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.
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
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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.
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
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.
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