Quick Equilibrium review
The Concept of Equilibrium
• As the substance warms it begins to
decompose:
N2O4(g)  2NO2(g)
• When enough NO2 is formed, it can react to
form N2O4:
2NO2(g)  N2O4(g).
• At equilibrium, as much N2O4 reacts to form
NO2 as NO2 reacts to re-form N2O4
N2O4(g)
2NO2(g)
• The double arrow implies the process is
dynamic.
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.
A System at Equilibrium
Once equilibrium is
achieved, the
amount of each
reactant and product
remains constant.
The Equilibrium Constant
 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
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
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
n = (moles of gaseous product) − (moles of gaseous reactant)
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.
Manipulating Equilibrium Constants
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)
2 N2O4 (g)
2 NO2 (g)
Kc =
4 NO2 (g) Kc =
[NO2]2
[N2O4]
[NO2]4
[N2O4]2
= 0.212 at 100C
= (0.212)2 at 100C
Manipulating Equilibrium Constants
The equilibrium constant of a reaction in the
reverse reaction is the reciprocal of the
equilibrium constant of the forward reaction.
N2O4 (g)
2 NO2 (g)
2 NO2 (g)
N2O4 (g)
Kc =
Kc =
[NO2]2
[N2O4]
[N2O4]
[NO2]2
= 0.212 at 100C
=
1
0.212
= 4.72 at 100C
Applications of Equilibrium Constants
Predicting the Direction of Reaction
• We define Q, the reaction quotient, for a reaction at
conditions NOT at equilibrium
aA + bB(g)
as
Q
pP + qQ
p
q
P  Q
a
b
A  B
where [A], [B], [P], and [Q] are molarities at any
time.
• Q = K only at equilibrium.
The Reaction Quotient (Q)
 To calculate Q, one substitutes the initial
concentrations on 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.
Applications of Equilibrium Constants
Predicting the Direction of Reaction
• If Q > K then the reverse reaction must
occur to reach equilibrium (go left)
• If Q < K then the forward reaction must
occur to reach equilibrium (go right)
Applications of Equilibrium Constants
Predicting the Direction of Reaction
• If Q > K then the reverse reaction must
occur to reach equilibrium (go left)
• If Q < K then the forward reaction must
occur to reach equilibrium (go right)
Le Châtelier’s Principle
Change in Reactant or Product
Concentrations
• Adding a reactant or product shifts the equilibrium
away from the increase.
• Removing a reactant or product shifts the
equilibrium towards the decrease.
• To optimize the amount of product at equilibrium,
we need to flood the reaction vessel with reactant
and continuously remove product (Le Châtelier).
• We illustrate the concept with the industrial
preparation of ammonia
N2(g) + 3H2(g)
2NH3(g)
Le Châtelier’s Principle
Effects of Volume and Pressure
• The system shifts to remove gases and
decrease pressure.
• An increase in pressure favors the direction
that has fewer moles of gas.
• In a reaction with the same number of
product and reactant moles of gas, pressure
has no effect.
• Consider
N2O4(g)
2NO2(g)
Le Châtelier’s Principle
Effect of Temperature Changes
• Removing heat (i.e. cooling the vessel), favors
towards the decrease:
– if H > 0, cooling favors the reverse reaction,
– if H < 0, cooling favors the forward reaction.
• Consider
Cr(H2O)6(aq) + 4Cl (aq)
2CoCl4 (aq)
+ 6H2O(l)
for which H > 0.
– Co(H2O)62+ is pale pink and CoCl42- is blue.
CATALYST—EQUILIBRIUM is achieved faster, but
the equilibrium composition remains unaltered.
Manipulating Equilibrium Constants
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 INVOLVING THE SOLUBILITY
AND PRECIPITATION OF COMPOUNDS
Equilibrium and Solubility
Saturated solutions
 A saturated solution is a solution that is in
equilibrium with undissolved solute
 Example:

BaSO4 (s) D Ba+2 (aq)+ SO4-2 (aq)
Solubility Products
The equilibrium constant expression for this
equilibrium is
Ksp = [Ba2+] [SO42-]
where the equilibrium constant, Ksp, is called the
solubility product.
Solubility Products
The equilibrium constant expression for this
equilibrium is
Ksp = [Ba2+] [SO42-]
where the equilibrium constant, Ksp, is called the
solubility product.
Solubility Products
 Ksp is not the same as solubility.
 Solubility is generally expressed as the mass of
solute dissolved in 1 L (g/L) or 100 mL (g/mL)
of solution, or in mol/L (M).
Will a Precipitate Form?
 In a solution,
 If Q = Ksp, the system is at equilibrium and the
solution is saturated.
 If Q < Ksp, more solid will dissolve until Q = Ksp.
 If Q > Ksp, the salt will precipitate until Q = Ksp.
Selective Precipitation of Ions
One can use
differences in
solubilities of
salts to separate
ions in a
mixture.
Common Ion Effect
 If a solution containing two dissolved substances
share a common ion, then the solubility of the salt
is more difficult to determine
 Adding “common ion” will cause the solubility to
be less in the presence of the common ion
 Causes less of the substance with the smaller Ksp
will dissolve.