Transcript Equilibrium for a general reaction

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2008
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59-241 Labs
• Find your lab partner within the next few
days.
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Expressions of the equilibrium constant K
• The equilibrium constant, K, (a dimensionless
quantity) can be expressed in terms of fugacities
for gas phase reactions or activities for aqueous
phase reactions.
• Fugacity ( a dimensionless quantity) is equal to
the numerical value of partial pressure, i.e. pj/pθ
where pθ = 1 bar).
• The activity, a, is equal to the numerical value of
the molality, i.e. bj/bθ where bθ = 1 mol kg-1.
For reactions occurring in electrolyte
solutions
• Effects of the interactions of ions on the reaction
process should be considered.
• Such a factor can be expressed with the activity
coefficient, γ, which denotes distance from the ideal
system where there is no ion-interaction.
• The activity shall now be calculated as αj = γj*bj/bθ
• For a reaction A + B ↔ C + D
K =
aC a D  C  D bC bD


a Aa B  A B bAbB
 K K b
The activities of solids are
equal to 1
• α(solid) = 1 (!!!)
• Illustration: Express the equilibrium constant for the
heterogeneous reaction
NH4Cl(s) ↔ NH3(g) + HCl(g)
• Solution:
In term of fugacity (i.e. partial pressure): Kp =
In term of molar fraction: Kx =
Estimate reaction compositions at equilibrium
•
Example 1: Given the standard Gibbs energy of reaction H2O(g) → H2(g) +
1/2O2(g) at 2000K is + 135.2 kJ mol-1, suppose that steam at 200k pa is passed
through a furnace tube at that temperature. Calculate the mole fraction of O2
present in the output gas stream.
•
Solution: (details will be discussed in class)
lnK = - (135.2 x 103 J mol-1)/(8.3145 JK-1mol-1 x 2000K)
= - 8.13037
K = 2.9446x10-4
K=
( PO2 / P  )1 / 2 ( PH 2 / P  )
PH 2O / P 
Ptotal = 200Kpa
assuming the mole fraction of O2 equals x
PO2 = x* Ptotal,
PH2 = 2(x*Ptotal)
PH2O = Ptotal – PO2 – PH2 = (1-3x)Ptotal
Equilibria in biological systems
• Biological standard state: pH = 7.
• For a reaction: A + vH+(aq) ↔ P
Δ rG = Δ r
Gθ
= Δr
Gθ
+ RT
ln(
+ RT ln(
bP
v
b A bH

)
1
 v
[H ]
)  RT ln
bp
bA
the first two terms of the above eq. form ΔrG‡
ΔrG‡ = ΔrGθ + 7vRTln10
Example: For a particular reaction of the form A → B
+ 2H+ in
aqueous solution, it was found that ΔrGθ = 20kJ mol-1 at 28oC. Estimate
the value of ΔrG‡.
• Solution: ΔrG‡ = ΔrGθ
+ 7vRTln10
here v = - 2 !!!
ΔrG‡ = 20 kJ mol-1 + 7(-2)(8.3145x10-3 kJ K-1mol-1)
x(273+ 28K)ln10
= 20 kJ mol-1 – 80.676 kJ mol-1
= -61 kJ mol-1
(Note that when measured with the biological standard, the
standard reaction Gibbs energy becomes negative!)
Molecular Interpretation of equilibrium
The response of equilibria to
reaction conditions
• Equilibria respond to changes in pressure,
temperature, and concentrations of
reactants and products.
• The equilibrium constant is not affected by
the presence of a catalyst.
How equilibria respond to pressure
• Equilibrium constant K is a function of the standard
reaction Gibbs energy, ΔrGθ .
• Standard reaction Gibbs energy ΔrGθ is defined at a
single standard pressure and thus is independent of
pressure.
• The equilibrium constant is therefore independent of
pressure: K
(
p
)T  0
• K is independent of pressure does NOT mean
that the equilibrium composition is independent
of the pressure!!!
consider the reaction 2A(g) ↔ B(g)
assuming that the mole fraction of A equals xA at quilibrium, then
xB = 1.0 – xA,
K=
(1.0  x A ) Ptotal / P
( x A Ptotal / P ) 2
(1.0  x A ) P

2
x A Ptotal
because K does not change, xA must change in response to any
variation in Ptotal!!!
Le Chatelier’s Principle
• A system at equilibrium, when
subject to a disturbance,
responds in a way that tends
to minimize the effect of the
disturbance.
Example: Predict the effect of an increase in pressure on the Haber
reaction, 3H2(g) + N2(g) ↔ 2NH3(g).
• Solution:
According to Le Chatelier’s Principle, an increase in pressure will
favor the product.
prove:
K=
2
p NH
3
3
p N pH
2
2


2
2
x NH
p
total
3
3
x N 2 ptotal x H
2
3
ptotal

2
x NH
3
3
xN2 xH
p2
2 total
Kx
2
ptotal
Therefore, to keep K unchanged, the equilibrium mole fractions Kx will
change by a factor of 4 if doubling the pressure ptotal.
The response of equilibria to
temperature
• According to Le Chatelier’s Principle:
Exothermic reactions: increased temperature favors the reactants.
Endothermic reactions: increased temperature favors the
products.
• The van’t Hoff equation:
(a)
(b)
d ln K
dT
d ln K
1
d( )
T

r H 
RT 2
r H 
 
R
(7.23a)
(7.23b)
Derivation of the van’t Hoff equation:
 r G
ln K  
RT
•
Differentiate lnK with respect to temperature
d ln K
1 d (  r G / T )

dT
R
dT
•
Using Gibbs-Helmholtz equation (eqn 3.53 8th edition)
d (  r G / T )
r H 

dT
T2
thus
•
d ln K
dT

r H 
RT 2
Because d(1/T)/dT = -1/T2:
d ln K
1
d( )
T
r H 
 
R
d ln K
 0 , suggesting
• For an exothermic reaction, ΔrHθ < 0, thus
dT
that increasing the reaction temperature will reduce the equilibrium
constant.