CHEM17 POST-LAB LECTURE - 1

Kurt W.E. Sy Piecco

List of Experiments

1. Colligative Properties of Solutions

2. Heat Effects and Calorimetry

3. Chemical Kinetics

4. Le Chatelier’s Principle

Colligative Properties of Solutions

True Solution (NaCl solution)

Colloidal Dispersion(Cooked starch

solution)

Coarse Mixture(Uncooked starch

solution)

Appearance Homogeneous, transparent or translucent

Dilute solutions will have

suspended tiny particles

Large particles or clumps settled on the bottom of the

container

Particle Visibility No particles visible Visible Visible

Settling Rate NA Very slow Very fast

Effect on Light Beam

No scattering Scattered No scattering (opaque)

Particle Size NA Very tiny Large

A. Dispersed Systems

� Answers to questions:

1. Light beams can only be used to differentiate colloidal dispersions from pure solutions

2. Solids in dispersed systems can be separated from the solution through filtration (using an appropriate filter, which depends on the size of the particles).

3. Within a typical cell, proteins and nucleic acids are colloidal sized particles dispersed in an aqueous solution of small molecules and ions. And the cleansing action of soaps works by converting oily dirt into an aqueous colloidal suspension.

4. The Tyndall effect is the scattering of a beam of light by suspended colloidal particles. As a result, we can clearly see the light beams with our naked eyes.

B. Boiling point elevation

� Both solutions will have elevated BP compared to pure water

� The NaCl solution will have a higher BP than the urea solution because of the greater number of solute particles in the NaCl solution (Na+ and Cl-).

C. Freezing point depression

� The PDB + “unknown” organic compound solution froze at a temperature lower then the FP of pure PDB

D. Osmotic pressure

� In every case, the more concentrated solution increased in volume:

� If the more concentrated solution is in the pig intestine bladder, it will expand.

� If the external solution is more concentrated, the bladder will shrink.

� If both concentrations are equal (isotonic) no change in the volumes is observed.

� Solutions will exhibit colligative properties only when the solute is non-volatile (e.g. salts of metal ions)

� Colligative properties and Entropy:

� Boiling Point Elevation (BPE)

� The solvent molecules would rather stay in the solution due to the greater entropy (between solvent molecules and solute molecules) in the solution than in the vapor (solvent molecules only).

� Freezing Point Depression (FPD)

� There is generally less entropy in solids (crystals) specially when only the solvent will freeze, therefore the solvent molecules will also prefer to stay in the solution where there is greater entropy.

� Osmotic Pressure

� The solvent molecules will move from the less dilute to the more dilute solution because of the greater entropy that will result due to the greater number of solute molecules to mix with.

� Molecular weight (MW) determination: solve for molality, then calculate MW.

� ∆Tf = -kfm ∆Tb = kb m ∆T = Tsolution – Tpure solvent

� Answers to Questions:

� BPE and FPD

1. Dissolving a solute in another substance lowers the freezing point of that substance

2. The solid should be finely powdered when determining its MP (melting point) so that heat will be uniformly distributed throughout the solid sample and to prevent the solid from getting burnt.

3. The solution has a lower melting point than the pure compound.

4. Calculation of FPD

� ∆Tf = Tsolution – Tpure solvent

5. Calculation of the molal concentration from FPD data:

6. Calculation of the molar mass (MW) of a compound from FPD data

� (see previous slide). From these mathematical relationships, one can calculate the molar mass of an unknown compound.

� Osmotic pressure

1. The more concentrated solution increased in volume.

2. The solution with equal concentration to the solution on the other side of the membrane (20% solution on both sides).

3. The solvent in the 10% solution moved to the 20% solution. The solvent in the 20% solution moved to the 30% solution.

4. In every case, the more concentrated solution increased in volume

Heat Effects and Calorimetry

A. Determination of heat capacity (of the calorimeter)

� To determine the amount of heat the calorimeter can absorb or release.

� Ideally, we want a calorimeter that doesn’t absorb or release heat so that it would not greatly affect our measurements of heat produced or absorbed during reactions.

� In other words, we need a very good heat insulator. However, for practical reasons, we can use styrofoam – which minimizesthe absorption/release of heat.

B. Heat of neutralization

� The heat released or absorbed during a reaction between an acid and base

� The reaction between HCl and NaOH is exothermic

C. Exothermic reaction

� A reaction that releases heat to the environment (like the dissolution of NaOH pellets in water).

� ∆H and q will be negative for this type of reaction.

D. Endothermic reaction

� A reaction that absorbs heat from the environment (e.g. dissolving Na2S2O3 in water).

� ∆H and q will be positive for this type of reaction.

E. Heat of combustion

� Heat released during combustion (explosive reaction of a substance with oxygen gas at high temperatures)

� Calculations:

� Heat capacity of the calorimeter

Assume that the calorimeter doesn’t allow heat to go in or out of it:

0 = qcalorimeter + qhotwater + qtapwater

qhotwater = weighthotwater × specific heat × ∆Th ∆Th = Tmixture – Thot water

qtapwater = weighttapwater × specific heat × ∆Tc ∆Tc = Tmixture – Ttap water

� Specific heat of water = 4.184 J/g°C

Heat capacity of the calorimeter = (qhotwater – qtapwater) ÷∆Tc

� Heat of neutralization

� The resulting NaCl solution is very dilute; assume its density and specific heat is equal to that of pure water

∆H = (total volume of mixed solution × density of water × specific heat × ∆T) ÷moles NaCl formed

� Heat of solution

� Exothermic (the NaOH solution is very dilute; same assumptions as above)

moles NaOH = weight NaOH ÷molar weight of NaOH

∆H = (volume of solvent × density of water × specific heat × ∆T) ÷moles NaOH

∆T = Tsolution – Tsolvent

� Endothermic (the Na2S2O3 solution is very dilute; same assumptions as above)

moles Na2S2O3 = weight Na2S2O3 ÷molar weight of Na2S2O3

∆H = (volume of solvent × density of water × specific heat × ∆T) ÷moles Na2S2O3

� Answers to questions

1. To be sure that the temperature is constant; that the system is really in thermal equilibrium.

2. The calorimeter also absorbs or releases some heat from or to the system.

3. For an adiabatic system, one component of a system has to absorb the heat released by another component. The signs are just conventions used to designate release (negative) or absorb (positive).

4. Possible source of error: leaks in the setup where heat and/or matter can enter or escape.

Collision and Transition State Theories

The Collision Model

� Rates of reactions are affected by concentration and temperature.

� We need to develop a model that explains this observation.

� An explanation is provided by the collision model, which is based on kinetic-molecular theory.

� In order for molecules to react they must collide.

� The greater the number of collisions the faster the rate.

� The more molecules present, the greater the probability of collision and the faster the rate.� Thus reaction rate should increase with an increase in the concentration of

reactant molecules.

� The higher the temperature, the more energy available to the molecules and the more frequently the molecules collide.� Thus reaction rate should increase with an increase in temperature.

� However, not all collisions lead to products. � In fact, only a small fraction of collisions lead to products.� In order for reaction to occur the reactant molecules must collide in the

correct orientation and with enough energy to form products.

The Orientation Factor

� The orientation of a molecule during collision can have a profound effect on whether or not a reaction occurs.

� Consider the reaction between Cl and NOCl: Cl + NOCl→ NO + Cl2� If the Cl collides with the Cl of NOCl, the products are Cl2 and NO.

� If the Cl collides with the O of NOCl, no products are formed.

NO + NO3 2 NO2

Molecular orientation and effective collisions

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

The proposed transition state in the reaction

between CH3Br and OH-

The TS is trigonal bipyramidal; note the elongated C-Brand C-O bonds

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Reaction energy diagram for the reaction between CH3Br and OH-

Activation Energy

� Arrhenius: molecules must posses a minimum amount of energy to react. Why?

� In order to form products, bonds must be broken in the reactants.

� Bond breakage requires energy.

� Molecules moving too slowly, with too little kinetic energy, do not react when they collide.

� Activation energy, Ea, is the minimum energy required to initiate a

chemical reaction.

� Eawill vary with the reaction.

Transition State Model

� Consider the rearrangement of methyl isonitrile to form acetonitrile:

� Energy is required to stretch the bond between the CH3 group and the N≡C group to allow the N≡C to rotate.

� The C–C bond begins to form.

� The energy associated with the molecule drops.

� The energy barrier between the starting molecule and the highest energy state found along the reaction pathway is the activation energy. � The species at the top of the barrier is called the activated complexor transition state.

� The change in energy for the reaction is the difference in energy between CH3NC and CH3CN.� ∆Erxn has no effect on reaction rate.

� The activation energy is the difference in energy between the reactants (CH3NC) and the transition state.� The rate depends on the magnitude of the Ea.

� In general, the lower the Ea, the faster the rate.

� Notice that if a forward reaction is exothermic (CH3NC → CH3CN), then the reverse reaction is endothermic (CH3CN → CH3NC).

� How does this relate to temperature?

� At any particular temperature, the molecules present have an average kinetic energy associated with the population.

� In the same distribution, some molecules have less energy than the average while others have more than the average value.

� The fraction of molecules with an energy equal to or greater than ∆Ea is given by:

� Molecules that have an energy equal to or greater than ∆Ea have sufficient energy to react.

� As we increase the temperature, the fraction of the population that has an energy equal to or greater than ∆Ea increases.

� Thus more molecules will react.

RTaEef

−=

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Reaction energy diagrams and possible

transition states for three reactions

endothermic

exothermic

exothermic

The Arrhenius Equation and the Effect of Temperature on the Reaction Rate

� Arrhenius discovered that most reaction-rate data obeyed an equation based on three factors:

� The number of collisions per unit time.

� The fraction of collisions that occur with the correct orientation.

� The fraction of the colliding molecules that have energy equal to or greater than ∆Ea.

� From these observations Arrhenius developed the Arrhenius equation:

� Where k is the rate constant, Ea is the activation energy, R is the ideal-gas constant (8.314 J/K·mol), and T is the temperature in K.

� A is called the frequency factor.

� It is related to the frequency of collisions and the probability that a collision will have a favorable orientation.

� Both A and Ea are specific to a given reaction.

� Ea and Amay be determined experimentally.

� We need to rearrange the Arrhenius equation:

� Alternatively, we can use:

RTaEAek

−=

ART

Ek a lnln +−=

−=

122

1 11ln

TTR

E

k

k a

Chemical Kinetics

� Nature of reactants

� The iron filings dissolved and produced bubbles faster in HCl than in HOAc

� Answers to questions:

1. Fe + HCl is comparatively more vigorous

2. H2 is produced, which is a flammable gas

3. H2

4. FeCl2 and Fe(OAc)25. Fe + 2 HCl → H2 + FeCl2 and Fe + 2 HOAc → H2 + Fe(OAc)2

� Effects of concentration

� The more concentrated NaOH solution dissolved the aluminum foil faster than the dilute solutions.

� Higher concentration of reactants results in a faster reaction

� Answers to questions:

1. Increasing the reactant concentration decreases the reaction time.

2. Increasing the reactant concentration increases the reaction rate.

3. Reaction rate depends on reactant concentration but the product concentration is dependent on the reaction rate.

� Effect of temperature

1. Hotter water dissolved and decomposed the alkaseltzer tablets faster than cooler water resulting in a more rapid evolution of CO2

gas for inflating the balloon. Hot water also lowered the solubility of CO2 in the solution.

2. Higher water temperatures also lower the solubility of gases.

3. Higher temperature increased the reaction rate.

4. Hotter water dissolved the alkaseltzer tablets faster resulting in a faster evolution of CO2 gas, which filled the balloon. Besides the rapid evolution of gas, the relationship between temperature and volume of gases are directly proportional (Charles’ and Gay-Lussac’s Law).

� Effect of surface area

1. Zinc dust dissolved faster in HCl and produced gas more vigorously than mossy zinc.

2. Increase in surface area results in the increase of the reaction rate.

3. Zn + 2HCl → ZnCl2 + H2

� Answers to questions:

1. According to the Collision and Transition-State theories of chemical kinetics, a reaction becomes faster

a) if the reactants are more chemically reactive,

b) when solid reactants have high surface area,

c) at high reactant concentrations,

d) at elevated temperatures

e) in the presence of a catalyst

2. Increasing the surface area is equivalent to increasing the concentration of the solid reactant by providing more surfaces on which molecules in the liquid can collide with atoms/molecules on the solid.

3. Osterized food (greater surface area) promotes its reaction (digestion) with (by) HCl (gastric juice).

� Effect of catalyst

� MnO2 accelerated the decomposition reaction of H2O2

� H2O2 → H2O + O2

� Answers to questions:

1. Catalysts hasten reactions. Inhibitors slow down reactions. Poisons are substances that deactivate (destroy) catalysts.

2. The catalyst concentration doesn’t necessarily have to appear in the rate law

3. The evolution of O2 is faster when MnO2 is added.

� Reaction order

� Reaction: 6I- + BrO3- + 6H+ → 3I2 + Br

- + 3H2O

� Rate law: rate = k [I-]x [BrO3-]y [H+]z

� Calculations: see examples on the board

� Answers to questions:

1. x = 1, y = 1, z = 2

2. Overall order of the reaction = 4

The Concept of Equilibrium

� Consider colorless frozen N2O4.

� At room temperature, it decomposes to brown NO2.

N2O4 (g) ↔ 2NO2 (g)

� At some time, the color stops changing and we have a mixture of N2O4 and NO2.

� Chemical equilibrium is the point at which the concentrations of all species are constant.

� Consider a simple reaction.

� A↔ B

� Assume that both the forward and reverse reactions are elementary processes.

� We can write rate expressions for each reaction.

� Forward reaction: A→ B

� Rate = kf[A] kf = rate constant (forward reaction)

� Reverse reaction: B→ A

� Rate = kr[B] kr = rate constant (reverse reaction)

� Place some pure compound A into a closed container.

� As A reacts to form B, the concentration of A will decrease and the concentration of B will increase.

� Thus we expect the forward reaction rate to slow and the reverse reaction rate to increase.

� Eventually we reach equilibrium: the forward and reverse rates are equal.

� At equilibrium: kf[A] = kr[B]

� Rearranging, we get:

� At equilibrium the concentrations of A and B do not change.

� This mixture is called an equilibrium mixture.

� This is an example of a dynamic equilibrium.

� A dynamic equilibrium exists when the forward and reverse rates of the reaction are equal.

� No further net change in reactant or product concentration occurs.

� The double arrow ↔ implies that the process is dynamic.

[ ][ ]

constant aA

B==

r

f

k

k

The Equilibrium Constant

� Consider the reaction: N2 (g) + 3H2 (g) ↔ 2NH3 (g)

� If we start with a mixture of nitrogen and hydrogen (in any proportions), the reaction will reach equilibrium with a constant concentration of nitrogen, hydrogen, and ammonia.

� However, if we start with just ammonia and no nitrogen or hydrogen, the reaction will proceed, and N2 and H2 will be produced until equilibrium is achieved.

� No matter what the starting composition of reactants and products is, the equilibrium mixture contains the same relative concentrations of reactants and products.

� Equilibrium can be reached from either direction.

� We can write an expression for the relationship between the concentration of the reactants and products at equilibrium.

� This expression is based on the law of mass action.

� For a general reaction:

aA + bB ↔ pP + qQ

� The equilibrium expression is given by:

� Where Kc is the equilibrium constant.

� The subscript c indicates that concentrations (molarities) are used in this expression.

� Note that the equilibrium constant expression has products in the numerator and reactants in the denominator.

[ ] [ ]

[ ] [ ]ba

qp

cKBA

QP=

� Consider the reaction:

N2O4 (g) ↔ 2NO2 (g)

� The equilibrium constant is given by:

� The value of this constant (at 100°C) is 0.211 (regardless of the initial N2O4 (g) or NO2 (g) concentrations.

� The equilibrium expression depends on stoichiometry.

� It does not depend on the reaction mechanism.

� The value of Kc varies with temperature.

� We generally omit the units of the equilibrium constant.

[ ]

[ ]42

2

2

ON

NO=cK

Equilibrium Constants in Terms of Pressure

� If the reactants and products of a reaction are gases we can write an equilibrium expression in terms of the partial pressures of the species.

� Where Kp is the equilibrium constant.

� Kp is based on partial pressures measured in atmospheres.

� Note: Kc and Kp are numerically different.

� We can use the ideal-gas equation to convert between partial pressures and molarity:

� For a substance A: PA=[A]RT

� A general relationship can be derived:

� Where ∆n is the sum of the stoichiometric coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants.

� For example, for the reaction: N2O4 (g) ↔ 2NO2 (g)

� ∆n = 2-1 = 1

[ ] [ ][ ] [ ]b

B

a

A

q

Q

p

pPP

PPK

p=

nRTPV = MRTRTV

nP ==

( ) n

cp RTKK∆

=

The Magnitude of Equilibrium Constants

� The equilibrium constant, K, is the ratio of products to reactants.

� Therefore, the larger K, the more products are present at equilibrium.

� Conversely, the smaller K, the more reactants are present at equilibrium.

� If K >> 1, then products dominate at equilibrium and equilibrium lies to the right.

� If K << 1, then reactants dominate at equilibrium and the equilibrium lies to the left.

The Direction of the Chemical Equation and K

� Equilibrium can be approached from any direction.

� Consider the reaction: N2O4 (g) ↔ 2NO2 (g)

� The equilibrium constant for this reaction (at 100°C) is:

� However, when we write the equilibrium expression for the reverse reaction,

2NO2 (g) ↔ N2O4 (g)

� The equilibrium constant for this reaction (at 100°C) is:

� The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.

212.0]ON[

]NO[

42

22 ==cK

[ ]

[ ]72.4

212.0

1

NO

ON2

2

42 ===cK

Applications of Equilibrium Constants

Predicting the Direction of Reaction

� For a general reaction: aA + bB ↔ pP + qQ

� We define Q, the reaction quotient, as:

� Where [A], [B], [P], and [Q] are molarities at any given time.

� Note: Q = K only at equilibrium.

� If Q < K, then the forward reaction must occur to reach equilibrium.

� If Q > K, then the reverse reaction must occur to reach equilibrium.

� Products are consumed, reactants are formed.

� Q decreases until it equals K.

Calculation of Equilibrium Concentrations

� The same steps used to calculate equilibrium constants are used to calculate equilibrium concentrations.

� Generally, we do not have a number for the change in concentration.

� Therefore, we need to assume that xmol/L of a species is produced (or used).

� The equilibrium concentrations are given as algebraic expressions.

[ ] [ ]

[ ] [ ]ba

qp

QBA

QP=

Le Châtelier’s Principle

� Consider the Haber process: N2 (g) + 3H2 (g) ↔ 2NH3 (g)

� As the pressure increases, the amount of ammonia present at equilibrium increases.

� As the temperature increases, the amount of ammonia at equilibrium decreases.

� Can this be predicted? Of course! Just read on…

� Le Châtelier’s principle:

� If a system at equilibrium is disturbed by a change in temperature, a change in pressure, or a change in the concentration of one or more components, the system will shift its equilibrium position in such a way as to counteract the effects of the disturbance.

Change in Reactant or Product Concentration

� If a chemical system is at equilibrium and we add or remove a product or reactant, the reaction will shift to reestablish equilibrium.

� For example, consider the Haber process again:

N2 (g) + 3H2 (g) ↔ 2NH3 (g)

� If H2 is added while the system is at equilibrium, Q < K.

� The system must respond to counteract the added H2 (by Le Châtelier’sprinciple).

� That is, the system must consume the H2 and produce products until a new equilibrium is established.

� Therefore, [H2] and [N2] will decrease and [NH3] increase until Q = K.

� We can exploit this industrially.

� Suppose that we wanted to optimize the amount of ammonia we formed from the Haber process.

� We might flood the reaction vessel with reactants and continuously remove the product.

� The amount of ammonia produced is optimized because the product (NH3) is continuously removed and the reactants (N2 and H2) are continuously added.

N2 (g) + 3H2 (g) ↔ 2NH3 (g)

Effects of Volume and Pressure Changes

� Consider a system at equilibrium.

� If the equilibrium involves gaseous products or reactants, the concentrations of these species will be changed if we change the volume of the container.

� For example, if we decrease the volume of the container, the partial pressures of each gaseous species will increase.

� Le Châtelier’s principle predicts that 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.

� Consider the following system: N2O4 (g) ↔ 2NO2 (g)

� An increase in pressure (by decreasing the volume) favors the formation of colorless N2O4.

� The instant the pressure increases, the concentration of both gases increases and system is not at equilibrium.

� The system moves to reduce the number of moles of gas.

� A new equilibrium is established.

� The mixture is lighter in color.

� Some of the brown NO2 has been converted into colorless N2O4(g).

� In a reaction with the same number of moles of gas in the products and reactants, changing the pressure has no effect on the equilibrium.

� In addition, no change will occur if we increase the total gas pressure by the addition of a gas that is not involved in the reaction.

Effect of Temperature Changes

� The equilibrium constant is temperature dependent.

� How will a change in temperature alter a system at equilibrium?

� It depends on the particular reaction.

� For example, consider the endothermic reaction:

Co(H2O)62+

(aq) + 4Cl–(aq) ↔ CoCl4

2–(aq) + 6H2O(l) ∆H > 0

� Co(H2O)62+ is pale pink and CoCl4

2– is a deep blue.

� At room temperature, an equilibrium mixture (light purple) is placed in a beaker of warm water.

� The mixture turns deep blue.

� This indicates a shift toward products (blue CoCl42–).

� This reaction is endothermic.

� For an endothermic reaction (∆H > 0), heat can be considered as a reactant.

� Thus adding heat causes a shift in the forward direction.

� The room-temperature equilibrium mixture is placed in a beaker of ice water.

� The mixture turns bright pink.

� This indicates a shift toward reactants (pink Co(H2O)62+).

� In this case, by cooling the system we are removing a reactant (heat).

� Thus the reaction is shifted in the reverse direction.

The Effect of Catalysts

� 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 affect the composition of the equilibrium mixture.

Le Chatelier’s Principle

� Gaseous equilibrium

� Reactions:

� Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2

� 2NO2 �N2O4

� NO2 is brown, N2O4 is colorless

� Effect of pressure: increase in pressure will result in a lighter colored gas (increase in concentration of N2O4)

� Effect of temperature: heating will make the gas mixture darker (increase in the concentration of NO2)

� Common-ion effect

� Reaction: HOAc� H+ + OAc-

� Effect of addition of NaOAc: the solution became more alkaline

NO2

� Complex-ion equilibrium

� Reaction: FeCl3 + KSCN � FeSCN2+ + K+ + Cl-

� Net reaction: Fe3+ + SCN- � FeSCN2+

� FeSCN2+ is a dark red complex

� Effect of Fe(NO3)3: darker red color

� Effect of NH4SCN: darker red color

� Effect of KCl: lighter red color

� Effect of Na2HPO4: no effect

� Chromate – Dichromate equilibrium

� Cr2O72- + H2O � 2CrO4

2- + 2H+

� Effect of pH: low pH (high H+ concentration) will turn the solution orange. High pH will turn the solution yellow.

� Answers to questions:

� Gaseous equilibrium

1. N2O4 is very toxic, corrosive and combustible. NO2 is also toxic. The equilibrium between NO2 and N2O4 is affected by temperature and pressure as shown in the previous slide.

2. See previous slide

� Common-ion effect

� Increase in the concentration of OAc- decreased the dissociation of HOAcleading to decreased H+ concentration

� Complex-ion equilibrium

1. (See previous slide)

2. Increase in the concentration of Fe3+ and SCN- will result in the increase of the concentration of the red colored complex. Increase of the concentration of KCl will result in the decrease of the concentration of the red complex. Na2HPO4 will not affect the reaction.

� Chromate-Dichromate equilibrium

1. Chromate is yellow. Dichromate is orange.

2. Cr2O72- + 2OH- � 2CrO4

2- + H2O

3. High concentrations of OH- will favor the formation of chromate

4. Dichromate

5. See previous slide

6. Keq1 = [chromate]2 ÷ ( [dichromate] × [OH-]2 )

Keq2 = [chromate]2 × [H+]2 ÷ [dichromate]

7. Keq1 = (3.2 × 10-7)2 × 34-1 × (1 × 10-14)-2

Keq2 = (3.2 × 10-7)2 × 34-1

References

� Brown, Le May and Bursten. Chemistry: the Central Science, 9th ed. 2004

� Silberberg. Chemistry: The Molecular Nature of Matter and Change, 4th ed. 2006