Entropy, free energy and equilibrium Spontaneity Entropy Free energy and equilibrium.
Spontaneity and Equilibrium
47
Spontaneity and Equilibrium isolated system: 0 S Isothermal process A w dA dw TS E d dw TS d dE dw TdS dE dw dw dE TdS dQ TdS dQ dQ rev A w TS E A Energy Helmholz Maximum work obtained in a process at constant temperature is equal to the decrease in Helmholtz energy of the system.
description
Spontaneity and Equilibrium. isolated system : . Isothermal process. Maximum work obtained in a process at constant temperature is equal to the decrease in Helmholtz energy of the system. Calculate the maximum work that can be obtained from the combustion of 1 mole of methane at 298 K. - PowerPoint PPT Presentation
Transcript of Spontaneity and Equilibrium
Spontaneity and Equilibrium
isolated system: 0S
Isothermal process
Aw
dAdw
TSEddw
TSddEdw
TdSdEdw
dwdETdS
dQTdS
dQdQrev
Aw
TSEA
EnergyHelmholz
Maximum work obtained in a process at constant temperature is equal to the decrease in Helmholtz energy of the system.
Calculate the maximum work that can be obtained from the combustion of 1 mole of methane at 298 K.
Given Ho and So of the the combustion of methane.
)(2)(2)(2)(4 22 gggg OHCOOCH
Aw
STRTnHA
STpVHA
STpVHA
STEA
TSEA
max
Transformation at constant temperature and pressure
Gw
dGdw
TSHddw
TdSdHdw
TdSVdppdVdHdwpdV
TdSVdppdVdHdwdw
TdSpVHddw
dwdETdS
dQTdS
dQdQ
Vpnon
Vpnon
Vpnon
Vpnon
Vpnon
VpnonVp
rev
Gw Vpnon
TSHG
EnergyGibbs
Maximum work, over and above pV-work, obtained in a process at constant temperature and pressure is equal to the decrease in the Gibbs energy of the system.
G
Gw Vpnon
0
Special case:No work over and above pV-work
wnon p-V=0
forcedG
mEquilibriuG
eoussponG
0
0
tan0
Calculate the maximum non-pV work that can be obtained from the combustion of 1 mole of methane at 298 K.
Fundamental equations of Thermodynamics
Maxwell Relations
pdVTdSdE
pdVdQdE
dwdQdE
rev
VS S
pVT
VdpTdSdH
VdppdVpdVTdSdH
VdppdVdEdH
pVEH
pS SV
pT
pVET
SE
dVVEdS
SEdE
VSfE
SV
SV
),(
VpHT
SH
dppHdS
SHdH
pSfH
Sp
Sp
),(
pdVSdTdA
SdTTdSpdVTdSdA
SdTTdSdEdA
TSEA
VT Tp
VS
VdpSdTdG
SdTTdSVdpTdSdG
SdTTdSdHdG
TSHG
pT TV
pS
pVAS
TA
dVVAdT
TAdA
VTfA
TV
TV
),(
VpGS
TG
dppGdT
TGdG
pTfG
Tp
Tp
),(
1212
12
2
1
2
1
2
1
:/
),(
ppVpGpG
dpVpGpGsolidsliquidsFor
VdpdG
VdpdppGdG
dppGdT
TGdG
pTfG
p
p
p
p
G
G
T
Tp
Transformation at constant temperature
pRTp
ppRTp
ppRT
nG
npG
ppnRTGpG
pp
ppnRTpGpG
dpp
nRTpGpGgasidealFor
o
oo
o
o
oo
o
p
p
ln
ln
ln
ln
ln
:
22
22
22
1
1
212
12
2
1
Chemical potential
121
1
2
2
121
1
2
2
121
1
2
2
2
/
/
2
22
22
11
11
11
/
./
/
11/
/11/
2
1
22
11
TTH
TTG
TTG
TTH
TTG
TTG
TTH
TTG
TTG
dTTHTGd
eqHelmholzGibbsdTTHTGd
TH
TS
TH
TS
TTG
TTSH
TS
TG
TS
TTG
TTG
TG
TTTG
T
T
TG
TG
p
p
ppp
The value of Gfo of Fe(g)
is 370 kJ/mol at 298 K. If Hf
o of Fe(g) is 416 kJ/mol (assumed to be constant in the range 250-400 K), calculate Gf
o of Fe(g) at 400 K.
Transformation at constant pressure
G dependence on n
H2O
H2O
nMwtmnG
Mwtnm
nG
GG
propertyextensivenondependsE
EE
cmE
21
21
2
22112211
2121
nMwtnMwtmnnG
mmmGGG
Given a system consisting of two substances:
ii
i nG
iiipT
iii
nTnp
nTnp
npTnpTnTnp
i
dndG
dndppGdT
TGdG
dndndppGdT
TGdG
dnnGdn
nGdp
pGdT
TGdG
nnnpTfG
ii
ii
iiii
,
,,
2211,,
2,,2
1,,1,,
22
...
...
),...,,,,(
21
j
npTjji
nG
,,
If there is no change in composition:
0
0
,
pT
i
dG
dn
System at constant T and p
a
dn1
Each subsystem is a mixture of substances.
b
Chemical Equilibrium
ba 11 substsubst
1111111
11
11
dndndndG
dndG
dndG
dGdGdG
tot
tot
abba
bb
aa
ba
mequilibriudGif
eousspondGif
tot
tot
0
tan0
11
11
ab
ab
Equilibrium is established if chemical potential of all substances in the system is equal in all parts of the system.
Matter flows from the part of system of higher chemical potential to that of lower chemical potential.
pTpT
xRTpTpT
xRTpTpT
xRTpRTT
xRTpRTT
pxRTTpRTT
pTpRTTpRTT
oHpure
mixtureH
io
ipurei
Ho
HpuremixtureH
HoH
mixtureHH
HoH
mixtureHH
HoHH
oH
mixtureHH
oHpure
oHH
oH
pureHH
,,
ln,,
ln,,
lnln
lnln
lnln
,lnln
22
222
2222
2222
222222
222222
ab
bb
bb
aa
a
Pure H2
b
N2 + H2
Pd membrane
Equilibrium never reached
constant T & p
ba pp
G and S of mixing of gases
iiimix
if
oof
oof
mixturemixturemixturei
iif
oopurei
iii
ifmix
xnRTxRTnxRTnG
xRTnxRTnGG
xRTnxRTnpTnpTnG
xRTpTnxRTpTnG
pTnpTnnG
pTnpTnnG
GGG
lnlnln
lnln
lnln,,
ln,ln,
,,
,,
2211
2211
22112211
222111
2211
2211
.0
0ln1
lnlnln
lnlnln
2211
22
11
2211
spontalwaysG
xx
xxRTnxRTxnxRTxnG
nnx
nnx
xnRTxRTnxRTnG
mix
ii
iiitottottotmix
tottot
iiimix
mixp
mix
pp
STGS
TGS
TG
0ln
lnln
iiitotmix
iiitot
iiitot
p
mix
xxRnS
xxRndT
xxRTnd
TG
0.0 0.2 0.4 0.6 0.8 1.00
2
4
6
Sm
ix (J
/mol
.K)
x1
0.0 0.2 0.4 0.6 0.8 1.0
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
x1
Gm
ix/nRT
STHG
TdSdHdG
dTconstT
SdTTdSdHdG
TSHG
0.
0
mixmixmix
mixmixmix
STGH
STHG
mixT
mix
TT
Vp
G
VpGV
pG
Chemical reactionsCH4(g) +2O2(g) → CO2(g) + 2H2O(g)
i
oii
o
iii
CHOCOOH
ii
iim
CHmOmCOmOHm
Rpif
G
G
G
nGG
GGGGG
GGGGG
4222
4222
22
22
,
,,,,
Heat of FormationFormation reaction: reaction of forming 1 mole of product from the elements in their stable form at 25ºC and 1 atm.Heat of formation = H of formation reaction = FH
Standard heat of formation = Hº of formation reaction = FHº
FHº(NO(g)): ½ N2(g)+½ O2(g) → NO(g) Hº
FHº(CO(g)): Cgraphite(s)+½ O2(g) → CO(g) Hº
FHº(O(g)): ½ O2(g) → O(g) Hº FHº(Cdiamond(s)): Cgraphite(s) → Cdiamond(s) Hº
FHº(O2(g)): O2(g) → O2(g) Hº=0 FHº(Cgraphite(s)): Cgraphite(s) → Cgraphite(s) Hº=0
G of FormationGibbs energy of formation = G of formation reaction = FG
Standard Gibbs energy of formation = Gº of formation reaction = FGº
FGº(NO(g)): ½ N2(g)+½ O2(g) → NO(g) Gº
FGº(CO(g)): Cgraphite(s)+½ O2(g) → CO(g) Gº
FGº(O(g)): ½ O2(g) → O(g) Gº FGº(Cdiamond(s)): Cgraphite(s) → Cdiamond(s) Gº
FGº(O2(g)): O2(g) → O2(g) Gº=0 FGº(Cgraphite(s)): Cgraphite(s) → Cgraphite(s) Gº=0
01,25
atmCelementsstable
G oo
F
oiFi
orct
oiFi
orct GGHH
Applying Hess’s Law
oiF
oi
i
oiFi
i
oii
o
G
GG
01,25
01,25
atmCelementsstable
atmCelementsstable
G
ooi
ooiF
Chemical reactions
CH4(g) +2O2(g) → CO2(g) + 2H2O(g)
., constpT
As the reaction proceeds: The number of moles of involved substances
changes. G of system will change:
ii
i
ii
i
dndG
nG
ioii nn
ddn ii
Extent of reactionReaction
advancementDegree of reaction
ii
ipT
ii
iii
i
ddG
ddndG
,
As the forward reaction proceeds: grows, d positive, d > 0
destablishemequilibriudGddGif
spontproceedtdoesnrctforwarddGddGif
spontproceedsrctforwarddGddGif
pT
pT
pT
00
.'00
.00
,
,
,
.spontproceedsrctreversed
even though G of products larger than G of reactants, the reaction proceeds!!!!!!!!!!
Reason: Gmix
0,
ii
ipTd
dG
mequilibriuat
For a mixture:
i
oii
oi
i
oi
oi
ii
oi
oi
iipure
mixpureoii
ii
oi
iitotal
oii
oi
ii
oi
oii
iitotal
ii
itotal
nnnG
GGnnG
nnG
nG
mixpuretotal
mixpuretotal
GGG
GGG
G more negative if Gpure is small Gmix largely negative
Equilibrium constant
DdCcBbAa
po
io
bB
aA
cC
dDo
aA
bB
cC
dD
o
aA
bB
cC
dD
oA
oB
oC
oD
AoAB
oB
CoCD
oD
AoAB
oB
CoCD
oD
ioii
ABCDi
ii
QRTGpRTGG
ppppRTGG
ppppRTGG
pRTpRTpRTpRT
TaTbTcTdG
pRTaTapRTbTb
pRTcTcpRTdTdG
pRTTapRTTb
pRTTcpRTTdG
pRTTT
abcdG
i lnln
ln
lnlnlnln
lnlnlnln
lnln
lnln
lnln
lnln
ln
atmpRTTT
ppRTTT
ioii
oio
ii
1ln
ln
R, P: Ideal gases
RTGK
KRTG
KQpp
pp
constpmequilibriuat
ppppRTGG
Gmequilibriuat
o
p
po
peqpeq
bB
aA
cC
dD
i
eqbB
aA
cC
dDo
ln
0ln
.:
0ln
0:
mequilibriuGKQif
spontrctforwardGKQif
spontnonrctforwardGKQif
KQ
RTG
QRTKRTG
QRTGG
pp
pp
pp
p
p
pp
po
0
.0
.0
ln
lnln
ln
eq
b
o
Ba
o
A
c
o
Cd
o
D
eqbB
aA
cC
dD
p
pp
pp
pp
pp
ppppK
totalii pxp
i
totalxp
bacd
o
totalxp
eq
b
o
totala
o
total
c
o
totald
o
total
eqbB
aA
cC
dD
eq
b
o
totalBa
o
totalA
c
o
totalCd
o
totalD
p
pKK
ppKK
pp
pp
pp
pp
xxxx
ppx
ppx
ppx
ppx
K
Kp relation to Kx
ptotal in atm
Kp relation to Kc
RTcpRTVnpRTnVp ii
iiii
iRTKK
RTcc
ccRTcRTcRTcRTc
ppppK
cp
bacd
eqbB
aA
cC
dD
eqb
Ba
A
cC
dD
eqbB
aA
cC
dD
p
10821.010821.01
1/1
KKmolatmL
LatmmolpRc
atmpLmolc
pRTc
cc
pcRTcc
pRTc
ppp
ppppK
o
o
oo
o
o
o
i
oo
oi
o
i
o
ii
eqbB
aA
cC
dD
p
eq
bB
aA
cC
dD
c ccccK c in mol/L
R=0.0821 atmL/mol.K
Consider the reaction N2O4(g) → 2 NO2(g)
FGº(NO2(g))=51.31 kJ/mol FGº(N2O4(g))=102.00 kJ/mol
Assume ideal behavior, calculate
242
42
0,1
?%50)4
1)3
)2
25)1
NOmolONmolmolesofnumberinitial
ddissociateONissystemtheinpressurewhatAt
atmatK
K
CatK
x
c
op
no neq Sni xeq pi
N2O4 1 1-x 1+x (1-x)/(1+x) (1-x)/(1+x)*Ptot
NO2 0 2x 2x/(1+x) 2x/(1+x)*Ptot
5.01
deg x
nnondissociatiofree o
reacted p
xxxK p
11
2 2
Temperature dependence of Kp
CTR
HK
TdT
RHKd
TTRHTKTK
TKTK
TdT
RHKd
dTRT
HKd
RTH
TH
RdTTGd
RdTKd
TG
RRTGK
o
p
o
p
o
ppp
p
T
T
oTK
TKp
o
p
oo
o
p
oo
p
p
p
1ln
ln
11lnlnln
ln
ln
11ln
1ln
2
1212
1
2
2
)(
)(
2
22
2
1
2
1
bxay
0 1000 2000 3000 4000 5000
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
.0 endothH
RH
o
o
500 1000 1500 2000 2500 3000
0
10
20
30
40
50
60
70
1/T
T
ln K
p
K p
0.0005 0.0010 0.0015
0
2
4
.0 exothH
RH
o
o
RSconstIntercept
RS
RTH
RTSTH
RTGK
o
ooooo
p
.
ln
For the reaction N2O4(g) → 2 NO2(g)
FHº(NO2(g))=51.31 kJ/mol FHº(N2O4(g))=102.00 kJ/mol
Kp(25oC)=0.78 atm, calculate Kp at 100oC?
For a given reaction, the equilibrium constant is 1.80x103 L/mol at 25oC and 3.45x103 L/mol at 40oC. Assuming Ho to be independent of temperature, calculate Ho and So.
Heterogeneous Equilibria
)(2)()(3 gss COCaOCaCO
i
oii
o
iii
G
G
eqCOp
COo
COoi
oi
oi
oi
oiCO
oi
oii
oii
COoii
ioii
iii
pK
pRTGG
pRTCaCOCaOCOG
CaCOCaOpRTCOG
CaOCaO
CaCOCaCO
pRTCOCO
pRTp
CaCOCaOCOG
2
2
2
2
2
ln
ln
ln
ln
ln
32
32
33
22
32
FGº kJ/mol
FHº kJ/mol
CaCO3(s) -1128.8 -1206.9CaO(s) -604.0 -635.1CO2(g) -394.4 -393.51
Calculate The pressure of CO2 at 25oC and at 827oC?
Gº =130.4 kJ
Hº =178.3 kJ
ln(Kp)=ln(pCO2)=-52.6
pCO2=1.43x10-23 atm
At 1100 K: ln(pCO2)=0.17
pCO2=0.84 atm
Vaporization Equilibria
)()( gl AA )(2)(2 gl OHOH
eqOHp
OHo
loiOHg
oi
ligi
g
g
g
pK
pRTGG
OHpRTOHG
OHOHG
)(2
)(2
)(2
ln
ln )(2)(2
)(2)(2
dTRTH
pd
dTRT
HKd
ovap
OH
o
p
g 2
2
)(2ln
ln
CTR
Hp
ovap
OH v
1ln)(2
Clausius-Clapeyron Equation
Derive the above relations for the sublimation phase transition!
Mass Action Expression (MAE)
• For reaction: aA + bB cC + dD
Reaction quotient– Numerical value of mass action expression– Equals “Q” at any time, and– Equals “K” only when reaction is known to be at
equilibrium
badc
[B][A][D][C]Q
41
Calculate [X]equilibrium from [X]initial and KC
Ex. 4 H2(g) + I2(g) 2HI(g) at 425 °CKC = 55.64 • If one mole each of H2 and I2 are placed in a
0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI?
• Step 1. Write Equilibrium Law
64.55]][[][
22
2
IHHIKc
42
Ex. 4 Step 2. Concentration Table
Conc (M) H2(g) + I2(g) 2HI (g)Initial 2.00 2.00 0.000ChangeEquil’m
• Initial [H2] = [I2] = 1.00 mol/0.500L =2.00M• Amt of H2 consumed = Amt of I2 consumed = x • Amt of HI formed = 2x
– x +2x– x+2x2.00 – x 2.00 – x
222
)00.2()2(
)00.2)(00.2()2(64.55
xx
xxx
43
Ex. 4 Step 3. Solve for x• Both sides are squared so we can take
square root of both sides to simplify
22
)00.2()2(64.55x
xK
)00.2(2459.7
xx
xx 2)00.2(459.7
xx 2459.7918.14
58.1459.9918.14 x
x459.9918.14
44
Ex. 4 Step 4. Equilibrium Concentrations
Conc (M) H2(g) + I2(g) 2HI (g)Initial 2.00 2.00 0.00ChangeEquil’m
• [H2]equil = [I2]equil = 2.00 – 1.58 = 0.42 M• [HI]equil = 2x = 2(1.58) = 3.16
– 1.58 +3.16– 1.58+3.160.42 0.42
45
Calculate [X]equilibrium from [X]initial and KC
Ex. 5 H2(g) + I2(g) 2HI(g) at 425 °CKC = 55.64 • If one mole each of H2, I2 and HI are placed in
a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI?
• Now have product as well as reactants initially• Step 1. Write Equilibrium Law
64.55]][[][
22
2
IHHIKc
46
Calculate [X]equilibrium from [X]initial and KC
Ex. 6 CH3CO2H(aq) + C2H5OH(aq) CH3CO2C2H5(aq) +
acetic acid ethanol ethyl acetate H2O(l)
KC = 0.11 • An aqueous solution of ethanol and acetic
acid, each with initial concentration of 0.810 M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium?
47
Calculating KC Given Initial Concentrations and One Final Concentration
Ex. 2aH2(g) + I2(g) 2HI(g) @ 450 °C• Initially H2 and I2 concentrations are 0.200
mol each in 2.00L (= 0.100M); no HI is present
• At equilibrium, HI concentration is 0.160 M• Calculate KC• To do this we need to know 3 sets of
concentrations: initial, change and equilibrium
