II.2- Saponification of ethyl ethanoate, a second-order kinetic?Data: R = 8.314 J mol·' K-'The kinetic study of the saponification reaction of ethyl ethanoate by caustic soda (sodium hydroxideNaOH):CH3COOCH2CH3 + OH → CH3CO2¯ + CH3CH2OHwas achieved at 298 K for initial concentrations a in ester and hydroxide that are identical.Experimental results, for different values of a, are reported in the following table:a (mol L-')0.17.5 × 10-25 x 10-24 x 10-2t1/2 (s)6.66 x1058.89 ×10$1.33 ×1061.66 ×106Recall the differential expression of the rate law for a second-order kinetic, with a partial orderof 1 to each of the individual reactants. When initial concentration of each of the reactant isthe same, deduce the expression of the integrated form of this equation. The rate constant willbe denoted k and the concentration of each of the compounds produced (products) will becalled x.

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II.2- Saponification of ethyl ethanoate, a second-order kinetic? Data: R = 8.314 J mol-l K- The kinetic study of the saponification reaction of ethyl ethanoate by caustic soda (sodium hydroxide NaOH): CH3COOCH2CH3 + OH → CH;CO2 + CH3CH2OH was achieved at 298 K for initial concentrations a in ester and hydroxide that are identical. Experimental results, for different values of a, are reported in the following table: a (mol L-') 0.1 7.5 x 10-2 5 x 10-2 4 x 10-2 t1/2 (s) 6.66 x105 8.89 x105 1.33 x106 1.66 x106 Recall the differential expression of the rate law for a second-order kinetic, with a partial order of 1 to each of the individual reactants. When initial concentration of each of the reactant is the same, deduce the expression of the integrated form of this equation. The rate constant will be denoted k and the concentration of each of the compounds produced (products) will be called x.

II.2- Saponification of ethyl ethanoate, a second-order kinetic? Data: R = 8.314 J mol-l K- The kinetic study of the saponification reaction of ethyl ethanoate by caustic soda (sodium hydroxide NaOH): CH3COOCH2CH3 + OH → CH;CO2 + CH3CH2OH was achieved at 298 K for initial concentrations a in ester and hydroxide that are identical. Experimental results, for different values of a, are reported in the following table: a (mol L-') 0.1 7.5 x 10-2 5 x 10-2 4 x 10-2 t1/2 (s) 6.66 x105 8.89 x105 1.33 x106 1.66 x106 Recall the differential expression of the rate law for a second-order kinetic, with a partial order of 1 to each of the individual reactants. When initial concentration of each of the reactant is the same, deduce the expression of the integrated form of this equation. The rate constant will be denoted k and the concentration of each of the compounds produced (products) will be called x.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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