• One of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction.
A
(g)

B
(g)

C
(g)

D
(g)
• The rates of the forward and reverse reactions can be represented as:
Rate
Rate f r

 k k r f
A B which
  
which represents represents
the
the forward reverse rate.
rate.
• When system is at equilibrium:
Rate f
= Rate r
Equilibrium constants are dimensionless because they actually involve a thermodynamic quantity called activity.
Activities are directly related to molarity

• K c is the equilibrium constant .
• K c is defined for a reversible reaction at a given temperature as the product of the equilibrium concentrations (in M) of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation, divided by the product of the equilibrium concentrations (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.

c
• The value of K c depends upon how the balanced equation is written.
PCl
5
• This reaction has a K c
=[PCl
PCl
3

3
][Cl
2
]/[PCl
5
Cl
]=0.53
2
PCl
3

Cl
2
PCl
5
• This reaction has a K c
=[PCl
5
]/=[PCl
3
][Cl
2
]=1.88

• The mass action expression or reaction quotient has the symbol Q.
– Q has the same form as Kc
• The major difference between Q and Kc is that the concentrations used in Q are not necessarily equilibrium values.
• Why do we need another “equilibrium constant” that does not use equilibrium concentrations?
• Q will help us predict how the equilibrium will respond to an applied stress.
• To make this prediction we compare Q with K c
.
• Q<K products favored
• Q>K reactants favored
• favored Q=K equilibrium

• LeChatelier’s Principle
- If a change of conditions (stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress in reaching a new state of equilibrium.
– We first encountered LeChatelier’s Principle in Chapter 14.
•
Some possible stresses to a system at equilibrium are:
1.
Changes in concentration of reactants or products.
2.
Changes in pressure or volume (for gaseous reactions)
3.
Changes in temperature.

p
c
• The relationship between K p
K p

K c and K c is:
 n or K c

K p
    n
 n = (# of moles of gaseous products) (# of moles of gaseous reactants)
• Heterogeneous equilibria have more than one phase present.
– For example, a gas and a solid or a liquid and a gas.
3
 


2
o
•
How does the equilibrium constant differ for heterogeneous equilibria?
– Pure solids and liquids have activities of unity.
– Solvents in very dilute solutions have activities that are essentially unity.
– The Kc and Kp for the reaction shown above are:
K c
= [CO
2
] K p
= P
CO
2


o rxn
 
G (notice no o indicating standard state) is the free energy change at nonstandard conditions
•
For example, concentrations other than 1 M or pressures other than 1 atm.
 
G is related to

G o by the following relationship.

G =

G o 
RT lnQ or

G =

G o 
2 .
303 RT log Q
R = universal gas constant
T = absolute temperatu re
Q = reaction quotient


o rxn
• The relationships among 
G o rxn
, K, and the spontaneity of a reaction are:

G o rxn
K
< 0 > 1
= 0 = 1
> 0 < 1
Spontaneity at unit concentration
Forward reaction spontaneous
System at equilibrium
Reverse reaction spontaneous

•
There are three classes of strong electrolytes.
1 Strong Water Soluble Acids
Remember the list of strong acids from Chapter 4.
HNO
3(  )

H
2
O
(  )
   
H
3
O

(aq)
 
NO
3(aq)
2 or
KOH
HNO

(s)
2

O

  
 

3(
H  
H
K

(aq)
NO
 3(aq)
OH
-
(aq)
3
Sr(OH)
2(s)
  O   
Sr
2

(aq)

2 OH
-
The solubility guidelines from Chapter 4 will help you remember these salts.
(s)
 H  O   

(aq)

-
(aq)
3
2
 
 H  O   
(aq)
HCl
2
 
3(aq)
NaOH
Arrhenius Produces H + Produces OH -
Brönsted-Lowery Donates H +
Lewis Accepts e pair
Accepts H +
Donates e pair

• We can define a new equilibrium constant for weak acid equilibria that uses the previous definition.
K a


H
3
O



CH
3
CH
3
COO

COOH



1 .
8

10

5 for acetic acid
– This equilibrium constant is called the acid ionization constant.

+

– The symbol for the ionization constant is K a
.

• Many weak acids contain two or more acidic hydrogens.
– Examples include H
3
PO
4 and H
3
AsO
4
.
• The calculation of equilibria for polyprotic acids is done in a stepwise fashion.
– There is an ionization constant for each step.
• Consider arsenic acid, H
3
AsO
4
, which has three ionization constants.
1 K a1
= 2.5 x 10 -4
2 K a2
= 5.6 x 10 -8
3 K a3
= 3.0 x 10 -13
K a1

K a2

K a3
• This is a general relationship.
– For weak polyprotic acids the K a1 is always > K a2
, etc.

• Calculate the concentration of all species in 0.100 M arsenic acid, H
3
AsO
4
, solution.
1 Write the first ionization step and represent the concentrations.
Approach this problem exactly as previously done.
2 Substitute the algebraic quantities into the expression for K a1
.
3.
Use the quadratic equation to solve for x , and obtain both values of x .
4 Next, write the equation for the second step ionization and represent the concentrations.
5 Substitute the algebraic expressions into the second step ionization expression.
6 Finally, repeat the entire procedure for the third ionization step.
7.
Substitute the algebraic representations into the third ionization expression.

• There are two common kinds of buffer solutions:
1 Solutions made from a weak acid plus a soluble ionic salt of the weak acid.
2 Solutions made from a weak base plus a soluble ionic salt of the weak base
1.
Solutions made of weak acids plus a soluble ionic salt of the weak acid
• One example of this type of buffer system is:
–
The weak acid - acetic acid CH
3
COOH
–
The soluble ionic salt - sodium acetate NaCH
3
COO

log
 log K a
 log
 
  multiply
 log pH
 pK a by 1
  log K a
 log
 
 
 log
 
 
The Henderson-Hasselbach equation is one method to calculate the pH of a buffer given the concentrations of the salt and acid. The Henderson-Hasselbach
Equation can be used for bases by substituting OH for H + and base for acid.

1 Calculate the pH of the original buffer solution.
2 Next, calculate the concentration of all species after the addition of the gaseous strong acid or strong base.
– This is another limiting reactant problem.
3 Using the concentrations of the salt and base and the Henderson-
Hassselbach equation, the pH can be calculated.
4 Finally, calculate the change in pH.

• We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

• We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

• Titration curves for Strong Acid/Weak Base Titration Curves look similar to Strong Base/Weak Acid Titration Curves but they are inverted.
• Weak Acid/Weak Base Titration curves have very short vertical sections.
• The solution is buffered both before and after the equivalence point.
• Visual indicators cannot be used.