# Esterification Equilibrium

IB Group 4   Internal Assessment                    Chemistry               Biology               Physics

Name:                                                         Date(s):

Candidate Number:                          Session:                 Level:

Investigation Title:

Internal Assessment Marking Form

Introduction:

The equilibrium constant could be calculated by using the formula. Where A + B -> C + D

In determining the equilibrium constant for an esterification reaction,

The equilibrium constant, KC , would be;

The research question would be; determine the KC values for different mixtures.

Table 1. Uncertainties of Equipments

Table 2. Mixtures and Its Properties

Table 3. Mass of Each of the Liquids

Results:

Table 4. Raw Data – Titres against 0.9531M of NaOH

Table 5. Qualitative Data

Table 6. Processed Data - Rearrangements of Data

Table 7. Processed Data – Average Titration Value for Each Mixture

Mixture 1; average and uncertainty calculation

Average = 44.65ml

Uncertainty = 0.60ml

Calculations:

Balanced equation for the esterification reaction is;

Calculate the number of moles of ethyl ethanoate, ethanoic acid, ethanol and water present in the         original mixture.

Mixture 1, 2, 3, 4 and 5;

Since there were no ethanoic acid present in the mixture, the amount in moles of ethanoic                         acid present in the original mixture is 0.

Mixture 1 contained 5ml of ethyl ethanoate, hence the amount in moles of ethyl ethanoate                         present in the original mixture is;

Moles of ethyl ethanoate;                calculating uncertainty

0.052 ± 2.3x10-4 mol of ethyl ethanoate was present

Moles of water are 0.212±0.0009mol.

Volume of water         = 5 – (0.032x36.45)

= 3.83g

Calculating uncertainty,

Uncertainty        = 0.9531 x (33.24x10-3±0.15x10-3)

= 0.032±0.451%

= 5±0.15 – (0.032±0.451% x 36.45)

= 5±0.15 – 1.1664±0.00526

= 3.83ml±0.16ml

Mixture 2 contained 1ml of water and 4ml of ethyl ethanoate; hence the amount in moles                         of ethyl ethanoate and water present in the original mixture are;

Moles of ethyl ethanoate;                calculating uncertainty

0.041 ± 2.3x10-4 mol of ethyl ethanoate was present.

Moles of water are 0.268±0.0017mols

Mixture 3 contained 2ml of water and 3ml of ethyl ethanoate; hence the amount in moles                         of ethyl ethanoate and water present in the original mixture are;

Moles of ethyl ethanoate;                calculating uncertainty

0.031 ± ...