The Equilibrium Constant

• For a general reaction

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 =

Kc =

Kp =

Equilibrium

concentrations

pressures

• 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:

– experimentally, the amount of CO2 does not seem to depend on the amounts of CaO and CaCO3. Why?

Heterogeneous Equilibria

• 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

• 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

Calculating Equilibrium Constants

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

Predicting the Direction of Reaction• We define Q, the reaction quotient, for a general

reaction

as

• Q = K only at equilibrium.

Applications of Equilibrium Constants

• 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.

• Consider the production of ammonia

• 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?

Le Châtelier’s Principle

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.

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.

Le Châtelier’s Principle

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?

• 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. C2O42-

4. SO32-

Determining Oxidation Numbers

Half Reactions• The half-reactions for

Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)are

Sn2+(aq) Sn4+(aq) +2e-

2Fe3+(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.

3. 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 OH-

6. Check!

Example:MnO4

- + C2O42- → Mn2+ + CO2

• The two incomplete half reactions areMnO4

- Mn2+

C2O42- CO2

Examples – Balance the following oxidation-reduction 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) + 2e-

– Cathode: 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, Ered 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 assign2H+(aq, 1M) + 2e- H2(g, 1 atm)

• Ered of zero.• The emf of a cell can be calculated from standard

reduction potentials:

anodecathode oxredcell EEE

• For any electrochemical process

• A positive E indicates a spontaneous process (galvanic cell).

• A negative E indicates a nonspontaneous process.

Spontaneity of Redox Reactions

processoxidation processreduction oxredcell EEE

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.

Electrolysis

• 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?