# 5.60 Thermodynamics &amp; Kinetics ```MIT OpenCourseWare
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5.60 Thermodynamics &amp; Kinetics
Spring 2008
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.
&frac12; 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 &micro;i (T , p ) for an ideal gas in a mixture
and we know that G = ∑ ni &micro;i
i
⇒
{
}
∆G (ε ) = ε ⎡⎣ν C &micro;C ( g,T , p ) + ν D &micro;D ( g,T , p ) ⎤⎦ − ⎡⎣ν A &micro;A ( g,T , p ) + ν B &micro;B ( 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.
We know that
⎡ pi
⎤
⎢⎣ 1 bar implied⎥⎦
&micro;i ( g,T , p ) = &micro;io (T ) + RT ln pi
where &micro;io (T ) is the standard chemical potential of species “i” at 1 bar
and in a pure (not mixed) state.
∴
⎧⎪
⎛ pCνC pDνD
o
o
o
o
⎡
⎤
⎡
⎤
∆G (ε ) = ε ⎨ ⎣ν C &micro;C (T ) + ν D &micro;D (T ) ⎦ − ⎣ν A &micro;A (T ) + ν B &micro;B (T ) ⎦ + RT ln ⎜ ν A ν B
⎝ pA pB
⎩⎪
⇒
∆G = ∆G &deg; + RT lnQ
⎞ ⎫⎪
⎟⎬
⎠ ⎭⎪
(taking ε=1)
where
∆G &deg; = ⎡⎣ν C &micro;Co (T ) + ν D &micro;Do (T ) ⎤⎦ − ⎡⎣ν A &micro;Ao (T ) + ν B &micro;Bo (T ) ⎤⎦
and
pCνC pCνD
Q = νA νB is the reaction quotient
pA pB
∆G &deg; is the standard change in free energy for taking pure reactants
into pure products.
o
o
o
∆G o = ∆Grxn
= ∆Hrxn
−T ∆Srxn
or
o
o
∆G o = ∆Gform
( reactants )
( products ) − ∆Gform
If
∆G (ε ) &lt; 0
then the reaction will proceed spontaneously to form
more products
∆G (ε ) &gt; 0
then the backward reaction is spontaneous
∆G (ε ) = 0
No spontaneous changes
⇒
Equilibrium
5.60 Spring 2008
Lecture #15
page 3
o
= −RT lnQeq
At Equilibrium ∆G (ε ) = 0 and this implies ∆Grxn
Define
Qeq = K p
the equilibrium constant
νD
νC
⎛ pCνC pCνD ⎞
∆ν ⎛ XC XC ⎞
K p = ⎜ ν A νB ⎟ = p ⎜ ν A ν B ⎟ = p ∆ν KX
⎝ pA pB ⎠eq
⎝ XA XB ⎠eq
and thus
o
∆Grxn
= −RT ln K p , K p = e −∆G
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.
________________________________________________
T = 298 K
p =1 bar
Example: H2(g) + CO2(g) = H2O(g) + CO(g)
H2(g)
CO2(g)
H2O(g)
CO(g)
Initial #
of moles
a
b
0
0
# moles
at Eq.
a-x
b-x
x
x
Total # moles at Eq. = (a – x) + (b – x) + 2x = a + b
Mole fraction
a −x
a +b
b −x
a +b
x
a +b
x
a +b
5.60 Spring 2008
Lecture #15
page 4
at Eq.
o
∆Gform
(kJ/mol)
∴
∆G
o
rxn
and
0
= 28.6 kJ/mol
Kp =
pH O pCO
2
pH pCO
2
2
=
-396.6
Kp = e
⇒
XH OXCO
2
XH XCO
2
2
−
-228.6
28,600 kJ/mol
(8.314 J/K-mol)(298 K )
-137.2
= e −11.54 = 9.7 x 10 −6
x2
=
( a − x ) (b − x )
Let’s take a = 1 mol and b = 2 mol
We need to solve
A)
x2
(1 − x )(2 − x )
= 9.7 x 10 −6
Using approximation method:
K &lt;&lt; 1, so we expect x &lt;&lt; 1 also.
Assume
1 − x ≈ 1, 2 − x ≈ 2
⇒
x2
(1 − x )(2 − x )
≈
x2
2
= 9.7 x 10 −6
x ≈ 0.0044 mol (indeed &lt;&lt; 1)
B)
Exactly:
x2
= K p = 9.7 x 10 −6
2
x − 3x + 2
x 2 (1 − 9.7x 10 −6 ) + 3x ( 9.7x 10 −6 ) − 2 ( 9.7x 10 −6 ) = 0
x =
−3 ( 9.7x 10 −6 )
2 ( 9.7x 10 −6 )
9 ( 9.7x 10 −6 ) + 4 (1 − 9.7x 10 −6 ) 2 ( 9.7x 10 −6 )
2
&plusmn;
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
Xi’s at Eq.
n −x
n +x
2x
n +x
Total # moles at Eq. =
n – x + 2x = n + x
2
Kp =
2
pNO
2
pN O
=
2 4
Kp = p
2
p 2XNO
pXN O
2 4
4α 2
1 − α2
where α = x n is the fraction reacted
K
(1 − α ) 4 pp = α 2
2
2
⎛ 2x ⎞
⎜
4x 2
n + x ⎟⎠
=p⎝
=p 2
n −x2
⎛n − x ⎞
⎜
⎟
⎝n + x ⎠
Kp ⎞ Kp
⎛
α ⎜1 +
⎟=
4
p
⎝
⎠ 4p
2
Kp
4p
1
=
α2 =
Kp ⎞ ⎛
⎛
4p ⎞
⎜1 +
⎟ ⎜⎜ 1 +
⎟
4p ⎠
Kp ⎟
⎝
⎝
∴
⎛
4p ⎞
α = ⎜1 +
⎟
⎜
K p ⎠⎟
⎝
−1 2
⎠
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)
Initial mol #
2
1
0
# at Eq.
2-2x
1-x
2x
Xi’s at Eq.
2 (1 − x )
3−x
1−x
3−x
2x
3−x
Kp =
2
pNO
2
2
pNO
pO
2
=
2
p 2XNO
=
2
2
p 2XNO
pXO
2
2
XNO
1
2
2
p XNO
XO
2
=
K p = 2.3x 1012 at 298 K
Total # moles at Eq.
= 2 – 2x + 1 – x + 2x
=3-x
2
1 x (3 − x )
p
3
(1 − x )
K p &gt;&gt; 1 so we expect x ≈ 1 ⇒ 3 - x ≈ 2
Kp ≈
1
2
p (1 − x )3
or
(1 − x )
3
≈
2
pK p
⎛ 2
x =1−⎜
⎜ pK p
⎝
13
⎞
⎟⎟
⎠
In this case, if p↑ then x↑ as expected from Le Chatelier’s principle.
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