材料化学(双语)第四章重点

11Chapter2StructureandpropertyofmaterialsChapter4

ChemicalThermodynamicsofMaterials材料化学热力学24.2

EllinghamDiagramsanditsapplication（埃灵罕姆图的应用）4.2.1EllinghamDiagramsΔG0-TlineLinearrelationshipB=－ΔS03stableunstable4Slope（斜率）ofG0-Tline:※Ifthenumberofgasdecreaseduringtheoxidationprocess,S0<0,(-S0)>0,slopispositive.（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）Metal(s)+O2(g)-----Metaloxide(s)※Ifthenumberofgasincreaseduringtheoxidationprocess,S0>0,(-S0)<0,slopisnegative※Ifthenumberofgasunchangeduringtheoxidationprocess,S0=0,(-S0)=0，slopiszero,thatistosay,

G0isunrelatedtotemperature.5

wecanstudythethermodynamicsandoxidativepropertiesofvariousmaterialsinalargetemperaturerangeaccordingtoEllinghamdiagrams,andthenprovideprovidenceanddataforthedevelopmentofnewmaterials.4.2ApplicationofEllinghamDiagrams（埃灵罕姆图的应用）61)EquilibriumandControlofoxideformation

（氧化物生成平衡及控制）

Underacertaintemperature,wecancontrolthedirectionoftheinteraction（相互作用）bymodifytheO2pressure.

72)Comparisonofstabilityofoxide

（氧化物稳定性比较）MetaloxidewithG0-Tlineinthebeloworthe-G0hasbigger

negativevalue,theoxideismorestable.曲线越在下方，G0

负值越大，越稳定Atagiventemperature,correspondingelementinthebelowlinemaymaketheelementreductionoftheaboveline.

在给定的温度，在相应元素线下面的可使上面的元素还原。MetaloxidelieabovetheH2OgeneratinglinewillbereducedbyH.

金属氧化物在水生成线上的可被H还原ThedistancebetweentwolinesofoxidativeandreductivereactionsstandfortheG0。8i.e.ComparisonofTiO2andMnOTiO2generatinglineliesbelowtheMnOgeneratingline,sotheTiO2ismorestable.

1000℃,thedistancebetweentwolines：

G0<0,MnOwillbereducedbyTiatthestandardcondition.93)Reverseofreducingcapacity

还原能力的相互反转Whenthetwogeneratinglinesconvergeatcertaintemperature（当两根氧化物生成线在特定温度相交时）,therelativereducingcapacityoftwoelementsmayreverse.SlopofCOgeneratinglineisnegative,soCOismorestablewiththeincreaseoftemperature.

Alltheoxidecouldbereducedaslongasthetemperatureishighenough.Bywhat?104.3PhasebalanceandphasediagramKeyterms:

Phasebalance,phasediagram,component,unitarysystem,binarysystem,ternarysystem,isomorphous,leverrule,eutecticphasediagram,peritecticphasediagram,monotectic11

f——isthenumberofdegreesoffreedom,whichmeansthenumberofpropertiessuchastemperatureorpressure,whichareindependentofothervariables.（自由度数）

c

——

isthenumberofcomponent（独立组元数）

p——

isthenumberofphasesinthermodynamicequilibriumwitheachother（平衡相的数目）

2

——

temperatureandpressure;ifitisasolidsystem,itshouldbe1.GibbsPhaseRule

wasproposedbyJosiahWillardGibbsinthe1870sastheequality

f=c-p+2

Binaryphasediagrams

二元相图c=2Condensedstatussystem：f=c-p+1=3-pMaximumnumberofdegreesoffreedom:f=3-1=2Componentandtemperature2Dplane

?132curves(liquidlineandsolidusline)2single-phaseregions1two-phaseregion1)Binaryisomorphousdiagramandleverrule

二元匀晶相图与杠杆规则P103ABTATBLαL+αwB/%※Twocomponentshaveanalogouschemicalpropertiesandsamecrystalstructure.※Theydissolveeachotheratbothliquidandsolidusstate.※Theyforminfinite(successive)solidsolution.14LeverruleTACuNiC0CLCαTBT1LαL+αacb0100.wNi/%15Deduceofleverrule16PhaseanalysisBinaryisomorphousdiagram液相线固相线液相区L固相区α两相共存区ABTATBL+αwB,%2curves(liquidlineandsolidusline)c=2，p=2，f=12single-phaseregionsc=2，p=1，f=21two-phaseregionc=2，p=2，f=1f=c-p+117m1818Atextremumpoint,itisnotaccordwithphaserule.Cshouldbeconsideredasaspecificcomponent.thephasediagramshouldbeconsideredasacombinationoftwobinaryisomorphousdiagram(ACandCB).Binaryisomorphousdiagramwithextremum极值MaximumpointMinimumpoint192)Eutecticphasediagram二元共晶相P105

20SoliduslinewB21Twosolidusphasewereprecipitatedsimultaneouslyfromaliquidphase.Accordingtophaserule,whenthreephaseinaequilibriumstate,f=c-p+1=2-3+1=0,sointhiscase,thecomponentandtemperaturearecertain,itappearaplateau（平台）.CEDthree-phaselineCharacteristicsofeutecticreaction（共晶反应）22eutecticpoint共晶点,thelocationisdefinedbyeutecticcomponentandtemperature(E).eutectictemperature共晶温度eutecticcomposition共晶成分Eutecticreactionline2324Pb-SnAlloy1ABEFP113/8.吉布斯相律通常为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

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P113/9.一合金之成分为90Pb－10Sn。（a）请问此合金在100℃、200℃、300℃时含有哪几种相？（b）在哪个温度范围内将只有一相存在？Compositionofanalloyis90Pb-10Sn.(a)Whichphasesiscontainedat100℃,200℃,300℃?(b)Inwhichtemperaturerangewillbeonlyonephaseexists?(a)Compositionofalloyis90Pb-10Snalloywhichcorrespondingtotheredverticalline,itintersectwiththetemperaturelevellineata,bandcpoints.So,therearetwophasesat100℃,αandβ;thereisonlyonephaseat200℃,α;pointcliesontheliquidusline,sotherearetwophasesat300℃,Landα.(b)Drawaperpendicularlineacrosstheingredientlineof90Pb-10Snalloy,wecangetthreepoints,d,eandf,thendrawhorizontallinestogetthecorrespondingtemperature.Sowhenthetemperatureishigherthan300℃,liquidphaseLexists.Whenthetemperatureisbetween148~268℃,thereisonlyαphase.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α＋β相区，所以有α和β两相。

200℃时，交点b位于α单相区，所有只有α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α两相。（b）过成分为90Pb-10Sn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148～268℃时，只有α相存在。

P113/10.固溶体合金的相图如下图所示，试根据相图确定：Phasediagramofsolidsolutionalloyisshowninthefigure,pleasedeterminethefollowingquestionsaccordingtoit.成分为40％B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分含60％B，合金的成分是什么？成分为70％B的合金最后凝固的液体成分是什么？合金成为为50％B，凝固到某温度时液相含有40％B，固体含有80％B，此时液体和固体各占多少分数？a)Whatisthecompositionofthesolidfirstfrozenfromalloycontain40%B?b)Ifthesolidfirstfrozencontaining60%B,pleasedeterminecompositionofthealloy.c)Whatisthecompositionoftheliquidfinalsolidifiedfromalloycontaining70%B?d)Analloycontaining50%B,whenitarriveatsometemperature,theliquidphasecontaining40%B,andthesolidphasecontaining80%B,pleasedeterminethefractionoftheliquidandsolid.Thecompositionofthesolidfirstfrozenfromalloycontain40%Bcorrespondtothepointa’,whichcontaining80％Band20％A.b)Ifthesolidfirstfrozencontaining60%B,wecangetpoin