### Unit 12 - Equilibrium

```Equilibrium
UNIT 12
Overview

Concept of Equilibrium




Equilibrium constant
Equilibrium expression
Heterogeneous vs
homogeneous equilibrium
Solving for equilibrium
constant



Solving for K
Manipulating equilibrium
constant
Using gases

Equilibrium Calculations



Reaction Quotient (Q)


ICE method
ICE Stoichiometry
Relationship K and Q
Le Chatlier’s Principle





Pressure
Volume
Concentration
Temperature
Catalysts
The Concept of Equilibrium

We’ve already used the phrase “equilibrium” when talking

In principle, every chemical reaction is reversible ...
capable of moving in the forward or backward direction.
2 H2 + O2
2 H 2O
Some reactions are easily reversible ...
Some not so easy ...
The Concept of Equilibrium

Chemical equilibrium occurs when a reaction and
its reverse reaction proceed at the same rate.
The Concept of Equilibrium


As a system approaches
equilibrium, both the
forward and reverse
reactions are occurring.
At equilibrium, the
forward and reverse
reactions are
proceeding at the same
rate.
A System at Equilibrium

Once equilibrium
is achieved, the
amount of each
reactant and
product remains
constant.
A System at Equilibrium
Rates become equal
Concentrations become constant
Depicting Equilibrium

In a system at equilibrium, both the forward and
reverse reactions are running simultaneously. We
write the chemical equation with a double arrow:
Equilibrium Constant
Forward reaction:
Rate law
Reverse reaction:
Rate Law
Equilibrium Constant
At equilibrium
r
Rearranging gives:
Equilibrium Constant

The ratio of the rate constants is a constant (as long as T
is constant).
The expression becomes
Equilibrium Expression
(Law of Mass Action)
To generalize, the reaction:
Has the equilibrium expression:
This expression is true even if you don’t know the elementary
reaction mechanism.
Equilibrium Expression

[A], [B], [C], [D] = molar concentrations or partial pressures
at equilibrium

Products = numerator and reactants = denominator

Coefficients in balanced equation = exponents
Equilibrium Expression

Equilibrium constant can have different subscripts

Kc = concentration
 Moles, liters, or molarity

Kp = partial pressure
 Atmospheres

Ksp = solubility product
 Reactant is a solid so only products are in expression
Equilibrium Expression

Does not depend on initial concentration of reactants and
products

Solids and pure liquids are not included in equilibrium
expression (only gases and solutions)

Keq does not have units (describes activity of reaction)

Depends only on the particular reaction and the temperature
Equilibrium Expression

The Concentrations of Solids and Liquids Are
Essentially Constant

Therefore, the concentrations of solids and liquids do not
appear in the equilibrium expression
PbCl2 (s)
Pb2+ (aq) + 2 Cl−(aq)
Kc = [Pb2+] [Cl−]2
Sample Exercise 1

Write the equilibrium expression for Kc for
the following reactions:
Homogeneous vs Heterogeneous

Homogeneous equilibrium

Reactants and products are all in the same phase


Example: 2NOBr(g) ↔ 2NO(g) + Br2(g)
Heterogeneous equilibrium

Reactants and products are occur in different phases

Example: PbCl2 (s) ↔ Pb2+(aq) + 2Cl-(aq)
Equilibrium - Summary

At equilibrium…

Concentrations of reactants and products no longer change
with time

Neither reactants nor products can escape from the system

The equilibrium constant comes from a ratio of concentration
Solving for Keq
Kc, the final ratio of [NO2]2 to [N2O4], reaches a
constant no matter what the initial concentrations of
NO2 and N2O4 are.
Solving for Keq
=
[0.0172]2
[0.00140] = 0.211
Solving for Keq
This graph shows data
from the last two trials
from the table.
What does the value of K mean?

If K >> 1, the reaction is
product-favored; contains
mostly products at
equilibrium
• If K << 1, the reaction is
reactant-favored; contains
mostly reactants at
equilibrium.
Manipulating Equilibrium Constants

The equilibrium constant of a reaction in the reverse
reaction is the reciprocal of the equilibrium constant
of the forward reaction.
Manipulating Equilibrium Constants

The equilibrium constant of a reaction that has been
multiplied by a number is the equilibrium constant
raised to a power that is equal to that number.
Manipulating Equilibrium Constants

If two reactions can be added together to create a
third reaction, then the Keq for the two reactions can
be multiplied together to get the Keq for the third
reaction.
If
and
then
A+B↔C
C↔D+E
A+B↔D+E
Keq = K1
Keq = K2
Keq = K1K2
Manipulating Equilibrium Constants
(Summary)

The equilibrium constant in the reverse direction is the
inverse of the equilibrium constant in the forward
direction.

The equilibrium constant of a reaction that has been
multiplied by a number is the equilibrium constant raised
to a power equal to that number.

The equilibrium constant for a net reaction made up of
multiple steps is the product of the equilibrium constants
for individual steps.
Equilibrium Constant and Gases

Because pressure is proportional to concentration
for gases, the equilibrium expression can also be
written in terms of partial pressures (instead of
concentration):
Mixed versions are also used sometimes:
Relationship between Kc and Kp
From the ideal gas law we know that
= Pressure in terms of concentration
Relationship between Kc and Kp

Substituting P=[A]RT into the expression for
Kp for each substance, the relationship
between Kc and Kp becomes
Kp = Kc (RT)n
Where:
n = (moles of gaseous product) – (moles of gaseous reactant)
R = 0.0821 L∙atm/mol∙K
GIVEN TO YOU ON AP CHEAT SHEET!!
Relationship between Kc and Kp

In the synthesis of ammonia from nitrogen and
hydrogen, N2(g) + 3H2(g) ↔ 2NH3(g). Given
Kc of 9.60 at 300°C, calculate Kp.
Kp = Kc (RT)n
n = (moles gaseous products – moles gaseous reactants)
= 2-4 = -2
Kp = 9.60(0.0821×573K)-2= 4.34×10-3
ICE
Br2 (g)

2Br (g)
At 12800C the equilibrium constant (Kc) for the
reaction is 1.1 x 10-3. If the initial concentrations are
[Br2] = 0.063 M and [Br] = 0.012 M, calculate the
concentrations of these species at equilibrium.
Comparing initial concentrations to
equilibrium concentrations
At 12800C the equilibrium constant (Kc) for the reaction is 1.1 x 10-3. If the
initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the
concentrations of these species at equilibrium.
Br2 (g)
2Br (g)
Let x be the change in concentration of Br2
Br2 (g)
2Br (g)
Initial (M)
0.063
0.012
Change (M)
-x
+2x
0.063 - x
0.012 + 2x
Equilibrium (M)
[Br]2
Kc =
[Br2]
(0.012 + 2x)2
= 1.1 x 10-3
Kc =
0.063 - x
Solve for x
14.4
Kc =
(0.012 + 2x)2
0.063 - x
= 1.1 x 10-3
4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x
4x2 + 0.0491x + 0.0000747 = 0
-b ±
2
x=
ax + bx + c =0
b
2a
x = -0.0105
Br2 (g)
2Br (g)
Initial (M)
0.063
0.012
Change (M)
-x
+2x
0.063 - x
0.012 + 2x
Equilibrium (M)
At equilibrium, [Br] = 0.012 + 2x = -0.009 M
At equilibrium, [Br2] = 0.062 – x = 0.0648 M
2
– 4ac
x = -0.00178
or 0.00844 M
14.4
Example Problem: Calculate Concentration
Note the moles into a 10.32 L vessel stuff ... calculate molarity. Starting concentration
of HI: 2.5 mol/10.32 L = 0.242 M
2 HI
H2 + I2
0.242 M
Initial:
Change: -2x
0.242-2x
Equil:
0
+x
x
Keq 
[ H 2 ][ I 2 ]
[ HI ]
Keq 
37
[ x ][ x ]
[ 0 . 242  2 x ]
2
0
+x
x

2
x
2
[ 0 . 242  2 x ]
2
 1 . 26 x10
3
Example Problem: Calculate Concentration
And yes, it’s a quadratic equation. Doing a bit of rearranging:
x
2
[ 0 . 242  2 x ]
2
 1 . 26 x10
0 . 995 x  1 . 22 x10
2
x
b
x  7 . 38 x10
b  4 ac
2
2a
38
3
3
5
0
x = 0.00802 or –0.00925
Since we are using this to model a real,
physical system, we reject the negative
Root. The [H2] at equil. is 0.00802 M.
Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M
Initial
change
equil:
I2
2I
0.20
-x
0.20-x
0
+2x
2x
Keq 
[I ]
2
 2 . 94 x10
 10
[I2 ]

[2 x]
2
[ 0 . 20  x ]
 2 . 94 x10
 10
Approximating
If Keq is really small the reaction will not proceed to the right
very far, meaning the equilibrium concentrations will be
nearly the same as the initial concentrations of your
reactants.
0.20 – x is just about 0.20 is x is really dinky.
(Example: A million dollars minus a nickel is still a million
dollars)
If the difference between Keq and initial concentrations is
around 3 orders of magnitude or more, go for it.
Otherwise, you have to use the quadratic.
40
Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M
I2
Initial
change
equil:
0.20
-x
0.20-x
2I
0
+2x
2x
Keq 
[I ]
2
 2 . 94 x10
 10
[I2 ]

[2 x]
2
[ 0 . 20  x ]
 2 . 94 x10
 10
More than 3
orders of mag.
between these
numbers. The
simplification will
work here.
Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M
I2
Initial
change
equil:
[2 x]
2I
0.20
-x
0.20-x
2
0 . 20
 2 . 94 x10
10
0
+2x
2x
Keq 
[I ]
2
 2 . 94 x10
 10
[I2 ]

[2 x]
2
[ 0 . 20  x ]
x = 3.83 x 10-6 M
 2 . 94 x10
 10
More than 3
orders of mag.
between these
numbers. The
simplification will
work here.
Equilibrium Stoichiometry
Given initial concentrations of reactants
AND
Equilibrium concentration of products
(or vice versa)
Equilibrium Stoichiometry
A closed system initially containing1.000 x 10-3 M H2 and 2.000 x
10-3 M I2 at 448C is allowed to reach equilibrium. Analysis of the
equilibrium mixture shows that the concentration of HI is 1.87 x
10-3 M. Calculate Kc at 448C for the reaction:
H2 (g) + I2 (g)
2 HI (g)
Equilibrium Stoichiometry
ICE method:
[H2], M
Initially
[I2], M
1.000 x 10-3 2.000 x 10-3
[HI], M
0
Change
At
Equilibrium
Write what we know.
1.87 x 10-3
Equilibrium Stoichiometry
[H2], M
Initially
[I2], M
1.000 x 10-3 2.000 x 10-3
[HI], M
0
Change
+1.87 x 10-3
At
equilibrium
1.87 x 10-3
Calculate the change in [HI]. [HI] Increases by 1.87 x 10-3 M
Equilibrium Stoichiometry
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3 2.000 x 10-3
Change
-9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3
At
equilibrium
0
1.87 x 10-3
Use stoichiometry to solve for the change in concentration of
[H2] and [I2] from the given change in [HI].
Equilibrium Stoichiometry
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3 2.000 x 10-3
Change
-9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3
At
equilibrium
6.5 x 10-5
1.065 x 10-3
Now calculate concentrations at equilibrium
0
1.87 x 10-3
…and, therefore, the equilibrium
constant
[HI]2
Kc =
[H2] [I2]
=
(1.87 x 10-3)2
(6.5 x 10-5)(1.065 x 10-3)
= 51
The Reaction Quotient (Q)

To calculate Q, one substitutes the initial concentrations
of reactants and products into the equilibrium expression.

Q gives the same ratio the equilibrium expression gives,
but for a system that is not at equilibrium.
Q
[A], [B], [C], [D] = initial
molar concentrations or
partial pressures
If Q = K,
The system is at
equilibrium.
If Q > K,
There is too much
product and the
reaction proceeds
backwards
(equilibrium shifts to
the left).
If Q < K,
There is too much
reactant, and the
reaction proceeds
forward
(equilibrium shifts
to the right).
Summary

Q uses same as equilibrium expression but with initial
concentrations or pressures

Q = K, reaction is at equilibrium
Q > K, reaction goes backwards (shifts left)
Q < K, reaction goes forward (shifts right)


Le Châtelier’s Principle
“If a system at equilibrium is disturbed by a change in
temperature, pressure, or the concentration of one of
the components, the system will shift its equilibrium
position so as to counteract the effect of the
disturbance.”
Le Chatlier’s Principle
1.
System starts at equilibrium.
2.
A change/stress is then made to system at
equilibrium.





3.
Change in concentration
Change in volume
Change in pressure
Change in Temperature
System responds by shifting to reactant or product
side to restore equilibrium.
Le Chatlier’s Principle – basic terms

When you take something away from a system at equilibrium,
the system shifts in such a way as to replace some what you’ve
taken away.

When you add something to a system at equilibrium, the
system shifts in such a way as to use up some of what you’ve
Le Chatlier’s Principle - Concentration

A system shifts to offset
the change in
concentration of reactants
or products
Le Chatlier’s Principle - Concentration

Adding reactants or products causes a shift in the other
direction to offset the added reactants or products
aA + bB
cC + dD
Shifts right
aA + bB
cC + dD
Shifts left
Le Chatlier’s Principle - Concentration

Removing reactants or products causes a shift toward
the removed reactants or products to offset the change
Remove
aA + bB
Shifts left
Remove
cC + dD
aA + bB
cC + dD
Shifts right
Le Chatlier’s Principle – Volume & Pressure

Only matters when
dealing with gases

Count the total moles
of gas on each side of
the equation to
determine shift
Le Chatlier’s Principle – Volume & Pressure
The system shifts to remove gases and decrease
pressure.
An increase in pressure favors the direction that has
fewer moles of gas.
In a reaction with the same number of product and
reactant moles of gas, pressure has no effect.



62
Le Chatlier’s Principle – Volume & Pressure
A (g) + B (g)
CHANGE
Increase pressure
Decrease volume
Decrease pressure
Increase volume
C (g)
SHIFT
Toward side with less moles of gas
Toward side with less moles of gas
Toward side with more moles of gas
Toward side with more moles of gas
Le Chatlier’s Principle – Temperature

Endothermic: heat can be considered as a reactant


Exothermic: heat can be considered as a product



A + heat  B + C
A + B C + heat
Adding heat (i.e. heating the vessel) favors away from
the increase
Removing heat (i.e. cooling the vessel), favors
towards the decrease
Le Chatlier’s Principle – Temperature


Equilibrium constant, K, is temperature dependant
Temperature is only factor that can change the value
of K
Change
Exothermic
Endothermic
Increase temperature
K decreases
K increases
Decrease temperature
K increases
K decreases
Le Chatlier’s Principle – Catalysts



Does not change K
Does not shift the position of an equilibrium system
System will reach equilibrium sooner
Catalysts increase the rate of both the
forward and reverse reactions.
Equilibrium is achieved faster, but the
equilibrium composition remains unaltered.
Le Chatelier Example #1
A closed container of ice and water is at equilibrium. Then,
the temperature is raised.
Ice + Energy  Water
right to restore
The system temporarily shifts to the _______
equilibrium.
Le Chatelier Example #2
A closed container of N2O4 and NO2 is at equilibrium.
NO2 is added to the container.
N2O4 (g) + Energy  2 NO2 (g)
left to
The system temporarily shifts to the _______
restore equilibrium.
Le Chatelier Example #3
A closed container of water and its vapor is at equilibrium.
Vapor is removed from the system.
water + Energy  vapor
right to
The system temporarily shifts to the _______
restore equilibrium.
Le Chatelier Example #4
A closed container of N2O4 and NO2 is at equilibrium.
The pressure is increased.
N2O4 (g) + Energy  2 NO2 (g)
left to restore
The system temporarily shifts to the _______
equilibrium, because there are fewer moles of gas on
that side of the equation.
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