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5.60 Thermodynamics & Kinetics
Spring 2008
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5.60 Spring 2008 Lecture #15 page 1
Chemical Equilibrium
Ideal Gases
Question: What is the composition of a reacting mixture of ideal
gases?
e.g. ½ N2(g, T, p) + 3/2 H2(g, T, p) = NH3(g, T, p)
What are pN , pH , and pNH
2 2 3
at equilibrium?
Let’s look at a more general case
νA A(g, T, p) + νB B(g, T, p) = νC C(g, T, p) + νD D(g, T, p)
The νi‘s are the stoichiometric coefficients.
Let’s take a mixture of A, B, C, and D with partial pressures
pA = XA p , pB = XA p , pC = XC p , and pD = XD p
Is this mixture in equilibrium?
We can answer by finding ∆G if we allow the reaction to proceed
further.
We know µi (T , p ) for an ideal gas in a mixture
and we know that G = ∑ ni µi
i
⇒ { }
∆G (ε ) = ε ⎡⎣ν C µC ( g,T , p ) + ν D µD ( g,T , p ) ⎤⎦ − ⎡⎣ν A µA ( g,T , p ) + ν B µB ( g,T , p ) ⎤⎦
5.60 Spring 2008 Lecture #15 page 2
where ε is an arbitrary small number that allows to let the reaction
proceed just a bit.
⎡ pi ⎤
We know that µi ( g,T , p ) = µio (T ) + RT ln pi ⎢⎣ 1 bar implied⎥⎦
where µio (T ) is the standard chemical potential of species “i” at 1 bar
and in a pure (not mixed) state.
⎧⎪ ⎛ pCνC pDνD ⎞ ⎫⎪
∴ ∆G (ε ) = ε ⎨ ⎣ν C µC (T ) + ν D µD (T ) ⎦ − ⎣ν A µA (T ) + ν B µB (T ) ⎦ + RT ln ⎜ ν A ν B
⎡ o o
⎤ ⎡ o o
⎤ ⎟⎬
⎩⎪ ⎝ pA pB ⎠ ⎭⎪
⇒ ∆G = ∆G ° + RT lnQ (taking ε=1)
where ∆G ° = ⎡⎣ν C µCo (T ) + ν D µDo (T ) ⎤⎦ − ⎡⎣ν A µAo (T ) + ν B µBo (T ) ⎤⎦
pCνC pCνD
and Q = νA νB is the reaction quotient
pA pB
∆G ° is the standard change in free energy for taking pure reactants
into pure products.
∆G o = ∆Grxn
o
= ∆Hrxn
o
−T ∆Srxn
o
or ∆G o = ∆Gform
o
( products ) − ∆Gform
o
( reactants )
If ∆G (ε ) < 0 then the reaction will proceed spontaneously to form
more products
∆G (ε ) > 0 then the backward reaction is spontaneous
∆G (ε ) = 0 No spontaneous changes ⇒ Equilibrium
5.60 Spring 2008 Lecture #15 page 3
At Equilibrium ∆G (ε ) = 0 and this implies ∆Grxn
o
= −RT lnQeq
Define Qeq = K p the equilibrium constant
⎛ pCνC pCνD ⎞ ∆ν ⎛ XC XC ⎞
νC νD
K p = ⎜ ν A νB ⎟ = p ⎜ ν A ν B ⎟ = p ∆ν KX
⎝ pA pB ⎠eq ⎝ XA XB ⎠eq
and thus ∆Grxn = −RT ln K p , K p = e −∆G
o
o RT
Note from this that K p (T ) is not a function of total pressure p.
It is KX = p −∆ν K p which is KX ( p ,T ) .
Recall that all pi values are divided by 1 bar, so Kp and KX are
both unitless.
________________________________________________
Example: H2(g) + CO2(g) = H2O(g) + CO(g) T = 298 K
p =1 bar
H2(g) CO2(g) H2O(g) CO(g)
Initial # a b 0 0
of moles
# moles a-x b-x x x
at Eq.
Total # moles at Eq. = (a – x) + (b – x) + 2x = a + b
a −x b −x x x
Mole fraction
a +b a +b a +b a +b
5.60 Spring 2008 Lecture #15 page 4
at Eq.
∆Gform
o
(kJ/mol) 0 -396.6 -228.6 -137.2
28,600 kJ/mol
−
(8.314 J/K-mol)(298 K )
∴ ∆G o
rxn = 28.6 kJ/mol ⇒ Kp = e = e −11.54 = 9.7 x 10 −6
pH O pCO XH OXCO x2
and Kp = 2
= 2
=
pH pCO
2 2
XH XCO
2 2
( a − x ) (b − x )
Let’s take a = 1 mol and b = 2 mol
x2
We need to solve = 9.7 x 10 −6
(1 − x )(2 − x )
A) Using approximation method:
K << 1, so we expect x << 1 also.
x2 x2
Assume 1 − x ≈ 1, 2 − x ≈ 2 ⇒ ≈ = 9.7 x 10 −6
(1 − x )(2 − x ) 2
x ≈ 0.0044 mol (indeed << 1)
x2
B) Exactly: = K p = 9.7 x 10 −6
x − 3x + 2
2
x 2 (1 − 9.7x 10 −6 ) + 3x ( 9.7x 10 −6 ) − 2 ( 9.7x 10 −6 ) = 0
−3 ( 9.7x 10 −6 ) 9 ( 9.7x 10 −6 ) + 4 (1 − 9.7x 10 −6 ) 2 ( 9.7x 10 −6 )
2
x = ±
2 ( 9.7x 10 −6 ) 2 (1 − 9.7x 10 −6 )
The “-“ sign gives a nonphysical result (negative x value)
Take the “+” sign only ⇒ x = 0.0044 mol (same)
5.60 Spring 2008 Lecture #15 page 5
Effect of total pressure: example
N2O4(g) = 2 NO2(g)
Initial mol # n 0
# at Eq. n-x 2x Total # moles at Eq. =
n – x + 2x = n + x
n −x 2x
Xi’s at Eq.
n +x n +x
2
⎛ 2x ⎞
⎜
pNO p 2XNO n + x ⎟⎠
2 2
4x 2
Kp = 2
= 2
=p⎝ =p 2
pN O pXN O ⎛n − x ⎞ n −x2
2 4 2 4
⎜ ⎟
⎝n + x ⎠
4α 2
Kp = p where α = x n is the fraction reacted
1 − α2
Kp
−1 2
K ⎛ Kp ⎞ Kp 4p 1 ⎛ 4p ⎞
(1 − α ) 4 pp = α 2
2
α ⎜1 +
2
p
⎟= α2 = =
Kp ⎞ ⎛
α = ⎜1 +
⎜ ⎟
K p ⎠⎟
⎝ 4 ⎠ 4p ⎛ 4p ⎞ ⎝
⎜1 + ⎟ ⎜⎜ 1 + ⎟
⎝ 4p ⎠ Kp ⎟
⎝ ⎠
∴ If p increases, α decreases
Le Chatelier’s Principle, for pressure:
An increase in pressure shifts the equilibrium so as to decrease the
total # of moles, reducing the volume.
In the example above, increasing p shifts the equilibrium toward the
reactants.
---------------
5.60 Spring 2008 Lecture #15 page 6
Another example:
2 NO(g) + O2(g) = 2 NO2(g)
K p = 2.3x 1012 at 298 K
Initial mol # 2 1 0
# at Eq. 2-2x 1-x 2x Total # moles at Eq.
= 2 – 2x + 1 – x + 2x
2 (1 − x ) 1−x 2x
Xi’s at Eq. =3-x
3−x 3−x 3−x
pNO
2
p 2XNO
2
1 XNO
2
1 x (3 − x )
2
Kp = 2
= 2
= 2
=
pNO
2
pO 2
p 2XNO
2
pXO 2
p XNO
2
XO 2
p 3
(1 − x )
K p >> 1 so we expect x ≈ 1 ⇒ 3 - x ≈ 2
13
1 2 2 ⎛ 2 ⎞
(1 − x )
3
Kp ≈ or ≈ x =1−⎜ ⎟⎟
p (1 − x )3 pK p ⎜ pK p
⎝ ⎠
In this case, if p↑ then x↑ as expected from Le Chatelier’s principle.
