CHAPTER 14
CHEMICAL EQUILIBRIUM
I. Chemical Equilibrium
A. Definition / Explanation
When a chemical reaction occurs, the conc. of
reactants decrease and the conc. of products
increase. Eventually chemical equilibrium is
reached, in which the concentration of the
products and reactants remain constant.
“No net change in reaction mixture”.
This does not mean [reactants] = [products] !!!
B. Achieving Equilibrium
1. Equilibrium is not reached instantaneously.
How fast or slow determined by kinetics.
2. How much product and reactants are present
at equilibrium is determined by
thermodynamics (energetics of reaction).
What is meant by “product-favored” reaction?
What is meant by “reactant-favored” reaction?
C. Dynamic Equilibrium
Chemical reaction reach dynamic equilibrium.
State of a reaction in which its forward and
reverse reactions occur at equal rates so that
the conc. of the reactants and products do
not change with time.
** Reactants and products formed at same rate.
** Does not mean [reactants] = [products].
D. Direction of Approach
Dynamic equilibrium is independent of
the direction of approach. For a given
temperature, the equilibrium state will be
the same whether the reaction is approached
from starting materials of pure reactants or
pure products.
Consider rxn: 2 N2(g) + 3 H2(g)  2 NH3(g)
Note following diagram
Explain idea
E. Catalysts Do No Effect Equilibrium
Conc.
They effect rate, but not equilibrium
state !!!
How Can We Quantitatively Describe Equilibrium?
Regardless of the
starting concentrations;
once equilibrium is
reached …
… the expression with products
in numerator, reactants in
denominator, where each
concentration is raised to the
power of its coefficient, appears
to give a constant.
II. Equilibrium Constant
A. Equilibrium Expression
“Mass Action Expression”
For rxn: aA + bB  cC + dD
c
d
C   D

Kc 
b
a
 A  B 
** No units
** Exponents always = reaction coefficients
B. Species Not in Equilibrium Expression
1. Pure solids – (NaCl(s), Na(s)
2. Pure liquids - H2O(l)
3. Solvents - may be involved in reaction, but
its concentration does not change.
(dilute solutions)
**General Rule – only include substances that
are (g) or (aq)
Write equilibrium expressions for the
following reactions:
1. CO(g) + 3 H2(g)  CH4(g) + H2O(g)
2. H2(g) + I2(g)  2 HI(g)
3. AgCl(s)  Ag+(aq) + Cl-(aq)
4. NH3(aq) + H2O(l)  NH4+(aq) + OH- (aq)
C. Kc for Related Reactions
1. Changing Stoichiometric Coefficients
N2(g) + 3 H2(g)  2 NH3(g)
Kc = 3.5 x 108
What if we alter the coefficients in equation?
2 N2(g) + 6 H2(g)  4 NH3(g) K’c = ?
When the coefficients of an equation are multiplied
by a common factor “n”, we raise the original Kc
value to the power “n” to obtain the new equilibrium
constant.
2. Changing Direction of Reaction
N2(g) + 3 H2(g)  2 NH3(g)
Kc = 3.5 x 108
What if we consider the reverse reaction?
2 NH3(g)  N2(g) + 3 H2(g)
K’c = ?
When we reverse the equation for a chemical
reaction for which the equilibrium constant is Kc,
we invert the equilibrium constant. The reverse
reaction has the equilibrium constant 1 / Kc.
D. Kc For Combining Reactions
A + BC
C +BF
Kc1 = 2.0 x 10-4
Kc2 = 1.0 x 10-6
What is the Kc for the reaction?
A + 2 B  F Kc = ?
When we add the equations for individual reactions
to obtain an overall equation, we multiply their
equilibrium constants to obtain the equilibrium constant
for the overall reaction.
E. Kc For Gas Phase Reactions
If all reactants and products are in gas
phase, the concentration terms are replaced
by partial pressures.
(Derived from Ideal Gas Law, PV=nRT )
What is the Kp expression for the rxn:
2 H2(g) + O2(g)  2 H2O(g)
What is the relationship between
Kp and Kc?
Kp  Kc   RT 
n
Δn = (sum of moles of gas products) – (sum of
moles of gas reactants)
As determined from coefficients in
chemical reaction.
F. Types of Equilibrium Expressions
Kc
Kc
Ka
Kb
Kw
Ksp
Type
Reference Table
general rxns
acid
Acid Table
base
Base Table
water
slight
Solubility Product
solubility
Table
III. Meaning of Kc
Kc Value
Reaction
very large
product favored
Kc > 1
Larger Kc: more product favored
very small
Kc < 1
reactant favored
Smaller Kc : more reactant favored
Kc = 1
significant amounts of both
reactants and products
1. If Kc = 1, does that mean that the concentration
of products equal the concentrations of reactants?
2. Which reaction has the greatest tendency to go
to completion (product favored)?
a. 2 NO2(g)  N2O4(g) Kc = 1.7 x 102
b. N2(g) + O2(g)  2 NO(g) Kc = 1.7 x 10-3
IV. Equilibrium Problems
A. Solving For Equilibrium Constant
1. Guidelines
a. Write equilibrium expression for
reaction.
b. Construct an “ICE” table with the
Initial, Change, and Equilibrium
concentrations of all reactants and
products.
When determining concentrations for table:
1) Units of conc. = Molarity = mol / L
(may use pressures if all gases and Kp)
2) You may need to use “x” to represent
unknown changes in concentrations in
table.
3) Remember changes in concentration of
reactants and products occur in same
ratio as equation coefficients.
c. Always plug in equilibrium concentrations
into the equilibrium expression to solve the
problem.
2. Problems – Solving for Kc Value
1. At equilibrium, a 2.0 L flask is found to
contain 0.10 mol of CO, 0.20 mol of CO2 ,
and 4.8 g of carbon.
C(s) + CO2(g)  2 CO(g)
What is Kc for this reaction?
2. A student placed 1.00 mol of CO and 2.00
mol of H2 in a 2.00 L flask at 7800C. After
reaching equilibrium, the CO
concentration is found to be 0.150 M.
Determine the value of the equilibrium
constant for this reaction.
CO(g) + 2 H2(g)  CH3OH (g)
Predicting Direction of a Reaction
1. Reaction Quotient, Q
“Q” has the same form as Kc expression,
but uses the actual concentrations in the
reaction mixture. “Q” is not a constant.
2. Significance of Q
By comparing the Reaction Quotient to the Kc
value, one can determine the direction a
reaction will occur as it proceeds towards
equilibrium.
If Q > Kc
[Product] greater than would be at equilibrium
Reaction would proceed to the left.
If Q < Kc
[Product] less than would be at equilibrium
Reaction would proceed to the right.
If Q = Kc
Reaction is at equilibrium.
Dynamic equilibrium. No net change occurs.
Example Problem
A 50.0 L reaction vessel contains 1.00 mol N2,
3.00 mol H2, and 0.500 mol NH3. Will more
ammonia, NH3, be formed or will it dissociate
when the mixture goes to equilibrium at 400oC?
The reaction is:
N2(g) + 3 H2(g)  2 NH3(g)
Kc = 0.500
B. Solving for Equilibrium Concentrations
1. Guidelines
a. Follow same guidelines as with Kc calculations.
b. Additional Comments
1) Some equilibrium concentrations may contain
a number (n) plus or minus “x”, (i.e., n+x, n-x).
2) (When Kc is very small or large, 10-5 > Kc > 105),
“x” may be very small compared to “n” such that
simplification can be made to problem. n ± x ≈n.
3) Verify assumption once “x” is determined to see
that it is less than (± 5%) of “n”. 5 % RULE
4) If assumption is not true, you will need to solve
problem using the quadratic equation (see later).
2. Problems
1. Problem Solved Since Perfect Square
(or Quadratic Equation)
A 0.20 mol sample of H2 and a 0.20 mol sample
of I2 were placed into a 1.0 L flask. What were
the equilibrium concentrations of H2, I2, and
HI? Reaction temperature was 500K.
H2(g) + I2(g)  2 HI(g)
Kc = 49.5
Another Example Problem:
Given the equilibrium reaction at 745 K:
H2(g) + I2(g)  2 HI(g)
Kc = 50.0.
Suppose that 0.75 mol HI(g), 0.025 mol H2(g), and 0.025
mol I2(g) are placed into a 20.0 L flask and heated to
745 K. Calculate the equilibrium concentrations of all
three substances.
2. Problems Solved Using Assumptions
(or Quadratic Equation)
A 2.0 L flask contained initial concentrations of
0.0330 M N2 and 0.00810 M O2. What is the
equilibrium concentration of NO in the flask?
N2(g) + O2(g)  2 NO(g)
Kc = 4.8 x 10-31
3. Problem Solved Using Quadratic Equation
(can only solve by using Quadratic Equation)
Consider the following reaction at 4480C:
I2(g) + Cl2(g)  2 ICl(g)
Kp = 50.5
The reaction mixture initially contains pressure of
I2 = 60.0 atm and pressure of Cl2 = 120.0 atm.
Determine the equilibrium pressure of ICl(g).
VI. Shifting Chemical Equilibriums
A. Le Chatelier’s Principle
1. Definition – When any change in
concentration, temperature, pressure, or volume
is imposed on a system at equilibrium, the system
responds by attaning a new equilibrium
condition that minimizes the impact of the
imposed change.
** Reaction tries to maintain equilibrium state.
** Significance
2. Factors (Stress) Affecting Equilibrium
a. Changing Concentrations
Reaction will try to maintain equilibrium or
maintain a similar Kc value.
A + BC + D
Adding reactants: shifts equilibrium to right
Adding products: shifts equilibrium to left
Removing reactants?
Removing products?
Will solids or pure liquids affect equilibrium?
Significance
Le Chatelier’s Principle Related to “Q” and “Kc”
b. Changing Pressure in Gas Equilibria
Pressure increases (or volume decreases)
for gaseous equilibrium cause shift in
equilibrium. Think about it like a
concentration factor.
An increase in pressure (or decrease in volume)
always drives the reaction in the direction of the
fewer # of gas molecules.
How will increase in pressure affect equilibrium?
N2(g) + 3 H2(g)  2 NH3(g)
H2(g) + I2(g)  2 HI(g)
N2O4(g)  2 NO2(g)
Initial
… two molecules of
NO2 combine …
When pressure
is increased …
… to give one molecule
of N2O4, reducing the
pressure increase.
c. Changes in Temperature
When considering exothermic or endothermic
reactions, consider heat as a product or
reactant. (Use same principles as changes in
concentration.)
Add heat to exothermic: shifts equilibrium to left
Add heat to endothermic: shifts right
N2(g) + 3 H2(g)  2 NH3(g) ΔH = -92.2 kJ
N2(g) + 3 H2(g)  2 NH3(g)
(+ heat on which side??)
B. Example Problem
How will the following factors affect the
final equilibrium amount of Cl2 ?
PCl5(g)  PCl3(g) + Cl2(g)
ΔH = 21 kcal/mol
a. Add PCl3
b. Remove PCl5
c. Increase temperature
d. Decrease volume
e. Decrease pressure
f. Adding inert gas to mixture