Beyond Nash Equilibrium - Correlated Equilibrium and Evolutionary Equilibrium
A + 2 → Chapter 15 Equilibrium - web.gccaz.eduweb.gccaz.edu/~jaszi38221/2014/Fall/Lectures/CHM 152...
Transcript of A + 2 → Chapter 15 Equilibrium - web.gccaz.eduweb.gccaz.edu/~jaszi38221/2014/Fall/Lectures/CHM 152...
9/22/2014
1
Chapter 15 – Equilibrium
Equilibrium in Chemical Reactions
time
Am
ou
nt
of
reac
tan
t/p
rod
uct
Lets’ look back at our hypothetical reaction from the kinetics chapter.
A + 2B → C
[A]
[C]
Why doesn’t the concentration of A ever go to zero?
What if we waited longer?
What is happening here?
In reality, a better was to represent our reaction would be…
A + 2B ⇌ C
Equilibrium – Some Reactions are Reversible
Tells us there is more than one reaction taking place.
N2O4 ⇌ 2NO2
2NO2 → N2O4 N2O4 → 2NO2
⇌ Reaction 1 Reaction 2
and “Forward” reaction “Reverse” reaction
Will there be more reactants or more products at equilibrium?
⇌
Reversible Reactions & Equilibrium
2 Rev worker takes the two pieces, puts back together. Sends back….
However, it takes REV worker twice as long to put the pieces
together….
How will the amounts of products and reactants change over time?
Products Reactants
1 Fwd worker breaks reactant in two, sends pieces down the line….
+
Reactants Products
3 Because the fwd and rev workers are working at different rates,
over time, there will be a build-up of products
4 Since the REV worker is being overwhelmed with more products,
the foremen sends in help!!!
As the workday progresses…..
Reversible Reactions & Equilibrium
Reactants
5 Because there are now two REV workers, they together they work
at the same rate as the FWD worker
Eventually…
Products
From this point on, the amounts of reactants and products will stay the same…they are in
dynamic equilibrium
Reversible Reactions & Equilibrium
Once dynamic equilibrium is reached, the amounts of reactants and products
DO NOT CHANGE.
The reaction appears to have stopped, but the forward and reverse reaction are taking
place at the same rate.
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2
Notice that
although each trial
started with a
different ratios of
reactant/products,
the “ratio” of
concentrations of
reactants/products
at equilibrium will
be the same
INITIAL FINAL
INITIAL FINAL
Experimental Evidence of Equilibrium
Trial 1
Trial 2
Equilibrium Constant – K
Do not confuse the two kay’s
K k • Upper-case / large K
• Denotes “how far” a reaction
proceeds
• Determine by equilibrium
amounts of reactants / products
• NO UNITS!!
• lower-case / small k
• Related to “how fast” a
reaction proceeds
• Determined based on the
rate law
• UNITS!!
Equilibrium Constant – K The quantitative measure of “how “far a reaction proceeds is represented by a constant, K.
K can be thought of as the result of two reactions: the forward reaction and the reverse reaction. Whether we end up with more reactant and more product in the end is determined by the rate of each reaction.
Let’s assume the following reaction is a one-step process. We can determine the rate laws since it is an elementary step:
N2O4⇌ 2NO2
“Forward” reaction “Reverse” reaction
Rate = krev[NO2]2 Rate = kfwd[N2O4]
c
42
22
rev
fwd
]O[N
][NO K
k
k
At equilibrium
Similar rxn rates Fwd faster than rev Rev faster than fwd
Reactant Favored Product Favored Reactant ~ Product
c22rev
42fwd
][NO
]O[NK
k
k mequi l ibriu at
[reactant]
[product]
tcoefficien
tcoefficien
c K
Equilibrium Constant
Re
acta
nt(
s)
Pro
du
ct(s
)
K ≈1 K <<1 K >>1
What does the numerical value of Kc actually mean?
“Reactant Favored” “Product Favored”
K <<<<1 K >>>>1
Negligible Complete
Pro
du
ct(s
)
Re
acta
nt(
s)
mequi l ibriu at [reactant]
[product]
tcoefficien
tcoefficien
c K
Substantial
amount of both
reactants and
products
Mostly reactants,,
very small
amount of
products
Mostly products,
very small
amount of
reactants
Equilibrium Constant and concentration – Kc
aA + bB ⇌ cC + dD mequi l ibriu at [B][A]
[D][C]
ba
dc
c K
For aqueous species and gases we can use concentration in our constant
Initial (M) Equilibrium (M) Experiment [N2O4] [NO2] [N2O4] [NO2]
1 0.000 0.0200 0.00140 0.0172 2 0.000 0.0300 0.00280 0.0243 3 0.000 0.0400 0.00452 0.0310
]O[N
][NO
42
22
c K 211.0(0.00140M)
(0.0172M)K
2
c
Since
We can determine Kc experimentally by measuring concentrations of reactants and products at equilibrium
Kc
0.211 0.211 0.213
N2O4 ⇌ 2NO2
K values and equilibrium concentration practice
Oxygen gas can be converted into ozone based on the following reaction:
3O2 (g) ⇌ 2O3 (g)
• If the equilibrium concentration of O2 is 0.50M and O3 is
2.2x10-7M, determine the value of the equilibrium constant, Kc
• If the equilibrium concentration of O2 is 0.10M, what would be the
equilibrium concentration of O3?
-8 M
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Heterogeneous Equilibria
How do we deal with reactions that have different states of matter within the same reaction? For example:
Fe(OH)3(s) → Fe3+(aq) + 3OH-(aq)
Although a gas can be represented by Kc or Kp, how do we represent solids? What is the concentration of a solid? Well, there is no need to calculate it since it will not change even if you change the amount do we. So, we take it out of the K expression.
Pure liquids and solids are not represented in the equilibrium
equation
3-33c ]][OH[Fe ][Fe(OH) ' K
CO2(g) +C(s) → 2CO(g)
][CO
[CO]
2
2
c K
So for the following equation…
…solid carbon will not be a part of the equilibrium expression
H+(aq) +OH-(aq) → H2O(l)
And for the following equation…
]][H[OH
1
-c K
…liquid H2O will not be a part of the equilibrium expression
][Fe(OH)
]][OH[Fe
3
3-3
c
KA solid or ,liquid does not change
concentration
3-3c ]][OH[Fe K
Equilibrium Constant and Pressure – Kp
Just as we can have an equilibrium constant based on pressure of products and reactants (Kc), we can also use partial pressures when dealing with gases
mequi l ibriu at Reactants)of Pressure (Partial
Products)of Pressure (Partial p K
aA + bB ⇌ cC + dD
mequi l ibriu at )()(
)()(
bB
aA
dD
cC
pPP
PPK Pgas = the partial pressure the gas
exerts in the reaction mixture
So, K’s for the reaction:
2O3⇌ 3O2
23
32
c][O
][OK
2O
3O
p)(
)(
3
2
P
PK
Based on partial pressures Based on concentrations
The Relationship between Kc and Kp
The relationship between Kc and Kp is determined by the following equation
nKK Δcp (RT)
Mol gas products – mol gas reactants from balanced equation
Kmolatm L0.08206
For example, the following reaction: 2SO2(g) + O2(g) ⇌ 2SO3(g) has a Kc of 4.09x10-2 at 1000.K. To determine Kp:
3)(2
KmolatmL2
p )(1000.K)])[(0.08206(4.09x10 K 4.98x10-4
The reaction: H2(g) + Cl2(g) ⇌ 2HCl(g) has a Kp of 4.0x1031 at 300.K. What is Kc?
2)-(2Kmol
atmLc
31 )(300.K)][(0.08206 )(K )(4.0x10 4.0x1031
When the number of moles of gas are the same on both sides of the equation, Kc = Kp
Relationships between related K’s
aA + bB ⇌ cC + dD ba
dc
1c[B][A]
[D][C] K
Remember that for the reaction….
But what about the reverse reaction?
cC + dD ⇌ aA + bB
Since we have written the equation in reverse, products and reactants are
switched. Thus, K is different. dc
ba
2c[D][C]
[B][A] K
1
1
1
2
1 K
KK
When we deal with reversible reactions, we can write it any direction we choose.
Relationships between related K’s
aA + bB ⇌ xX ba
x
1c[B][A]
[X] K
xX ⇌ cC
213 KKK
x
c
2c[X]
[C] K
aA + bB ⇌ cC
Let’s say we have two reactions that have a reactant/product in common:
We can add the two reaction together
x
c
ba
x
c21c[X]
C][
[B][A]
[X] KK
ba
c
3[B][A]
[C] K
How do we combine K expressions?
Reaction Quotient and Equilibrium Constant What do we do when we know the concentration of reactants before equilibrium (or before
we mix them to perform a reaction)? Do we have an expression that can quantify this?
K Q
Q
Forward
More Products
Formed
Reverse
More Reactants
Formed
mequi l ibriu at NOT anytime at [B][A]
[D][C]
ba
dc
c Q
Equilibrium Q<K Q>K
ba
dc
c[B][A]
[D][C] Q
Ratio of reactants higher than at
equilibrium ba
dc
c[B][A]
[D][C] Q
Ratio of products higher than at
equilibrium
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Le Chatelier’s Principle
When an equilibrium is disturbed, the reaction will proceed in the direction that reestablishes an equilibrium .
Equilibrium mixture of NO2 and N2O4 gives a yellowish color.
More NO2 added
The addition of more NO2 turns the mixture orange (NO2 is
orange in color).
Wait 10 minutes
After 10 minutes, the color returns as before, suggesting that the
equilibrium mixture has returned.
N2O4← 2NO2
Which way did the reaction shift to reach equilibrium?
N2O4 + 2NO2
colorless
Dark orange
N2O4 ⇌ 2NO2 N2O4 ⇌ 2NO2 N2O4 ? 2NO2
Le Chatelier’s Principle
A + B ⇌ C + D
What happens when a equilibrium is “disturbed”? Once it is disturbed, it will shift toward reactants or products to return to equilibrium
Increase the amount of a reactant
Q<K
Reaction shifts toward products
A + B → C + D
K is reestablished
A + B ⇌ C + D
A + B ⇌ C + D
Decrease the amount of a reactant
Q>K
Reaction shifts toward reactants
A + B ← C + D
K is reestablished
A + B ⇌ C + D
A + B ⇌ C + D
Increase the amount of a product
Q>K
Reaction shifts toward reactants
A + B ← C + D
K is reestablished
A + B ⇌ C + D
A + B ⇌ C + D
Decrease the amount of a product
Q<K
Reaction shifts toward products
A + B → C + D
K is reestablished
A + B ⇌ C + D
To make a reaction shift toward products •Decrease the amount of product •Increase the amount of reactant
To make a reaction shift toward reactants •Decrease the amount of reactant •Increase the amount of product
Le Chatelier’s Principle Only reactants and/or products that are part of the equilibrium equation, will affect equilibrium.
Fe(OH)3
Fe3+
Fe3+ OH-
OH-
OH-
OH- OH- OH-
OH-
Fe(OH)3
Fe3+
Fe3+ OH-
OH-
OH-
OH-
OH-
OH-
OH-
Add more
Fe(OH)3 Adding more Fe(OH)3 will not cause more ions to dissolve.
No shift in equilibrium
3-3c ]][OH[Fe K
Fe(OH)3 is not part of the equilibrium expression
Ca(CO3)2
CO2 CO2
CO2
CO2
CO2 CO2
CaCO3
CaO
CO2 CO2
CO2
CO2
CO2 CO2
CaCO3
Add more
CaCO3
][CO 2c K
Concentration of Fe(OH)3 ions
does not change
Concentration of CO2 does not
change
Adding more CaCO3 will not cause more CO2 or CaO to be
created. No shift in equilibrium
Fe(OH)3(s) → Fe3+(aq) + 3OH-(aq)
CaCO3(s) → CO2(g) + CaO(s)
Le Chatelier’s Principle – Change in Temp In order to determine the effect of temperature on equilibria, it is convenient to view heat either
as a product or reactant.
A + B ⇌ C + HEAT A + B ⇌ C + HEAT
Reaction shifts toward
reactants
Reaction shifts toward products
A + B ← C + HEAT
A + B ⇌ C + HEAT
Equilibrium is established w/ smaller K
A + B → C + HEAT
A + B ⇌ C + HEAT
Equilibrium is established w/ larger K
+ A + B ⇌ C HEAT
Reaction shifts toward products
Equilibrium is established w/ larger K
+ A + B ⇌ C HEAT
Reaction shifts toward
reactants
+ A + B → C HEAT
+ A + B ⇌ C HEAT
Equilibrium is established w/ smaller K
+ A + B ← C HEAT
+ A + B ⇌ C HEAT
Exothermic reactions (-ΔH) Endothermic reactions (+ΔH)
Le Chatelier’s Principle
N2O4 ⇌ 2NO2
N2O4 ← 2NO2 N2O4 ⇌ 2NO2
Equilibrium shifts to the side with the fewest moles of gas.
More of that gas (product or reactant depending on the balanced equation) is
produced
2 atm 2 atm 1 atm
Decrease in volume An increase in total pressure will shift a reaction toward the side with the fewest moles of gas.
Ca(CO3)2
CO2 CO2
CO2 CO2
CaCO3
CO2
CO2
CaCO3
CO2 CO2
CO2
CO2
CaCO3
2 atm 2 atm
CO2
CO2
CO2
CO2
CO2
CaCO3(s) ⇌ CO2(g) + CaO(s) CaCO3(s) ← CO2(g) + CaO(s) CaCO3(s) ⇌ CO2(g) + CaO(s)
The “I.C.E” Table
A certain reaction has a balanced equation of A ⇌ 2B + 3C. After placing 6.00M A in reaction vessel @ 1000.K, the reaction begins. During the reaction, the reactant decreased by1.33M . What are the equilibrium concentrations and what is Kc?
I
E C
nitial
hange
quilibirum
6.00M 0M
-1.33M +(2)1.33
(6.00M – 1.33M) 4.67M
(0M+2.66M) 2.66M
A ⇌ 2B + 3C
0M
+(3)1.33
(0M+3.99M) 3.99M
•We plug everything we know into the table.
•Write the balanced equation
•Since A changed -1.33M, how should the concentration of B and C change in comparison?
Look at the balanced equation
•B should change 2 times as much and C should change 3
times as much a A.
[A]
[C][B] 32
c K 96.24.67
(3.99)(2.66) 32
c K
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Determining Change from Initial and Equilibrium Values
I
E C
nitial
hange
quilibirum
1.00atm 0atm
-3x +2x
3A ⇌ 2B
0.40atm
Gas “A” is put in a reaction flask and heated to 40.0°C until it changes to gas “B” in a 3-to-2 ratio. Before the reaction, gas “A” had a partial pressure of 1.00atm. At the end of the reaction, after it has cooled gas “A” had a partial pressure of 0.40atm. Determine the
equilibrium pressure of gas “B” and determine Kp for the reaction.
•Write the balanced equation
•Plug in what we know
•Since we don’t know the change, we, can substitute with a variable
“x” (temporarily).
•We can actually work backward from equilibrium.
Therefore, we know that -3x = -0.60
x = 0.20 atm
-0.60atm
So 2x = 0.40atm
+0.40atm
0.40atm
Change = Initial – Equilibrium = 1.00atm-0.40atm = 0.60atm
Change = -3x
2.5(0.40)
(0.40)3
2
p K
and
3A
2B
p)(
)(
P
PK
Don’t forget your coefficients
Calculating Equilibrium Concentration using K
I
E C
nitial
hange
quilibirum
A ⇌ B
3.33M 0M
-x +x
3.33M - x 0M + x
A certain reactant “A” rearranges into a product “B” at room temperature in a 1-to-1 ration. If you started with 3.33M “A”, what will be the concentration of both “A” and
“B” at equilibrium if Kc = 1.11x10-1
•Write the balanced equation
•Plug in what we know
•Since we don’t know the change, we, can substitute with a variable x.
•Equilibrium is just the initial plus the change
•Substitute equilibrium concentrations
0.111[A]
[B]Kc 0.111
x-3.33
x 0.111x0.370x 0.3701.111x
0.333x [A]eq = 3.33M - x
[B]eq = x
[A]eq = 3.00M
[B]eq = 0.333M
Double-check by working backward
0.1113.00
0.333c K
Since… Then…
Calculating Equilibrium Concentration using K – Perfect Squares
I
E C
nitial
hange
quilibirum
0.100M 0.100M
-x -x
0.100M - x 0.100M - x
H2 + I2 ⇌ 2HI
The reaction between H2 and I2 produces HI. Assume 0.100mols of H2 and 0.100 moles of I2 are placed into a 1.00L container. The gas react and are allowed to reach
equilibrium. Calculate the equilibrium concentration of all the gases. Kc for the reaction at 443°C is 50.5.
•Write the balanced equation
•Plug in what we know
•Since we don’t know the change, we, can substitute with a variable x.
•Equilibrium is just the initial plus the change
•Substitute equilibrium concentrations
0M
+2x
0M + 2x
50.5]][I[H
[HI]
12
2
c KSince…
50.5x)-x)(0.100-(0.100
(2x)2
Perfect Square
50.5x-0.100
2x2
50.5
x-0.100
2x2
or
Calculating Equilibrium Concentration using K – Perfect Squares
I
E C
nitial
hange
quilibirum
0.100M 0.100M
-x -x
0.100M - x 0.100M - x
H2 + I2 ⇌ 2HI
0M
+2x
0M + 2x
50.5x-0.100
2x2
7.11x)-(0.100
2x
7.11x-0.7112x
9.11
0.7119.11x
x = 0.0780
Since [H2]eq and [I2]eq = 0.100-x [HI]eq = 2x
[H2]eq and [I2]eq = 0.022M [HI]eq = 0.156M Double-check by working backward
50.3022)(0.022)(0.
(0.156)2
c K
Calculating Equilibrium Concentration using K – Quadratic Formula
I
E C
nitial
hange
quilibirum
We can calculate equilibrium concentrations if we know initial concentrations and the equilibrium constant, Kc.
•Write the balanced equation N2O4 ⇌ 2NO2
•Plug in what we know
If we start with an initial concentration of N2O4 of 5.0M before decomposition , how
much NO2 will there be at equilibrium? Kc for the reaction is 4.6x10-3.
5.0M 0M •Since we don’t know the change, we, can
substitute with a variable x. -x +2x
5.0M - x 2x •Equilibrium is just the initial plus
the change
3
42
22
c 4.6x10]O[N
][NO K
•Substitute equilibrium concentrations
Since… 32
c 4.6x10x)-(5.0
(2x) K Solve for x…
Calculating Equilibrium Concentration using K – Quadratic Formula
I
E C
nitial
hange
quilibirum
N2O4 ⇌ 2NO2
5.0M 0M
-x +2x
5.0M - x 2x
32
c 4.6x10x)-(5.0
(2x) K
(2x)2 = (5.0-x)(0.0046)
4x2 = -0.0046x + 0.023
4x2 + 0.0046x - 0.023 = 0
2a
4acbb 2
Quadratic Formula
This is in the form of a quadratic
ax2 + bx + c = 0
General form of a Quadratic
4x2 + 0.0046x - 0.023 = 0
2(4)
3)4(4)(-0.02(0.0046)0.0046- 2
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6
2(4)
3)4(4)(-0.02(0.0046)0.0046 2
x = 0.075 and x = -0.076
[NO2]eq = 0.15M 9150.004
4.9
(0.15)2
c K
Double-check by working backward
This value for x does not make “chemical sense” because you cannot
have negative concentrations
[NO2]eq = 2x [N2O4]eq = 5.0M-x
[N2O4]eq = 4.9M
Calculating Equilibrium Concentration using K – Quadratic Formula
7
2
2
1.67x102x)-(0.0250
(x)(2x)
Calculating Equilibrium Concentration using K – Simplifying Assumptions
I
E C
nitial
hange
quilibirum
2H2S ⇌ 2H2 + S2
0.0250M 0M
-2x +2x
0.0250M - 2x 2x
At this point, you had better know which steps to take!!!
0M
+x
+x
7
22
22
2c 1.67x10
S][H
][S][H K
Consider the decomposition of H2S, to give H2 and S2. If Kc = 1.67x10-7 at 800°C, determine the equilibrium concentrations for all gases if the initial amount of H2S is
0.0250M.
7
2
2
1.67x102x)-(0.0250
(x)(2x)
This looks like it is going to be a cubic function; 3rd degree polynomial!!! That is: ax3 + bx2 + cx +d = 0. I don’t feel like solving this!!! Although you can do this via a graphing calculator (which you can’t use
on an exam, of course…..) or using successive approximations.
Let’s make some simplifying assumptions. For example, since K is so small, I’m going to assume that the change in x will be small also. So…..
…Let’s get rid of this..since it should
small
7
2
2
1.67x10(0.0250)
(x)(2x) HHMMM….Better……
Calculating Equilibrium Concentration using K – Simplifying Assumptions
I
E C
nitial
hange
quilibirum
2H2S ⇌ 2H2 + S2
0.0250M 0M
-2x +2x
0.0250M - 2x 2x
0M
+x
+x
7
2
2
1.67x10(0.0250)
(x)(2x)
7
4-
3
1.67x106.25x10
4x
4
))(6.25x10(1.67x10x
473
42.97x10x
3 113 3 2.61x10x
[H2S]eq = 0.0250M – 2x
[H2]eq = 2x
[S2]eq = x
[H2S]eq = 0.0244M
[H2]eq = 5.94x10-4M
[S2]eq = 2.97x10-4M 7-
2
-42-4
c 1.76x10 (0.0244)
)(2.97x10)(5.94x10K
Double-check by working backward
Slightly off because of
approximation. But O.K.
Is our approximation appropriate? Calculate what percentage x is of your
original []’s. If it is 5% or smaller, it is OK.
1.19%x1000.0250
2.97x10-4
Our approximation is
appropriate