Ethyl Acetate – NaOH Reaction Kinetics Experiment
Martin Novick Group 14, Chemical Engineering Laboratory Submitted to Prof. David B. Henthorn September 25, 2012 Summary The goal of this project was to determine the pre-exponential factor, k o , the activation energy, E, and the reaction rate constants, k, of the saponification process of ethyl acetate using sodium hydroxide (NaOH) at 5 temperature between 15 and 25 degrees Celsius. Two trails were performed at temperatures 16, 18, 20, 22, and 24 degrees Celsius. The main equipment of the project were the jacketed beaker batch reactor and the LabPro conductivity probe. The solution’s conductivity throughout the reaction was collected and plotted in a linearized plot against time to retrieve ������ value for each trial. The rate law was assumed to be ������������������������ = −������������ ������������ ������������������ , where ������������������ and ������������������ are the concentrations of sodium hydroxide and ethyl acetate respectively. The ln k) values were plotted against the inverse temperatures to ( linearize the Arrhenius equation. The k o value and E value from the linearized Arrhenius plot were found to be 15 ± 3M −1 s −1 and −36402 ± 8191j × mol−1 respectively. The E value being negative suggests the reaction is exothermic. The large standard errors of the ������������ and ������ values were probably caused by the low number of data points collected or the assumed rate law was wrong.
Introduction The objective of this project was to determine the pre-exponential factor, k o , the activation energy, E, and the reaction rate constants, k, of the saponification process of ethyl acetate using sodium hydroxide (NaOH). Saponification is a chemical process heavily used in industry, especially in soap production. Knowing the effects of temperature on the reaction rate allows better control over the reaction process and find the optimizing point of production. The right temperature maximizes production and minimizes reactants and heating or cooling energy, finding the maximum possible profit.
The saponification of ethyl acetate with sodium hydroxide (NaOH) is an equi-molar reaction given as NaOH + CH3 COOCH2 CH3 → CH3 COONa + CH3 CH2 OH.  The rate was assumed to be second order overall, but first order relative to either reactant, with the disappearance rate of sodium hydroxide given as: dCOH = −kCOH Cac , dt (1)
where COH is the NaOH concentration, Cac is the concentration of ethyl acetate, and k is the rate constant. COH is represented with x, and Cac is assumed to be in excess with a starting concentration of a. The concentration of ethyl acetate throughout the reaction was given as Cac = a − xo + x, where xo is the initial starting concentration of NaOH. Substituting the expressions for COH and Cac into Equation 1 results in: dx = −kx(a − xo + x). dt Equation 2 is then separated and integrated shown as the following: ∫ t dx = −k ∫ dt. xo x(a − xo + x) 0 x
The result of equation 3 would become: ln ( xo (a − xo + x) ) ax = kt. a − xo
Since a 20% excess ethyl acetate solution was used, the initial ethyl acetate concentration was 1.25 times of the initial starting concentration of NaOH, so a = 1.25xo . After the substitution of a = 1.25xo , Equation 4 was simplified to:
0.2x ln ( x o + 0.8) = kt. 0.25xo The relationship between
and conductivity of the solution was given as: ������������ ������������ − ������∞ = , ������ ������ − ������∞ (6)
where ������������ is the initial conductivity of NaOH before adding ethyl acetate, ������∞ is the conductivity after the reaction has reached completion, and ������ is the conductivity at any point of time during the reaction process. Substituting Equation 6 into Equation 5 gives the following: ������������ − ������ ln (0.2 ������ − ������ ∞ + 0.8) ∞ = kt. 0.25xo
Since ������������ , ������������ and ������∞ are constants, the only...
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