第03章-化學(xué)反應(yīng)熱力學(xué)以及相變

1122aAbByYzZ0aAbByYzZ0vBBvBB的化學(xué)計量數(shù)反應(yīng)物vB0,產(chǎn)物vB

aAbByYzZt0,tt,

nA,0

nB(t)

d 0vBBddnB(mold

nB(

nB(0 0

nB()nB(0v

2

N2(g)3H2(g)2NH3(g)NvN2

11N(g)3H(g)NH( NvN2

1

d

dnBdnYdnZ

N2(g)3H2(g)2NH3(g)Δ＝1mol的意思是：1molN2和1mol(3H2反應(yīng),1N(g)3H(g)NH( 1mol的意思是：1

2和

3 反應(yīng)生成1molNH3

2dXXBdnBvBXBd X T,

vBXB

rXm rGrGm（平衡（DeDonder）首先引進(jìn)了化學(xué)反A

德（DeA

0BBdGSdTVdpB,

T,pBB

當(dāng)1 r BB

(

,

BT, BAA(GT BA

AA

用(G

G~圖上曲線的斜率,AAAAA

T,p

(G

T,p

，，0 θRT B p θRT pθ pθB

pRT RTln(pB RTln(pB)p

vp

v

B BrmGθ RTlnrmp

B RTln(fBf f θ

RT B

pθ pθfRT RTln(fB RTln(fBf

v

B fBpθf Gθ RTln 平衡時 Gθ RTlnJ peqlpeq L MKθJ

(

/pθ)

peq

peq A Bpθpθ GθRTln θθp mG RTlnKRTln RTln(J/KθθpppprmRTp/K)θp（平衡當(dāng) K 時, 0,A

θJ當(dāng)pKpθJ

0A0, Kθ時,

K K

+1.5H

K

rrKK1(Kθ2C(s)

Gθ(1)RTln CO(g)1O(g)CO C(s)1O(g) (1)(2)得

Gθ(2)RTln Gθ(3)RTln Gθ(3)Gθ(1)Gθ KθppKθ(3)pp

Kθ (θRTlnBRTlnxp p pθ θRTln BRTlnx pθBB

(

pθRTlnBpθB

G*RTln rmx 將

GθRTlnrmx

GθRTlnJeq GθRTln expθθx m (θRTlnBRTlnap p )令:)

BB

RT pθBpθJaG G*llnJ

G*RTln rma 將

GθRTlnrma

GθRTlnJeq GθRTlnK KJKJ (eqxeq (eq)vB(xeq 0AA p (θRTln ARTlnx)(θRT B

θ

θ A

RTln A pθ pθ

Bp

G*RTln rm 將

GθRTlnrm

GθRTlnJeq GθRTlnKθ pK Jp

(xeq)vA(

A

HHθrmBv (B,,Tθ mGθHθTSθ rm K( )K( )BGθrmRTlnKt

b

l mBB l x) x) x xf(aaa pB pB pnLpBpB

BRTBRTlnKmrGθnB,0＝該反應(yīng)物的原始數(shù)量nBnB,0某反應(yīng)物消耗掉的數(shù)量

/pθ) xAxBxL1124.54gCl2(g)和4.19gSO2(g),在463K下，進(jìn)行tnB

Cl2(g)+SO2(g)＝SO2Cl200x(0.1295pVn

0.1295 分別求得：pSOCl pSO2pCl2SO2Cl2(g)＝Cl2(g)t

B (1B y Kθ B ( B

p(Cl p(SO) K

p p(SO2Cl2

dXXBdnBvBXBd X T,

vBXB

rXm

vAvBvYv rH B nY

r r v B

H(T)vH

B (B, 1 1 mHe,Ne,H2,O2,N2,Br2(lS(斜方)2H2

H m,C(s)O2(g)CO2(g)H H m,CO2(g1H(g,p)1Cl(g,p)HCl(g,p H mH

Hf

m maA(g)bB(g)yY(g)zZ(s)aA(g)+yY(g)+aA(g)+yY(g)+H HrHmy(298.15K)=H2－H=yfHmy(Y,g,298.15K)＋zfH－afHmy(A,g,298.15K)－bfHmy(B,H

vH

2C2H2(g)+5O2(g)=4CO2(g)+2H2O rH (298.15KH =4fH (COg)+2H (H =（–393.51×4–241.82×2–226.7×2）kJmol=–2511.08kJmolHθ(B,,T CCO(2SSO(2HH2O(l)Cl1完全燃燒反應(yīng)的H1

H m,CH4(g)O2(g)CO2(g)2H2O(l)H H m,CH4(g22H H H m,C石墨CHCOOH(l)2O(g)2CO(g)2H H m,CHCOOH(l則H m,CHCOOH(l3H

H

Hθ(T)vH

BaAbByYHθ(T)aH bH yH m( m( m(HθH

θzz (COOH)2(g)+2CH3OH(l)=(COOCH3)2(s)+ 3c y((COOCH)，s)=–1678.0kJmol 3cc

y(CHOH，l)=–726.5kJmol y((COOH)，g)=–120.2kJ mrH (298.15K)=–120.2–2×726.5–(–1678.0 JlmkJmol例已知例已知CO2(g)的ΔfHm?(298.15K)=–393.51kJ.mol-1，?(298.15K)=aA()bB()yY()zZ(T標(biāo)準(zhǔn)態(tài)aA(bB()298.15K標(biāo)準(zhǔn)態(tài)

Hθ(T H

T標(biāo)準(zhǔn)態(tài)yY(zZ()298.15K，標(biāo)準(zhǔn)態(tài)Hθ(T)

Hθ(298.15K) H1

[aCpm,A()bCpm,B()

298.15K[yCpm,Y()zCpm,Z()Hθ(298.15K查表 (Hθ

vv

vH 即 vBCpm,BTBTHθ(T)Hθ(298.15K)

rCp,m §3-3 limST0K limS limS limST

S(T)S(0K)

Q S*(0Ｋ)=

0K QSm(B,T)Sm

0K Smθ(B,,T Sm 0=ΣBB

θ(T)

vSθ

aA(g)bB(g)yY(g)zZ(s)Sθ(T)ySθ(Y,g,T)zSθ(Z,s,T aSθ(A,g,T)bSθ(B,g,T 例4已知在25℃,H2O(l)和H2O(g)的標(biāo)準(zhǔn)摩爾生成焓分別為-285.830KJmol-1和-241.818KJmol-1，在此溫度下,水的飽和蒸汽壓為3.166KPa，69.91Jmol1KHm,Hm,m,2

m

m, m, Sθ(HO,g)Sθ(HO,l)

m,

m,

m,

m,m,1

Hθ(HO,g)Hθ(HO,l)m,

H,T

m,1 m,1m, m,

Smm,m

Rp

m,aAbByYS vS

S ySθ(Y)zSθ(Z)aSθ S Sθ(298.15K)S vAv 11

TSθ(T

vYv vAv

TSθ(T

vYv Sθ(T)SθSθ(T) SθSθSθ

T1vACp,m(A)dTT

T1vBCp,mT SθSθSθ

T2vYCp,m(Y)dTT

T2vZCp,mT SθT查表

(SθvS

Sθ(T)Sθ(T)

T2rCp,mdTT rCp,m vBCp,mB 第一問中T升高，說明反應(yīng)在恒壓，非體積功為零的條件應(yīng)放熱。當(dāng)反應(yīng)是在恒溫、恒壓條0vBBdef

nxeq nxAA θpB(BB)1. RTlnKrmHθrmBv (B,,Tθ mGθHθTSθ rm 習(xí)題2:下列反應(yīng)在1000K時，Ky=1.39CO(g)+H2O(g)=CO2(g)+H2(g)①+=+110α0α

K 2/(1)2nyCO(1)/2 CO(g)+H2O(g)=CO2(g)+H2

4

∑nB=K =Kn=α2/（1–α）（4–α）=得：α0.840yCO（1–α）/5的平衡轉(zhuǎn)化率為α=0.688,C2H6(g)==C2H4(g)+1000K的Ky

C2H6(g)==C2H4(g)+

nB nB1BB Kn(npθBBK

C2H6(g)==C2H4(g)+

nB開始：

ppB200K

(p/pθ)2(200p)/pθ

pyC

2

p/pB96.5/296.5yC

120.3251122rmrUHHUpVdHdHdUd(pVdXXB vBXB XXXXX vB rBT, rBdHHBdnBvBHB dUUBdnBvBUB dVVBdnBvBVB ：dHdUd(pV：vBHBdvBUBdpvBVB vBHB vBUB pvBVB 即： p X vBXB T, p p v( B RTv( B例實(shí)驗(yàn)測得0.5320gC6H6（l）在298.15K，定1）C6H6(l)+7.5O2(g)=6CO2(g)+2）2C6H6(l)+15O2(g)=12CO2(g)+ξ=nB/νB=-0.5320×10-3kg/(-3kg.mol-1)=6.811×10-3rU=QV=-rUm=rU/ξ=-22.23kJ/6.811×10-=-3.264×103kJ.mol-rHm=rUm+RT=-3.264×103kJ.mol-1+（-1.5）×8.314×298.15J.mol-=-3.628×103kJ.mol-對反應(yīng)2）ν苯-2ΣB（g）-3ξ=nB/νB=3.406×10-rUm=rU/ξ=-6.527×103kJ.mol-rHm=rUm+RTΣB（g）=-6.536×103kJ.mol-解：C2H2(g2.5O2(g)2CO2(g)

Qp=ΔH=0

2CO2(g)+H2O2CO2(g)+H2OT2 p2C2H2(g)+2.5O2(g)T1=298.15K10N2p1=101.325kPaC2H2(g)+2.5O2T1=298.15Kp2CO2(g)+H2OT1=298.15Kp

H=H1＋r ΔH1=

y(298.15K)＋H=r y(298.15K)= y(B,T ΔH2=(2Cp,mCO2(g)+Cp,mH2O(g)+10Cp,mN2(g))(mdlnKθrHm m mK ,對正反應(yīng)有利m m升高溫度K,θ RlnθTrmRTlnK((θ )T(lnKθ) THθ(Gθ m)Tv BlnKθTT2( m)θTvB( B)TBTGθBvθ Gθ m)TrHvθBlnKθ BT2lnKlnKθrHθmpdlnKrHHθrmHθf(TrHprm p,m p,mvBlnKθ1作圖,是一條直線lnKθ1作圖,是一條直線T斜率mH Rθθ R1Cm0Hθf(Tr vBm

0rHθ1TRln 21TRln 2 mK1H K

rC

vBCp,m

0,rHθ2T 2T1

rCp,m

mvBCp,mm

0,rHθf(TmHθ(T)f(mHθ(T)f(TrmrHθ(TdlnKθln 2 )mK )m1

1.rCp,m1.rCp,mvBCp,mB2.H , , TH m(298.15K4.H m(TBpeq BK3. m(298.15K5.H f(TK2θKTHθ(T2rT11m2 f(T RTln vp B Bpθ yeq pK ( ) ( B )B(yeq)p pθ pθ K (yeq) KKK( pθ)KpypθvB0K一定，pBθyθvB0,K一定，p,Ky θp

K

p

B(yeq)

)B

nnBBnBBB

Bpθn BKn pθn BKKK(p)yKpnpθBKKpθpBnKnB若ΣvB＞0,(p/py∑nB)∑B，Kn，平衡右移;若ΣvB0p/py∑nB)B↑Kn，平衡左移。3H2(g)+N2(g)=常從循環(huán)氣(主要是H2N2NH3CH4)中排除CH4氣體。GRTlnKRTlnJeqθ pp

K rGm rGm

Kf(TKf(Trθmln 2 mKH1K1RT1)2T1 1.T1.T2.p3.nBB4.rnBa成4f(TK Gv BGv BθBB §5-11.

蒸發(fā)焓vapHm，熔化焓fusHm,升華焓:subHm,晶型轉(zhuǎn)變焓trsHH(T)：kJmol1 有 def (g) (l)g

Hm(l)Hm(s)l l HdefH(g)H(s)g

HdefH(C,2)H(C,1)C,2 -ΔsubHm。mpHn mpT及該溫度 n (T n nB(g)n (T B(g)n B(g)n B(g)n B(g)n ΔH2表示在恒溫下壓力從p2變至p1時液體的焓變，壓差不 T1 (l)dT

p,m

T

1p1 H(T)

H(T)T1

H(T) H(T) (g)

，上式表明，若知道ΔvapHm(T1)及Cp,m(l)和Cp,m(g)的數(shù)值，則不難求得另一溫度T2下B的摩爾蒸發(fā)焓ΔvapHm(T2)。上式還表明，ΔvapHm隨溫度而變的原例；已知水在100℃、101.325kPa下其摩爾蒸發(fā)焓ΔvapHm(100℃)=40.63kJmol-1，水與水蒸氣的平均摩爾定壓熱容分別為Cp,m(l)=76.56Jmol-1K-1 Cp,m(g)=34.56Jmol-1K-1。設(shè)水蒸氣為理想氣體，試求水在