more stable. negative value, the oxide is more stable. 曲线越在下方

7、， G0 负值越大，越稳定 At a given temperature, corresponding element in the At a given temperature, corresponding element in the below line may make the element reduction of the below line may make the element reduction of the above line.above line. 在给定的温度，在相应元素线下面的可使上面的元素还原。 Metal oxide lie above the HMetal

8、 oxide lie above the H2 2O generating line will be O generating line will be reduced by H.reduced by H. 金属氧化物在水生成线上的可被H还原 The distance between two lines of oxidative and The distance between two lines of oxidative and reductive reactions stand for the reductive reactions stand for the G G0 0。 8i. e.

9、 Comparison of TiO2 and MnO TiOTiO2 2 generating line lies below the MnO generating generating line lies below the MnO generating line, so the line, so the TiOTiO2 2 is more stableis more stable. . 1000, the distance between two lines1000, the distance between two lines： o0221000 CTi(s) + O (g) = Ti

10、O (s) 674.11kJGo021000 CMn(s) + O (g) = 2MnO(s) 586.18kJGo021000 CTi(s) + 2MnO(s) = 2Mn(s) + TiO (s) 87.93kJG G G00, 00, MnO will be reduced by Ti at the MnO will be reduced by Ti at the standard condition.standard condition.93) Reverse of reducing capacity还原能力的相互反转还原能力的相互反转 When the two generating

11、lines When the two generating lines convergeconverge at at certain temperaturecertain temperature（当两根氧化物生成线在特定温度相交（当两根氧化物生成线在特定温度相交时）时）, the relative reducing capacity of two , the relative reducing capacity of two elements may reverse.elements may reverse. Slop of CO generating line is negative, so

12、 Slop of CO generating line is negative, so CO is CO is more stable with the increase of temperaturemore stable with the increase of temperature. . All the oxide could be reduced as long as the All the oxide could be reduced as long as the temperature is high enough. temperature is high enough. By w

13、hat?By what?104.3 Phase balance and phase diagramKey terms: Phase balance, phase diagram, component, unitary system, binary system, ternary system, isomorphous, lever rule, eutectic phase diagram, peritectic phase diagram, monotectic11f f is the is the number of degrees of freedomnumber of degrees o

14、f freedom, which means the , which means the number of properties such as number of properties such as temperature or pressuretemperature or pressure, which , which are independent of other variables.are independent of other variables.（自由度数）（自由度数） c c is the number of is the number of componentcompo

15、nent（独立组元数）（独立组元数） p p is the number of is the number of phases phases in thermodynamic in thermodynamic equilibrium with each other equilibrium with each other （平衡相的数目）（平衡相的数目） 2 2 temperature and pressuretemperature and pressure; if it is a solid system, it ; if it is a solid system, it should be

16、1.should be 1.Gibbs Phase Rule was proposed by Josiah Willard Gibbs in the 1870s as the equality f = c-p+2 Binary phase diagrams二元相图二元相图c c=2=2 Condensed status systemCondensed status system：f f=c c- -p p+ +1 1=3-=3-p p Maximum number of degrees of freedom: Maximum number of degrees of free

17、dom: f f=3-1=2 =3-1=2 Component and temperatureComponent and temperature2 D plane2 D plane ?132 curves (liquid line and solidus line)2 single-phase regions1 two-phase region1) Binary isomorphous diagram and lever rule二元匀晶相图与杠杆规则二元匀晶相图与杠杆规则 P103ABTATBLL+wB/% Two components have analogous chemical pro

18、perties and same crystal structure. They dissolve each other at both liquid and solidus state. They form infinite (successive) solid solution.1400LLCCWbcWabCCLever ruleLever ruleTACuNiC0CLCTBT1LL+acb0100.wNi/% 15Deduce of lever rule000LLLWWWW CW CWC00LLCCWbcWabCC16Phase analysisBinary isomorphous di

19、agram液相线液相线固相线固相线液相区液相区 L固相区固相区两相共存区两相共存区ABTATBL+wB, %2 curves (liquid line and solidus line)c=2，p=2，f=12 single-phase regionsc=2，p=1，f=21 two-phase regionc=2，p=2，f=1f = c-p+117m1818At extremum point, it is not accord with phase ruleAt extremum point, it is not accord with phase rule. C should . C s

20、hould be considered as a specific component. the phase diagram be considered as a specific component. the phase diagram should be considered as should be considered as a combination of two binary a combination of two binary isomorphous diagram (AC and CB).isomorphous diagram (AC and CB).Binary isomo

21、rphous diagram with extremumBinary isomorphous diagram with extremum极值极值Maximum pointMaximum pointMinimum pointMinimum point192) Eutectic phase diagram二元共晶相二元共晶相P10520Solidus linewB21DCEL Two solidus phase were precipitated simultaneously Two solidus phase were precipitated simultaneously from a liq

22、uid phase.from a liquid phase. According to phase rule, when three phase in a According to phase rule, when three phase in a equilibrium state, equilibrium state, f f=c c- -p p+1=2-3+1=0, so in this case, +1=2-3+1=0, so in this case, the component and temperature are certain, it the component and te

23、mperature are certain, it appear a plateauappear a plateau（平台）（平台）. . CED three-phase lineCED three-phase lineCharacteristics of eutectic reaction（共（共晶反应）晶反应）22 eutectic pointeutectic point共晶点共晶点, the location is defined by , the location is defined by eutectic component and temperature (eutectic co

24、mponent and temperature (E E). ). eutectic temperatureeutectic temperature共晶温度共晶温度 eutectic compositioneutectic composition共晶成分共晶成分 Eutectic reaction lineEutectic reaction line23DEECWCCWCC100%DEDCCCwCC100%ECDCCCwCCDCEL24Pb-Sn97.561.9100%45.4%97.5 19w61.9 19100%54.6%97.5 19wAlloy 1Alloy 1ABEFP113/8.吉

25、布斯相律通常为f=c-p+2，为什么在固体材料的研究中，相律一般可表达成f=c-p+1？Gibbs phase rule is usually expressed as f = c-p+2, why in the study of solid materials, the phase rule be expressed as f = c-p+1? 答：在固体材料的研究中，压力对固相反应的影响很小，通常可以忽略，所以非成分的变量只有温度这一项，所以相律一般可表达成f=c-p+1。In the study of solid materials, pressure has little effect

26、 on the solid phase reaction, usually it can be ignored, so temperature is the only non-composition variable, so the phase rule generally be expressed as f = c-p+1.重点题目重点题目 P113/9. 一合金之成分为一合金之成分为90Pb10Sn。（。（a）请）请问此合金在问此合金在100、200、300时含有哪几种时含有哪几种相？（相？（b）在哪个温度范围内将只有一相存在？）在哪个温度范围内将只有一相存在？ Composition o

27、f an alloy is 90Pb-10Sn. (a) Which phases is contained at 100, 200, 300? (b) In which temperature range will be only one phase exists? (a) Composition of alloy is 90Pb-10Sn alloy which corresponding to the red vertical line, it intersect with the temperature level line at a, b and c points. So, ther

28、e are two phases at 100, and ; there is only one phase at 200, ; point c lies on the liquidus line, so there are two phases at 300, L and . (b) Draw a perpendicular line across the ingredient line of 90Pb-10Sn alloy, we can get three points, d, e and f, then draw horizontal lines to get the correspo

29、nding temperature. So when the temperature is higher than 300, liquid phase L exists. When the temperature is between 148 268, there is only phase.答：（a）成分为90Pb-10Sn的合金对应的为红色的垂直线，分别在100、200、300作水平线得到交点a，b，c，所以 100时，交点a位于相区，所以有和两相。 200时，交点b位于单相区，所有只有一相。 300时，交点c位于液相线上，此为两相共存线，所以有液相L和两相。（b）过成分为90Pb-10S

30、n的线作垂直线，可得到交点d和e，然后作水平线得到相应的温度。所以在温度大于300时，只有液相L存在。在温度为148268时，只有相存在。 P113/10. 固溶体固溶体合金的相图如下图所示，试根据相图确定：合金的相图如下图所示，试根据相图确定：Phase diagram of solid solution alloy is shown in the figure, please determine the following questions according to it. 成分为成分为40B的合金首先凝固出来的固体成分是什么？的合金首先凝固出来的固体成分是什么？ 若首先凝固出来的固体成

31、分含若首先凝固出来的固体成分含60B，合金的成分是什么？，合金的成分是什么？ 成分为成分为70B的合金最后凝固的液体成分是什么？的合金最后凝固的液体成分是什么？ 合金成为为合金成为为50B，凝固到某温度时液相含有，凝固到某温度时液相含有40B，固体含有，固体含有80B，此时液体和固体各占多少分数，此时液体和固体各占多少分数？ a) What is the composition of the solid first frozen from alloy contain 40% B? b) If the solid first frozen containing 60% B, please determine composition of the alloy. c) What is the composition of the liquid final solidified from alloy containing 70% B? d) An alloy containing 50% B, when it a