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© 2008 Brooks/Cole 1
Chapter 14: Chemical Equilibrium
© 2008 Brooks/Cole 2
Characteristics of Chemical Equilibrium
© 2008 Brooks/Cole 3
Species do not stop forming OR being destroyed
Rate of formation = rate of removal
Concentrations are constant.
Reactants convert to products
a A + b B c C + d D
Equilibrium is Dynamic
© 2008 Brooks/Cole 4
N2(g) + 3 H2(g) 2 NH3(g)
Equilibrium is Independent of Direction of Approach
© 2008 Brooks/Cole 5
Equilibrium and Catalysts
© 2008 Brooks/Cole 6
For the 2-butene isomerization:
At equilibrium:
rate forward = rate in reverse
An elementary reaction, so:
kforward[cis] = kreverse[trans]
C=C
CH3
H H
H3C
C=C CH3
H
H
H3C
The Equilibrium Constant
2
© 2008 Brooks/Cole 7
At equilibrium the concentrations become constant.
The Equilibrium Constant
© 2008 Brooks/Cole 8
or = [trans]
[cis] kforward
kreverse
We had:
kforward[cis] = kreverse[trans]
“c” for concentration based
Kc = = = 1.65 (at 500 K) kforward
kreverse
[trans]
[cis]
The Equilibrium Constant
© 2008 Brooks/Cole 9
The Equilibrium Constant
For a general reaction:
a A + b B c C + d D
Kc = = [C]c [D]d
[A]a [B]b
kforward
kreverse
Products raised to
stoichiometric powers…
…divided by reactants
raised to their stoichiometric powers
© 2008 Brooks/Cole 10
The Equilibrium Constant
N2(g) + O2(g) 2 NO(g) Kc = [NO]2
[N2] [O2]
S8(s) + O2(g) SO2(g) 1 8 Kc =
[SO2]
[O2]
© 2008 Brooks/Cole 11
[Solid] is constant throughout a reaction.
• pure solid concentration =
• [S8] = dS8 / mol. wt S8
• d and mol. wt. are constants, so [S8] is constant.
• This constant factor is “absorbed” into Kc.
Equilibria Involving Pure Liquids and Solids
Pure liquids are omitted for the same reason.
density
mol. wt
g / L
g / mol
© 2008 Brooks/Cole 12
Equilibria in Dilute Solutions
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
Kc = [NH4
+][OH-]
[NH3] = 1.8 x 10-5 (at 25°C)
(Units are customarily omitted from Kc)
3
© 2008 Brooks/Cole 13
The Equilibrium Constant What is the equilibrium constant for:
SiH4(g) + 2 O2(g) SiO2(s) + 2 H2O(g) ?
Write down Kc for:
2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O(l)
[Na2SO4]
[H2SO4][NaOH]2
Omit SiO2 (a solid)
Include H2O (a gas)
Omit H2O (aq. solution)
[H2O]2
[SiH4][O2]2
© 2008 Brooks/Cole 14
Kc for Related Reactions
Change the stoichiometry… …change Kc
This new Kc is the square root of the original Kc. • Multiply an equation by a factor… • … Raise Kc to the power of that factor.
Kc = [NH3]
2
[N2][H2]3 N2(g) + 3 H2(g) 2 NH3(g)
3 2 N2(g) + H2(g) NH3(g) Kc =
[NH3]
[N2] [H2] 3
2
reaction divided by 2
© 2008 Brooks/Cole 15
Kc for Related Reactions
N2(g) + 3 H2(g) 2 NH3(g)
Kc = [NH3]
2
[N2][H2]3
2 NH3(g) N2(g) + 3 H2(g)
Kc = [N2][H2]
3
[NH3]2
Reverse a reaction… … Invert Kc:
© 2008 Brooks/Cole 16
Kc(step1) x Kc(step2) = [NO]2
[N2][O2]
[NO2]2
[NO]2[O2]
The overall Kc is the product of the steps:
If a reaction can be written as a series of steps:
Kc = [NO2]
2
[NO]2[O2] 2 NO(g) + O2(g) 2 NO2(g)
Kc = [NO2]
2
[N2][O2]2 N2(g) + 2 O2(g) 2 NO2(g)
Kc = [NO]2
[N2][O2] N2(g) + O2(g) 2NO(g)
= [NO2]
2
[N2][O2]2
Kc for a Reaction that Combines Reactions
© 2008 Brooks/Cole 17
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: PV = nRT
and for gas A: PAV = nART
PA= RT nA
V = [A]RT
P is easily measured, so another K is used…
© 2008 Brooks/Cole 18
Kp = PC
c PDd
PAa PB
b “p” for pressure
based
In general, Kp Kc. Your text shows:
Kp = Kc(RT) ngas
ngas = (c + d) – (a + b)
=(moles of gaseous products) (moles of gaseous reactants)
Equilibrium Constants in Terms of Pressure
4
© 2008 Brooks/Cole 19
Calculate Kp from Kc for the reaction:
N2(g) + 3 H2(g) 2 NH3(g) Kc = 5.8 x 105 at 25°C.
Kp = Kc(RT) ngas
T = 25 + 273 = 298 K
ngas = 2 – (3 + 1) = 2
R must have L/mol units.
([ ] has mol/L units).
Kp = 5.8 x 105 0.0821 L atm mol-1 K-1(298 K) -2
= 9.7 x 102 mol2 L-2 atm-2
mol and L units are assumed to cancel (units omitted from Kc).
Kp = 9.7 x 102 atm-2
Equilibrium Constants in Terms of Pressure
© 2008 Brooks/Cole 20
If all the equilibrium concentrations are known it’s
easy to calculate Kc for a reaction….
Kc = = [C]c [D]d
[A]a [B]b
kforward
kreverse
For:
a A + b B c C + d D
Determining Equilibrium Constants
Kc = [B]/[A]2 = 4.0/(2.0)2 = 1.0
At equilibrium [A] = 2.0 M & [B] = 4.0 M. What is Kc?
© 2008 Brooks/Cole 21
… In other cases stoichiometry is used to find some
concentrations.
Determining Equilibrium Constants
© 2008 Brooks/Cole 22
H2(g) + I2(g) 2 HI(g)
[ ]initial 0.01000 0.00800 0
Concentrations change – use stoichiometry:
[ ]change
Let HI made
= 2x
2x
Stoichiometry: 1H2 2HI If
HI made = 2x, H2 lost = x
-x
1I2 2HI
I2 lost = x
-x
Determining Equilibrium Constants
© 2008 Brooks/Cole 23
H2(g) + I2(g) 2 HI(g)
[ ]equilib 0.01000 - x 0.00800 - x 2x
Add [ ]initial and [ ]change to get [ ]equilib
Since [I2] = 0.00560 M at equilibrium
0.00560 = 0.00800 - x
-0.00240 M = - x
x = 0.00240 M
Determining Equilibrium Constants
© 2008 Brooks/Cole 24
Calculate [ ]equilib and then Kc
Kc = = [HI]2
[H2][I2]
(0.00480)2
(0.00760)(0.00560)
Kc = 0.541
[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
Determining Equilibrium Constants
5
© 2008 Brooks/Cole 25
When:
Meaning of the Equilibrium Constant
© 2008 Brooks/Cole 26
Predicting the Direction of a Reaction
Q = [C]c [D]d
[A]a [B]b
For the reaction: aA + bB cC + dD
© 2008 Brooks/Cole 27
Reactants Products
Q = [Products]
[Reactants] Kc=
[Products]equilib
[Reactants]equilib
Kc is a constant (only changes if T changes). So:
Predicting the Direction of a Reaction
© 2008 Brooks/Cole 28
Predicting the Direction of a Reaction
© 2008 Brooks/Cole 29
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.
Q = [ SO3]
2
[SO2]2[O2]
(0.250)2
(0.085)2(0.100) = = 86.5
Predicting the Direction of a Reaction
© 2008 Brooks/Cole 30
If Kc is known, [ ]equilib can be calculated.
Calculating Equilibrium Concentrations
6
© 2008 Brooks/Cole 31
H2(g) + I2(g) 2 HI(g)
[ ]initial 0 0 0.0500
Change on reaching equilibrium:
[ ]change +x +x -2x
Amount lost
1H2 2HI
1I2 2HI
[ ]equilib +x +x 0.0500 – 2x
Kc = = [HI]2
[H2][I2]
(0.0500 – 2x)2
(x)(x)
Calculating Equilibrium Concentrations
© 2008 Brooks/Cole 32
Kc = = [HI]2
[H2][I2]
(0.0500 – 2x)2
x2
Since Kc = 76
76 x2 = (0.0500 – 2x)2
Take the square root of both sides:
8.718 x = ±(0.0500 – 2x)
Ignore the negative root (x cannot be negative).
Calculating Equilibrium Concentrations
© 2008 Brooks/Cole 33
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 = = = 76. [HI]2
[H2][I2]
(0.0407 M)2
(0.00467 M)(0.00467 M)
Matches Kc given
in the problem
Calculating Equilibrium Concentrations
© 2008 Brooks/Cole 34
Consider:
A(g) B(g) + C(g) Kc = 2.500
A is added to an empty container. [A]initial = 0.1000 M.
What are [A], [B] and [C] at equilibrium?
Calculating Equilibrium Concentrations
© 2008 Brooks/Cole 35
A B + C
[ ]Initial 0.100 0 0
[ ]change -x x x
[ ]equilib (0.100 – x) x x
Kc = = [B][C]
[A] (x)(x)
(0.1000 – x)
x2
(0.1000 – x) 2.500 =
x2 + 2.500 x – 0.2500 = 0 A quadratic
equation
Calculating Equilibrium Concentrations
© 2008 Brooks/Cole 36
ax2 + bx + c = 0
-b ± b2 - 4ac
2a has two roots (solutions): x =
x = +0.0963 M or -2.596 M
Here:
x =
Calculating Equilibrium Concentrations
7
© 2008 Brooks/Cole 37
Ignore the negative root (negative concentration!)
Then x = 0.0963 M, and:
[ A ] = 0.1000 - x = 0.0037 M
[ B ] = x = 0.0963 M
[ C ] = x = 0.0963 M
Most of the A is converted to products.
Check: (0.0963)(0.0963)/0.0037 = 2.5
Calculating Equilibrium Concentrations
© 2008 Brooks/Cole 38
Le Chatelier’s Principle
“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 system at equilibrium resists change.
• If “pushed”, it “pushes back”
© 2008 Brooks/Cole 39
Changing Concentrations
a A + b B c C + d D
Add A (or B) and the system adjusts to remove it.
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.
© 2008 Brooks/Cole 40
Changing Concentrations
C=C
CH3
H H
H3C
C=C CH3
H
H
H3C
© 2008 Brooks/Cole 41
Changing V or P in Gaseous Equilibria
Kc=Q=
At equilibrium
[ NO2 ]2
[ N2O4]
Consider: N2O4(g) 2 NO2(g)
Q= = = = Kc
After V change, Q < Kc equilibrium shifts to products
( [ NO2 ])2
[ N2O4] [ NO2 ]
2
[ N2O4]
[ NO2 ]2
[ N2O4]
© 2008 Brooks/Cole 42
Consistent with Le Chatelier
N2O4(g) 2 NO2(g)
V doubled = lower concentration. Minimize change by:
• Making more molecules (increase concentration).
• Shifting toward products - convert 1 molecule into 2.
P doubled. Minimize the change by:
• Removing molecules (decrease P).
• Shift toward reactants - convert 2 molecules into 1.
Changing V or P in Gaseous Equilibria
8
© 2008 Brooks/Cole 43
P and V changes have no effect if ngas = 0:
H2(g) + I2(g) 2 HI(g)
Kc=Q=
At equilibrium
[HI]2
[H2][I2] Q = = = = Kc
After V change, Q = Kc still at equilibrium!
( [HI])2
[H2] [I2] [HI]2
[H2][I2]
[HI]2
[H2][I2]
If V is doubled, the system cannot adjust – 2 gas
molecules each side.
Changing V or P in Gaseous Equilibria
© 2008 Brooks/Cole 44
PA= RT nA
V = [A]RT
Changing V or P in Gaseous Equilibria
© 2008 Brooks/Cole 45
N2(g) + O2(g) 2 NO(g) H° = 180.5 kJ
Changing Temperature
T (°C) Kc
298 4.5 x 10-31
900 6.3 x 10-10
2300 8.2 x 10-3
Kc increases as T increases.
Explained by LeChatelier…
© 2008 Brooks/Cole 46
T increased? Minimize by removing heat.
Forward reaction is endothermic
Equilibrium shifts forward, removing heat.
T reduced? Add heat.
Reverse reaction is exothermic
Equilibrium shifts in reverse, releasing heat
N2(g) + O2(g) 2 NO(g) H° = 180.5 kJ
Changing Temperature
© 2008 Brooks/Cole 47
Changing Temperature
© 2008 Brooks/Cole 48
The Haber-Bosch Process
Production of NH3 for fertilizer from atmospheric N2
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…
N2(g) + 3 H2(g) 2 NH3(g)
9
© 2008 Brooks/Cole 49
The Haber-Bosch Process
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.
N2(g) + 3 H2(g) 2 NH3(g) H° = -92.2 kJ
Run at 450°C; NH3 is continually liquefied and
removed.
© 2008 Brooks/Cole 50
The Haber-Bosch Process
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