Characteristics of Chemical Equilibrium
Chapter 14: Chemical Equilibrium
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Equilibrium is Dynamic
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Equilibrium is Independent of Direction of Approach
Reactants convert to products
aA+bB
cC+dD
N2(g) + 3 H2(g)
2 NH3(g)
Species do not stop forming OR being destroyed
Rate of formation = rate of removal
Concentrations are constant.
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Equilibrium and Catalysts
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The Equilibrium Constant
For the 2-butene isomerization:
CH3
H 3C
H
H
H 3C
C=C
C=C
H
H
CH3
At equilibrium:
rate forward = rate in reverse
An elementary reaction, so:
kforward[cis] = kreverse[trans]
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1
The Equilibrium Constant
The Equilibrium Constant
At equilibrium the concentrations become constant.
We had:
kforward[cis] = kreverse[trans]
[trans]
kforward
= [cis]
kreverse
or
Kc =
kforward
=
kreverse
[trans]
= 1.65 (at 500 K)
[cis]
“c” for concentration based
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The Equilibrium Constant
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The Equilibrium Constant
For a general reaction:
aA+bB
cC+dD
N2(g) + O2(g)
Products raised to
stoichiometric
powers…
Kc =
kforward
=
kreverse
[C]c [D]d
[A]a [B]b
…divided by reactants
raised to their stoichiometric
powers
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S8(s) + O2(g)
2 NO(g)
SO2(g)
[NO]2
Kc = [N ] [O ]
2
2
[SO2]
Kc = [O ]
2
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Equilibria in Dilute Solutions
Equilibria Involving Pure Liquids and Solids
[Solid] is constant throughout a reaction.
• pure solid concentration =
density
mol. wt
g/L
g / mol
• [S8] = dS / mol. wt S8
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NH3(aq) + H2O(l)
• d and mol. wt. are constants, so [S8] is constant.
• This constant factor is “absorbed” into Kc.
Kc =
NH4+(aq) + OH-(aq)
[NH4+][OH-]
= 1.8 x 10-5
[NH3]
(at 25°C)
Pure liquids are omitted for the same reason.
(Units are customarily omitted from Kc)
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The Equilibrium Constant
Kc for Related Reactions
What is the equilibrium constant for:
SiO2(s) + 2 H2O(g) ?
SiH4(g) + 2 O2(g)
N2(g) + 3 H2(g)
[H2O]2
[SiH4][O2]2
Omit SiO2 (a solid)
Include H2O (a gas)
Write down Kc for:
2 NaOH(aq) + H2SO4(aq)
Na2SO4(aq) + 2 H2O(l)
[Na2SO4]
[H2SO4][NaOH]2
Omit H2O (aq. solution)
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Change the stoichiometry…
…change Kc
N2(g) + 3 2 H2(g)
Kc =
Kc =
NH3(g)
[NH3]
3
[N2][H2] 2
reaction divided by 2
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Kc for a Reaction that Combines Reactions
If a reaction can be written as a series of steps:
2 NH3(g)
2NO(g)
Kc =
[NO]2
[N2][O2]
2 NO(g) + O2(g)
2 NO2(g)
Kc =
[NO2]2
[NO]2[O2]
N2(g) + 2 O2(g)
2 NO2(g)
Kc =
N2(g) + O2(g)
[NH3]2
Kc =
[N2][H2]3
2 NH3(g)
[NH3]2
[N2][H2]3
This new Kc is the square root of the original Kc.
Kc for Related Reactions
Reverse a reaction…
Kc =
• Multiply an equation by a factor…
• … Raise Kc to the power of that factor.
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N2(g) + 3 H2(g)
2 NH3(g)
… Invert Kc:
N2(g) + 3 H2(g)
[N2][H2]3
[NH3]2
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[NO2]2
[N2][O2]2
The overall Kc is the product of the steps:
2
[NO2]2
[NO2]2
Kc(step1) x Kc(step2) = [NO]
=
2
2
[N
[N2][O2] [NO] [O2]
2][O2]
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Equilibrium Constants in Terms of Pressure
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Equilibrium Constants in Terms of Pressure
In a constant-V reaction, partial pressures change as
concentrations change.
At constant T, P is proportional to molar conc.:
Ideal gas:
and for gas A:
“p” for pressure
based
PV = nRT
PCc PDd
PAa PBb
In general, Kp Kc. Your text shows:
Kp = Kc(RT)ngas
ngas = (c + d) – (a + b)
=(moles of gaseous products) (moles of gaseous reactants)
PAV = nART
PA= nA RT = [A]RT
V
P is easily measured, so another K is used…
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Kp =
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Equilibrium Constants in Terms of Pressure
Determining Equilibrium Constants
Calculate Kp from Kc for the reaction:
If all the equilibrium concentrations are known it’s
easy to calculate Kc for a reaction….
N2(g) + 3 H2(g)
Kc = 5.8 x 105 at 25°C.
2 NH3(g)
Kp = Kc(RT)ngas
T = 25 + 273 = 298 K
ngas = 2 – (3 + 1) = 2
For:
aA+bB
R must have L/mol units.
([ ] has mol/L units).
Kc =
Kp = 5.8 x 105 0.0821 L atm mol-1 K-1(298 K)
-2
cC+dD
kforward
=
kreverse
[C]c [D]d
[A]a [B]b
= 9.7 x 102 mol2 L-2 atm-2
At equilibrium [A] = 2.0 M & [B] = 4.0 M. What is Kc?
mol and L units are assumed to cancel (units omitted from Kc).
Kp = 9.7 x
102
atm-2
Kc = [B]/[A]2 = 4.0/(2.0)2 = 1.0
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Determining Equilibrium Constants
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Determining Equilibrium Constants
… In other cases stoichiometry is used to find some
concentrations.
H2(g) +
[ ]initial
0.01000
I2(g)
0.00800
2 HI(g)
0
Concentrations change – use stoichiometry:
[ ]change
-x
-x
Stoichiometry: 1H2 2HI If
HI made = 2x, H2 lost = x
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Determining Equilibrium Constants
H2(g) +
0.01000 - x
I2(g)
0.00800 - x
Let HI made
= 2x
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Determining Equilibrium Constants
Add [ ]initial and [ ]change to get [ ]equilib
[ ]equilib
1I2 2HI
I2 lost = x
2x
Calculate [ ]equilib and then Kc
[H2]equilib = 0.01000 - (0.00240) = 0.00760 M
[I2]equilib = 0.00800 - (0.00240) = 0.00560 M
[HI]equilib = 2(0.00240) = 0.00480 M
2 HI(g)
2x
Since [I2] = 0.00560 M at equilibrium
Kc =
0.00560 = 0.00800 - x
-0.00240 M = - x
x = 0.00240 M
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[HI]2
(0.00480)2
=
[H2][I2]
(0.00760)(0.00560)
Kc = 0.541
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Meaning of the Equilibrium Constant
Predicting the Direction of a Reaction
When:
For the reaction: aA + bB
cC + dD
[C]c [D]d
Q = [A]a [B]b
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Predicting the Direction of a Reaction
Reactants
[Products]
Q = [Reactants]
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Predicting the Direction of a Reaction
Products
[Products]equilib
Kc= [Reactants]
equilib
Kc is a constant (only changes if T changes). So:
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Predicting the Direction of a Reaction
Calculating Equilibrium Concentrations
2 SO2(g) + O2(g)
2 SO3(g) Kc = 245 at 1000 K
Predict the direction of the reaction if SO2 (0.085 M),
O2 (0.100 M) and SO3 (0.250 M) are mixed in a
reactor at 1000 K.
If Kc is known, [ ]equilib can be calculated.
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[ SO3]2
(0.250)2
Q = [SO ]2[O ] = (0.085)2(0.100) = 86.5
2
2
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Calculating Equilibrium Concentrations
[ ]initial
H2(g) +
I2(g)
2 HI(g)
0
0
0.0500
+x
+x
+x
+x
Kc =
1I2 2HI
76 x2 = (0.0500 – 2x)2
-2x
Take the square root of both sides:
8.718 x = ±(0.0500 – 2x)
1H2 2HI
[ ]equilib
0.0500 – 2x
[HI]2
(0.0500 – 2x)2
=
[H2][I2]
(x)(x)
Kc =
[HI]2
(0.0500 – 2x)2
=
[H2][I2]
x2
Since Kc = 76
Amount lost
Change on reaching equilibrium:
[ ]change
Calculating Equilibrium Concentrations
Ignore the negative root (x cannot be negative).
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Calculating Equilibrium Concentrations
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Calculating Equilibrium Concentrations
So:
[H2]equilib = x = 0.00467 M
[I2]equilib = x = 0.00467 M
[HI]equilib = 0.0500 – 2x = 0.0407 M
It’s a good idea to check your work:
Kc =
Consider:
A(g)
Matches Kc given
in the problem
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Calculating Equilibrium Concentrations
[ ]Initial
A
B
0.100
0
0
x
x
x
x
[ ]change
[ ]equilib
-x
(0.100 – x)
+
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Calculating Equilibrium Concentrations
ax2 + bx + c = 0
C
has two roots (solutions):
x = -b ±
b2 - 4ac
2a
Here:
Kc =
(x)(x)
[B][C]
= (0.1000 – x)
[A]
x2
2.500 =
(0.1000 – x)
x2 + 2.500 x – 0.2500 = 0
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Kc = 2.500
A is added to an empty container. [A]initial = 0.1000 M.
What are [A], [B] and [C] at equilibrium?
[HI]2
(0.0407 M)2
=
= 76.
[H2][I2]
(0.00467 M)(0.00467 M)
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B(g) + C(g)
x=
A quadratic
equation
x = +0.0963 M
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or
-2.596 M
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Calculating Equilibrium Concentrations
Le Chatelier’s Principle
Ignore the negative root (negative concentration!)
Then x = 0.0963 M, and:
“If a system is at equilibrium and the conditions are
changed so that it is no longer at equilibrium, the
system will react to reach a new equilibrium in a way
that partially counteracts the change”
[ A ] = 0.1000 - x = 0.0037 M
[ B ] = x = 0.0963 M
[ C ] = x = 0.0963 M
• A system at equilibrium resists change.
• If “pushed”, it “pushes back”
Most of the A is converted to products.
Check: (0.0963)(0.0963)/0.0037 = 2.5
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Changing Concentrations
aA + bB
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Changing Concentrations
C=C
C=C
H
Add A (or B) and the system adjusts to remove it.
H
H 3C
CH3
H 3C
cC + dD
H
H
CH3
more product is created.
the reaction shifts forward.
Remove A (or B) the system adjusts to add more.
more reactant is created.
the reaction shifts backward.
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Changing V or P in Gaseous Equilibria
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Changing V or P in Gaseous Equilibria
Consistent with Le Chatelier
Consider:
N2O4(g)
N2O4(g)
2 NO2(g)
2 NO2(g)
V doubled = lower concentration. Minimize change by:
• Making more molecules (increase concentration).
• Shifting toward products - convert 1 molecule into 2.
Kc=Q=
[ NO2 ]2
[ N2O4]
At equilibrium
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P doubled. Minimize the change by:
([ NO2 ])2 [ NO2 ]2 [ NO2 ]2
Q= [ N O ] =
=
= Kc
[ N2O4]
[ N2O4]
2 4
• Removing molecules (decrease P).
• Shift toward reactants - convert 2 molecules into 1.
After V change, Q < Kc equilibrium shifts to products
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Changing V or P in Gaseous Equilibria
Changing V or P in Gaseous Equilibria
P and V changes have no effect if ngas = 0:
H2(g) + I2(g)
2 HI(g)
If V is doubled, the system cannot adjust – 2 gas
molecules each side.
Kc=Q=
[HI]2
[H2][I2]
At equilibrium
([HI])2
[HI]2
[HI]2
Q = [H ][I ] =
=
= Kc
[H2][I2] [H2][I2]
2
2
PA= nA RT = [A]RT
V
After V change, Q = Kc still at equilibrium!
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Changing Temperature
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Changing Temperature
N2(g) + O2(g)
N2(g) + O2(g)
T (°C)
298
900
2300
Kc
4.5 x 10-31
6.3 x 10-10
8.2 x 10-3
2 NO(g)
T increased? Minimize by removing heat.
H° = 180.5 kJ
Forward reaction is endothermic
Equilibrium shifts forward, removing heat.
Kc increases as T increases.
T reduced? Add heat.
Reverse reaction is exothermic
Equilibrium shifts in reverse, releasing heat
Explained by LeChatelier…
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H° = 180.5 kJ
2 NO(g)
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Changing Temperature
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The Haber-Bosch Process
Production of NH3 for fertilizer from atmospheric N2
N2(g) + 3 H2(g)
2 NH3(g)
Process developed by Fritz Haber (chemist) and Carl
Bosch (engineer). Perfected in 1914.
Germany could not import guano from S. America
and needed to make explosives…
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The Haber-Bosch Process
N2(g) + 3 H2(g)
2 NH3(g)
The Haber-Bosch Process
H° = -92.2 kJ
It is:
• Exothermic – product favored at low-T, but…
• … very slow at room-T, and must be heated.
• Carried out at high-P (200 atm), to favor product.
• Uses a catalyst to speed the reaction.
Run at 450°C; NH3 is continually liquefied and
removed.
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