Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and
Bruce E. Bursten
Chapter 15
Chemical Equilibrium
Directions: Carefully read all slides.
Copy into your notes only the slides written in BLACK font!
Also – Please proofread and correct any errors (if needed).
Equilibrium
Chemical
Equilibrium
Equilibrium
Chemical Equilibrium
1. Equilibrium is a state of balance between
two opposing reactions.
Some examples of opposing reactions:
Evaporation & Condensation
Melting & Freezing
Dissolving & Re-crystallization
2. The reactions must occur in a closed
system so that neither product nor reactant
can escape.
Equilibrium
Chemical Equilibrium (con’t)
3. When equilibrium is achieved, the two
opposing reactions are occurring at the
same rate or speed.
4. As a result, at equilibrium, the
concentrations of the reactants and
products are constant. (NOT EQUAL)
Equilibrium
Chemical Equilibrium (con’t)
5. Equilibrium is a dynamic process. It
looks like nothing is happening, but the
reactions have not stopped.
6. Double arrows are used to represent a
system at equilibrium.
Example:
Equilibrium
N2O4  2 NO2
The Concept of Equilibrium
N2O4 gas is nearly colorless. As it is heated, it
dissociated into brown NO2 gas. This reaction is
reversible. NO2 can also reform N2O4.
Equilibrium
Eventually, equilibrium is reached.
The Concept of Equilibrium
Chemical equilibrium occurs when a
reaction and its reverse reaction proceed at
the same rate.
Equilibrium
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.
Equilibrium
A System at Equilibrium
Once equilibrium is
achieved, the
amount of each
reactant and product
remains constant.
Equilibrium
Depicting Equilibrium
In a system at equilibrium, both the
forward and reverse reactions are being
carried out; as a result, we write its
equation with a double arrow
N2O4 (g)
2 NO2 (g)
Equilibrium
Check
Yourself
by Answering
the Questions
that Appear
Throughout
the Slideshow
Equilibrium
When a reaction is at a state of
equilibrium, the rate of the
forward reaction __________
the rate of the reverse reaction.
a.
b.
c.
d.
is equal to
is slower than
is faster than
is the reverse of
Equilibrium
When a reaction is at a state of
equilibrium, the rate of the
forward reaction __________
the rate of the reverse reaction.
a.
b.
c.
d.
is equal to
is slower than
is faster than
is the reverse of
Equilibrium
When a reaction is at a state of
equilibrium, the concentration of
the reactants and the
concentration of the products
are _____________.
a.
b.
c.
d.
equal to each other
constant, but not usually equal
constantly increasing
constantly decreasing
Equilibrium
When a reaction is at a state of
equilibrium, the concentration of
the reactants and the
concentration of the products
are _____________.
a.
b.
c.
d.
equal to each other
constant, but not usually equal
constantly increasing
constantly decreasing
Equilibrium
The
Equilibrium
Constant
Kc
Equilibrium
The Equilibrium Constant
• Forward reaction:
N2O4 (g)  2 NO2 (g)
• Rate law:
Rate = kf [N2O4]
Equilibrium
The Equilibrium Constant
• Reverse reaction:
2 NO2 (g)  N2O4 (g)
• Rate law:
Rate = kr [NO2]2
Equilibrium
The Equilibrium Constant
• Therefore, at equilibrium
Ratef = Rater
kf [N2O4] = kr [NO2]2
• Rewriting this, it becomes
kf
kr
[NO2]2
=
[N2O4]
Equilibrium
The Equilibrium Constant
The ratio of the rate constants is a
constant at that temperature, and the
expression becomes
kf
Keq =
kr
[NO2]2
=
[N2O4]
Equilibrium
The Equilibrium Constant (Kc)
(K means constant, c means concentration)
• To generalize this expression, consider
the reaction
aA + bB
cC + dD
• The equilibrium expression for this
reaction would be
[C]c[D]d
Kc =
[A]a[B]b
Equilibrium
The Equilibrium Constant (Kc)
con’t
• Note: K expressions are written as
“products over reactants”.
• Include only gases & aqueous
substances in the expression.
• Omit all solids and liquids.
Equilibrium
The Equilibrium Constant (Kc)
con’t
• Example - For the equilibrium:
• N2(g) + 3 H2 (g)  2 NH3 (g)
Kc =
[NH3]2
[N2] [H2]3
Equilibrium
The Equilibrium Constant (Kc)
con’t
• You try this equilibrium:
• 2 Cl2(g) + 2 H2O(g)  4 HCl(g) + O2 (g)
•
Kc = ?
Equilibrium
2 Cl2(g) + 2 H2O(g)  4 HCl(g) + O2 (g)
• Answer:
Kc = [HCl]4 [O2 ]
[Cl2]2 [H2O]2
Equilibrium
What Are the Equilibrium
Expressions for These Equilibria?
SnO2(s) + 2 CO(g)  Sn(s) + 2 CO2(g)
CaO(s) + CO2(g)  CaCO3(s)
Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)
Copy answers to your notes!
Equilibrium
Answers
Kc = [CO2]2
[CO]2
Kc =
1
[CO2]
Kc = [Zn2+]
[Cu2+]
Equilibrium
At equilibrium, the equilibrium
constant is equal to:
a.
b.
c.
d.
[reactants]/[products]
[products]/[reactants]
[reactants] x [products]
[reactants] + [products]
Equilibrium
At equilibrium, the equilibrium
constant is equal to:
a.
b.
c.
d.
[reactants]/[products]
[products]/[reactants]
[reactants] x [products]
[reactants] + [products]
Equilibrium
The Equilibrium Constant
Because pressure is proportional to
concentration for gases in a closed
system, the equilibrium expression can
also be written
(PC)c (PD)d
Kp =
(PA)a (PB)b
Equilibrium
Relationship between Kc and Kp
• From the ideal gas law we know that
PV = nRT
• Rearranging it, we get
n
P=
RT
V
Equilibrium
Relationship between Kc and Kp
Plugging this into the expression for Kp
for each substance, the relationship
between Kc and Kp becomes
Kp = Kc (RT)n
Where R = 0.0821 L-atm/mol-K and T = Kelvin temp and
n = (moles of gaseous product) − (moles of gaseous reactant)
Your teacher will do problems like this with you later.
Equilibrium
More
About
Equilibrium
Systems
Equilibrium
Equilibrium Can Be Reached from
Either Direction
As you can see, the ratio of [NO2]2 to [N2O4]
remains constant at this temperature no matter
what the initial concentrations of NO2 and N2O4
are. Note: K values are temp dependent.
Equilibrium
Equilibrium Can Be Reached from
Either Direction
This is the data from
the last two trials from
the table on the
previous slide.
See how concentration
becomes constant at
equilibrium.
Equilibrium
Equilibrium Can Be Reached from
Either Direction
Here is another example. It does not matter
whether we start with N2 and H2 or whether
we start with NH3. We will have the same
proportions of all three substances at
Equilibrium
equilibrium.
What Does the Value of K Mean?
• If K >> 1, the reaction
is product-favored;
product predominates
at equilibrium.
Equilibrium
What Does the Value of K Mean?
• If K >> 1, the reaction
is product-favored;
product predominates
at equilibrium.
• If K << 1, the reaction is
reactant-favored;
reactant predominates
at equilibrium.
Equilibrium
What Does the Value of K Mean?
• Please note: The values for K change
with changes in temperature!
Equilibrium
A large value of the equilibrium
constant indicates that when the
reaction reaches equilibrium,
mostly ______ will be present.
a.
b.
c.
d.
reactants
products
catalysts
shrapnel
Equilibrium
A large value of the equilibrium
constant indicates that when the
reaction reaches equilibrium,
mostly ______ will be present.
a.
b.
c.
d.
reactants
products
catalysts
shrapnel
Equilibrium
The equilibrium constant for a
reaction is 375. This means the
equilibrium mixture is mainly:
a. reactant, because 375 is more than 1
b. product, because 375 is more than 1
c. cannot be determined.
Equilibrium
The equilibrium constant for a
reaction is 375. This means the
equilibrium mixture is mainly:
a. reactant, because 375 is more than 1
b. product, because 375 is more than 1
c. cannot be determined.
Equilibrium
Rules for Manipulating K Values
1. Reverse the Reaction?
Take the Reciprocal of K
In other words, 1/K
Equilibrium
Example:
The equilibrium constant of a reverse
reaction is the reciprocal of the
equilibrium constant of the forward
reaction.
N2O4 (g)
2 NO2 (g)
[NO2]2
2 NO2 (g) Kc = [N O ] = 0.212 at 100C
2 4
[N2O4]
1
K
=
N2O4 (g) c
=
[NO2]2
0.212
= 4.72 at 100C
Equilibrium
You try:
• If K = 3.5 x 10-4 for:
H2(g) + Br2(g)  2 HBr(g)
• What is K for:
2 HBr(g)  H2(g) + Br2(g) ?
• Answer: Use 1/K = 1/ 3.5 x 10-4
Answer: 2857 or 2.9 x
103
Equilibrium
(no units)
Rules for Manipulating K Values
(con’t)
2. Multiply a Reaction by a factor of ‘n’ ?
Use: Kn
Equilibrium
Example:
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.
N2O4 (g)
[NO2]2
2 NO2 (g) Kc = [N O ]
2 4
2 N2O4 (g)
[NO2]4
4 NO2 (g)Kc = [N O ]2 = (0.212)2 at 100C
2 4
= 0.212 at 100C
Equilibrium
You try:
• If K = 3.5 x 10-4 for:
H2(g) + Br2(g)  2 HBr(g)
• What is K for:
3 H2(g) + 3 Br2(g)  6 HBr(g)
• Answer: Use K3 = (3.5 x 10-4)3
Answer: 4.3 x 10-11
Equilibrium
You try:
• If K = 3.5 x 10-4 for:
H2(g) + Br2(g)  2 HBr(g)
• What is K for:
½ H2(g) + ½ Br2(g)  HBr(g)
• Answer: Use K1/2 = (3.5 x 10-4)1/2
Answer:
(Take square root)
1.9 x10-2
Equilibrium
Rules for Manipulating K Values
3. Add several reactions together?
Multiply the K values together
Equilibrium
In other words
The equilibrium constant for a net
reaction made up of two or more steps
is the product of the equilibrium
constants for the individual steps.
Equilibrium
You try:
• If K1 = 1 x 10-3 for:
AB2(g)  2 B(g) + A(g)
and
• If K2 = 1 x 10-5 for:
A(g) + B(g)  AB(g)
• What is the overall equation when these 2 reactions
are added together and what is the K value for the
new reaction?
•
Answers:
AB2(g)  B(g) + AB(g)
(Muliply K values) 1 x 10-8
Equilibrium
Equilibrium
Calculations
Equilibrium
Simple Equilibrium Calculations
• Given: N2O4 (g)  2 NO2 (g)
• Write the Kc expression and calculate its
value if the the equilibrium
concentrations are: 0.0172 M for NO2
and 0.00140 for N2O4
• Answers on the next slide.
Equilibrium
Answer
Kc = [NO2]2
[N2O4]
Kc = [0.0172]2
[0.00140]
Kc = 0.211
Equilibrium
Simple Equilibrium Calculations
• Given: N2(g) + 3 H2 (g)  2 NH3 (g)
• What is the partial pressure of the
ammonia in the equilibrium mixture if
the equilibrium mixture contains 0.982
atm hydrogen gas and 0.432 atm
nitrogen gas when Kp = 1.45 x 10-5 at
500 Celsius.
• Answers on the next slide.
Equilibrium
Answer
Kp = [NH3]2
[N2] [H2]3
1.45 x 10-5 =
[NH3]2
[0.432] [0.982] 3
(1.45 x 10-5 )(0.432) (0.982) 3 = [NH3]2
Answer: [NH3] = 2.48 x 10-3 M
Equilibrium
Stop Here!
Start Homework
Textbook Assignment
Due Tuesday
• Questions #15.13ab, 15.14ab,
15.15, 15.16, 15.19, 15.22ab,
15.27, 15.28, 15.29, 15.30
Equilibrium
Warm Up
Equilibrium
Warm-Up
1. Write the equilibrium expression for the
reaction:
Ti(s) + 2Cl2(g)  TiCl4(l)
Answer: Kc =
1
[Cl2]2
Equilibrium
Warm-Up
(con’t)
2. Write the equilibrium expression for
the reaction:
CH4(g) + 2 H2S (g)  CS2(g) + 4 H2(g)
Answer: Kc = [CS2] [H2]4
[CH4] [H2S]2
Equilibrium
Warm-Up
(con’t)
3. What are the rules for manipulating K
values when a Rx is
a) reversed?
b) doubled?
c) tripled?
d) halved?
e) added to a 2nd Rx?
Answers:
Equilibrium
2
3
1/2
a) 1/K b) K c) K d) K e) multiply K values
Warm-Up
(con’t)
4. Given: Kp = 1.5 x 10-10 for
N2(g) + O2(g)  2 NO(g)
What is the Kp for
NO(g)  ½ N2(g) + ½ O2(g)
Answer: 1/ Kp1/2 = 1/(1.5 x 10-10)1/2 =
8.2 x 104
Equilibrium
Warm-Up
(con’t)
5. Given: K1 = 67 for
CoO(s) + H2(g)  Co(s) + H2O(g)
And, K2 = 490 for
CoO(s) + CO(g)  Co(s) + CO2(g)
What is K3 for:
H2(g) + CO2(g)  CO(g) + H2O(g) ?
Equilibrium
Warm-Up
(con’t)
5. Given: K1 = 67 for
CoO(s) + H2(g)  Co(s) + H2O(g)
And, K2 = 490 for
CoO(s) + CO(g)  Co(s) + CO2(g)
What is K3 for:
H2(g) + CO2(g)  CO(g) + H2O(g) ?
Answer: Reverse the Rx for K2 and add it to the Rx for K1
So, take the reciprocal of K2 and multiply it times K1
So, use (1/ K2) x K1 = (1/490) (67) = 0.14
Equilibrium
Warm-Up
(con’t)
6. Define the term ‘equilibrium’.
Answer: A state of balance between 2
opposing reactions that are occurring at
the same rate. As a result, the
concentrations of the reactants and
products are constant.
Equilibrium
Warm-Up
(con’t)
7. A reaction has an equilibrium constant of
8.6 x 10-3. Does the equilibrium mixture
favor the production of reactants or
products?
Answer: Whenever K is << 1, the
equilibrium mixture consists primarily of
reactants.
Equilibrium
Kp = Kc
n
(RT)
Equilibrium
Kp = Kc (RT)n
Where R = 0.0821 L-atm/mol-K and T = Kelvin temp and
n = (moles of gaseous product) − (moles of gaseous reactant)
If Kc = 0.042 for PCl3(g) + Cl2(g)  PCl5(g) at
500 K, what is the value of Kp for this reaction
at this temperature?
Kp = Kc (RT)n
Kp = (0.042)[(0.0821 L-atom/mol-K)(500 K)]1-2
Kp = 1.0 x 10-3
Equilibrium
Kp = Kc (RT)n
Where R = 0.0821 L-atm/mol-K and T = Kelvin temp and
n = (moles of gaseous product) − (moles of gaseous reactant)
Calculate Kc at 30.0oC for
SO2(g) + Cl2(g)  SO2Cl2(g)
if Kp = 34.5 for this reaction at this temperature?
Kp = Kc (RT)n
34.5 = (Kc)[(0.0821 L-atom/mol-K)(303 K)]1-2
34.5 = (Kc) (0.04019)
Equilibrium
Kc = 858
The
Reaction
Quotient
(Q)
Equilibrium
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.
Equilibrium
If Q = K,
the system is at equilibrium.
Equilibrium
If Q > K,
there is too much product and the reaction
shifts to the left to reach equilibrium.
Equilibrium
If Q < K,
there is too much reactant, and the reaction
shifts to the right to reach equilibrium.
Equilibrium
Reaction Quotient Problem
N2(g) + 3 H2 (g)  2 NH3 (g)
At 723 K, the Kc = 6.0 x 10-2 for ammonia production.
If the system contains 61.2 M nitrogen, 122 M
hydrogen, and 122 M ammonia. Is the system at
equilibrium? If not, which way must it shift to reach
equilibrium? Prove your answer.
Solution: Q = [NH3]2 =
(122) 2
[N2] [H2]3
(61.2)(122) 3
Q = 1.3 x
10-4
Q<K
Equilibrium
Must shift to right to reach equilibrium
ICE
Problems
Equilibrium
ICE Problems
I = Initial Concentration
C = Change in Concentration
E = Equilibrium Concentration
Where: E = the difference between I & C
Equilibrium
ICE Problems
We often do not know the equilibrium
concentrations of all the chemicals in an
equilibrium mixture.
If we know the eq conc of at least one
chemical, we can use the stoichiometry
of the reaction to deduce the eq conc of
the other chemicals.
Equilibrium
ICE Problems
Steps to follow:
1)Calculate all the known initial conc’s and
equilibrium conc’s
Molarity = moles/L moles = mass/gfm
2) Calculate the change in conc for the chemical
whose initial conc’s and eq conc’s are known
3) Use the stoichiometry (coefficients) to calculate
the changes in conc for the other chemicals
4) Use the change in conc’s & the initial conc’s to
determine the equilibrium conc’s
Equilibrium
5) Solve the problem
ICE Problem
A closed system initially containing
1.000 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 taking place, which is
H2 (g) + I2 (g)
2 HI (g)
Equilibrium
What Do We Know?
[H2], M
Initially
[I2], M
1.000 x 10-3 2.000 x 10-3
[HI], M
0
Change
At
equilibrium
1.87 x 10-3
Equilibrium
[HI] Increases by 1.87 x 10-3 M
[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
Equilibrium
Stoichiometry tells us [H2] and [I2]
decrease by half as much
[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
Equilibrium
We can now calculate the equilibrium
concentrations of all three compounds…
[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
0
1.87 x 10-3
Equilibrium
…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
Equilibrium
Complex ICE Problems
Ask Mrs. Amuso for the worksheet of
sample problems using the quadratic
equation. For ax2 + bx + c = 0, the value
of x is given by:
Equilibrium
Le Châtelier’s
Principle
Equilibrium
Recall: Equilibrium
1. Equilibrium can only occur in a closed system.
2. Equilibrium is a state of balance between two
opposing reactions that are occurring at the same
reaction rate or speed.
3. The amounts or concentrations of the reactants and
products are constant, but not equal.
4. Equilibrium is a dynamic process. It looks like nothing
is changing, but the reactions have not stopped.
Equilibrium
Le Châtelier’s Principle
1. When a stress is applied to a system at
equilibrium, the system shifts to relieve the
stress and restore the equilibrium to new
conditions.
2. Stresses include changes in:
a) Concentration of products or reactants
b) Pressure (gases)
c) Temperature
3. When a system shifts, either the forward or
reverse reaction speeds up and a new
equilibrium point is established.
Equilibrium
When the Forward Reaction
Speeds Up
The equilibrium point shifts to the Right
Reactants  Products
When equilibrium is reestablished,
there will be more product and less reactant.
Equilibrium
When the Reverse Reaction
Speeds Up
The equilibrium point shifts to the Left
Reactants  Products
When equilibrium is reestablished,
there will be more reactant and less product.
Equilibrium
Rules for Concentration Changes
Stress
Inc Product(s)
Dec Product(s)
Inc Reactant(s)
Dec Reactant(s)
aA + bB
cC + dD
Reactants
Products
Relief
Dec Product(s)
Inc Product(s)
Dec Reactant(s)
Inc Reactant(s)
Shift
Left
Right
Right
Left
Equilibrium
Concentration Changes
N2 (g) + 3H2 (g)
Stress
Increase N2
2NH3 (g)
Relief
Shift
Decrease N2
Final Result : Less N2 than before
To the Right
Less H2
More NH3
Equilibrium
Concentration Changes
If H2 is added to the
system, N2 will be
consumed and the
two reagents will
form more NH3.
Equilibrium
Concentration Changes
N2 (g) + 3H2 (g)
2NH3 (g)
Equilibrium
shifts left to
offset stress
Add
NH3
Equilibrium
Pressure Changes
1. Affect only gases.
2. Pressure is increased by decreasing volume.
3. To relieve the stress, the system shifts in the
direction which will decrease the number of
molecules of gas (fewer moles).
4. Pressure can also be increased by adding an
inert gas. This does not cause a shift.
Equilibrium
Pressure Changes
CaCO3(s)
Stress
The piston is pushed in,
decreasing the volume
and increasing the
pressure.
CaO(s) + CO2 (g)
Relief
Decrease pressure
Shift
In the direction that
consumes CO2(g),
lowering the pressure
again.
Equilibrium
Rules for Pressure Changes
2 A (g) + B (g)
3 moles
Stress
4 C (g)
4 moles
Shift
1. Increase pressure
(reduce volume)
To fewest moles of gas
2. Decrease pressure
(increase volume)
To most moles of gas
(In this case, to the left)
(In this case, to the right)
Equilibrium
Practice
N2 (g) + 3H2 (g)
Stress
Increase
pressure
Relief
Shift to fewer
moles of gas
2NH3 (g)
Shift
Equilibrium
Practice
H2 (g) + Cl2 (g)
Stress
Decrease pressure
2 CO2 (g)
Stress
2 HCl (g)
Shift
No shift
CO2 (g) + O2 (g)
Shift
Increase pressure
Equilibrium
Rules for Temperature Changes
2 A (g) + heat
Stress
Increase temperature
B (g)
Shift
In endothermic direction
(In this case, to the right)
Decrease temperature
In exothermic direction
(In this case, to the left)
Equilibrium
Temperature Changes
Co(H2O)62+(aq) + 4 Cl(aq) + heat
Pink
CoCl4 (aq) + 6 H2O (l)
Blue
When heated
When cooled
Equilibrium
Temperature Changes
Ways to represent an exothermic reaction:
N2 (g) + 3H2 (g)
N2 (g) + 3H2 (g)
2NH3 (g) + 92 kJ
2NH3 (g)  H = -92 kJ
 H value = forward reaction
Stress
Decrease Temp
(+ = endo
Relief
Increase Temp
(Exothermic)
- = exo)
Shift
Equilibrium
Temperature & Equilibrium Constants
1. For Endothermic Reactions:
Increasing T results in an increase in K
2. For Exothermic Reactions:
Increasing T results in a decrease in K
Equilibrium
Adding a Catalyst
1. Catalysts do not change the equilibrium constant (K
value) or shift the equilibrium.
2. Catalysts increase the rate of both the forward and
reverse reactions equally.
3. Catalyst lower the Ea for both the forward and reverse
reactions.
4. The system reaches equilibrium sooner.
.
uncatalyzed
catalyzed
Equilibrium
The Haber Process
N2 (g) + 3H2 (g)
2NH3 (g) + 92 kJ
Under normal conditions the amount of ammonia produced by this reversible reaction is not sufficient
for our needs. The Haber Process refers to the economically feasible method of producing ammonia
which was developed by German chemist Fritz Haber during WWI. When the Allies blocked off their
trade routes, the Germans lost all source of sodium nitrate and potassium nitrate, their source of
nitrogen for making explosives. The Germans were forced to use ammonia to make their explosives.
Because ammonia can also be used in fertilizer production, Fritz Haber received a Noble Prize for
developing this process. This decision was controversial because it is believed that WWI would have
ended before 1918 if it weren’t for the Germans being able to use ammonia to make their explosives.
Ironically, Haber was expelled from Germany in 1933 because he was Jewish.
Under what conditions will the production of ammonia be maximized?
Did you guess use a Catalyst at High Pressure and Low Temperature?
Actually it’s a catalyst at about 200 atm and 450 Celsius
(At a lower temp the reaction is too slow)
Equilibrium
The Haber Process
This apparatus,
developed by German
Maz Born, helps push
the equilibrium to the
right by removing the
ammonia (NH3) from
the system as a liquid.
Equilibrium
Equilibrium Shift Song
http://www.youtube.com/watch?v=qsR5lA_T2B
0&feature=related
Equilibrium