## Tuesday, April 22, 2014

### Kinetic Studies on Saponification of Ethyl Acetate by the Conductance Method - Physical Chemistry

Propose

1.     To know the characteristics of a second-order reaction by a graphical method.
2.     To determine the effect of temperature on the reaction rate of ethyl acetate with dilute sodium hydroxide.
3.     To be familiarized with the operation of a digital conductometer.

Principles

The saponification of ethyl acetate with sodium hydroxide is a second-order, irreversible reaction which can be represented by the following equation:

If the initial concentration of the reactants are equal (both a) and that converted concentration is x at reaction time t, then the concentration of ethyl acetate and NaOH is C0-x. Supposing that the reverse reaction can be ignored, the reactant and product concentrations at different time are

 t = 0 C0 C0 0 0 t = t C0 - x C0 - x x X t -> ∞ -> 0 -> 0 -> C0 -> C0

The rate equation for the above second-order reaction can be expressed as

Where k2 is second-order rate constant,  mol-1 L min-1
The equation can be integrated to give:

From the concentrations of the reactant and product in the reaction vessel and time of reaction, rate constant k2 can be calculated.

In this reaction, OH- ion is the most highly conductive species therefore the conductivites of the ethyl acetate and ethyl alcohol may be ignored. Since the reaction solution is dilute aqueous, it can be assumed that sodium acetate is completely ionized. The concentration of Na+ remains invariable before and after reaction. As the reaction time increases, the number of OH- ions decreases continuously, and the conductance of the system declines continuously.

 t = 0 к0 = A1C0 t = t кt = A1(C0 - x) + A2x t = ∞ к∞ = A2a

Then

Where к0 and кt are the conductivity at beginning and time t, respectly, к is the conductivity at the end of reaction, and A is the proportionality constant. Substituting the equation into below

or

A plot of 0 - кt)/(кt – к) against t sjould yield a straight line with a slpoe of k2C0 AND k2 can be calculated from the slope. The rate of reaction as characterized by its rate constant k is strongly temperature dependent. This is generally express as the Arrhenius equation:

Where  Ea : activation energy, kJ/mol
T : the reaction temperature, K
R : gas constant, J/(mol K)
Therefore,
Then

From the equation Ea can be obtained based on the determination of kT2 and kT1.

Chemicals

1.     NaOH(aq) (0.01852 M, standardized )
2.     NaOAc(aq) (0.00926 M)

3.     Ethyl acetate (A.R.)
4.     Distilled water

Apparatus

1.     Digital conductometer and computer

2.     Platinum electrode

3.     Glass reactor

5.     Thermostatic water bath
6.     50-mL test tube
7.     Pipette
8.     Washing ear ball

Procedures

Measurement of к0 ~ кt
Add 20-mL NaOH(aq) in the tube 1 and 20-mL ethyl acetate in tube 2. Set up the platinum electrode on the reactor and put the reactor in the thermostatic water bath for at least 10 minutes until the temperature is constant. Turn on the computer and the recorder. Use a washing ear ball to force the solution in tube 1 to remove to tube 2 and quickly mix for 3~5 times. Let the recorder works for about 20 minutes. Raise the temperature high for 3 and repeat the same steps above until the temperature is above 30.

Measurement of к
Also suck a 50-mL test tube in the thermostatic water bath and hold with 0.00926M NaOAc(aq). After each temperature of measurement of к0 ~ кt , wash the platinum electrode with distilled water and then suck the electrode in to the sodium acetate solution. Record the reading on the conductometer.

Experimental Record

Raw Data

NaOH(aq): 0.01852M ; Room Temperature.20.0 ; EtOAc(aq): 0.181 mL
Temperature 1 20.08 ; Conductivity of 0.00926M NaOAc(aq): 420 uS/cm

Temperature 2 22.90; Conductivity of 0.00926M NaOAc(aq): 448 uS/cm

Temperature 3 26.30 ; Conductivity of 0.00926M NaOAc(aq): 478 uS/cm

Analysis

Take the points after the 100th point

According to the slope of these three diagrams, the k2 can be easily figured out as follow:

 Temperature(℃) slope k2 20.08 -5.8324 x10-4 6.2984 x10-2 22.90 -6.8120 x10-4 7.3564 x10-2 26.30 -8.2301 x10-4 8.8878 x10-2

Draw a ln(k2)-(1/T) figure and do a linear fitting with these points. The slope of the diagram could be expressed as: slope = -Ea /R

For the slope = -4863.73, R = 8.314, and then the is Ea 40.52 kJ/mol
To compare with the literature value 39.9 kJ/mol , it is very close.

 Ea (experimental) 40.52 kJ/mol Ea (literature) 39.9  kJ/mol Percentage error 1.6%

References

  傅獻彩, 沈文霞, 姚天揚. 物理化學, 上冊歐4 . 北京:高等教育出版社, 1990:144.
  清華大學化學系物理化學實驗編寫組. 物理化學實驗. 北京：清華大學出版社, 1991.
  Robert C. Wcast Handbook of Chemistry and Physics. Physics. 58th ed. Ohio: CRC Press, 1977.
  朱文濤. 物理化學. 北京：清華大學出版社，1995.
  W. T. Gooch, J. Am. Chem. Soc., 1927, 49 (9), pp 2257–2257
  ADELIO M. MENDES, LUIS M. MADEIRA, FERNAo D. MAGALHAES, J 0513 M. SOUSA. Universidade do Porto - Porto, Portugal

#### 1 comment:

1. Could you please be so kind as to explain to me how you got the second set of graphs. Thank you!