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2-01 Equilibrium
We have hinted that reactions can proceed in the forward or
reverse direction
• CH3-CH(OH)-CH3 with an H2SO4 catalyst  CH3-CH=CH2
• But CH3-CH=CH2 with an H2SO4 catalyst  CH3-CH(OH)-CH3
• This makes sense from our PE diagrams
• The forward and reverse reactions just have different
activation energies and opposite ∆H values
A reversible reaction is said to be at equilibrium when the rate of
the forward reaction is equal to the rate of the reverse reaction
• If the products are able to escape from the reaction vessel
they won’t be available to reform reactants
o The reaction must be in a closed system
Closed system
• Nothing can enter or leave
Assign 1-2
• The temperature affects equilibrium
o A new equilibrium is attained at a new temperature
• The same equilibrium will exist regardless of whether we
start with excess reactants or products
• When a system is at equilibrium no macroscopic changes
occur
Dynamic equilibrium (chemistry)
• Equilibrium in which microscopic changes but not
macroscopic changes occur
• Constant microscopic back and forth between reactants and
products but the rates cancel out and nothing is visible on the
large scale
• http://www.mhhe.com/physsci/chemistry/animations/chang_
7e_esp/kim2s2_5.swf
reversible-reactions.jar
Static equilibrium (Physics)
• No changes occur at all
Assign 3-5
2-02 Characteristics of Equilibrium
Since equilibrium reactions are reversible there is both a forward
and reverse rate.
REACTANTS ↔ PRODUCTS
The rate of a reaction increases when the [reactants] increases
• The rate must be proportional to the [reactants]
• Rateforward = kforward [reactants]
• Ratereverse = kreverse [products]
• K is some constant for each reaction…a number
Try 6 and 7 on graph paper
Should have found that at equilibrium:
• Rate of consumption of reactants = rate of
production of reactants
• [reactants] is not = [products] in general
• [reactants] is constant and [products] is constant
• forward and reverse rates do not change
• a system which is not at equilibrium will tend to
move towards equilibrium
 once equilibrium is reached, no further change
occurs
reversible-reactions.jnlp
For H2 + Cl2 <--> 2HCl
H2
+
Cl2
=
HCl
The large boxes represent the total amounts at some equilibrium
• the number of moles of each gas present is not evident
just from the equation
• the ratio of [product]/[reactant] is a constant at
equilibrium
 more later…
The small boxes are the amounts that react every second at
equilibrium
• total reactants changing = total products changing
back= no net change
• 1H2 and 1Cl2 react to make 2HCl while at the same
time 2HCl react to make 1H2 and 1Cl2
• The rate that something is reacted is equal to the rate
that it is produced
Assign 8-13
2-03 Predicting if a reaction is spontaneous or not
Spontaneous change
• Occurs by itself without outside assistance
• Some conditions allow while other forbid spontaneous
reactions
Exothermic reactions should be spontaneous provided the
activation energy is not too high.
PE
Reaction proceeds
If you give the reaction a little nudge it goes to completion
• Match to paper
• Ignite explosives…
But endothermic reactions require constant energy input to proceed
and are not expected to be spontaneous unless sufficient heat is
available.
• Molecules on the downhill side would be expected to stay
there unless energy is supplied to them.
• The side having minimum energy (minimum enthalpy) is
favored
• lazy
There is more to spontaneity
• Some endothermic reactions are spontaneous
o Ammonium nitrate and water react spontaneously and
absorb energy from the surroundings
• Some exothermic reactions don’t go to completion
o Equilibrium is established as the reverse endothermic
reaction occurs
Randomness is a second factor that comes into play for spontaneity
• Most arrangements in space are disordered
• Only a few special arrangements are ordered
o Throw a deck of cards in the air
• There are a few situations where order could
develop
• Card house
• Back in sequence
• All Spades together
• Most situations from that random event are
random (disordered)
• A disordered result from a random event is more
probable than an ordered result.
As a crystal of salt dissolves water molecules strike the ions and
they bounce out of the lattice
• To get back in another molecule of water will have to hit it
perfectly to go back into its ordered position
o There are many other possible ways the ion could be hit
that won’t take it back to its crystal
o Disorder is favored and the crystal dissolves
Entropy is the amount of randomness in a system.
• Entropy tends to a maximum value
• Things tend to get jumbled up over time
• Slob
Ba2+(aq) + 2SCN-(aq) 2NH3(g) +
10H2O(l)
2NH4SCN (s) +
Ba(OH)2.8H2O(s)
The reactants are highly ordered crystals and the products are very
random ions so entropy favors the reaction even if enthalpy does
not.
The reactants absorb energy and then become so random that they
can’t reorganize to reform.
• The reaction is spontaneous and can drop the surrounding
temperature by 55o C
All reactions are driven towards maximum randomness (entropy)
and minimum energy (enthalpy)
• The universe favors the sloppy and the lazy
Tendency to minimum enthalpy
Exothermic
AB + heat
Endothermic
C + heat  D
The arrow points to the tendency for minimum enthalpy
• the down hill side
• the side with heat as a reactant
Tendency to maximum entropy
• look at the relative randomness of phases
• gases >> solutions > liquids >> solids
H2O (s) ↔ H2O(l)
• the liquid is more random and the products are favored
CaC(s) + 2H2O(l) ↔ C2H2(g) + Ca(OH)2(aq)
• the most random phase present is gas and the products are
favored
A(g) + B(s) ↔ 2C(g) + D(s)
• if the reactants and products are in the same phase then the
side with more molecules is more random (more possible
disordered arrangements)
Consider
• C2H2(g) + 2Cl2(g)  C2H2Cl4(l) + 386kJ
o Enthalpy favors the products
o Entropy favors the reactants
o Opposite tendencies lead to equilibrium
• CH4(g) + O2(g)  CO2(g) + 2H2O(g) + 394 kJ
o Enthalpy favors the products
o Entropy favors the products
o Reaction goes “100%”
• 4Au(s) + 3O2(g) + 162kJ  2Au2O3(s)
o enthalpy favors the reactants
o entropy favors the reactants
o the reaction goes “0%”
 no reaction
o gold doesn’t oxidize!
Try 14-16
Notes 2-04
Le Chatelier’s Principle
Le Chatelier’s Principle
• If a closed system at equilibrium is subjected to a
change, processes will occur that tend to counteract that
change
• Whatever we do, nature tries to undo
This principle is crucial to quickly predicting what will
happen in equilibrium systems and will be used throughout
the course.
Fe3+ + SCN-   FeSCN2+
• If we increase [Fe3+] or [SCN-] then the [FeSCN2+]
increases.
• If we decrease [Fe3+] or [SCN-] then the [FeSCN2+]
decreases.
Think of the water in a bathtub representing a system at
equilibrium with the reactants at one side of the tub and the
products at the other.
• If you dump some water at one end of the tub the water
flows towards the other end of the tub to reestablish
equilibrium
• If you scoop out some water at one end of the tub the
water flows back towards the “hole” to reestablish
equilibrium.
The effect of temperature changes:
• If we decrease the temperature in a reaction like:
2NO(g) + Cl2 (g) ↔ 2NOCl(g) + 76kJ
• We removed energy (a product)
• It’s like we scooped out some water on the product
side
• The reaction shifts to the right to compensate
• More reactants become products
Cl2
NO
[…]
NOCl
Time at which temp
changed
[Cl2] only changes half
as much as [NO] since
they react in a 1:2 ratio
Time
The effect of concentration changes:
• If we increase the [Cl2] in a reaction like:
2NO(g) + Cl2 (g) ↔ 2NOCl(g) + 76kJ
• We added a reactant
• It’s like we dumped some water on the reactant side
• The reaction shifts to the right to compensate
• More reactants become products
Cl2
Time at which Cl2 was
injected quickly
NO
Not all the Cl2 added is
used up… the change
can’t be cancelled
completely because the
other parts changed
[…]
NOCl
Time
The effect of pressure changes:
• A decrease in volume increases the pressure and the
concentration of all the gases.
2NO(g) + Cl2 (g) ↔ 2NOCl(g) + 76kJ
• Figure out which side has the most gases (3 moles reactants and
2 moles products)
• It’s like we dumped some water on the high gas side
(reactants)
• The reaction shifts to the right to compensate (want fewer
gases at high pressure)
• More reactants become products
Cl2
Time at which pressure
changed
NO
All gases increase in
concentration at first and
then the equilibrium
shifts to make products
[…]
NOCl
Time
The effect of adding a catalyst
• Since a catalyst speeds up the forward and reverse rate
the reaction remains at equilibrium
• Just gets there faster if it wasn’t already there.
Graphing notes:
• Temperature changes
• Concentrations change slowly to a new value
• Changing the concentration of a species
• One species suddenly jumps up or down then all
adjust slowly to compensate
• Changing the pressure
• The concentration of all gasses simultaneously jump
up or down then slowly adjust.
Assign 17-28
PS It helps to draw a little mound or hole over the
substance that increases or decreases to visualize which
way the equilibrium will shift.
Notes 02-05
Equilibrium in industry
In WWll Germany couldn’t get nitrates from the main global
source, Chile. Nitrates are used to make Trinitrotoluene (TNT).
Fritz Haber figured out how to make ammonia on a huge scale,
which can easily be oxidized to nitrates.
The Haber Process for making ammonia (important to
remember)
• N2(g) + 3H2(g) ↔ 2NH3(g) + 92kJ
Figure out what conditions will maximize the yield of ammonia
• Pressure…high
• Temperature?
o Reaction rate = high vs yield = low
• Other…iron oxide catalyst
http://www.youtube.com/watch?v=c4BmmcuXMu8
The Making of Cement from Limestone
• CaCO3(s) + 175kJ ↔ CaO(s) + CO2(g)
• Best conditions are
o Pressure….low
o Temperature…high
 Is there a conflict between rate and yield?…no
Notes 2-06
The equilibrium expression and the equilibrium constant
Pull out the two graphs we drew for questions 6 and 7. At the first
equilibrium there was 5 times as much B as A. At the second
equilibrium there was 5 times as much B as A.
• The constant ratio is not a fluke!
For the equilibrium equation:
A+B⇔C+D
it has been found that
[C ]× [D] = a constant
[A]× [B]
the expression is called the equilibrium expression and Keq is a
number called the equilibrium constant.
K eq =
Examples
• H2O(g) + CO(g) ↔ H2(g) + CO2(g)
o K eq = [H2(g) ]× [CO2(g)]
[H2O(g) ]× [CO(g)]
• PCl5(g) ↔ PCl3(g) + Cl2(g)
o K eq = [PCl3(g) ]× [Cl2(g)]
[PCl5(g) ]
• H2(g) + F2(g) ↔ HF(g) + HF(g)
or H2(g) + F2(g) ↔ 2HF(g)
2
o K eq = [HF(g) ]× [HF(g)] = [HF(g) ]
[H2(g) ]× [F2(g) ] [H2(g) ]× [F2(g) ]
o The coefficient becomes the exponent
In general:
• aA + bB + cC +… ↔ pP + qQ + rR +… gives
p
q
r
[
P ] × [Q] × [R ] ...
• K eq = a
or
[A] × [B]B × [C ]c ...
• K eq = [Products ]
[Reactants]
But there is a little wrinkle:
• for CaF2(s) ↔Ca2+(aq) + 2F-(aq)
•
[Ca
=
2+
]× [F
−
]
2
= 8.4 ×10 −13 (measured experimentally)
[CaF2(s)]
• but for a pure solid the concentration is a constant because its
density is a constant
o so [CaF2(s)] = 40.7M from 3.18 g/L and 78.1 g/mol
o if the concentration of the solid never changes the
expression is really:
K eq
( aq )

K eq
( aq )
[Ca
=
2+
( aq )
]× [F
−
]
2
( aq )
40.7
= 8.4 ×10 −13
 Why bother having 2 constants in the same
expression?
 Combine the constants to get

K eq ( new)
[Ca
=
2+
( aq )
]× [F
−
]
2
( aq )
= 3.4 ×10 −11
Instead of having to calculate the concentration of a solid every
time, just combine it into the Keq.
• Eliminate any concentrations that have a constant value in the
equilibrium expression
o Solids…the concentration comes from the density
which doesn’t change
o Pure liquids…the concentration can come from density
Only counts if it is the only liquid that exists in

the entire equilibrium expression
If there is more than one then dilution occurs and

concentration will vary
2
[
HBr ( g ) ]
• Br2(l) + H2(g) ↔ 2HBr(g) gives K eq =
[H2(g)]
• CH3COCH3(l) +Cl2(g) ↔ CH3COCH2Cl(l) + HCl(g) gives
• Cl2(g) + 8H2O(l) ↔ Cl2.8H2O(s) gives
Adding a reactant or a product which is a solid or pure liquid will
have no effect on the equilibrium
Assign 31-35
02-07 Le Chatelier’s Principle and the Equilibrium Constant
When the concentration pressure or temperature are changed the reaction tends to
counteract these changes.
• One shot change
 The system regains equilibrium
• Continuous change
 You have an open system
 Does not reach equilibrium
When temperature is decreased and held
2NO(g) + Cl2(g) ↔ 2NOCl(g) + 76kJ
shifts to the product side. This results in an increase in [products]
and a decrease in [reactants] so that
increases.
To figure out the direction of the change in Keq draw an arrow in
the Keq expression that shows the shift.
• An upward arrow indicates a higher Keq and vice versa
•
Only changes in temperature change Keq
Large Keq means that [products] > [reactants]
• lots of products at equilibrium
Small Keq means that [products] < [reactants]
• lots of reactants at equilibrium
Assign 36-46
02-08 Equilibrium Calculations
To determine the value of Keq for any reaction you need to
accurately know the concentrations of all of the species present.
• Must be experimentally determined
Example A
For 2NO(g) + O2(g) ↔2NO2(g)
A 2.0 L bulb contains 6.00 mol of NO2(g), 3.0 mol of NO(g) and
0.20 mol of O2(g) at equilibrium. What is the Keq?
First determine the equilibrium expression:
Use the data to calculate each of the concentrations:
Substitute the values into the expression:
[NO 2 ]2
[3.0]2
=
= 40
K eq =
[NO]2 [O 2 ] [1.5]2 [0.10]
Example B
4.00 mol of NO2 is introduced into a 2.00L bulb. After a while
equilibrium is attained. At equilibrium 0.50 mol of NO is found.
What is Keq?
The question implies that time had to pass for equilibrium to pass. The initial
concentration is not an equilibrium concentration.
∴ you need to figure out the equilibrium [ ]’s.
[NO 2 ]2
The Keq expression is K eq =
[NO]2 [O 2 ]
Calculate the concentrations from the questions:
4.00mol
= 2.0M
2.00L
Set up an “ice” box
[NO 2 ]start =
2NO(g)
[NO]eq =
0.500mol
= 0.250M
2.00L
O2(g)
↔
2NO2(g)
Initial
0
0
2.00
Change
Equilibrium
0.250
Units are omitted but they must all be the same (usually all M but
it can work with mol, contrary to what your book says)
Fill in the rest of the box. The values in the change row obey the
rules of stoichiometry. 2:1:2 ratio
∆ NO = + 0.250 (a shift left)
∆ O2 = ½ ∆ NO = + 0.125 (the shift added to the reactants)
∆ΝΟ2 = − ∆ NO = - 0.250 (the shift took away NO2)
Initial
Change
Equilibrium
2NO(g)
O2(g)
0
+0.250
0.250
0
+0.125
0.125
↔
2NO2(g)
2.00
-0.250
1.75
Substitute the equilibrium concentrations into the equilibrium
expression:
[NO 2 ]2
[1.75]2
K eq =
=
= 392
[NO]2 [O 2 ] [0.250]2 [0.125]
Example C
An unknown amount of NO2 was introduced to a 5.00L bulb.
When equilibrium was attained according to the equation
2NO(g) + O2(g) ↔2NO2(g)
the concentration of NO was 0.800M. If Keq is 24.0, how many
moles of NO2 were originally put in the bulb?
There is a start and then time passing before equilibrium. We
know [NO]eq = 0.800M and we want [NO2]start = x.
Set up the ice box
Initial
Change
Equilibrium
2NO(g)
O2(g)
0
0
↔
2NO2(g)
x
0.800
[NO] went from 0 to 0.800 so the shift was left. Fill in the box
with changes that match the stoichiometry of 2:1:2.
∆ NO = + 0.800
∆ O2 = ½ ∆ NO = + 0.400
∆ΝΟ2 = − ∆ NO = - 0.800
Initial
Change
Equilibrium
2NO(g)
O2(g)
0
+0.800
0.800
0
+0.400
0.400
↔
2NO2(g)
x
-0.800
x-0.800
Substitute the values into the equilibrium expression:
[NO 2 ]2
[x - 0.800]2
=
= 24.0
K eq =
[NO]2 [O 2 ] [0.800]2 [0.400]
You won’t have to use the quadratic equation in chem. 12. There
will always be a way around it.
(x-0.800)2=(24.0)(0.800)2(0.400)=6.144 take the square root of
both sides
x-0.800=2.479
x=3.279M = [NO2]start
We needed moles at the start so:
mol
3.279
× 5.00L = 16.4mol
# moles NO2 =
L
Example D
Keq= 49 for 2NO(g) + O2(g) ↔2NO2(g). if 2.0mol of NO , 0.2mol O2
and 0.40mol of NO2 are put into a 2.0L bulb, which way will the
reaction shift in order to reach equilibrium?
This question is asking for a decision not a numerical answer. You
need to compare two numbers
[NO 2 ]2
K eq =
= Q , where Q is a trial value for Keq
[NO]2 [O 2 ]
if Q = Keq then the system is at equilibrium
if Q < Keq then [products] is too small and the reaction will shift to
[reactants]
products to raise Keq
if Q > Keq then [products] is too big and the reaction will shift to
[reactants]
reactants to lower Keq
First find the equilibrium expression:
K eq =
[products]
[NO 2 ]2
=
[reactants] [NO]2 [O 2 ]
Calculate the starting concentrations:
[NO]start =
2.0mol
= 1.0M
2.0L
[NO 2 ]start =
0.40mol
= 0.20M
2.0L
Calculate the reaction quotient
∴ the reaction must shift right to the products
Example E
Keq = 3.5 for SO2(g) + NO2(g) ↔ SO3(g) + NO(g). If 4.0 mol of SO2(g)
and 4.0 mol of NO2(g) are placed in a 5.0 L bulb and allowed to
come to equilibrium. What will be the equilibrium concentrations
of all the species?
First write the equilibrium expression
K eq =
[SO 3 ][NO]
[SO 2 ][NO 2 ]
Set up an ice box
Initial
Change
Equilibrium
SO2(g)
0.80
NO2(g)
0.80
SO3(g)
0
NO(g)
0
We will need to enter variables for the rest of the spaces. It’s
easiest to assign x as the change for one of the species:
SO2(g)
NO2(g)
SO3(g)
NO(g)
Initial
0.80
0.80
0
0
Change
-x
-x
+x
+x
Equilibrium
0.80-x
0.80-x
x
x
Substitute the equilibrium concentrations into the Keq expression:
[SO 3 ][NO]
x2
K eq =
=
= 3.5 take the square root of both
[SO 2 ][NO 2 ] (0.80 - x) 2
sides
∴[SO3]=[NO]= 0.52M; [SO2]=[NO2]=0.28M
Example F
A 1.0 L reaction vessel contained 1.0 mol of SO2, 4.0 mol of
NO2, 4.0 mol of SO3 and 4.0 mol of NO at equilibrium
according to SO2(g) + NO2(g) ↔ SO3(g) + NO(g). If 3.0 mol of SO2
is added to the mixture, what will the new concentration of NO be
when equilibrium is re attained?
You have all the information needed to calculate Keq.
Set up an ice box (any additions or removals are put in the initial
line)
SO2(g)
NO2(g)
SO3(g)
NO(g)
Initial
1.0 +3.0
4.0
4.0
4.0
Change
-x
-x
+x
+x
Equilibrium
4.0-x
4.0-x
4.0+x
4.0+x
Take the root of both sides
∴ [NO] = 4.0 + x = 5.3M
Assign 47-65 odd and 66.