CHEMICAL ENGINEERING LABORATORY CHEG 4137W/4139W

Reaction Kinetics

Saponification of Isopropyl Acetate with Sodium Hydroxide

Objective: The purpose of this experiment is to examine and determine the reaction kinetics of a simple homogenous liquid-phase system and to model this reaction in a continuously stirred-tank reactor (CSTR). Simple batch (flask) reactions will be run initially to determine initial rate data from concentration versus time profiles before using the continuous reactor. Major Topics Covered: Chemical reaction kinetics, reaction order, Arrhenius-style rate law. Theory: The reaction:

Isopropyl Acetate + Sodium Hydroxide → Sodium Acetate + Isopropyl Alcohol is an example of a saponification reaction (the reverse reaction would be esterification). This reaction may be either reversible or irreversible. For the irreversible case, the rate equation for a batch reactor may be written:

rA = -d[A]/dt = k[A][B] where k is the rate constant and [A] and [B] are the concentrations of the reactants in the appropriate units. If strictly equimolar concentrations of reactants are used, the rate equation can be simplified to a general nth-order reaction. rA = kn [A]n where n ≅ 2. With good experimental data, this model allows the determination of the reaction order for an irreversible reaction. The effect of temperature on the rate constant can be compared with that predicted by the Arrhenius expression: k = Ae(-E/RT), where A is the frequency factor, E is the activation energy of the reaction, R is the gas constant, and T is the absolute temperature. Rate data at various temperatures can be used to determine the frequency factor and activation energy of the reaction. Once basic rate data is obtained, it can be used to size or design reactors, or determine the maximum conversion for a reactor of a given size. Design equations for different types of reactors are provided in the references. Safety Precautions:

1. This lab uses both acids and bases. Use caution when handling these reagents, as well as when sampling from flasks or the reactor. Appropriate personal protective equipment should be used at all times.

2. Always make sure that the pump reservoirs are filled prior to turning on the pumps.

3. The ester solution should be mixed in a well-ventilated area.

Available Variables: Batch Experiment: Temperature. CSTR Experiment: Temperature, reactor residence time. Procedure: See Kinetics Operational Checklist Analysis: Your analysis must include:

1. A graphical determination of the reaction order as determined by the batch reaction experiment.

2. Determination of the activation energy and frequency factor for the reaction, and a comparison with appropriate literature values.

3. A predictive model for CSTR results based on the batch results and intended CSTR

operating conditions.

4. A comparison of CSTR experimental results with the predictive model, and a discussion surrounding the agreement/disagreement between theory and experiment.

Report: Describe the design of your experiments and the results obtained, including an error analysis. Provide thoughtful and quantitative discussion of results, explain trends using physical principles and relate your experimental observations to predicted results (e.g. discrepancies between predicted CSTR results and experimental data). Express any discrepancies between observed and predicted results in terms of quantified experimental uncertainties or limitations of the correlations or computational software used. Pro Tips:

1. Because of variations in density of the solutions, generate your calibration curves using the actual reagents, not pure water. The setup allows you to continuously recycle the reagents during the calibration phase.

2. For ease of taking data, it is recommended to start at lower temperatures and work incrementally upward.

3. Flush both pumps with pure water at the end of the experiment; this will prolong the life of the pump internals.

4. Carefully monitor the level in the reactor, and adjust the inlet or outlet flow rates to keep the level steady. Only take data when the system is at steady state.

References: 1. Fogler, H. S., Elements of Chemical Reaction Engineering, Prentice-Hall, Inc.,

Englewood Cliffs, NJ, 2nd Ed., 1992. 2. C. H. Bamford and C. F. H. Tipper, eds., Comprehensive Chemical Kinetics, vol. 10,

3. "Ester Formation and Hydrolysis," Elsevier, New York, NY, 1972. International Critical Tables, 7,129 (1930).

APPENDIX AOn Curved Kinetic Data and

the Straight Scoop on Straight Lines

Why straight lines?

Many experimental results are analyzed in a fashion that produces a straight-line plot. Forexample, the Arrhenius theory appears as

k Ae E RT= − / (1)

and yet is plotted as

ln ln /k A E RT= − (2)

to get, obviously, a straight line.

Why is this? Can one simply analyze Eq. 1?

The historical reason for the straight lines is simple: it was virtually impossible for theexperimentalist to do anything else until about the 1960’s when digital computers becameavailable. Even a straight-line least squares (LS) analysis was tedious. Thus engineers andscientists sought straight lines as the final result in their analyses, and these analyses became a partof the literature. You still find them in textbooks and probably always will. “Everyone”understands that an Arrhenius plot is ln k vs. 1/T, and not k vs. T. However, you can analyze Eq.1 directly and forget about taking logs. The key is nonlinear least squares (NLLS), available onmost spread sheets, numerical packages and plotting programs (e.g., Polymath, Excel).

What’s good about straight lines?

Aside from tradition, straight lines allow the eye to see slight deviations from the expectedbehavior, because the eye is very sensitive to straight vs. curved.

What’s bad about straight lines?

In the case of the Arrhenius plot, probably nothing. However, it is important to realize that one ismaking several assumptions in any least-squares analysis; some are:

1. Every data point is independent; that is, there is no more connection between any twopoints than any other two points.

2. There is no error in the independent variable (x axis).3. The errors in the data are normally distributed and have equal variances.

It’s the latter that can cause trouble with transforms (e.g., ln k) vs. the real data (k), but thetransform can also improve the agreement with item 3. In the case of the Arrhenius plot, it often

helps, because the errors in rate data tend to be relative rather than absolute. If you have enoughdata, you can tell from the plot; the "scatter" should be about the same everywhere.

The big problem with the urge to get straight lines as the bottom line of an analysis is one ofhidden parameters, so

BEWARE OF HIDDEN PARAMETERS!

Watch the following:

xC CCA A

A=

−0

0 (3)

Eq. 3 looks like a perfectly correct conversion of concentration CA to conversion x using theinitial concentration CA0. After all, we know that the latter is, say, 0.05 mole/L. Or do we? Infact, in the same sense as we have error in each CA , we do not really know what CA0 is; it’s ahidden parameter and is not a constant like π. If the CA0 assumed is off only slightly, high errorsin x will result, especially at low x.

So what is one to do? The solution is simple and accessible to all. Cast the theory you are testingin terms of actual observations and unknown parameters. For example, second-order kinetics forthe disappearance of species A is classically analyzed using the equation:

1 10C C

ktA A− = (4)

assuming stoichiometric, irreversible reaction with rate constant k. No problem here--it comesstraight from integration of the mass-action theory. But wait! Are there any hidden parameters inour observations of the concentration of A, i.e., CA? Good question. With direct titration ofproduct resulting from CA , don’t we get CA free from worry? Unfortunately, there are hiddenparameters, and a careful look at the mass balance with no assumptions about the concentrationof anything gives the nonlinear equation

s

t

VNkt

VV

VV+

−

−=

∞

∞

0

11 (5)

where Vt (the observation) is the titer in mL at time t, V∞ is the titer at t = ∞ ( a parameter), V0 isthe titer needed at t = 0 (another parameter, and not an observation or a constant), N is thenormality of the titration solution, and Vs is the volume of the sample. Of course, k is the rateconstant that we want to find. Unfortunately, the rate constant is tied up with N and Vs.; there islittle we can do about separating them and they therefore must be accurately known. If N and Vsare in error, the errors will appear in k and we will never know it. But note that all the worry

about the zero-time titer is gone. That’s the trick: expose the hidden parameters, and let the datatell you what their values are. The penalty is more parameters and no nice straight lines (usually).

An example is shown in Figure 1. This is real data from a previous year. The points labeled“conventional analysis” are plotted according to a measured zero-time titer of 0.5 mL andassumed to be free of error. The points labeled NLLS are simply the observations straight fromthe burette at times other than zero. The line is Eq. 5.

Note that the changes are accurately described, whereas the conventional analysis is not straight,mainly because of the too-high zero-time titer. The rate constant from the NLLS is about thesame as the steep slope drawn on the conventional analysis. However, unlike the practice ofusing just the first few data from the conventional analysis to get the slope, the NLLS uses all thepoints and therefore has far greater precision. The truly dedicated will note that there are otherpossibilities, including reversible reaction and lack of stoichiometry, to explain curved data on theconventional plot. It may be possible with very accurate and plentiful data to distinguish these,but probably not with any great confidence.

The underlying message is that all experimental observations will have error. If you treat anobservation as a constant, then you are asking for systematic error in the end result.

?

Conventional analysis

NLLS on Vt (t)

?

0 500 1000 1500 2000 2500 3000

Time, s

0

10

20

30

40

50

60

70

80

90

1/(C

A, m

ol/

L) a

nd

10.

( V t

, mL)

Figure 1. Saponification of isopropyl acetate using 0.05 M NaOH.

There are exceptions to the rule that all named variable are either variables or parameters. Anexample is the Arrhenius expression given in the Kinetics handout:

ln k = ln k0 – E(1/T − 1/T0)/R (6)

In this version, T0 is an arbitrary temperature and k0 is the rate constant at T0. As the originalArrhenius equation (Equation 1) contains only two parameters, A and E, one of the extra symbolsintroduced in Equation 6 is arbitrary and can be set at any value. It can thus be regarded as aconstant.