材料化學(雙語)第四章重點

11Chapter2StructureandpropertyofmaterialsChapter4

ChemicalThermodynamicsofMaterials材料化學熱力學24.2

EllinghamDiagramsanditsapplication（埃靈罕姆圖的應用）4.2.1EllinghamDiagramsΔG0-TlineLinearrelationshipB=－ΔS03stableunstable4Slope（斜率）ofG0-Tline:※Ifthenumberofgasdecreaseduringtheoxidationprocess,S0<0,(-S0)>0,slopispositive.（金屬氧化物形成過程中，反應中有氣體，而生成物為固體，反應結果是熵減少，斜率為正。即在高溫下）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（埃靈罕姆圖的應用）61)EquilibriumandControlofoxideformation

（氧化物生成平衡及控制）

Underacertaintemperature,wecancontrolthedirectionoftheinteraction（相互作用）bymodifytheO2pressure.

72)Comparisonofstabilityofoxide

（氧化物穩(wěn)定性比較）MetaloxidewithG0-Tlineinthebeloworthe-G0hasbigger

negativevalue,theoxideismorestable.曲線越在下方，G0

負值越大，越穩(wěn)定Atagiventemperature,correspondingelementinthebelowlinemaymaketheelementreductionoftheaboveline.

在給定的溫度，在相應元素線下面的可使上面的元素還原。MetaloxidelieabovetheH2OgeneratinglinewillbereducedbyH.

金屬氧化物在水生成線上的可被H還原ThedistancebetweentwolinesofoxidativeandreductivereactionsstandfortheG0。8i.e.ComparisonofTiO2andMnOTiO2generatinglineliesbelowtheMnOgeneratingline,sotheTiO2ismorestable.

1000℃,thedistancebetweentwolines：

G0<0,MnOwillbereducedbyTiatthestandardcondition.93)Reverseofreducingcapacity

還原能力的相互反轉Whenthetwogeneratinglinesconvergeatcertaintemperature（當兩根氧化物生成線在特定溫度相交時）,therelativereducingcapacityoftwoelementsmayreverse.SlopofCOgeneratinglineisnegative,soCOismorestablewiththeincreaseoftemperature.

Alltheoxidecouldbereducedaslongasthetemperatureishighenough.Bywhat?104.3PhasebalanceandphasediagramKeyterms:

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

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

c

——

isthenumberofcomponent（獨立組元數）

p——

isthenumberofphasesinthermodynamicequilibriumwitheachother（平衡相的數目）

2

——

temperatureandpressure;ifitisasolidsystem,itshouldbe1.GibbsPhaseRule

wasproposedbyJosiahWillardGibbsinthe1870sastheequality

f=c-p+2

Binaryphasediagrams

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

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

二元勻晶相圖與杠桿規(guī)則P103ABTATBLαL+αwB/%※Twocomponentshaveanalogouschemicalpropertiesandsamecrystalstructure.※Theydissolveeachotheratbothliquidandsolidusstate.※Theyforminfinite(successive)solidsolution.14LeverruleTACuNiC0CLCαTBT1LαL+αacb0100.wNi/%15Deduceofleverrule16PhaseanalysisBinaryisomorphousdiagram液相線固相線液相區(qū)L固相區(qū)α兩相共存區(qū)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極值MaximumpointMinimumpoint192)Eutecticphasediagram二元共晶相P105

20SoliduslinewB21Twosolidusphasewereprecipitatedsimultaneouslyfromaliquidphase.Accordingtophaserule,whenthreephaseinaequilibriumstate,f=c-p+1=2-3+1=0,sointhiscase,thecomponentandtemperaturearecertain,itappearaplateau（平臺）.CEDthree-phaselineCharacteristicsofeutecticreaction（共晶反應）22eutecticpoint共晶點,thelocationisdefinedbyeutecticcomponentandtemperature(E).eutectictemperature共晶溫度eutecticcomposition共晶成分Eutecticreactionline2324Pb-SnAlloy1ABEFP113/8.吉布斯相律通常為f=c-p+2，為什么在固體材料的研究中，相律一般可表達成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：在固體材料的研究中，壓力對固相反應的影響很小，通?？梢院雎?，所以非成分的變量只有溫度這一項，所以相律一般可表達成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.重點題目P113/9.一合金之成分為90Pb－10Sn。（a）請問此合金在100℃、200℃、300℃時含有哪幾種相？（b）在哪個溫度范圍內將只有一相存在？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）成分為90Pb-10Sn的合金對應的為紅色的垂直線，分別在100℃、200℃、300℃作水平線得到交點a，b，c，所以

100℃時，交點a位于α＋β相區(qū)，所以有α和β兩相。

200℃時，交點b位于α單相區(qū)，所有只有α一相。

300℃時，交點c位于液相線上，此為兩相共存線，所以有液相L和α兩相。（b）過成分為90Pb-10Sn的線作垂直線，可得到交點d和e，然后作水平線得到相應的溫度。所以在溫度大于300℃時，只有液相L存在。在溫度為148～268℃時，只有α相存在。

P113/10.固溶體合金的相圖如下圖所示，試根據相圖確定：Phasediagramofsolidsolutionalloyisshowninthefigure,pleasedeterminethefollowingquestionsaccordingtoit.成分為40％B的合金首先凝固出來的固體成分是什么？若首先凝固出來的固體成分含60％B，合金的成分是什么？成分為70％B的合金最后凝固的液體成分是什么？合金成為為50％B，凝固到某溫度時液相含有40％B，固體含有80％B，此時液體和固體各占多少分數？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