Equilibrium: Lewis Adducts of Iodine

1 lab period; work in groups. Complete the Preparation page before laboratory.

Goals

Background

Dynamic equilibrium is one of the four or five central concepts in chemistry. It is dynamic equilibrium that must be outsmarted in the synthesis of ammonia from nitrogen and hydrogen in chemical reaction systems; that determines the maximum level of humidity in the air on a summer day; and that maintains balance in the complex coupled reactions of biological systems. A synthetic chemist must be cognizant of chemical equilibrium in attempting to create a target molecule, such as a potential new drug.

Dynamic equilibrium is manifested at the macroscopic scale, but its origins are at the molecular level, where motions and collisions of reacting molecules stimulate their conversion to products, and vice versa. Although possible, it is surprisingly difficult to cite everyday examples of dynamic equilibrium; for this reason, equilibrium concepts are often difficult to grasp. This is unfortunate, because an understanding of equilibrium is crucial to an appreciation for chemistry.

In this experiment, we will study the dynamic equilibrium established in reactions of type (1):

(1): A + :B <===> A-B

In this Lewis acid-base reaction (or electron pair donation), the base, :B, donates a free electron pair to an empty orbital on the acid, A, to form an adduct, A-B, in which acid and base are joined by a normal covalent bond. This is a very fundamental and widespread reaction process that we will encounter again and again. It will behoove us to develop an appreciation for dynamic equilibrium in these simple processes. The expression for the equilibrium constant, Keq for this reaction is in (2).

(2): Keq = [AB]/[A][B]

where [AB] signifies the molarity (concentration in moles AB per liter of solution) of species AB. The following statements can be made about the equilibrium constant expression:

  1. In a solution containing arbitrary initial concentrations of A, B, and AB, reaction will take place until these concentrations have adjusted to values that satisfy (2).
  2. For some reactions, attainment of equilibrium is very rapid; for others it is very slow.
  3. The magnitude of Keq depends on temperature and solvent. Thus when reporting a value for Keq for a reaction, the T and solvent must be specified.
  4. Minor rearrangement of (2) gives (3):

    (3): Keq[B] = [AB]/[A]

    In this form, the equilibrium constant expression shows us that the distribution of A between the two forms, A and AB, is a function of the magnitude of the equilibrium constant and the concentration of the base B. The larger the concentration of B and/or the larger the equilibrium constant, the larger the amount of AB relative to free A.

    For example, suppose Keq for (1) has the value 0.01 M-1. This tells us that in a solution containing a total of 0.1 moles of A per liter, the ratio [AB]/[A] will be only 0.01 when [B] = 1.0 M. Thus even though the [B] is relatively large, very little A is converted to AB because the equilibrium constant is small. In contrast, suppose Keq = 100 M-1. In this case, [B] = 1 M is sufficient to make the ratio [AB]/[A] = 100, so essentially all A has been converted to AB.

Suppose we want to experimentally determine a value for Keq for reaction (1). In order to do this, we would begin by mixing known initial amounts of the participating species. We denote these initial concentrations [A]o, [B]o, and [AB]o. After mixing, the reaction will then proceed in one direction or the other at a characteristic rate until it reaches equilibrium. To assign a value to Keq, it is necessary to measure the concentration of one of the three species, A, B, and AB, at equilibrium. If we can do this, the other two can be obtained from mass balance and the equilibrium constant can readily be calculated by substituting concentrations into (2) or (3). For example, suppose that we start with [A]o = 0.1 M, [B]o = 0.2 M, and [AB]o = 0 in the solution, and that we are able to independently measure the concentration of AB at equilibrium. We allow the solution to come to equilibrium, measure AB, and calculate A and B from the relationships

[A] = [A]o - [AB] = 0.1 M - [AB]

[B] = [B]o - [AB] = 0.2 M - [AB].

These equations are mass balance equations. The first states mathematically that the total amount of A distributed among the forms A and AB must be the same after the reaction as before. The second makes a similar statement for B. Once we have [A], [B], and [AB], we can readily calculate Keq. Continuing our example, suppose that at equilibrium we find [AB] = 0.06 M. It follows that [B] = 0.2-0.06 = 0.14 M, and that [A] = 0.04 M. Thus

Keq = [AB]/[A][B] = (0.06)/(0.04)(0.14) = 10.7 M-1

If we were able to directly measure the concentration of one of the substances at equilibrium, the experimental determination of equilibrium constants for reactions of type (1) (and in fact for any reaction) would be a simple matter, because we can always control the initial (starting) concentrations. Unfortunately, this is often not possible. More commonly, we are able to measure some equilibrium property of the reaction solution (for example, the amount of light absorbed at a certain wavelength, or the conductivity of the solution) to which two substances contribute. Let us call this property P, and assume that both A and AB contribute to the observed value of this property. Further, we will assume that the contribution that A makes to P is directly proportional to the concentration of A, and similarly for AB. Then

(4) PA = kA[A]
(5) PAB = kAB[AB]
(6) Pobs = kA[A] + kAB[AB] = kA([A]o - [AB]) + kAB[AB]

Alternately,
(7) Pobs = kA[A] + kAB[AB] = kA[A] + kAB([A]o - [A])

where Pobs signifies the observed value of the property, P, in a solution containing both A and AB at equilibrium, and [A]o = [A] + [AB]. This is simply the total amount of A in the solution in both forms. If we start with a solution containing only A, and measure the property P, we obtain the value in (7)

(8): PoA = kA[A]o

We then begin to add B to the solution, causing conversion of some A to the form AB, and resulting in a change in the observed value of P to that in (6). If we add enough B, we convert essentially all A to AB, and the value of P will now be

(9): PoAB = kAB[A]o

We now want to obtain an expression for the ratio [AB]/[A] in terms of the observables Pobs, PoA, and PoAB. Subtracting (6) from (8) gives

(10): PoA - Pobs = [AB](kA - kAB)

Subtracting (9) from (7) gives

(11): Pobs - PoAB = [A](kA - kAB)

Dividing (10) by (11) gives the desired result:

(12): (PoA - Pobs)/(Pobs - PoAB) = [AB]/[A]

Finally, substituting (12) into (3) gives equation (13), the desired result:

(13) (PoA - Pobs)/(Pobs - PoAB) = Keq[B]

From this equation we can determine the value of Keq by measuring the value of the property, P, before reaction (PoA), at various stages during the reaction (Pobs), and when reaction is complete (PoAB).

In this experiment we will study the reaction of the Lewis acid, iodine, with a number of Lewis bases:

(14): I2 + :B ---> I2(B)

The measured property will be the amount of light absorbed by the solution (the absorbance, A), to which both I2 and I2(B) contribute. We will measure the absorbance of a solution containing only I2. This absorbance is signified Ao. We will then add aliquots of B to this solution, measuring the absorbance after each addition (Aobs). For large values of Keq, reaction (14) will proceed to completion and we can directly measure the absorbance at the end due entirely to the adduct (Afinal). We can then obtain Keq from equation (15):

(15): (Ao - Aobs)/(Aobs - Afinal) = Keq[B]

A plot of the left side of this equation versus [B] will provide Keq as the slope.

In the event that we can not measure Afinal directly, we can work with a rearranged version of (15):

(16): Aobs = Afinal + (Ao - Aobs)/Keq[B]

In this case a plot of Aobs versus the quantity (Ao - Aobs)/[B] will provide Keq from the slope.

Focus Questions

  1. Is the observed behavior of your system consistent with the principle of Le Chatelier? Explain how.
  2. As you continue to add aliquots of B, you change the total volume of the solution in the cell. What effect does aliquot addition have on the concentration of iodine in the solution? What effect does it have on the observed absorbance at 504 nm?
  3. Is the total volume of B added large enough to affect concentration significantly? Explain.
  4. If the answer to the previous question is yes, how would you "correct" the concentrations and observed absorbances for the dilution effect?
  5. Enter you data into a spreadsheet and determine the equilibrium constant for reaction of your assigned base with iodine. Suggested column headings are Volume B added, [B], A at 504, Corrected A at 504, and additional columns suggested by equation (16). Plot the data according to equation (16) to obtain Keq for formation of the adduct of I2 with each base.
  6. Compare you Keq with those determined by other groups studying different bases, B. Arrange the bases studied in order of increasing affinity for I2. Do you find any patterns? Can you explain them?

Equipment and Materials

Safety

Safety glasses must be worn at all times in the laboratory. In general, organic solvents and reagents are toxic in varying degrees. Avoid ingestion and contact with the skin. Do not open a container of any organic liquid unless it is in a fume hood. It is recommended that you wear latex gloves when handling these materials or their solutions. DMSO readily penetrates the skin; avoid skin contact. Pyridine should be handled with care IN THE HOOD because it has an obnoxious odor. In case of skin contact by any of these materials, flush with copious quantities of water.

Experimental

Record all data in your notebook. Obtain the necessary equipment and clean the glassware thoroughly using brushes and Alconox detergent. Rinse with distilled water and dry thoroughly, inside and out, via aspiration.

You will find in the lab a stock solution containing approximately 0.01 M iodine in dichloromethane solvent. From this solution you will prepare 10 mL of a working solution that is about 5.0 x 10-4 M in iodine. Plan how to do this and carry out your plan. The instructor will assign you two Lewis bases, B, to study. One will be either acetone, dimethylsulfoxide, or pyridine. The other will be either imidazole, triethylamine, or triphenylphosphine.

As the experiment progresses you will record a number of electronic absorption spectra. Save each one with a different file name so that all can be recalled when you have finished.

Transfer exactly 2.2 mL of working solution to a glass spectrometer cell, and record the electronic absorption spectrum between 700 and 300 nm. You will now add a succession of aliquots of B (or solution of B) to the cell, recording the spectrum after each addition. Because you do not know in advance how strongly the base will interact with iodine, your first aliquot should be the smallest volume that you can measure accurately with your syringe: 1 microliter. Add a 1-microliter aliquot of your assigned Lewis base (or solution of a Lewis base) to the cell, stopper, shake, and record the spectrum again. Adjust the aliquot size depending on the size of the absorbance change produced by the first aliquot (if DA small, use a larger aliquot). Add at least 4 more aliquots of B. After the addition of each aliquot, record the spectrum.

Retrieve all spectra from memory and display them overlayed on the screen. Measure the absorbance at about 504 nm for each spectrum and record in your notebook. Then print the overlayed spectra.

Carry out a similar study for the second Lewis base.

Clean-up. When you have finished all of your work:

Disposal Methods

All dichloromethane solutions should be disposed of in the chlorinated-solvent waste jar in the hood. Put broken glass in the receptacle provided for this purpose.

References

  1. Benesi, Hildebrand
  2. Ramette, J. Chem. Ed.


Preparation
Equilibrium: Lewis Adducts of Iodine


Read

  1. This experiment.
  2. The appropriate sections of your textbook.

Problems

  1. The following reaction is carried out at 25 oC by adding successive aliquots of reagent B to a solution containing a known amount of reagent A.

    A + B <===> AB

    After each addition of B to the solution, the absorbance of the solution is measured at a wavelength where both A and AB have some absorbance.

    The initial concentrations, the volumes of the B aliquots, and the measured absorbance after each aliquot addition are given below.

    Initial volume of reaction solution = 5.00 mL
    Initial [A] in reaction solution = 0.00120 M
    [B] in titrant solution = 2.00 M

    Total Volume B Solution Added, mLAbsorbance Measured at Equilibrium, M
    00
    0.1
    0.2

    Use equation (16) to determine the equilibrium constant for the reaction.