Download Chemical Equilibrium

Chemical Equilibrium
Reversible Reactions
• A chemical reaction in which products can reform reactants is called a reversible reaction
• A reversible reaction is in chemical equilibrium
when the rate of its forward reaction equals
the rate of its reverse reaction and the
concentration of its products and reactants
remain unchanged
The Equilibrium Expression
• Suppose the following reaction takes place:
– nA + mB  xC + yD [lower case letters represent
coefficients; capital letters represent substances]
• Initially, the concentrations of C & D are zero and
A & B are maximum. Over time the forward
reaction rate decreases as A & B are used up. The
reverse reaction rate increases as C & D are
formed. When the rates become equal,
equilibrium is established.
The Equilibrium Expression
• After equilibrium is reached, the
concentrations of products and reactants
remain constant, so a ratio of their
concentrations should remain constant
• K = [C]x[D]y
[A]n[B]m
Equilibrium Constant
• The numerical value for K is determined
experimentally
• Only substances that can change are included
in K (so no pure solids or liquids)
Equilibrium Constant
• The equilibrium constant (K) is the ratio of the
mathematical product of the concentrations of
substances formed at equilibrium to the
mathematical product of the concentrations of
reacting substances. Each concentration is raised
to a power equal to the coefficient of that
substance in the chemical equation.
• The expression for K is sometimes referred to as
the chemical equilibrium expression
Equilibrium Constant
• Example: An equilibrium mixture of N2, O2, and
NO gases at 1500 K is determine to consist of
6.4 x 10 -3 M of N2, 1.7 x 10-3 M of O2, and 1.1 x
10-5 M of NO. What is the equilibrium
constant for the system at this temperature?
• The balanced chemical equation is:
N2 (g) + O2 (g)  2NO (g)
Equilibrium Constant
• K = [NO]2
[N2][O2]
• K = [1.1 x 10-5]2
[6.4 x 10-3][1.7 x 10-3]
• K = 1.1 x 10-5
Le Chatelier’s Principle
• Le Chatelier’s Principle states that if a system
at equilibrium is subjected to stress, the
equilibrium is shifted in the direction that
tends to relieve the stress
Le Chatelier’s Principle
• Concentration Changes:
– An increase in the concentration of a component will
shift the equilibrium position away from the increased
component. A decrease in the concentration of a
component will shift the equilibrium position towards
the decreased component
• Changes in concentration have no effect on the
equilibrium constant (K) although the equilibrium
concentrations do change
Le Chatelier’s Principle
• Temperature Changes
– View temperature/heat as a reactant for
endothermic reactions and as a product for
exothermic reactions
• Shift will then follow as in concentration
changes
Le Chatelier’s Principle
• Pressure Changes
– Only affects reactions in which gases are involved
– In order for it to have an affect total moles of gas on
the left must be different than total moles of gas on
the right
• Increasing pressure/decreasing volume will
increases the concentration of all species – so
equilibrium will shift away from the side with
more moles of gas. The opposite will happen for
a decrease in pressure/increase in volume
Le Chatelier’s Principle
• Adding an inert substance, such as a noble
gas, will not cause a shift in equilibrium
position
Common Ion Effect
• The addition of an ion common to two solutes to
bring about precipitation or reduced ionization is
called the common ion effect.
• Suppose HCl (g) is bubbled into a saturated NaCl
(aq) solution. As the HCl dissolves, it increases
the concentration of Cl- which is a “common ion”
in the dissociation of NaCl in solution. This
increased Cl- concentration will shift the
equilibrium position away from the Cl-
Equilibria of Acids, Bases, and Salts
• When weak acids are put into water, there is only
partial dissociation, which leads to an equilibrium
state
• The equilibrium constant is called Ka, or the acid
ionization constant. It is constant for a specified
temperature
• The same thing happens for weak bases. Their
equilibrium constant is Kb, or the base ionization
constant.
Equilibria of Acids, Bases, and Salts
• Buffers are solutions that contain either a
weak acid & its conjugate salt or a weak base
and its conjugate salt
• Buffers resist changes in pH as small amounts
of acids or bases are added
Equilibria of Acids, Bases, and Salts
• A reaction between water molecules and ions
of a dissolved salt is hydrolysis
• If the anions react with water, the process is
anion hydrolysis and results in a basic solution
• If the cations react with water, the process is
cation hydrolysis and results in an acidic
solution
Equilibria of Acids, Bases, and Salts
• The equilibrium equation for a typical weak
acid (HA) in water is:
HA (aq) + H2O (l)  H3O+ (aq) + A- (aq)
Ka = [H3 O+][A-]
[HA]
Equilibria of Acids, Bases, and Salts
• The equilibrium equation for a typical weak
base (B) in water is:
B (aq) + H2O (l)  BH+ (aq) + OH- (aq)
Kb = [BH+][OH-]
[B]
Equilibria of Acids, Bases, and Salts
• Hydrolysis helps explain why the end point of
titrations can occur at a pH other than 7 (which
happens in strong acid-strong base titration)
• When a weak acid is titrated with a strong base,
the end point is basic
• When a weak base is titrated with a strong acid,
the end point is acidic
Solubility Equilibrium
• The degree of solubility of a substance can be
described by the solubility product constant
(Ksp)
• This is the product of the molar
concentrations of its ions in a saturated
solution, each raised to the power that is the
coefficient of that ion in the balanced
chemical equation
Solubility Equilibrium
• The solubility product constant is an equilibrium
constant representing the product of the molar
concentrations of its ions in a saturated solution.
It has only one value for a given solid at a given
temperature
• The solubility of a solid is an equilibrium position
that represents the amount of solid required to
form a saturated solution with a specific amount
of solvent
Solubility Equilibrium
• Example – Calculate the solubility of AgBr (Ksp
is 5.0 x 10-13)
• AgBr (s)  Ag+ (aq) + Br- (aq)
Ksp = [Ag+][Br-]
[Ag+] = [Br-] so let [Ag+] = x and [Br-] = x
Ksp = [Ag+][Br-] = x2
5.0 x 10-13 = x2
X = 7.1 x 10-7 M
Solubility Equilibrium
• If the ion product is greater than Ksp , the solid
will precipitate
• If the ion product is less than Ksp , the solution
is unsaturated and precipitation will not occur
Solubility Equilibrium
• Example – Will a precipitate form if 20.0 mL of
0.010 M BaCl2 is mixed with 20.0 mL of 0.0050
M Na2SO4?
• BaSO4 (s)  Ba2+ (aq) + SO4 2- (aq)
• Ksp = [Ba2+][[SO4 2-] = 1.1 x 10-10
• Calculate mole quantities of Ba2+ and SO42– 0.00020 mol Ba and 0.00010 mol SO4
Solubility Equilibrium
• Calculate the total volume  0.040 L
• Calculate the Ba and SO4 ion concentration in
the combined solution
– 5.0 x 10-3 M Ba and 2.5 x 10-3 M SO4
• Calculate the ion product
– [Ba2+][[SO4 2-] = (5.0 x 10-3)(2.5 x 10-3) = 1.2 x 10-5
• The ion product is greater than Ksp so
precipitation occurs
Solubility Equilibrium
• It is important to note that Ksp cannot be used
for soluble substances.
• It is also sensitive to changes in solution
temperature