The Equilibrium Constant of an Ester Hydrolysis Reaction
A limited time offer! Get a custom sample essay written according to your requirements urgent 3h delivery guaranteedOrder Now
Abstract: This experiment was conducted in order to discover the Kc, equilibrium constant, of a hydrolysis reaction of an unknown ester #2, unknown acid, and alcohol #2 products. The first week consisted of creating the reaction mixtures in bottles, next was preparing a NaOH solution while neutralizing with KHP. The final week consisted of titrating the bottles with the NaOH solution prepared previously. After calculations the Kc average was equivalent to 0.1031
Introduction: Chemical equilibrium in a reaction occurs when the rate of the forward reaction is equivalent to the rate of the reaction going backwards. Once a chemical reaction has attained equilibrium, the collisions are still continuing, it is just that the reaction is occurring the same in both directions at the same rate. Both the products and reactants are being formed in an equal amount of time. The concentration of both reactants and products are expressed by the equilibrium constant Kc.
Procedure: (Reference Lab Manual for Procedures)
The following table displays the reaction mixtures within the bottles Table 1.
Density: 0.8878 g/mL Molar Mass: 102.13 g/mol
Density: 0.7997 g/mL Molar Mass: 60.10 g/mol
Table 2 shows the NaOH solution standardized with the addition of KHP Table 2.
Table 3 is the titration of the reaction mixtures with the NaOH solution prepared. Table 3.
Average NaOH after 3 trials was 0.6578 M NaOH
Moles of HCL
Molarity of HCL (average between bottle 1 & 1A)
Grams of HCL
Grams of Water
4.319g HCL = 4.7181 g H20
Moles of Water in HCL Solution
Moles of Water added
Moles of ester initially present moles of ester
Moles of alcohol initially present
Moles of acid present
Moles of carboxylic acid present, moles of alcohol formed, moles of ester disappeared, moles of water disappeared 0.0350 – 0.01185 = 0.0232
Average Kc = 0.1031
The results of the experiment, using Esther and Alcohol unknown number 2, gave a Kc value average equivalent to 0.1031.