CHEMICAL EQUILIBRIUM
My faith helps me overcome such negative emotions and find my equilibrium.
- Dalai Lama
Reactions are Reversible





A + B  C + D (forward)
Initially there is only A and B so only the forward
reaction is possible
As C and D build up, the reverse reaction speeds
up while the forward reaction slows down.
C + D  A + B (reverse)
Eventually the rates are equal
Reaction Rate
Forward Reaction
Equilibrium
Reverse reaction
Time
What is equal at Equilibrium?




The Forward and Reverse
Reaction Rates are equal.
Forward and Reverse
Reactions are still occurring.
Concentrations are NOT
necessarily equal.
The concentrations of
reactants and products do
not change at equilibrium.
Ratio of Reactant to Products at EQ




At equilibrium, there is a set ratio of product
concentrations to reactant concentrations
This ratio is given by K, the equilibrium constant
K is unique for a particular temperature.
Since an increase in temperature usually favors
either the forward or reverse reaction (whichever is
endothermic), K value changes as the temperature
changes.
Law of Mass Action




For any reaction
jA + kB
lC + mD
Keq = [C]l[D]m
PRODUCTSpower
[A]j[B]k
REACTANTSpower
Keq is called the equilibrium constant.
Keq is commonly written as Kc since concentrations
are used to determine the Keq.
Mass Action Expression
2 HCl (g)
H2 (g) + Cl2 (g)
Keq or Kc =
N2 (g) + 3 H2 (g)
Keq or Kc =
2 NH3 (g)
K Values



K > 1: products are "favored"
K = 1: neither reactants nor products favored
K < 1: reactants are "favored“

T
This reaction has a large K value.
Keq is constant at a certain T

Regardless of what we begin with, the ratio of
products to reactants is constant at equilibrium
K of the reverse reaction



Write the reaction in reverse.
lC + mD
jA + kB
Then the new equilibrium constant is
Krev =
[A]j[B]k
= 1/Kfor
[C]l[D]m

If the Keq is 5.0 x 103, what is the Keq of the reverse
reaction?
The units for K




There is no set or common unit for K.
The units are dependent on the reaction.
There may be no units, M, M2, 1/M, etc.
Since units may vary, the standard method is to not
include any units.
SOLVING PROBLEMS AT
EQUILIBRIUM
Using the Mass Action Expression
Calculate Kc
N2 + 3H2
Initial
[N2] = 1.000 M
[H2] = 1.000 M
[NH3] = 0.0 M
2 NH3
At Equilibrium
[N2] = 0.921M
[H2] = 0.763M
[NH3] = 0.157M
Calculate Kc
N2 + 3H2
Initial
[N2] = 0.0 M
[H2] = 0.0 M
[NH3] = 1.000 M
2 NH3
At Equilibrium
[N2] = 0.399 M
[H2] = 1.197 M
[NH3] = 0.157M
K is the same regardless of starting concentrations as long
as temperature remains constant.
A Slightly Tougher K Problem

3.00 moles of pure SO3(g) are introduced into an 8.00 dm3
container at 1105 K. At equilibrium, 0.58 mol of O2(g) has
been formed.
2SO3(g)


2SO2(g) + O2(g)
Calculate Kc
What other information must we first gather?
ICE Table
SO3(g)
SO2(g)
O2(g)
Initial moles
3.00
0
0
Change moles
-2y
+2y
+y
(3.00 - 1.16)
1.16
0.58
1.84/8.00
1.16/8.00
0.58/8.00
Equilibrium moles
Equilibrium
concentrations
2SO3(g)
2SO2(g) + O2(g)
Practice Problem 1

An aqueous solution is prepared that is initially 0.100 M CdI42After equilibrium is established the solution is found to be
0.013 M in Cd2+.
CdI42- (aq)

Cd2+ (aq) + 4 I- (aq)
Derive the expression for the equilibrium expression and
calculate the value of the constant?
Practice Problem 2

In a saturated solution of MgF2 at 18°C, the concentration of
the Mg2+ is 1.21 x 10-3 molar. The equilibrium is represented
by the equation below.
MgF2 (s)

Mg2+ (aq) + 2 F- (aq)
Derive the expression for the equilibrium expression and
calculate the value of the constant at 18ºC.
Practice Problem 3
CH3COOH
CH3COO- + H+
In a solution of acetic acid, the equilibrium
concentrations are found to be [CH3COOH] = 1.000,
[CH3COO-] = 0.0042, and [H+] = 0.0042.
Calculate the Ka of acetic acid.
What is the pH of this solution? (pH = -log[H+])
Practice Problem 4

When H2 (g) is mixed with CO2 (g) at 2000 K, equilibrium is
achieved according to the equation.
CO2 (g) + H2(g)



H2O (g) + CO (g)
Determine the value of the equilibrium expression if, at
equilibrium, [H2] = 0.20 M, [CO2] = 0.30 M, [H2O] = 0.55 M,
and [CO] = 0.55 M.
If the temperature is lowered, the [CO] becomes 0.40 M.
Calculate the value of Kc at this lower temperature.
What is the Kp of this reaction at 2000K?
Converting from Kc to Kp
Kp = Kc(RT)Δn





Kp = Equilibrium constant based on pressure
Kc = Equilibrium constant based on molar conc.
R = Gas constant, use 0.0821 L atm mol-1 K-1
T = Temperature, use Kelvin temp
Δn = moles product gas – moles reactant gas
Tough Problem

4.00 moles of HI are placed in an evacuated 5.00 L flask and
then heated a 800K. The system is allowed to reach
equilibrium.
2 HI(g)

H2(g) + I2(g)
What will be the equilibrium concentration of each species if
Kc = 0.00016 at 800K?
ICE, ICE, baby!
2 HI(g)
H2(g) + I2(g)
HI
H2
I2
Initial
.80
0
0
Change
- 2x
+x
+x
.80 - 2x
+x
+x
EQ
.00016 =
(x) (x)
(.8 - x)2
x = 0.010 via the quadratic equation