Chapter 6: Chemical Equilibria

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Chapter 6 Chemical Equilibria

• Chemical Potential of mixtures
• Reaction Gibbs energy, ?Gr
• ? Variation of ?Gr with composition
• Equilibrium constant
• Effect of temperature
• Effect of pressure

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? Chemical potential of mixtures
at constant T and single component ideal gas
m G/n
For gas mixture
Pi ? partial pressure of gas xi P
xi ? mole fraction of gas A P ? total pressure of
gas mixtures
P
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Pure liquid in equilibrium with its vapor
A (g)
A (l)
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Liquid mixture in equilibrium with vapors
A (g) B (g)
A (l) B (l)
Amount of A in vapor corresponds to amount of A
in mixtures if amount of A in mixtures is xA
Raoults law
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Raoults law is correct for ideal solution for
real solution,
activity of solvent A
as
as
aA gAxA
for solvent
for solute
aB gBxB
gB ? 1 as xB ? 0
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For dilute solution
mB molality
Similarly one obtains
CB molality
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? Reaction Gibbs energy (?Gr )
dP
For general case
At constant T, P
In this case
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Since the increase in amount of B corresponds to
the reduce in amount of A. If d? is the amount
change or extent of reaction, we can write (in
this case).
Law of conservation of mass
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lt
0
dG
reverse reaction spontaneous
gt
gt 0
0
dG
equilibrium
0
dG

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reaction Gibbs energy
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Variation of ?Gr with composition
From
From
reaction quotient
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For
at equilibrium
equilibrium constant
equilibrium activity of component i
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• no achievement of 100P
• reaction carries on until equilibrium
• is reached
• at equilibrium more P than R
• K gt 1

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• reaction carries on until equilibrium
• is reached.
• at equilibrium more R than P
• K lt 1

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• reaction carries on until equilibrium
• is reached
• at equilibrium R P
• K 1

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Example 6.1 Calculate equilibrium constant at 298
K for the reaction
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Equilibrium constant
N2(g) 3H2(g) 2NH3(g)
From aA PA/PA for liquids aA
fA/P0 for gas fA fugacity (real gas
pressure) PA for ideal gas
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P0 1 atm For ideal gas
Pi xiP
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• For general case
• aA bB cC dD
• fi ? Pi as P ? 0 (ideal gas)

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• aBgBxBgBmB

m0 1 mol/kg solvent
Km as Kg ? 1 for very dilute solution
C0 1 mol/L
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Example 6.2 In an industrial process, N2 at 1.00
bar is mixed with H2 at 3.00 bar and two gases
are allowed to come to equilibrium with the
product NH3 (in presence of a catalyst) in a
reactor of constant volume. At temperature of
the reaction, it has been deter- mined
experimatally that K 977. What are the
equilibrium partial pressure of the three gases?
Find also the equilibrium constant in terms of
concentration. 3H2(g) N2(g) 2NH3(g) 3.00-3x
1.00-x 2x
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0.895, 1.12
gt1.00
(PNH3)eq 2x0.895 1.79 bar (PH2)eq 3.00
3x0.895 0.31 bar (PN2)eq 1.00 0.895
0.10 bar
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From PV nRT ? P (n/V)RT CRT
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Effect of Temperature
Gibbs-Helmholtz equation
DGr0 -RTlnK
vant Hoff equation
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lnK
DHr0/R
1/T
Alternative derivation of vant Hoff equation
intercept
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Example 6.3 The equilibrium constant for ammonia
synthesis at 298 K is 6.0x105. Estimate its value
at 500 K.
From
N2(g) 3H2(g) 2NH3 (g) DHr0
2DHf0(NH3,g) -92.2 kJ
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Effect of Pressure
From DGr0 -RTlnK since DGr0 is measured at
constant P then K is independent of P
However equilibrium composition depends on P
followed Le Chateliers principle
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Example 6.4 Find equilibrium composition of HI in
the reaction H2(g) I2(s) 2HI(g)