ch.12 reactions involving pure condensed phase and a
TRANSCRIPT
Ch.12 Reactions Involving Pure Condensed
Phase and a Gaseous Phase
§12-1. Introduction
§12-2. Reaction Equilibrium in a system containing pure
condensed phases and a gas phase
§12-3. Variation of “Standard Gibbs Free Energy Change”
with T
§12-4. Ellingham Diagrams (Stability of Metals and Metal-Oxides.)
§12-5. Effect of Phase Transition
§12-6. Stability of Oxides in CO/ 2CO and 2H / OH2 Gas
Mixtures
§12-7. Upper limit of (2CO
CO
pp ) at a fixed T
§12-1. Introduction
.etc,H,OH,Cl,S,O:GasMetalPure:PhaseCondensed
22222
e.g. 1. Oxidation of metals.
2. At a given T, what is the maximum 2Op , which can be tolerated in a gas
atmosphere without oxidation of a metal. “Bright” annealing of copper.
§12-2. Reaction Equilibrium in a system containing pure
condensed phases and a gas phase
Oxidation of a pure metal M at T, P=1atm
)s()g(2)s( MOO21
M =+
assumptions:
.tricstoichiomeisMO.Minlelubinsoisoxygen
Three Equil.
)g()g(2)g(
)g()s(
)g()s(
MOO21
M)3(
MOMO)2(MM)1(
=+
==
∴
2/1OM
MO)g(O)g(M)g(MO
MOMO
MM
2
2 ppp
lnRTG21
GGG)3(
)solid(G)gas(G)2()solid(G)gas(G)1(
⋅−=−−=∆
==
οοοο
∵
+=
+=
∫=
=
)g(MpP
1P )S(M)S(MM
)g(M)g(MM
dPVG)solid(G
plnRTG)gas(Gο
ο
∵ )s(MV = molar volume of solid M, it is insensitive to pressure change.
i.e. )g(MpP
1P )s(M plnRTdPV)g(M <<∫=
=
e.g. T=1000 Cο , Fe
≅×≅ −
3)s(Fe
10)g(Fe
cm34.7Vatm106p
∴
−=⋅−=
−=
∫ =J74.0cmatm34.7V
J224750plnRT3p
1P )s(Fe
)g(Fe
∴ οο)s(M)g(M)g(M GplnRTG ≅+
or )g(M)s(M)g(M plnRTGG −= οο
Similarly, )g(MO)s(MO)g(MO plnRTGG −= οο
Substitute into(3),
−=−−=∆
21
O
)g(O)s(M)s(MO
2
2 P
1lnRTG
21
GGG οοοο
∵ οG∆ ≡－RT lnK
∴K=2
1
O 2p
1
and οG∆ =[ οοο)g(O)s(M)s(MO 2
G21
GG −− ] is Standard Gibbs Free Energy Change for
reaction : )s()g(2(s) MOO21
M =+
Note: 1. Equilibrium constant K can be expressed only with species in gas phase.
2. οG∆ is f (T), and K=K(T) only.
3.∴At fixed T, ( ).eqO2
p = )T,eq(O2p
∴when 2Op > .)eq(O 2
p ⇒ Spontaneous oxidation of metal will occur.
⇒ consuming )g(2O , 2Op ↓
⇒ finally, 2Op = .)eq(O2
p , oxidation ceases.
e.g. Extraction metallurgy process: Reduction of oxide ores by 2Op .)eq(O 2
p<
4C )s(u + )g(2O =2C )s(2Ou
οG∆ =－333000+141.3T (J)
∴ οG∆ =－RT lnK=－RT ln(2Op
1)=RT ln
2Op
∴ln2Op =
RTGο∆
=－ 38.7T
17390+
Figure 12.1
§12-3. Variation of “Standard Gibbs Free Energy Change”
with T
)s(2)g(2)s( MOOM =+ , οG∆ (T)=?
＊∵ οG∆ =－RT lnK=RT ln2Op
∴ .)eq(O2p (T) is determined by οG∆ (T)
Exact calculation of οG∆ (T) can be done with )T(CP
＊ οTG∆ = ο
TH∆ －T οTS∆ at P=1atm
= [ ο298H∆ + ∫ ∆
T
298 PdTC ]－T[ ο298S∆ + ∫
∆T
298
P dTTC
]
For M(s)+ )g(O2 =M 2O (s)
Given
++=++=++=
∆∆
−
−
−
2333)s(MO,P
2222)g(O,P
2111)s(M,P
298298
TcbaCTcbaCTcTbaC
S,H
2
2
οο
∴ )aaa(C 123P −−=∆ + T)bbb( 123 −− + 2123 T)ccc( −−−
= 2cTbTa −∆+∆+∆
∴ οTH∆ = ο
298H∆ + ∫ ⋅∆T
298 P dTC = οH∆ + aT∆ +Tc
T2b 2 ∆
−∆
οH∆ =
∆
−∆
+∆+∆298
2298 T
CT
2b
aT)egrationintofitlimLower(Hο
Similarly, calculation of οοTT GS ∆⇒∆ can be obtained.
Or, ( )
P
T
TT/G
∂
∂∆∂ ο
=－ 2T
THο∆
∴ ∫∆T
298T )
TG
(dο
= ∫∆
−T
298 2 dT)TH
(ο
= ∫∆
+∆
−∆
−∆
−T
298 32dT)
Tc
2b
Ta
TH
( ο
∴ =∆
TGT
ο
I+TH0∆－ Tlna∆ － 2T2
cT
2b ∆
−∆
I =integration constant=
×∆
+×∆
+∆+∆
−∆
2298
2982c
2982b
298lna298H
298G ο
ο
∴ οTG∆ = I T+ οH∆
T2c
T2b
TlnaT 2 ∆−
∆−∆−
＊ e.g. for 4C )s(u + )g(2O =2C )s(2Ou
(1) calculate from : ο298H∆ , ο
298S∆ , ]OCu,O,Cu[C )s(2)g(2)s(P
T4.174T1085.0T103.9TlnT28.4333200G 1523T +×−×−−−=∆ −−ο
(2) ∵ οG∆ =－RT lnK=RT ln .)eq(O2p
Measure .)eq(O 2p (T) ⇒ )T(Gο∆ TCTlnTBA ′+′+′≅
οTG∆ = T247TlnT2.14338900 +−−
(3) Approximate (2) to a linear form οTG∆ =A+BT
οTG∆ =－333000+141.3T
Compare (3) with (1), error is small !!
∴Usually, it is convenient to use approximation, “ for oxidation of pure metals.”
BTAGT +=∆ ο
§12-4. Ellingham Diagrams (Stability of Metals and Metal-Oxides.)
1. Ellingham: Plot οG∆ vs. T for oxidation of many metals.
οG∆ BTA +≅ , for oxidation of pure metals.
c.p. οG∆ = οH∆ －T οS∆ , at constant P
∴
∆−=∆=
o
o
SBHA
, but oo S,H ∆∆ is almost independent of T.
)s(2)g(2)s( MOOM =+
Figure 12.2 T=0 Ko , οG∆ = οH∆ (Heat formation of oxide at 0 Ko )
οS∆ = ο)s(MO2
S ο)s(MS− o
O2S− o
O 2S−≅
>>
≅οοο
οο
)s(M)s(MOO
)s(M)s(MO
SorSS
SS
22
2
∴( >∆− )Sο 0
(1) T= οT , 0G =∆ ο , ∴ )s(M and )s(2MO are in equilibrium under =2Op 1atm
∵ 0plnRTKlnRTG2O ==−=∆ ο , ∴ ( ) atm1p
.eqO2≡
(2) T= 01 TT < , ∵ οG∆ < 0 (oxidation reaction occurs under =2Op 1atm.)
∴ ο)s(MO2
G is smaller than ο)s(MG
)s(M is spontaneous ly oxidized.
∵ ο1TG∆ =R 1T ln )T,eq(o 12
p < 0 , ∴ )T,eq(o 12p < 1atm
(3) T= 02 TT > , οG∆ > 0, decompose of )s(2MO occurs under =2Op 1atm.
)T,eq(o 22p > 1atm
2. ∵ οο2OSS −≅∆
∴ οS∆ corresponds with disappearance of 1 mole of )g(2O initially at 1atm.
Therefore, if all oxidation reactions for pure metals are expressed in terms of
1 mole of )g(2O , then lines in Ellingham diagram have similar slopes.
slope= οο2OSS ≅∆−
∴all lines are more or less parallel.
e.g.
+−=∆=++−=∆=+
T6.145769400G,MnO2OMn2T7.143467800G,CoO2OCo2
)s()g(2)s(
)s()g(2)s(ο
ο
∴Relative stability of oxides CoO and MnO are determined by their οH∆ , i.e.
More negative οH∆ ⇔ More negative οG∆ ⇔ More stable the oxide, or metal is more easily oxidized.
3. Consider: Two oxidation reactions with Ellingham lines intersecting one another.
2 )s(A + 2O =2 )s(AO … … (1)
)(B λ + 2O = )s(BO … … (2)
Assume:
∆>∆∆<∆
∆<∆οοοο
οο
)1()2()1()2(
)1()2(
SSorSS
HH
∴(2)-(1) ⇒ 2)s()( BOA2AO2B +=+λ … … (3)
Figure 12.3 , Figure 12.4
∴ ETT < , 0G <∆ ο ,
++
unstableare)AOB(stableare)BOA( 2
ETT > , 0G >∆ ο ,
++
stableare)AOB(ustableare)BOA( 2
∴If pure A were to be used as a reducing agent to reduce 2BO to form B and AO, reduction should be conducted at ETT > .
Note: In order to compare stabilities of oxides, Ellingham diagram must be
drawn for oxidation reactions with consumption of one mole of )g(2O .
4. Richardson nomographic scale of )T,eq(O2p
οTG∆ =－RT lnK=RT ln )T,eq(O2
p : for oxidation
∵For )g(2O , ideal gas model G= PlnRTG +ο
οG : Standard Gibbs Free Energy of )g(2O at P=1atm
∴ οTG∆ is the Gibbs Free Energy change of )g(2O from 1atm to )T,eq(O2
p at T.
∴For a given .)eq(O2p , ο
TG∆ =[R ln .)eq(O2p ] T⋅
a straight line of οTG∆ vs. T can be drawn.
∴For different .)eq(O2p , a series of lines of ο
TG∆ vs. T can be drawn .
These lines radiate from the point οG∆ =0, at T= K0ο .
Figure 12.5
＊ When .)eq(O2p =1atm ln .)eq(O2
p =0
∴ οTG∆ =R ln .)eq(O2
p T⋅ , slope=0
when .)eq(O2p < 1atm, ln .)eq(O2
p < 0
∴ οG∆ vs. T line, slope < 0
∴ )T,eq(O2p nomographic scale is added to Ellingham diagram along right-hand
edge and bottom edge.
Figure 12.6 e.g. at T= 1T
ab = ο1TG∆ =Gibbs Free Energy change of )g(2O
from 1atm→ 2010− atm
∴if 2Op < )T,eq(O 12
p = 2010− atm
oxidation of )s(M will not occur.
Figure 12.7 If 1T < ET
)T,eq(O 12p (2) < )T,eq(O 12
p (1)
∴B oxidized more easily than A.
＊ Starting with 2Op =1atm at 1T , (A+B+ 2O )
A and B oxidized simultaneously.
When 2Op ↓ to ≤ )T,eq(O 12
p (1)
Oxidation of A ceases and oxidation of B continues + part of )s(OA
decomposed.
Final equilibrium: ++ )s(2)s( BOA )T,eq(O 12p (2)
§12-5. Effect of Phase Transition
)s()g(2)s( AOOA =+ … … (1)
)(A λ = )s(A … … (2) at T= (A)m,T
(3)=(1)+(2) ∴ )s(2)g(2)( AOOA =+λ … … (3)
∴ ο)3(H∆ = ο
)s(AO2H － ο
λ)(AH － ο)g(O2
H
= ο)s(AO2
H － ο)s(AH － ο
)g(O2H －[ ο
λ)(AH － ο)s(AH ]
∴ ο)3(H∆ = ο
)1(H∆ － οA,mH∆
Similarly, ο)3(S∆ = ο
)1(S∆ － οA,mS∆ = ο
A,mO SS2
∆−− = )SS( A,mO2
ο∆+−
∵ οA,mH∆ > 0 and ο
A,mS∆ > 0
∴
∆<∆
∆<∆οο
οο
)1()3(
)1()3(
SS
HH ⇒＊ slope of Ellingham line=( οS∆− ) > 0
∴(slope) ↑ slope= οA,mO SS
2∆+
Figure 12.8 ＊ intercept at T= K0ο is lower.
)s(2)g(2)s( AOOA =+ … … (1)
=)s(2AO )2(AO λ … … (4) at T=2AOm,T
(5)=(1)+(4): )(2)g(2)s( AOOA λ=+ … … (5)
ο)5(H∆ = ο
λ)(AO2H － ο
)s(AH － ο)g(O2
H
= ο)s(AO2
H － ο)s(AH － ο
)g(O2H +[ ο
λ)(AO2H － ο
)s(AO2H ]
ο)5(H∆ = ο
)1(H∆ + ο2AO,mH∆
Similarly, ο)5(S∆ = ο
)1(S∆ + ο2AO,mS∆ = ο
22 AO,mO SS ∆+−
∵ ο2AO,mH∆ > 0, ο
2AO,mS∆ > 0
∴
∆<∆
∆<∆οο
οο
)1()5(
)1()5(
SS
HH
∴
∆−<∆−==
.decreases)slope(),S()S(slope*.higherisK0Taterceptint*
)1()5()5(οο
ο
Figure 12.8 , Figure 12.9
e.g. 2Fecl,mT =969K,
2Fecl,bT =1298K, Fe,mT =1809K
Given
−+−=∆+−=∆
+−−=∆=+=+=+
)J(T1.375TlnT8.41105600G)J(T68.63286400G
)J(T9.212TlnT68.12346300G)3......(FeClClFe)2......(FeClClFe)1......(FeClClFe
)3(
)2(
)1(
)g(2)g(2)s(
)(2)g(2)s(
)s(2)g(2)s(
ο
ο
ο
λ
∴(2)－(1), FeC )s(2l = FeC )(2l λ … … (4)
ο)4(G∆ = ο
2FeCl,mG∆ = ο)2(G∆ ο
)1(G∆− =59900+12.68T lnT－149T
+−−=∂
∆∂−=∆
−=∂
∆∂−=∆
14968.12Tln68.12]T
G[S
T68.1259900]T
)T/G([TH
2
2
2
2
FeCl,mFeCl,m
FeCl,m2FeCl,m
οο
οο
∴T=2Fecl,mT =969K,
=∆
=∆
K/J13.49S
J47610H
2
2
FeCl,m
FeCl,mο
ο
(3)－(2), FeC )(2l λ = FeC )g(2l … ...(5)
ο)5(G∆ = ο
2FeCl,bG∆ = ο)3(G∆ ο
)2(G∆− =180800+41.8T lnT－438.8T
+−−=∂
∆∂−=∆
−=∂
∆∂−=∆
8.4388.41Tln8.41]T
G[S
T8.41180800]T
)T/G([TH
2
2
2
2
FeCl,bFeCl,b
FeCl,b2FeCl,b
οο
οο
∴T=2Fecl,bT =1298K,
=∆
=∆
K/J46.97S
J126500H
2
2
FeCl,b
FeCl,bο
ο
Figure 12.10
§12-6. Stability of Oxides in CO/ 2CO and 2H / OH2 Gas
Mixtures
＊ In order to avoid oxidation of a metal, or reducing a metal oxide, a low 2Op is
usually needed.
＊ A vaccum system, p 1010−≅ atm ( Torr10 7− ), is not enough to get a low 2Op .
＊ Low 2Op can be obtained by CO/ 2CO or 2H / OH2 mixture with high
( COp /2COp ) or (
2Hp / OH2p ) ratio. Nomographic scales of ( COp /
2COp ) and
(2Hp / OH2
p ) are added to Ellingham diagram.
1. Consider reducing of 2MO with (CO/ 2CO ) gas mixture.
∵2 )g(2)g2)g( CO2OCO =+ … … (*1)
∴ οG∆ =－564800+173.62T (J)
οG∆ =－RT ln PK =－RT ln(2
2
O2CO
2CO
pp
p
⋅)
οG∆ =2RT ln(2CO
CO
pp
)+RT ln2Op
at T= οT
∴ln )1(*.)eq(O2
p =ο
οο
RT
G )T(∆－2 ln(
2CO
CO
pp
)
For oxidation of some metal: at T= οT
)s(2)g(2)s( MOOM =+ … … (*2)
οG∆ =－R 0T lnK=R 0T ln )2(*.)eq(O2
p
∴If )1(*.)eq(O2
p ＜ )2(*.)eq(O2
p , metal will not be oxidized (or oxide will be reduced)
2. Nomographic scale for ( COp /2COp )
2 )g(2)g2)g( CO2OCO =+ … … (*1)
[1atm] [1atm] [1atm]
ο1G∆ =－564800+173.62T
or ο1G∆ = ο
]CO[ 2H∆ －T ο
]CO[ 2S∆
2C )g(2O [1atm]= 2 )g(2CO [P atm]… … (*3)
∴ ο3G∆ =2RT lnP
∴(*1) + (*3)=(*4) 2 )g(2)g2)g( CO2OCO =+
[1atm] [1atm] [P atm]
∴ ο4G∆ = ο
1G∆ + ο3G∆
∴ ο4G∆ = ο
]CO[ 2H∆ －T ο
]CO[ 2S∆ +2RT lnP
ο4G∆ = ο
]CO[ 2H∆ +T[2R lnP+( ο
]CO[ 2S∆− )]
∴For any other pressure of )g(2CO : P atm, or (2CO
CO
pp
)=(P1
) ≠ 1
ο4G∆ vs. T is linear, originating from ο
]CO[ 2H∆ “C” point,
with a slope ≅ ( ο]CO[ 2
S∆− )+2R lnP
∴
∆−<><
∆−><>=
)S(slope,1)p
p(,1P
)S(slope,1)p
p(,1pP
]CO[CO
]CO[CO
COCO
2
2
CO
2
2
2
ο
ο
Figure 12.11
3. Using CO/ 2CO gas mixture to reduce )s(2MO
)s(M + )atm1(O2 = )s(2MO … … (*5)
(*1)－(*5)=(ix) )s(2MO +2CO(1atm)= )s(M +2 2CO (1atm)… … (ix)
∴ ο)ix(G∆ =( −∆ ο
(1)G ο(5)G∆ )=－RT ln
2
CO
CO
p
p2
ο)ix(G∆ =2RT ln(
2CO
CO
pp
)
∴at T= sT , (2CO
CO
pp
)=1, ο)ix(G∆ =0, (equil.)
(a) Fix (2CO
CO
pp
)=1, >rT sT , (r)point lies below Ellingham line(*5)
∴
.reducedisMO.oxidizedbenotwillM
2
(b) AT <uT sT
If (2CO
CO
pp
)=1, point(u) lies above line(*5) M will be oxidized.
Only when (2CO
CO
pp
)≥ 10, M will not be oxidized. (or 2MO can be reduced)
e.g. 2 2CO (1atm)= 2 2CO (0.1atm)… … (6)
ο6G∆ =2RT ln(0.1)
∴(ix)+(6)=(x) 2MO +2CO(1atm)= M +2 2CO (0.1atm)… … (x)
∴ ο)x(G∆ = ο
)ix(G∆ +2RT ln(0.1)=2RT ln(2CO
CO
pp
)+2RT ln(0.1)
at T= uT , (2CO
CO
pp
)=10 ο)x(G∆ =0, (equil.)
Ex.: For a given (2CO
CO
pp
) point A= 610 , and T=1000 Cο , then a point P in diagram
can be specified.
(1) ( ).eqO2
p can be obtained by drawing OP and its intercept at 2Op scale
(point B= atm10 26− ) (2) For an Ellingham line of some metal, if point P lies below line, metal will not
be oxidized. Figure 12.12
§12-7. Upper limit of (2CO
CO
pp ) at a fixed T
−−=∆=+
−−=∆=+
T3.175223400G,CO2OC2)2(
T84.0394100G,COOC)1(
)3()g()g(2)s(
)1()g(2)g(2)s(ο
ο
(2)－2 )1(⋅ =(3) 2 )g(CO + )g(2O =2 )g(2CO , ο)3(G∆ =－564800+173.62T
(2)－(1)=(4) )s(C + )g(2CO =2 )g(CO , ο)4(G∆ =170700－174.5T
1. Reaction Equilibrium of (3) (CO- 2CO - 2O )
∴ ο)3(G∆ =－RT ln PK =－RT ln(
2
2
O2CO
2CO
pp
p
⋅)
∴2 ln(CO
CO
p
p2 )= ln
2Op －RT
G )3(ο∆
∴log(CO
CO
p
p2 )=
21
log2Op
RT3.22
G )3(
×
∆−
ο
∴log(CO
CO
p
p2 )=
21
log2Op +
R3.2262.173
RT3.22564800
×−
×
For a fixed 2Op , log(
CO
CO
p
p2 ) vs. (
T1
) is linear with same slope of (RT3.22
564800×
)
Figure 12.13
Fix p2O , as T ↑ , (
T1
) ↓ ⇒
CO
CO
p
p2 ↓ ,
2CO
CO
p
p ↑
i.e. T ↑ ⇒
2CO
CO
p
p ↑, more reducing gas.
See Figure.12.14.
2. At a fixed T, (T1
), there is a lower limit of
CO
CO
p
p2 ,
[ upper limit of
2CO
CO
p
p] , because reaction (4) sets an equilibrium ,
as
2CO
CO
p
p ↑ ⇒ C )(S precipitates
(4) C )(S + CO )(g2 ＝ 2 CO )(g
∆G o4 = －RT ln K = －RT ln
2CO
2CO
pp
At a total pressure : P = 1 atm , p2CO = 1－ p CO
K = CO
2CO
p1p
−= exp(－
RTG o
4∆) = (1.3 × 10 9 ) exp(－
T20532
) = x
∴ p 2CO + p CO xx −⋅ = 0
∴ p CO = 2
x4xx 2 ++−
and p2CO = 1－ p CO =
2x4xx2 2 +−+
∴
CO
CO
p
p2
itlim =x4xx
x4xx22
2
++−
+−+= f(T)
T ↓, (T1
) ↑ ⇒ x ↓ ⇒
CO
CO
p
p2
itlim ↑, see Figure 12-13.
3. MO )(s reduced in CO/ CO 2 gas mixture
＊ MO )(s + CO )(g = M )(s + CO )g(2
Equilibrium line can be added into Figure12-13.
e.g. (A) FeO )(s + CO )(g = Fe )(s + CO )(g2
∆G o)A( = －22800 + 24.26T
∴ ∆G o)(A = －RT ln K = －RT ln
CO
CO
p
p2
∴ log
CO
CO
p
p2 = －
RT3.2
Go)A(∆
= T
1192－1.27 ( line AB )
Note : ＊ any gas state , which lies above AB is oxidizing with respect to Fe )(s
＊ Point A is the lowest temperature T A , at which FeO can be reduced by
C )(s .
Example 1 :
(1) CO )g( + 21
O )g(2 = CO )g(2 ∆G o1 =－282400+86.81T
Given (2) H )g(2 +21
O )g(2 = H 2 O )g( ∆G o2 =－246400+54.8T
(3) Co )s( +21
O )g(2 = CoO )s( ∆G o3 =－282400+71.85T
Ask : 1. Which ( H )(g2 or CO )(g ) is a more efficient reducing gas ?
2. One mole H )(g2 , how many moles of CoO )(s can be reduced at
T = 1673 K and T = 873 K ?
3. One mole CO )(g , how many moles of CoO )(s can be reduced at
T = 1673 K and T = 873 K ? Sol :
1. ∵ ∆G o1 = ∆G o
2 , T o = 1125 K = 852℃
When T＞T o , ∆G o1＜∆G o
2 , H )(g2 is a more efficient reducing gas.
When T＜T o , CO )(g is a more efficient reducing gas.
2. CoO )s( + H )g(2 = Co )s( + H 2 O )g( … … .(4)
(4) = (2)－(3), ∆G o4 = －12500＋17.05 T
CoO )s( + H )g(2 = Co )s( + H 2 O )g(
Initial 1 1 0 0 Final 1-x 1-x x x
∆G o4 = －RT ln K 4 = －RT ln
2
2
H
OH
p
p
∴ K 4 = exp(－RTG o
4∆) =
2
2
H
OH
p
p
T = 1673 K , K 4 = 19.1 = ( )x1
x
−
∴ x = 1K
K
4
4
+ =
1.201.19
= 0.95 (mole)
T = 873 K , K '4 = 43.5
x ' = 1K
K'4
'4
+ =
5.445.43
= 0.978 (mole )
3. CoO )s( + CO )g( = Co )s( + CO )g(2 .(5)
(5) = (1)－(3), ∆G o5 = －48500＋14.96 T
CoO )s( + CO )g( = Co )s( + CO )g(2
Initial 1 1 0 0 Final 1-y 1-y y y
∴ ∆G o5 = －RT ln K 5 = －RT ln
CO
CO
p
p2
∴ K 5 = exp(－RTG o
5∆) =
CO
CO
p
p2 = ( )y1
y
−
T = 1673 K , K 5 = 0.52 y = 1K
K
5
5
+ =
52.152.0
= 0.34 (mole)
T = 873 K , K '5 = 132 y ' =
133132
= 0.992 (mole)
∴ T = 1673 K , x ＞ y , H )g(2 is more efficient
T = 873 K , x ＜ y , CO )g( is more efficient
Example 2 : NiO )s( + Cl 2 )g( = NiCl2 )s( + 21
O )g(2
At T = 900 K ∆G o = －15490 J
If 90 % Cl2 )g( must be converted into O )g(2 ,
What is the total pressure ? P =?
Sol: NiO )s( + Cl 2 )g( = NiCl2 )s( + 21
O )g(2
Initial 1 1 0 0
Final 1-x 1-x x 21
x
∴ ∆G o = －RT ln K= －RT ln
2
2
Cl
21
O
p
p
∴
2
2
Cl
21
O
p
p = exp(－
RT
Go∆) = 7.925
For gas phase : n T = (1-x) + 21
x = 1－21
x
∴
⋅
−−
⋅
−
P)2/x(1
x1
P)2/x(1
2/x 21
= 7.925
But x = 90 % = 0.9 ∴ ( )
P182.0818.0 2
1
= 7.925 ∴ P = 0.393 atm
Example 3 : (1) ZnO )s( + CO )g( = Zn )g( + CO )g(2 , ∆G o1 =－177800－111.2 T
Q –1 : Reaction (1) : T = 1223 K , P = 1atm , final composition of gas phase p i =?
Sol–1: T = 1223 K , ∆G o1 = 41800
∵ ∆G o1 = －RT ln K 1= －RT ln
⋅
CO
ZnCO
p
pp2
∴
⋅
CO
ZnCO
p
pp2 = exp (－
RTG o
1∆) = 0.0164
P = p Zn + p2CO + p CO but p Zn = p
2CO
∴ p CO = 1－2 p Zn
∴ Zn
2Zn
p21p
−= 0.0164
∴ p Zn = 0.113 atm
p2CO = 0.113 atm
p CO = 0.774 atm
Q-2 : When P ↑ (p Zn ) increases , condensation of Zn )g( occurs .
T = 1223 K , p Zn = ? P C = ?
Given : Zn )l( = Zn )g(
ln p o)l(Zn =
T15250−
－1.255 ln T + 21.58 (atm)
Sol –2 : T=1223 K , p o)l(Zn = 1.49 atm
∵ K 1=
⋅
CO
ZnCO
p
pp2 = 0.0164 , p Zn = p
2CO =1.49 atm
∴ 0.0164 = )49.12(P
49.149.1
c ×−×
∴ P C = 138.35 atm , p CO = 135.37 atm
Q-3 : T=1223 K , P = 150 atm equilibrium p i = ?
Sol –3 : ∵ Zn )l( = Zn )g( equilibrium must be maintained .
∴ p Zn =1.49 atm
But p2CO ≠ p Zn ( ∵ P ＞ 138 atm )
∴ K 1=
⋅
CO
CO
p
49.1p2 = 0.0164 … … .(I)
P = 150 = 1.49 + p2CO + p CO … … ..(II)
Solve (I) (II)
∴ p2CO = 1.61 atm , p CO = 146.9 atm
Example 4 :
Given , CO )g(2 + H )g(2 = CO )g( + H 2 O )g( … (1) ∆G o1 = 36000－32 T
CO－CO 2－H 2－H 2 O gas mixture produced by mixing with
CO )g(2 + H )g(2 1:1 moles , T = 1000 K, P = 1 atm
Q-1 p )eq(i = ?
Sol : ∆G o1 = －RT ln K 1P K 1P = exp(－
RTG o
1∆)= 0.618
K 1P =
⋅
⋅
22
2
HCO
COOH
pp
pp
∵ p CO = p OH2 , p
2CO = p2H
∴ P = 1 = p OH2 + p CO + p
2CO + p2H = 2 (p
2H + p OH2 )
∴ p OH2+ p
2H = 0.5 ∴ p OH2= 0.5－p
2H
∴ p CO = p OH2 = 0.5 － p
2H
p2CO = p
2H
∴ K 1P =
⋅
−⋅−
22
22
HH
HH
pp
)p5.0()p5.0(=
−2
H
2H
2
2
p
)p5.0(
∴ p2H = p
2CO = 0.28 atm
p OH2 = 0.5 － p
2H = 0.22 atm = p CO
Q-2 : Equilibrium gas mixture at initial P = 1 atm , T = 1000 K
Contained in a constant volume vessel . What are final p 'i if CaO )s(
is introduced ?
CaO )s( + 5CO )g( = CaC )s(2 + 3 CO )g(2 … ..(2)
CaO )s( + H 2 O )g( = Ca(OH) )s(2 … … … … … … .(3)
CaO )s( + CO )g(2 = CaCO )s(3 … … … … … … … .(4)
∆G o2 = －37480+300.7 T
∆G o3 = －117600+ 145 T
∆G o4 = －168400+ 144 T
Sol : T = 1000 K
If : Equilibrium (I) K 2 = exp(－RTG o
2∆) =
5'CO
3'CO
p
p2 = 1.78 × 10 14−
substitute p 'CO2
= 0.053 atm
p 'CO = 152.9. atm impossible !!
(II) K 3 = 0.037 = '
OH2p
1
p 'OH2
= 27.027 atm
P ≤ 1 atm impossible !!
(III) K 4 = 18.82 = 'CO2
p1
p 'CO2
= 0.053 atm ＜ p2CO = 0.28 atm Yes !!
∴ only reaction (4) is equilibriated .
∴ p 'CO2
= 0.053 atm (＜ p2CO = 0.28 atm )
∴ CO )g(2 is consumed and (1) P ↓ , (2) reaction (1) ←
at new equilibrium p 'CO = p '
OH2
p 'OH2
+ p 'H2
= 0.5 atm (not changed )
∴ K 1P = 0.618 =
⋅⋅
22
2
H'
CO'
CO'
OH'
pppp
=
⋅− 053.0)p5.0(p
OH'
2OH
'
2
2
∴ p 'OH2
= 0.113 atm = p 'CO
p 'H2
= ( 0.5 － 0.113 ) = 0.387 atm
p 'CO2
= 0.053 atm
P ' = 0.666 atm
Q-3 : If C )s( , graphite is introduced , what is the final p ''i ?
Given , C )s( + CO )g(2 = 2CO )g( … … … … .(5)
∆G o5 = 170700 － 174.5 T
Sol-3 : K 5 = exp(－RTG o
5∆) = 1.579 =
2CO''
2CO
''
pp
In order to maintain equilibrium of (4)
∴ p ''CO2
= p 'CO2
= 0.053 atm
∴ p ''CO = (K 5 p ''
CO2) 2
1
= (1.579 × 0.053 ) 21
= 0.289 atm
However , p ''OH2
+ p ''H2
= 0.5 atm ( not changed !!)
∴ K 1P = 0.618 =
⋅⋅
22
2
H''
CO''
CO''
OH''
pppp
=
⋅
⋅−053.0p
289.0)p5.0(
2
2
H''
H''
∴ p ''H2
= 0.449 atm
p ''OH2
= ( 0.5 － 0.449 ) = 0.051 atm
p ''CO2
= 0.053 atm
p ''CO = 0.289 atm
P = 0.842 atm Note : If process 2 , 3 , Q2 , Q3 are reversed , the final equilibrium p ''
i is the same .