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.
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Spontaneity and Equilibrium isolated system : Isothermal process Maximum work obtained in a process at constant temperature is equal to the decrease.
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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 DHo and DSo 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
For a given reaction, the equilibrium constant is 1.80x103 L/mol at 25oC and 3.45x103 L/mol at 40oC. Assuming DHo to be independent of temperature, calculate DHo and DSo.
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
DFGº kJ/mol
DFHº kJ/mol
CaCO3(s) -1128.8 -1206.9
CaO(s) -604.0 -635.1
CO2(g) -394.4 -393.51
Calculate The pressure of CO2 at 25oC and at 827oC?
DGº =130.4 kJ
DHº =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
dTRT
Hpd
dTRT
HKd
ovap
OH
o
p
g 2
2
)(2ln
ln
CTR
Hp
ovap
OH v
1ln
)(2Clausius-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
ba
dc
[B][A][D][C]
Q
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Calculate [X]equilibrium from [X]initial and KC
Ex. 4 H2(g) + I2(g) 2HI(g) at 425 °C
KC = 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
IHHI
Kc
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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
2
22
)00.2(
)2()00.2)(00.2(
)2(64.55
x
xxx
x
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Ex. 4 Step 3. Solve for x• Both sides are squared so we can take
square root of both sides to simplify
2
2
)00.2(
)2(64.55
x
xK
)00.2(2
459.7x
x
xx 2)00.2(459.7
xx 2459.7918.14
58.1459.9918.14 x
x459.9918.14
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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
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Calculate [X]equilibrium from [X]initial and KC
Ex. 5 H2(g) + I2(g) 2HI(g) at 425 °C
KC = 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
IHHI
Kc
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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?
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Calculating KC Given Initial Concentrations and One Final Concentration
Ex. 2a
H2(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