Chemical Equilibrium
Chapter 6
pages 190 - 217
Reversible Reactions- most
chemical reactions are reversible
under the correct conditions
Equilibrium=
H2O(g) + CO(g)  H2(g) + CO2(g)
The graph represents changes in concentrations with time for the reaction
H2O(g) + CO(g)H2(g) + CO2(g) when equimolar quantities of H2O(g) and
CO(g) are mixed.
(a) H2O and CO are mixed in equal numbers and begin
to react (b) to form CO2 and H2. After time has passed,
equilibrium is reached (c) and the numbers of reactant
and product molecules then remain constant over time
(d).
H2O(g) + CO(g)  H2(g) + CO2(g)
or
X  Y
• At equilibrium, X forms Y at the same rate
as Y forms X.
• Therefore, the concentration of [X] and [Y]
will remain constant.
• Also, when equilibrium is reached, the rates
of the forward reaction will equal the rates
of the reverse reaction.
The Equilibrium Constant, Keq
(also called the mass action expression)
•
Two methods for describing the
equilibrium of a reaction.
1. Kc- describes the equilibrium of a reaction
where the concentrations of the materials
is known.
2. Kp- describes the equilibrium of a gaseous
reaction using partial pressures instead of
concentrations.
Kc
c = concentration
aA + bBcC + cD
Coefficients in the chemical equation become
exponents in the equilibrium constant expression.
c
d
[C] [D]
Kc 
a
b
[A] [B]
Include only substances in the gas or aqueous phase.
Solid’s and liquid’s concentrations do not change
during a chemical reaction.

Write the equilibrium expression for the
reaction below.
Ni (s) + 4CO (g)  Ni(CO)4(g)
Write the equilibrium expressions
for the following reactions:
a. H2 (g) + I2 (g)  2HI (g)
b.
2NH3 (g) + 3 CuO(s)  3H2O (g) + N2 (g) + 3Cu (s)
c. NH4Cl (s)  NH3 (g) + HCl (g)
Kp
p= pressure
• Equilibrium is described in terms of the
partial pressures of the reactants and
products.
2CO2 (g) 2CO (g) + O2 (g)
2
Kp 
(PCO ) (PO2 )
(PCO 2 )
2
Write the equilibrium constant for the
following mixture of gases at equilibrium.
2NO (g) + O2 (g)  2NO2 (g)
The Relationship Between Kc and
Kp
Where n = (total moles of gaseous products)-(total moles
of gaseous reactants)
R= ideal gas law constant (.08206 L atm / mol K
T in Kelvin
Example: A 3:1 starting mixture of hydrogen,
H2, and nitrogen, N2, comes to equilibrium
at 500oC. The mixture at equilibrium is
35.06% NH3, 96.143% N2 and 0.3506% H2
by volume. The total pressure in the
reaction vessel was 50.0 atm. What is the
value of Kp and Kc for this reaction?
Kp =
Kc =
• Example:
2NO (g) + Cl2(g)  2NOCl (g)
In the above reaction, the total pressure of the
mixture of gases at equilibrium is 1.55 atm.
The percentages of each gas in the mixture
areas follows:
NOCl = 77.4%, NO = 3.20%, and
Cl2 = 19.4%.
Calculate Kp and Kc for the reaction.
Interpretation of the Equilibrium
Constant
[Products]
K
[Reactants]
For K>1, we can state that the amount of
product is greater than the amount of
 reactant. The equilibrium is favored to the
right.
Likewise, for K<1, the amount of reactants is
much greater than the products. The
equilibrium favored to the left.
The Reaction Quotient- Q
• There are times when a chemist may be given
information about a chemical reaction that may or
may not be at equilibrium.
• A Q calculation can be used to determine the
direction the reaction is heading to reach
equilibrium.
• If Q > K, then the reaction has too many products
and is heading towards the reactant side.
• If Q < K, then the reaction has too many reactants
and is heading towards the product side.
• If Q = K, then the reaction is at equilibrium.
• Example:
N2 (g) + 3H2 (g)  2NH3 (g)
In the reaction shown above, the value of Kc
at 5000C is 6.0 X 10-2. At some point
during the reaction, the concentrations of
each material were measured. At this point,
the concentrations of each substance were
[N2] = 1.0 X 10-5 M, [H2] = 1.5 X 10-3M,
and [NH3] = 1.5 X 10-3 M. Calculate the
value of Q, and determine the direction that
the reaction was most likely to proceed
when the measurements were taken.
Solving Problems When Not All
Equilibrium Concentrations are Known
H2 (g) + I2 (g)  2HI (g)
When 2.00 mole each of hydrogen and iodine are
mixed in an evacuated vessel, 3.50 mole of HI are
produced. What is the value of the equilibrium
constant, Kc?
It is important to realize that not all of the reactants
are converted to products in reversible reactions.
Some reactants will remain at equilibrium.
Solution: Careful Stoichiometry!
H2 (mol)
Start
Change
(∆)
Final
2.0
I2 (mol)
2.0
2HI (mol)
0
+3.50
3.50
Use the mole ratios to determine the
change concentrations of the reactants.
1 mol H2
3.50 mole HI X
 1.75 mol H2
2 mol HI
H2 (mol)
Start
Change
(∆)
Final
2.0
I2 (mol)
2.0
2HI (mol)
0
+3.50
3.50
Since we are in 1.0 L, 0.25 mol = 0.25 M
Use the final concentrations to
calculate the value of Kc
• Example:
Nitric oxide gas, NO, and oxygen gas, O2, react to
form the poisonout gas nitrogen dioxide, NO2, in
the reaction shown below:
2NO (g) + O2 (g)  2NO2 (g)
10.0 moles of NO and 6.00 moles of O2 are placed
into an evacuated 1.00 L vessel, where they begin
to react. At equilibrium, there are 8.80 moles of
NO2 present. Calculate the value of Kc, assuming
that the temperature remains constant throughout
the reaction.
Kc =
Determining K When Only Initial
Concentrations Are Known
H2 (g) + I2(g)  2HI
0.500 mol of hydrogen and 0.500 mol of iodine are
added to a 1.00 liter reaction vessel. The mixture
is heated to 4980C and allowed to reach
equilibrium according to the reaction shown
above. At this temperature, Kc = 49.7. What is
the composition of the reaction mixture at
equilibrium in this system?
H2 (g) + I2(g)  2HI
Start
Change
(∆)
Final
H2 (mol)
I2 (mol)
2HI (mol)
0.50
0.50
0
Enter the Quadratic Formula
2
[HI ]
Kc 
[H 2 ][I 2 ]
(2 X ) 2
49.7 
(0.50  X ) 2
12.4  49.7 X  49.7 X 2  4 X 2
X 
b 
b  4 ac
2a
2
• Example:
0.500 moles of NO gas are placed into a 1.00
L reaction vessel. The gas is heated to an
extremely high temperature, where it
decomposes according to the reaction
shown below:
2NO (g) N2 (g) + O2 (g)
At equilibrium, Kc = 2.4 X 103. Determine
the equilibrium composition of the mixture.
5.1 X 10-3 M NO, 0.25 M N2, 0.25 M O2
Le Chatelier’s Principle
• When a system at equilibrium is disturbed, it will
react in a way that minimizes the disturbance.
• Factors that can influence the position of an
equilibrium.
–
–
–
–
1. Concentration (aq or g)
2. Temperature
3. Pressure (gases)
4. Addition of a catalyst
•
Example: Describe how the reaction
below adjusts its equilibrium in each of
the scenarios.
CO (g) + 2H2 (g)  CH3OH (g) H0 = -21.7 Kcal
a. Some of the methanol vapor is condensed
and removed from the reaction vessel.
b. The pressure is increased by decreasing
the volume of the reaction vessel.
c. The temperature is increased.