CHEMISTRY 2500 – EXERCISE K2

 

KINETICS OF SAPONIFICATION OF AN ESTER

 

Objective

 

To confirm that the total order of the reaction:

 

 

O              O

 

 

CH3

CH CH3

CH3

–

+   HO

 

H3C

O

 

 

+

–

O                      H3C

OH

CH

CH3             (1)

 

is two, to show that the individual orders of the two reactants are each unity, and to determine the reaction’s rate constant at 25 oC. Because of the large difference between the mobility of the hydroxide ion (20.52 x 10-8 m2 s-1 V-1 at infinite dilution in water at 25oC) and that of the acetate ion (4.24 x 10-8 m2 s-1 V-1 in similar circumstances) which replaces it during the reaction, conductimetry is the most convenient way of following this reaction. The method of treating the raw data to obtain significant results illustrates the use of “smoothed” data.

 

Note: there is a required calculation for Part 3 of the experimental procedure, and it is recommended that you perform this calculation before lab, so you can verify with your teaching assistant and prepare the solution in the time allotted.

 

Background Reading

 

Review the subject of the electrical conductivity of solutions, (see any standard Physical Chemistry text)1, making sure that you understand the meaning of such quantities as cell constant, specific resistivity, equivalent conductivity, transport number, etc.. Find out why alternating current is almost always used to measure conductivities of electrolyte solutions. In the first part of the experiment the equalization method is employed to simplify the handling of the kinetic data. In the second part, the initial concentrations are not equal. A slightly more elaborate treatment of the data is necessary. Ensure that you understand the difference. A similar experiment is described by Daniels et al.2

 

Theory

 

In Part 12, hydroxide ion and ester are mixed in equal concentrations and, therefore, from the stoichiometry of Reaction (1), their concentrations will remain equal thereafter.

Hence, assuming second-order kinetics (order one in each reactant species), i.e.

 

d éëOH- ùû

 

ë        û

–                = k éOH- ù [E]

dt

(2)

 

 

 

(where we represent the ester by E) We may write [OH-] = [E] = c and

 

and integrating,

 

• dc = kc 2

dt

 

 

 

(3)

 

1 = kt + 1

c            co

(4)

 

where co denotes the initial hydroxide (and ester) concentration. If a straight line results from plotting the reciprocal of the hydroxide ion concentration against time, second-order kinetics are confirmed, and the slope directly yields the rate constant.

 

In Part 2, the concentrations are not initially equal. Rather [E]o = 2[OH-]o. From the stoichiometry, we can write:

 

[E]o – [E] = [OH-]o – [OH-]

so [E] = [E]o – ([OH-]o – [OH-])                                       (5)

= 2[OH-]o – [OH-]o + [OH-]

= [OH-]o + [OH-]

= co + c                                                                  (6)

 

where co and c have the same significance as before, that is the initial and changing concentrations of OH-. We can now substitute for [E] and [OH-] in Equation (2) and obtain

 

 

 

which can be integrated to yield:

dc

= kc(c + co )

dt

(7)

 

 

 

ln(1 + co ) = c kt + ln 2

(8)

 

c         o

 

 

Thus, if the data for Part 2 are plotted as ln(1+co/c) vs. t, and if the order is 2 and the individual orders each unity (as has been assumed in the derivations), a straight line should again be obtained, with a slope equal to cok.

 

In both parts, we obtain the concentrations by the following argument:

 

By the Kohlrausch law of independent ion migration, the specific conductivity (k, ?-1

cm2) of a solution containing Na+, OH- and CH3COO- as the only ionic species is

k = ?Na[Na+] + ?OH[OH-] + ?Ac[CH3COO-]                                       (9)

where ?Na, ?OH and ?Ac are the equivalent conductivities of the sodium, hydroxide and acetate ions.

 

Sodium and hydroxide ions are initially present in equal concentration and the latter are progressively replaced by acetate ions; therefore

 

[Na+] = [OH-]o                                                                               (10)

 

 

and

[OH-]o = [OH-] + [CH3COO-]

 

 

Using these relations, we can substitute for [Na+] and [CH3COO-] in equation (9), to derive an expression for the specific conductivity, in terms of only [OH-]o and [OH-]:

k = ?Na[OH-]o + ?OH[OH-] + ?Ac([OH-]o – [OH-])

 

 

or, collecting coefficients:

k = (?Na + ?Ac) [OH-]o + (?OH – ?Ac) [OH-]                                      (11)

 

 

 

In principle, equation (11) can be used to determine the hydroxide ion concentration, since all the other terms are either known or can be measured. A more accurate and convenient method, however, makes particular use of the initial and final specific conductance ko and k?. Since the saponification proceeds virtually to completion, after

infinite time [OH-] = 0, and therefore from (11)

 

k? = (?Na + ?Ac) [OH-]o                                                                      (12)

Similarly, inserting initial conditions into equation (11) yields

 

ko = (?Na + ?Ac) [OH-]o + (?OH – ?Ac) [OH-]o = (?Na + ?OH) [OH-]o

(13)

 

Now, taking advantage of the fact that

 

ko – k? = (?Na + ?OH) [OH-]o – (?Na + ?Ac) [OH-]o = (?OH – ?Ac) [OH-]o

or

(?OH – ?Ac) = (ko – k? ) / [OH-]o

 

 

Equations (11), (12) and (13) may be combined into

 

k = k? + (?OH – ?Ac) [OH-]

k = k? + ((ko – k? ) / [OH-]o) * [OH-] or, finally, rearranging, we obtain:

 

 

[OH

– ] = æ k – k ¥

ö

–

÷÷[OH  ]o

 

çç

è k o  – k ¥ ø

or

 

 

çç

c = æ k – k¥

ö

÷÷ co

(14)

 

è ko – k¥ ø

 

This is the basic equation of the exercise, enabling the hydroxide ion concentration variation to be measured from current readings alone.

 

Values of ko and k? could be found by suitably extrapolating a graph of k versus t. However, the rate of the reaction under the conditions of this experiment is such that k does not approach k? until about 20 hours after the mixing of reactants.

 

On the other hand, the reaction is initially so rapid that extrapolation of k back to t = 0 is also insufficiently accurate. Because use of equation (14) requires accurate values of ko and k?, they are measured in this experiment with “synthetic solutions”, i.e. ones prepared by the experimenter in such a way as to have unchanging conductivities, and which emulate the initial and final conductivities of the reaction mixture.

 

From the table of values below, it is possible to calculate what the ratio k?/ko should be. How closely do your results agree with the calculated ratio?

 

Equivalent conductivities of ions at 25oC

(These apply to infinitely dilute solutions)  Ion ?/?-1 cm2 mol-1

Na+                                              50.11

K+                                               73.52

OH-                                            198.0

CH3COO-                                       40.9

 

5

 

 

 

 

Description

 

1. Digital Display
2. Control Range Switch
3. Function Switch
4. Probe
5. Probe Cap

 

 

 

 

Figure 1: Schematic representation of the Conductivity Meter.

 

 

Apparatus

 

1. Conductivity Meter (as shown in Figure 1). Obtained from and return to the instructor

 

1. Thermometer

 

• Stock watch or clock with second hand

 

1. 50 ml pipettes (two)

 

1. 25 ml pipettes (two)

 

1. 150 ml conical flask (to use as a conductivity cell)

 

• 250 ml conical flasks (three)

 

• 250 ml volumetric flask

 

1. Thermostat bath set at 25°C

 

1. Wash bottle containing distilled water

 

1. Bottle of isopropyl acetate

 

• Bottle of 0500 M dilute sodium hydroxide solution (note: this bottle should be carefully stopper when not in used to prevent ingress of atmospheric carbon dioxide)

 

• Bottle of isopropyl alcohol

 

• Bottle of sodium acetate

 

Procedure

 

Part 1: 1:1 [OH–]:[Ester] mixture kinetics

 

Note: Once you begin making measurements, you will want to complete Part 3 of this experiment, during the measurement phase, to use your time effectively.

 

Before beginning the exercise, check to ensure that the thermostat bath is maintaining the temperature at 25°C ± 0.01°C, and that the indicator light is going on and off regularly. It should not be necessary to adjust the controller.

 

To prepare a solution of 0.100 M isopropyl acetate, dilute the pure isopropyl acetate as follows: calculate the mass of isopropyl acetate required (note: Check the volume of your available volumetric flask to determine how much you will need to prepare that concentration) and using a disposable pipette, introduce this mass of the pure liquid into a pre-weighed beaker or small flask present in your caIDy. Introduce a small volume of deionized water into the container before aIDition of isopropyl acetate to minimize  fumes from evaporation. Be sure to tare the container and the small amount of water on the balance, before you aID the isopropyl acetate.  Transfer the contents to your available volumetric flask, and dilute to the line with deionized water.

 

Pour approximately 150 ml of each of the sodium hydroxide, ester solution, and distilled water into separate, labelled 250 ml conical flasks, and place these in the thermostat bath to come to the temperature of the bath. Record the temperature of the bath.

 

Pipette 50.0 ml of the 0.0500 M sodium hydroxide solution into a clean 150 ml conical flask, followed by 50.0 ml of distilled water. This is your “synthetic” t = 0 solution, whose conductivity is essentially the same as your reaction mixture at the beginning of the reaction. Mix by shaking. Place the flask into the thermostat bath. Rinse the probe of the conductivity meter with distilled water and place it into the t = 0 solution. Turn the control range switch at 20,000; your readings will now be in micromho (?mho) units. When the reading stops changing (indicating the flask contents have come to the bath temperature), record the conductivity reading while VERY CAREFULLY stirring the probe in the solution. The value is ko.

 

Now empty this flask, rinse it with distilled water and place the empty flask into the bath. From your previously heated solutions, pipette 50.0 ml of the dilute 0.0500 M sodium hydroxide solution into the 150 ml flask, followed by 25.0 ml of distilled water and

25.0 ml of the 0.100 M isopropyl acetate ester solution. Start the stopwatch (or note the time on your watch) when half the ester solution has been introduced (t = 0). Continue aIDing the ester solution and then mix the solution by shaking. Measure the conductivity (k) and the corresponding time (t); repeat this conductivity measurement every 2 minutes until you have readings for a total of 75 minutes. All conductivity measurements should be made as follows: switch the function switch to micromhos for about 5 seconds, read k and  t  while  carefully  stirring  with  the  probe,  then  switch  off.  At  the  end  of  the

 

experiment, re-determine the temperature of solution, which should be a little different from the initial value; the mean temperature may be taken to apply to the entire experiment. Do not discard the solution in the flask until the end of the period. Pour the solution into a clean dry flask or bottle, stopper it well, and set it aside to make one final measurement immediately before your clean up.

 

Following study of the theory of this experiment, it will be appreciated that the values of

ko and k¥ are very crucial.

Part 2: 1:2 [OH–]:[Ester] mixture kinetics

 

Rinse the 125 ml flask with distilled water, allowing it to drain well, and place it in the thermostat bath. Prepare the reaction mixture as before, but this time use 50.0 ml of dilute 0.0500 M sodium hydroxide solution (as before) and 50.0 ml of the 0.100 M ester solution. Make measurements of conductivity and time as in Part 12, for 75 minutes, as in Part 1.

 

Before cleaning up, make one final measurement of your solution you retained from Part

1. You should compare this value to the last values measured in Part 1 (as an estimate of changes near the end), and also compare that value to the “synthetic infinity” value for Part 12, in your final report.

 

Once you have completed all measurements, ensure all glassware and the conductivity probe have been thoroughly rinsed with distilled water.

 

Part 3: Synthetic “t = ?” solutions for parts 1 and 2

 

Measurement of the final “infinity” conductivity using synthetic solutions

 

To allow accurate measurement of k¥, you are to prepare synthetic solutions that will have the same composition that the actual reaction solutions would have at the end of the reaction. Infinity conductivity values (k¥) are obtained by preparing a solution of sodium acetate, isopropyl alcohol, and isopropyl acetate in water, in the same concentrations of each as the final solution would be expected to have. Note: You can determine what this should be, by considering the initial concentrations of reactants, and performing the necessary stoichiometric calculations to determine what would be present in the solutions assuming 100% reaction completion.

 

Prepare such a synthetic solution for the kinetic runs from Parts 1 and 2 – but recall, since the initial starting conditions of each part are not the same, these synthetic solutions will not be identical. It would be a good idea to calculate the concentrations of each reagent needed for both solutions before coming to the laboratory. Assume you will be preparing your solutions from solid sodium acetate, pure isopropyl alcohol, and pure isopropyl acetate.

 

Having made sure that the thermostat is maintaining the same temperature as before, place the containers of the two “synthetic infinity solutions” in the bath. Also place the empty 125 ml flak in the flask to be used as your conductivity cell. Pour the first infinity solution into the 125 ml flask and measure the conductivity as before, three or four times at intervals of a minute or two. If no apparent trend is apparent, calculate the average of the last three readings, and take this as k¥ for Part 1. Repeat with the other solution. The two values should not differ by a great deal, but will not be exactly equal.

 

Results

 

Your notebook should contain all the raw data, it is easiest to keep track of if it is recorded in chart form.

 

Treatment of Results

 

For each run, enter your ?-values into the Excel template provided. Use Equation (14) to calculate the [OH-] at each of these times, you can do this in the Excel template – a column is provided for the results. The value of ?? that should be used for the [OH-] calculation is that acquired from the “synthetic” infinity solutions. Next, fill in the column that will be graphed to determine the rate constant. For run 12, this is 1/[OH-], and for run 2 it is ln (1+[OH-]o/[OH-]).

 

The Excel template will prepare a total of four graphs, two of these are the rate constant graphs. The line for Run #1 should pass through the point (0, 1/[OH-]o) and the line for Run #2 should pass through the point (0, ln 2). If they do not, it is possible that the ester solution has undergone some hydrolysis since it was prepared. To determine the extent to which this has occurred, extrapolate each line until you reach the point at which the vertical axis value is equal to the value expected at t = 0 with Equations (4) or (8) (i.e. 1/[OH-]o or ln 2), and read the time at this point. This time value is equal to the extent to which the reaction has already occurred at the time at which you declared t = 0. If this correction is significant you should indicate this in your report. If this is the case, you can estimate the percentage of the ester that had been hydrolyzed in storage, and include this in your discussion.

 

From the two plots discussed above, it is possible to calculate the rate constant. Equations (4) and (8) are the equations of the best fit lines for Runs 1 and 2, respectively. From them, we can see that the slope of the graph for Run 1 is equal to k, and the slope of the graph for Run 2 is equal to c0k.

 

Report

 

After the usual introduction and notes on method followed by a brief discussion of the theory of the method, your report should contain a “Results” section, in which the graphs, sample calculations, and the final values for k are recorded.

 

In the “Discussion and Conclusions” section you should comment on:

 

 

• Whether the ratio of k¥ to ko in each case is reasonably close to that calculated from the tabulated equivalent conductivities given above (See Equations (12) and (13)).

 

• Whether the graphs of 1/[OH-]oand ln (1+[OH-]o/[OH-]) are sufficiently close to straight If they deviate by more than a little, what are the possible causes of the deviations?

 

• Whether the two calculated values of the rate constant agree closely enough to justify the assumption that the order of the reaction is unity in each

 

• Any sources of error present in the procedure or introduced during the performance of the

 

• A qualitative comparison of the value of the synthetic ?? value for Part 12, with the final value of ? measured for the Part 1 solution (which is measured after you have completed Part 2, before your cleanup). Using these values, include some comments describing the similarity of these two values – what might this suggest about the length of time it would take for this reaction to come to completion?

 

…and of course, any comments or observations you think relevant.

 

References

 

1. See for example M. Barrow, “Physical Chemistry”, 5thed., McGraw-Hill, New York, 1993.
2. Daniels, et al., Cit., pp. 144-149.
3. Smith and Lorenson, Am. Chem. Soc. 1939, 61, p. 117.