The Equilibrium Constant
• For a general reaction
aA + bB
cC + dD
the equilibrium constant expression for everything in
solution is
where Keq is the equilibrium constant.
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
1. Write an expression for Keq
2. Calculate K at a given temperature if [CH4] =
0.020 M, [O2] = 0.042 M, [CO2] = 0.012 M, and
[H2O] = 0.030 M at equilibrium. (include units)
• If K << 1, then reactants dominate at equilibrium and the
equilibrium lies to the left.
Types of Equilibrium Constant
Keq = Equilibrium
Kc =
concentrations
Kp =
pressures
Heterogeneous Equilibria
• When all reactants and products are in one phase, the
equilibrium is homogeneous.
• If one or more reactants or products are in a different
phase, the equilibrium is heterogeneous.
• Consider:
CaCO3(s)
CaO(s) + CO2(g)
– experimentally, the amount of CO2 does not seem to depend on the amounts of
CaO and CaCO3. Why?
• The concentration of a solid or pure liquid is its
density divided by molar mass.
• Neither density nor molar mass is a variable, the
concentrations of solids and pure liquids are constant.
• For the decomposition of CaCO3:
• We ignore the concentrations of pure liquids and pure
solids in equilibrium constant expressions.
• The amount of CO2 formed will not depend greatly on
the amounts of CaO and CaCO3 present.
Example – Write the Keq expression for the
following equations:
NH4Cl (s)
NH3 (g) + HCl (g)
NH3 (g) + HCl (g)
NH4Cl (s)
Example – Write the Keq expression for the following equations:
NH4Cl (s) ↔ NH3 (g) + HCl (g)
Keq = [NH3][HCl]
NH3 (g) + HCl (g) ↔ NH4Cl (s)
Keq = ([NH3][HCl])-1
Calculating Equilibrium
Constants
• Proceed as follows:
– Tabulate initial and equilibrium concentrations (or partial pressures) given.
– If an initial and equilibrium concentration is given for a species, calculate the change
in concentration.
– Use stoichiometry on the change in concentration line only to calculate the changes
in concentration of all species.
– Deduce the equilibrium concentrations of all species.
• Use “ICE” Charts
Example – N2 (g) + 3 H2 (g) 2 NH3 (g)
The initial concentration of N2 is 0.25 M and of H2 is
0.60 M. The equilibrium concentration of H2 is
0.45 M. What are the equilibrium concentrations
of N2 and NH3? What is the value of Keq?
Example – N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)
The initial concentration of N2 is 0.25 M and of H2 is 0.60 M. The
equilibrium concentration of H2 is 0.45 M. What are the
equilibrium concentrations of N2 and NH3? What is the value of
Keq?
1. [N2] = .2 M
[NH3] = .1 M
2. Keq = .549 M-2
Applications of Equilibrium
Constants
Predicting the Direction of Reaction
• We define Q, the reaction quotient, for a general
reaction
aA + bB
cC + dD
as
• Q = K only at equilibrium.
• If Q > K then the reverse reaction must occur to reach
equilibrium (i.e., products are consumed, reactants are
formed, the numerator in the equilibrium constant
expression decreases and Q decreases until it equals K).
• If Q < K then the forward reaction must occur to reach
equilibrium.
Le Châtelier’s Principle
• Consider the production of ammonia
N2(g) + 3H2(g)
2NH3(g)
• As the pressure increases, the amount of ammonia
present at equilibrium increases.
• As the temperature decreases, the amount of ammonia
at equilibrium increases.
• Can this be predicted?
Effects of Volume and Pressure Changes
• As volume is decreased pressure increases.
• Le Châtelier’s Principle: if pressure is increased the
system will shift to counteract the increase.
• That is, 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.
• Increasing total pressure by adding an inert gas has no
effect on the partial pressures of reactants and products,
therefore it has no effect on the equilibrium.
Effect of Temperature Changes
• The equilibrium constant is temperature dependent.
• For an endothermic reaction, H > 0 and heat can be
considered as a reactant.
• For an exothermic reaction, H < 0 and heat can be
considered as a product.
• Adding heat (i.e. heating the vessel) favors away from
the increase:
– if H > 0, adding heat favors the forward reaction,
– if H < 0, adding heat favors the reverse reaction.
• 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.
Le Châtelier’s Principle
The Effect of Catalysis
• A catalyst lowers the activation energy barrier for the
reaction.
• Therefore, a catalyst will decrease the time taken to
reach equilibrium.
• A catalyst does not effect the composition of the
equilibrium mixture.
SO3 (g)
SO2 (g) + 1/2 O2 (g)
• ΔH = +98.9 kJ
• Determine the effect of each of the following
on the equilibrium (direction of shift)
– What happens to the concentration of SO3 after
each of the changes?
A) Addition of pure oxygen gas.
B) Compression at Constant Temperature
C) Addition of Argon gas
D) Decrease temperature
E) Remove sulfur dioxide gas
F) Addition of a catalyst
•
•
•
•
•
•
Calculating Equilibrium Concentrations
The same steps used to calculate equilibrium constants are
used.
K is given.
Generally, we do not have a number for the change in
concentration line.
Therefore, we need to assume that x mol/L of a species is
produced (or used).
The equilibrium concentrations are given as algebraic
expressions.
We solve for x, and plug it into the equilibrium
concentration expressions.
Example – H2 (g) + FeO (s) H2O (g) + Fe (s)Kc
= 5.20
If the initial concentration of H2 is 0.50 M and the
inintial concentration of H2O is 6.50 M, what
will the equilibrium concentrations be?
If the initial concentration of H2 is 1.00 M (no H2O
present), what will the equilibrium
concentrations be?
Determining Oxidation
Numbers
• The oxidation numbers of all atoms in a molecule or ion
add up to the total charge (0 for a molecule, the ion’s
charge for an ion)
• Elements by themselves are the charge shown
• O is almost always -2 (exception H2O2)
• F is almost always -1
• Group 1 elements are +1
• Group 2 elements are +2
Determining Oxidation
Numbers
• Examples: Determine the oxidation numbers of each
atom in the following:
1. NaC2H3O2
2. Na2SO4
3. C2O424. SO32-
Half Reactions
• The half-reactions for
Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe2+(aq)
are
Sn2+(aq)  Sn4+(aq) +2e2Fe3+(aq) + 2e-  2Fe2+(aq)
• Oxidation: electrons are products.
• Reduction: electrons are reactants.
Loss of
Gain of
Electrons is
Electrons is
Oxidation
Reduction
Balancing Equations by the Method of Half Reactions
1. Write down the two half reactions.
2. Balance each half reaction:
a. First with elements other than H and O.
b. Then balance O by adding water.
c. Then balance H by adding H+.
d. Finish by balancing charge by adding electrons.
Multiply each half reaction to make the number of
electrons equal.
4. Add the reactions and simplify.
5. If it is in basic solution, remove H+ by adding OH6. Check!
3.
Example:
MnO4- + C2O42- → Mn2+ + CO2
• The two incomplete half reactions are
MnO4-  Mn2+
C2O42-  CO2
Examples – Balance the following oxidationreduction reactions:
1. Cr (s) + NO3- (aq)  Cr3+ (aq) + NO (g) (acidic)
2. Al (s) + MnO4- (aq)  Al3+ (aq) + Mn2+ (aq)
(acidic)
3. PO33- (aq) + MnO4- (aq)  PO43- (aq) + MnO2 (s)
(basic)
4. H2CO (aq) + Ag(NH3)2+ (aq) 
HCO3- (aq) + Ag (s) + NH3 (aq)
(basic)
• Voltaic cells consist of
–
–
–
Anode: Zn(s)  Zn2+(aq) + 2eCathode: Cu2+(aq) + 2e-  Cu(s)
Salt bridge (used to complete the electrical circuit): cations move from anode to
cathode, anions move from cathode to anode. Necessary to balance charge.
• The two solid metals are the electrodes (cathode and
anode).
• Anode = oxidation
Cathode = reduction
• As oxidation occurs, Zn is converted to Zn2+ and 2e-. The
electrons flow towards the cathode (through the wire)
where they are used in the reduction reaction.
• We expect the Zn electrode to lose mass and the Cu
electrode to gain mass.
• “Rules” of voltaic cells:
1. At the anode electrons are products. (Oxidation)
2. At the cathode electrons are reactants. (Reduction)
3. Electrons cannot swim.
• Electrons flow from the anode to the cathode.
• Therefore, the anode is negative and the cathode is
positive.
• Electrons cannot flow through the solution; they have
to be transported through an external wire. (Rule 3.)
Standard Reduction (Half-Cell) Potentials
• Convenient tabulation of electrochemical data.
• Standard reduction potentials, Ered are measured
relative to the standard hydrogen electrode (SHE).
• Oxidation Potentials are obtained by reversing a
reduction half-reaction (sign of E° is changed)
• The SHE is the cathode. It consists of a Pt electrode in a
tube placed in 1 M H+ solution. H2 is bubbled through
the tube.
• For the SHE, we assign
2H+(aq, 1M) + 2e-  H2(g, 1 atm)
• Ered of zero.
• The emf of a cell can be calculated from standard
reduction potentials:
Spontaneity of Redox
Reactions
• For any electrochemical process
• A positive E indicates a spontaneous process (galvanic
cell).
• A negative E indicates a nonspontaneous process.
Electrolysis
Electrolysis of Aqueous Solutions
• Nonspontaneous reactions require an external current in
order to force the reaction to proceed.
• Electrolysis reactions are nonspontaneous.
• In voltaic and electrolytic cells:
–
–
–
reduction occurs at the cathode, and
oxidation occurs at the anode.
In electrolytic cells, electrons are forced to flow from the anode to cathode.
•
•
•
•
Example, decomposition of molten NaCl.
Cathode: 2Na+(l) + 2e-  2Na(l)
Anode: 2Cl-(l)  Cl2(g) + 2e-.
Industrially, electrolysis is used to produce metals like Al.
Electroplating
• Active electrodes: electrodes that take part in
electrolysis.
• Example: electrolytic plating.
• Consider an active Ni electrode and another metallic
electrode placed in an aqueous solution of NiSO4:
• Anode: Ni(s)  Ni2+(aq) + 2e• Cathode: Ni2+(aq) + 2e-  Ni(s).
• Ni plates on the inert electrode.
• Electroplating is important in protecting objects from
corrosion.
•
•
•
•
•
Quantitative Aspects of Electrolysis
We want to know how much material we obtain with
electrolysis.
1 Ampere is 1 Coulomb per second (A= C/s)
1 mole of electrons = 96,485 C = 1 Faraday
Use balanced half-equation to equate moles of
substance to moles of electrons
Molar mass (in g) = 1 mole of substance
Examples
1. A car bumper is to be electroplated with Cr from a
solution of Cr3+. What mass of Cr will be applied
to the bumper if a current of 0.50 amperes is
allowed to run through the solution for 4.20 hours?
2. What volume of H2 gas (at STP) will be produced
from the SHE after 2.56 minutes at a current of 0.98
amperes?
3. What volume of F2 gas, at 25°C and 1.00 atm, is
produced when molten KF is electrolyzed by a
current of 10.0 A for 2.00 hours? What mass of
potassium metal is produced? At which electrode
does each reaction occur?