Propose
1.
To know the characteristics of
a secondorder 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 secondorder, 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 C_{0}x. Supposing that the
reverse reaction can be ignored, the reactant and product concentrations at
different time are
The rate equation for the above
secondorder reaction can be expressed as
Where k_{2}
is secondorder 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 k_{2} 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

к_{0} = A_{1}C_{0}

t = t

к_{t} = A_{1}(C_{0 }
x) + A_{2}x

t = ∞

к_{∞} = A_{2}a

Then
Where к_{0}_{ }and к_{t} 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 (к_{0}  к_{t})/(к_{t}
– к_{∞}) against t sjould yield a straight line with a slpoe of k_{2}C_{0}
AND k_{2} 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 E_{a} : activation energy, kJ/mol
T : the reaction temperature, K
R : gas constant, J/(mol K)
Therefore,
From
the equation E_{a }can be obtained based on the determination of k_{T2
}and k_{T1}.
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
4.
Volumetric flask
5.
Thermostatic water bath
6.
50mL test tube
7.
Pipette
8.
Washing ear ball
Procedures
Measurement of к_{0} ~ к_{t}
Add
20mL NaOH_{(aq)} in the tube 1
and 20mL 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 50mL test tube in the thermostatic water bath and hold with 0.00926M
NaOAc_{(aq)}. After each temperature of measurement of к_{0} ~ к_{t }, 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 100^{th} point：
According to the slope
of these three diagrams, the k_{2 }can be easily figured out as follow:
Temperature(℃)

slope

k_{2}

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(k_{2})(1/T)
figure and do a linear fitting with these points. The slope of the diagram
could be expressed as: slope = E_{a }/R
For
the slope = 4863.73, R = 8.314, and then the is E_{a}： 40.52 kJ/mol
To compare with the literature value^{[6]}： 39.9
kJ/mol , it is very close.
E_{a }(experimental)

40.52 kJ/mol

E_{a }(literature)

39.9 kJ/mol

Percentage error

1.6%

References
[1] 傅獻彩,
沈文霞, 姚天揚. 物理化學, 上冊歐4 版. 北京:高等教育出版社, 1990:144.
[2] 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.
[3] Robert C. Wcast Handbook of Chemistry and
Physics. Physics. 58^{th} ed. Ohio: CRC Press, 1977.
[4] 朱文濤.
物理化學. 北京：清華大學出版社，1995.
[5] W. T. Gooch, J. Am. Chem. Soc.,
1927, 49 (9), pp 2257–2257
[6]
ADELIO M. MENDES, LUIS M. MADEIRA, FERNAo D. MAGALHAES, J 0513 M. SOUSA.
Universidade do Porto  Porto, Portugal
Could you please be so kind as to explain to me how you got the second set of graphs. Thank you!
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