Chapter 15:
Chemical Equilibrium
By: Ms. Buroker
What is Chemical Equilibrium?
The state where the concentrations of all
reactants and products remain constant with
time.
1.) Reactions in closed vessels will reach
equilibrium
2.) Reactions which are “far to the right”
3.) Reactions which are “far to the left”
Chemical Equilibrium: “Big K”
kinetics: rate constant “little k”
kinetics “little k” told us how fast a reaction
proceeds and is used to indicate a possible
mechanism.
Eq. tells us to what extent a RXN proceeds to
completion
react.  prod. @ Eq: rate forward = rate reverse
The Equilibrium Condition
1.) Upon addition of reactants and/or products, they react until a
constant amount of reactants and products are present =
equilibrium.
2.) Equilibrium is dynamic since product is constantly made
(forward reaction), but at the same rate it is consumed (reverse
reaction).
The Equilibrium State
• Not all reactants are completely converted to product.
• Reaction equilibria deal with the extent of reaction.
• Arrows between reactants and products separate them and
qualitatively indicate the extent of reaction.
– Single arrow points to dominant side: H2(g) + O2(g)  H2O(g)
– Double arrow indicates both reactants and products present
after equilibrium obtained: N2O4 (g)  2NO2(g).
• Equilibrium exists when rates of forward and reverse reaction are
the same.
E.g. When rate of N2O4 decomposition equal the rate of formation
of N2O4, reaction at equilibrium
N2O4 (g)  2NO2(g).
The Equilibrium Constant
The Link Between Chemical Equilibrium and
Chemical Kinetics:
Reactions with one elementary step:
At equilibrium: Rforward = Rreverse
E.g. Decomposition of N2O4:
kf[N2O4] = kr[NO2]2. or
2
k f [NO 2 ]
Kc 

k r [N2O 4 ]
Equilibrium Constant
The Equilibrium Constant
• At equilibrium the ratio of concentrations
equals a constant.
• A generalized form of this expression that
describes the equilibrium condition:
aA + bB + cC + ...  mM + nN + oO ....
Kc 
m
n
a
b
o
[M] [N] [O]
c
[A] [B] [C]
Equilibrium
Concentrations
Kc 
m
n
a
b
o
[M] [N] [O]
c
[A] [B] [C]
Law of Mass Action: Values of Kc are constant for a RXN
at a given temperature. Any equilibrium mixture of the
above system at that temperature should give the
SAME Kc value.
For the reverse reaction:
K’ = 1/K
If the original reaction is
multiplied by a factor, then:
K” = Kn
Equilibrium Constants are
Temperature Dependent
and Mechanism
Independent!!
What Does “K” Really Mean???
When K is much larger than 1 …
A.) The reaction will consist mostly of products
B.) The Equilibrium lies to the right
When K is a very small value ….
A.) The reaction will consist mainly of reactants
B.) The Equilibrium condition is far left
C.) The reaction does NOT occur to any significant
event
Let’s Calculate a K Value …
The following equilibrium concentrations were
observed for the Haber process at 127○C:
[NH3] = 3.1 x 10-2 mol/L
[N2] = 8.5 x 10-1 mol/L
[H2] = 3.1 x 10-3 mol/L
a.) Calculate the value of K at 127○C for this reaction.
Answer: 3.8 x 104
b.) Calculate K for the reverse reaction at 127○C.
Answer: 2.6 x 10-5
c.) Calculate the value of the K for the reaction: Answer:
1/2N2(g) + 3/2H2(g) ↔ NH3(g)
1.9 x 102
Okay …
There is a term we refer to as the equilibrium
position, which is a set of equilibrium
concentrations.
*** It’s important to note that this is
different than the equilibrium constant, K.
The equilibrium constant remains the same
at a given temperature no matter what the
concentration because it is a ratio.
There is only one equilibrium constant for a particular system
at a particular temperature, but there are an infinite number
of equilibrium positions.
The following results were collected for two experiments involving the
reaction at 600○C between gaseous sulfur dioxide and oxygen to for
gaseous sulfur trioxide:
Experiment 2
Experiment 1
Initial
[SO2] = 2.00M
Equilibrium
[SO2] = 1.50M
Initial
[SO2] = 0.500M
Equilibrium
[SO2] = 0.590M
[O2] = 1.50M
[SO3] = 3.00M
[O2] = 1.25M
[SO3] = 3.50M
[O2] = 0
[O2] = 0.0450M
[SO3] = 0.350M [SO3] = 0.260M
Show that the equilibrium constant is the same in both cases.
Step 1: Write a balanced chemical equation.
2SO2(g) + O2(g) ↔ 2SO3(g)
From the law of mass action:
[SO 3 ]2
c  [SO ]2 [O2]
K
2
Equilibrium Expressions Involving Pressures
We also have Kp which is sometimes used when dealing with
gases with the “P” referring to the pressure of the gases.
aA(g) + bB(g)  dD(g) + eE(g)
Molar Concentration of Gas
n
Since: PV = nRT or P    RT
V
then P = CRT
Note: Pressure is proportional to Concentration!
For the General Reaction: aA(g) + bB(g)  dD(g) + eE(g)
K P  K c RT
n
(d + e) – (a + b)
0.0821L•atm
mol•K
Let’s Try One …
PCl3(g) + Cl2(g)  PCl5(g)
In a 5.00 L vessel (@ 230oC) an equilibrium mixture is found
to contain: 0.0185 mol PCl3, 0.0158 mol PCl5 and 0.0870 mol
of Cl2.
Determine Kc and Kp
 0.0158mol
 5.00L 
[PCl 5 ]



Kc 
= Kc = 49.08
 0.0185mol  0.0870mol
[PCl 3 ][Cl 2 ]
 5.00L   5.00L 



K P  Kc (RT) n  49.08(0.08
21 503)-1  1.19
Heterogeneous Equilibria
Equilibria involving more than one phase
Ex: CaCO3(s) ↔ CaO(s) + CO2(g)
Experimental results show that the position of a
heterogeneous equilibria does not depend on the
amounts of pure solids or liquids present.
Concentrations of pure solids and liquids can not
change; therefore:
K C  [CO2 ]
What is the Kc expression?
CaCO3(s)  CaO(s) + CO2(g)
Answer: Kc = [CO2]
What is the Kp expression?
Answer: Kp = PCO2
Using I.C.E. (Initial, Change, Equilibrium)
Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was
filled with 0.1500 mol of CO and 0.300 mol of H2. @ Eq.
@500K, 0.1187 mol of CO were present. How many moles of
each species were present @ Eq. and what is the value of Kc?
CO(g)
+ 2H2(g)

CH3OH(g)
I.
0.1500
0.300
0
C.
-X
- 2X
+X
E.
0.1187
0.2374
0.0313
If 0.1500 – X = 0.1187, Then X = 0.0313, So ….
Kc 
[C H3O H]
2
[C O ][H2 ]

 0.0313
 1.500 


 0.1187  0.2374
 1.500   1.500 



2
= 10.52
More I.C.E.
Problem: @ 77oC, 2.00 mol of NOBr (nitrosyl
bromide) placed in a 1.000L flask dissociates to
the extent of 9.4%. Find Kc.
2NOBr(g)

2NO(g) +
Br2(g)
I.
2.00
0
C.
- 2X
+ 2X
+X
0.188
+ 2X
0.094
+X
1.812
E. 2.00
– 2X
0
Since 9.4% dissociates, the Change in NOBr,
-2x = 2.00(0.094), So x = 0.094
2
Kc 
[NO ]2[Br2 ]
[NO Br]2
 0.188  0.0940
 1   1 
 


= 1.01 x 10-3
2
 1.812
 1 


Applications of the Equilibrium Constant
Consider This … What about a system that may not yet
be at equilibrium?
Reaction Quotient (Q) is determined using the law of
mass action and initial concentrations instead of
equilibrium concentrations.
N2(g) + 3H2(g) ↔ 2NH3(g)
2
[ NH 3 ]0
Q
3
[ N 2 ]0 [ H 2 ]0
1.) Q = K: The system is @ equilibrium … no shift will occur.
2.) Q > K: Shift to the left will occur
3.) Q < K: Shift to the right will occur
Equilibrium Quotient
Example:
PCl5(g)  PCl3(g) + Cl2(g)
-2
@ 250oC Kc= 4.0 x 10
If: [Cl2] and [PCl3] = 0.30M and [PCl5] = 3.0M, is the system at Equilibrium? If
not, which direction will it proceed?
Qc =
[PC l3 ][C l2 ] [0.30][0.30]
Find Qc and 
compare to Kc to decide.
= 3.0 x 10 2
[PC l5 ]
[3.0]
Qc < Kc (not @ Eq.)
Which way must the RXN go to achieve Equilbrium?
Remember ratio is prod./React
more products makes the number bigger
RXN goes
21
Le Chatelier’s Principle
States: If a change is
imposed on a
system at
equilibrium, the
position of the
equilibrium will shift
in a direction that
tends to reduce that
change.
Le Chatelier Another Way …
If a component (reactant or
product) is added to a reaction
system at equilibrium, the
equilibrium concentration will shift
in the direction that lowers the
concentration of that component. If
a compound is removed, the
opposite effect occurs.
Le Chatelier Example #1
A closed container of ice and water is
at equilibrium. Then, the temperature
is raised.
Ice + Energy  Water
The system temporarily shifts to the
_______
right to restore 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)
The system temporarily shifts to
Left to restore
the _______
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
The system temporarily shifts to the
right to 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)
The system temporarily shifts to the
_______
Left to restore equilibrium, because
there are fewer moles of gas on that side
of the equation.
Le Chatelier and Changes in Temp.
Changes
in
concentrations
or
total
Whether Kc increases or decreases
pressure
causetwo
shifts
in equilibrium
depends upon
factors,
T and Hrxn.
without changing the value of the
equilibrium
constant.
In contrast,causes
1.) An increase
in temperature
the equilibrium
to shift
towards
the
changes
in temperature
cause
the value
side.
of endothermic
Kc to change.
2.) A decrease in temperature causes
the equilibrium to shift in the opposite
direction.
Le Chatelier and Changes in Temp.
Whether Kc increases or decreases
depends upon two factors, T and Hrxn.
1.)
An increase
in temperature
causes
Reactants
+ heat
↔ Products
the equilibrium to shift towards the
Anside.
increase in heat will
endothermicEndothermic
Kc
cause the reaction to shift
to the right.
2.)
A decrease
temperature
causes
Reactants
↔inProducts
+ heat
the equilibrium to shift in the opposite
Exothermic
direction.
An increase in heat will
K
c
cause the reaction to shift
to the left.
Le Chatelier and Catalysts
1.) It has no effect on the
equilibrium.
2.) It speeds up attainment of
equilibrium. Kc related to the
H and not Ea