Equilibrium
Forward and Backward Reactions
Hydrogen reacts with iodine to make hydrogen iodide:
H2 (g) + I2 (g) −→ 2HI(g)
forward rate = kf [H2 ][I2 ]
2HI(g) −→ H2 (g) + I2 (g)
backward rate = kb [HI]2
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Equilibrium Constant
kf [H2 ][I2 ] = kb [HI]2
All the constants together make the equilibrium constant, K:
kf [H2 ][I2 ]
= kb [HI]2
kf
kb
=
K
=
[HI]2
[H2 ][I2 ]
[HI]2
[H2 ][I2 ]
K is a property of the reaction, and is something you could look
up in a table. K will also (typically) change with temperature.
Equilibrium Constant Expressions
We extend the above idea to reactions with two products:
aA + bB cC + dD
for which the equilibrium constant is given by
K=
[C]c [D]d
[A]a [B]b
so, for example:
6CO2 + 6H2 O C6 H12 O6 + 6O2
2CH3 Cl + 3O2
3H2 O + CO2 + COCl2
2
gives
K
=
K
=
[C6 H12 O6 ][O2 ]6
[CO2 ]6 [H2 O]6
[H2 O]3 [CO2 ][COCl2 ]
[CH3 Cl]2 [O2 ]3
2H2 + O2 2H2 O
K=
[H2 O]2
= 2.4 × 1047
[H2 ]2 [O2 ]
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Gas Equilibria
When a reaction just involves gases
aA + bB cC + dD
with
KP =
PCc PDd
PAa PBb
For
H2 (g) + I2 (g) 2HI(g)
2
PHI
= 2.9 × 10−1
PH2 PI2
If, at equilibrium, we have partial pressures of H2 and I2 of 0.1 atm,
we get
KP =
2
PHI
PH2 PI2
=
2.9 × 10−1
2
PHI
=
0.29PH2 PI2
=
0.29(0.1 atm)(0.1 atm)
=
2.9 × 10−3 atm2
=
0.054 atm
PHI
4
Homogeneous and Heterogeneous Equilibria
We define
Homogeneous equilibria are those where all reagents and all
products are in the same phase: solid, liquid, or gas.
Heterogeneous equilibria are everything else: at least one reagent
or product is in a different state from the others.
For
CO2 (g) + C(s) 2CO(g)
we can write
K
=
K · [C]
=
K0
=
[CO]2
[CO2 ][C]
[CO]2
[CO2 ]
[CO]2
[CO2 ]
or
KP =
2
PCO
PCO2
Generally, we assume that pure solid and liquid components have
been eliminated from an equilibrium expression.
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Sequential Reactions
Combustion of nitrogen occurs stepwise:
N2 (g) + O2 (g)
2NO(g) + O2 (g)
and so
K1 K2 =
2NO(g)
2NO2 (g)
K1
=
K2
=
K3
=
K1 = 2.0 × 10−25
K2 = 6.4 × 109
2
PNO
PN2 PO2
2
PNO
2
2
PNO PO2
2
PNO
2
PN2 PO2 2
2
2
2
PNO
PNO
PNO
2
2
=
= K3
2 P
PN2 PO2 PNO
PN2 PO2 2
O2
and
K3 = K1 K2 = (2.0 × 10−25 )(6.4 × 109 ) = 1.3 × 10−15
And, in general:
Adding two reactions (reagents and products) to make
a new reaction gives an equilibrium constant that is the
product of the two equilibrium constants.
Here are the tricks we could use to get from equilibrium constants
we know to ones we don’t:
1. If the forward reaction is K, the backward is 1/K.
2. If we multiply the
√ moles of all species by n, then K goes to
K n . (e.g., K 2 , K)
3. If we add two reactions, we multiply their equilibrium constants.
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Concentrations → Equilibrium Constant
2SO2 (g) + O2 (g) 2SO3 (g)
at 800 K, we measure an equilibrium mixture of 5.0 ×10−2 M SO3 ,
3.5 ×10−3 M O2 , and 3.0 ×10−3 M SO2 .
K
=
=
=
[SO3 ]2
[SO2 ]2 [O2 ]
(5.0 × 10−2 )2
(3.0 × 10−3 )2 (3.5 × 10−3 )
7.9 × 104
Let’s say we start with 0.05 M SO3 alone at 800 K, and we allow
it to reach equilibrium. Once it does, we measure a concentration
of 1.89 × 10−3 M O2 . Can we calculate K now?
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Initial:
Change:
Final:
[SO3 ]
0.05
−3.78 × 10−3
4.62 × 10−2
[SO2 ]
0
+3.78 × 10−3
3.78 × 10−3
[O2 ]
0
+1.89 × 10−3
1.89 × 10−3
and
K
[SO3 ]2
[SO2 ]2 [O2 ]
(4.62 × 10−2 )2
(3.78 × 10−3 )2 (1.89 × 10−3 )
=
=
7.9 × 104
=
Equilibrium Constant → Concentrations
H2 (g) + I2 (g) 2HI(g)
[HI]2
= 54
[H2 ][I2 ]
Initial:
Change:
Final:
[H2 ]
0.05
−x
0.05 − x
8
[I2 ]
0.05
−x
0.05 − x
[HI]
0
+2x
2x
[HI]2
[H2 ][I2 ]
(2x)2
(0.05 − x)(0.05 − x)
(2x)2
(0.05 − x)2
=
54
=
54
=
54
This gives us
[HI]2
[H2 ][I2 ]
=
54
(2x)2
(0.05 − x)(0.05 − x)
=
54
4x2
=
54(0.05 − x)(0.05 − x)
2
=
27(0.0025 − 0.1x + x2 )
0
=
2x
x
x
25x2 − 2.7x + 0.061
√
−b ± b2 − 4ac
=
√2a
2.7 ± 0.54
=
50
= 0.039 or 0.069
So x = 0.039
[H2 ]
[I2 ]
[HI]
=
=
=
0.05 − x
0.05 − x
2x
9
=
=
=
0.011
0.011
0.078
Same problem, with [H2 ] = 0.5 and [I2 ] = 0.05.
[H2 ]
0.5
−x
0.5 − x
Initial:
Change:
Final:
[I2 ]
0.05
−x
0.05 − x
[HI]
0
+2x
2x
and so
[HI]2
[H2 ][I2 ]
=
54
(2x)2
(0.5 − x)(0.05 − x)
=
54
4x2
=
54(0.025 − 0.55x + x2 )
0
=
x
[H2 ]
[I2 ]
[HI]
=
=
=
25x2 − 14.85x + 0.675
√
14.85 ± 220.52 − 67.5
=
50
1
=
(14.85 ± 12.37)
50
= 0.0496 or 0.544
0.5 − x
0.05 − x
2x
10
=
=
=
0.45
0.004
0.0992
The Reaction Quotient
aA + bB cC + dD
K=
[C]c [D]d
[A]a [B]b
eq
We define the reaction quotient:
Q=
[C]c [D]d
[A]a [B]b
which is exactly the same thing as K, but the system does not have
to be at equilibrium:
At equilibrium, Q = K. Away from equilibrium, Q can
take any (non-negative) value.
• If Q < K, this means that
[C]c [D]d
<
[A]a [B]b
[C]c [D]d
[A]a [B]b
eq
and the reaction proceeds, as written, to the right.
• If Q > K, the reaction proceeds, as written, to the left.
c d
[C]c [D]d
[C] [D]
>
[A]a [B]b
[A]a [B]b eq
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N2 (g) + 3H2 (g) 2NH3 (g)
and
KP = 2.14 × 10−2 @ 540 K
If we introduce a equal mixture of all three gases—0.333 atm each
of H2 , N2 , and NH3 —what happens?
QP
=
2
PNH
3
PN2 PH3 2
1 2
3
=
=
1
3
1 3
3
9
QP = 9 > 2.14 × 10−2 = KP
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LeChâtelier’s Principle
If a system is at equilibrium, the concentrations (by definition) of
reactants and products will no longer change. Changes in conditions—
stresses—can alter the position of equilibrium, however:
1. Adding or removing reactants or products changes the composition of the system, taking it away from equilibrium.
2. Changing the pressure or volume changes the concentration
of gas-phase components, but not liquid or solid components;
this can change relative concentrations for some reactions.
3. Changing temperature will usually change K.
LeChâtelier’s principle tells us
• If there are more moles of gas in the reactants than the products, the reaction decreases pressure, and increasing pressure
will favor reaction.
• If there are more moles of gas in the products than the reactants, the reaction increases pressure, and increasing pressure
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will inhibit reaction.
• If there are the same number of moles of gas in reagents and
products, changing pressure or volume is unlikely to affect
the reaction.
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