Determining an equilibrium constant. The aim of this experiment is to calculate the equilibrium constant (kc) of the reaction: CH3CO2C2H5(l) + H2O(l) C2H5OH(l) + CH3CO2H(l)

AS and A Level Science

Date: 4/3/2011

Exp. No.: 14

Title: Determining an equilibrium constant

Aim: The aim of this experiment is to calculate the equilibrium constant (kc) of the reaction:

CH3CO2C2H5(l) + H2O(l) C2H5OH(l) + CH3CO2H(l)

Ethyl ethanoate Ethanol Ethanoic acid

Introduction: The above reaction is very slow, so it takes a lot of time to attain equilibrium.

However, by using dilute hydrochloric acid as a catalyst, the reaction rate will become much fast and equilibrium can be attained in a few days.

This experiment is separated into 2 parts.

The first part of the experiment is to prepare mixture of different proportions of the reactants in sealed containers using dilute hydrochloric acid as a catalyst. These mixtures will be allowed to react for a week at room temperature before doing the second part to ensure that the reaction mixtures have attained equilibrium.

The second part of the experiment, the mixtures will be titrated one by one using standard sodium hydroxide. Sodium hydroxide reacts with acid in the mixture including ethanoic acid(which is a product of the reaction) and also hydrochloric acid(catalyst of the reaction).

Finally, using the amounts of ethanoic acid produced, we can calculate the equilibrium concentration of all four components and they can be used to determine the equilibrium constant of the reaction.

Apparatus and chemicals:

For part A:

5 boiling tubes with stoppers, electronic balance, air displacement pipette, about 30 cm3 of 2M HCl, two 10 cm3 measuring cylinder, about 15 cm3 of ethyl ethanoate, deionized water, plastic films(wrap)

For part B:

5 conical flasks, deionized water, a few cm3 of phenolphthalein, burette with stand, about 150 cm3 of approximately 1M standard sodium hydroxide solution

Procedure:

Part A

Step

5 boiling tubes were labeled with 1A, 1B, 2, 3 and 4 respectively and their corresponding stoppers were also labeled as the same as their boiling tubes.
Each stopper was wrapped with plastic films.
Each tube with its stopper was weighed using an electronic balance.
5.0 cm3 of 2M hydrochloric acid were added to each tube.
The tubes were weighed again using an electronic balance with their stoppers after adding the HCl.
Different volumes of ethyl ethanoate were added to each tube using a measuring cylinder according to the table below.

Tubes 2, 3 and 4 were then weighed with their stoppers after adding ethyl ethanoate.
Different volumes of deionized water were added to each tube using another measuring cylinder according to the table below.

Tubes 3 and 4 were weighed with their stoppers after adding deionized water.
All measured data were collected and recorded.
All 5 tubes were shook gently and put aside for a week.
After 3 days, the tubes were shook gently again.

Part B (After 1 week)

A burette was rinsed and filled with standard sodium hydroxide ...

This is a preview of the whole essay

Tubes 2, 3 and 4 were then weighed with their stoppers after adding ethyl ethanoate.
Different volumes of deionized water were added to each tube using another measuring cylinder according to the table below.

Tubes 3 and 4 were weighed with their stoppers after adding deionized water.
All measured data were collected and recorded.
All 5 tubes were shook gently and put aside for a week.
After 3 days, the tubes were shook gently again.

Part B (After 1 week)

A burette was rinsed and filled with standard sodium hydroxide solution.
The whole content in tube 1A was poured into a conical flask with a few washings.
2-3 drops of phenolphthalein were added to the conical flask and the solution was titrated against the standard sodium hydroxide solution.
The titre of titration was recorded.
Step 14-17 were repeated 4 times with tube 1B, 2, 3 and 4.

Observations:

Part A:

The liquid mixtures in the boiling tubes 2, 3 and 4 were in 2 immiscible layers and the mixtures were colourless.
Ethyl ethanoate and hydrochloric acid were colourless.
Ethyl ethanoate had a “glue” smell.

Part B:

The colour of the end-point of the titrations was from colourless to pink.
The liquid mixtures were miscible and remained colourless after a week.

Data:

Part A:

Part B:

Molarity of the standard sodium hydroxide = 1.0086 mol dm-3

Calculations:

Average titre for tube 1A and 1B = = 9.73 cm3

Since same amount of HCl catalyst were added to all 5 tubes, so the amount of HCl in tube 1A and 1B were the same of that in tubes 2, 3 and 4.

The amount of HCl present in each tube = × 1.0086 = 9.81×10-3

For tube 2

Total amount of acid at eqm (hydrochloric acid and ethanol acid) in tube 2 (Calculated using the titre)

= = 3.74×10-2 mol

Eqm amount of ethanoic acid in the mixture = 3.74×10-2 −9.81×10-3 = 2.76×10-2 mol

From the equation of this reaction, we know that the mole ratio between ethanol and ethanoic acid = 1:1, so the eqm amount of ethanol produced = eqm amount of ethanoic acid produced

= 2.76×10-2 mol

Number of moles of ethyl ethanoate added to tube 2 =

= = 5.04×10-2 mol

Since from the equation, the mole ratio between ethyl ethanoate and ethanoic acid = 1:1, so the amount of ethyl ethanoate used = amount of ethanoic acid produced

Eqm amount of ethyl ethanoate = 5.04×10-2 – 2.76×10-2 = 2.28×10-2 mol

Mass of pure HCl in HCl(aq) = Amount of HCl × Molar mass of HCl = 9.81×10-3×36.5 = 0.358g

Mass of water in HCl(aq) = 5.0 – mass of pure HCl = 5.0 – 0.358 = 4.64g

Initial amount of water =

= = 0.254 mol

Since from the equation, the mole ratio between water and ethanoic acid = 1:1, so the amount of water used = amount of ethanoic acid produced

= Initial amount of water – Eqm amount of ethanoic acid = 0.254 – 2.76×10-2 = 0.226 mol

By equilibrium law, the equilibrium constant (Kc) of the reaction

= = 0.148

By similar calculations, we can calculate the Kc of the reaction in tube 3 and 4 and the calculated results are shown in the table below.

The three calculated Kc values are 0.148, 0.129 and 0.113.

Discussion:

Why do the exact molarity of HCl(aq) used need not to be found out?

This is because the amount of HCl(aq) can be found by titrating tube 1A and 1B using sodium

hydroxide. The exact amount of HCl added can be found out directly using this method, so

there is no need to know the exact molarity of HCl.

This is also the reason why we need tube 1A and 1B which only contain HCl(aq) but not other

reactants – to find out the amount of HCl added.

Why were there 2 immiscible layers just after mixing but became miscible after a week?

There were 2 immiscible layers just after mixing because water and ethyl ethanoate are

immiscible, so they existed as 2 layers.

However, after a week, ethyl ethanoate and water reacted and formed ethanol, which can

dissolve ethyl ethanoate and also dissolve in water. So, the mixture became miscible after

enough amount of ethanol was produced.

Why do we need to wrap plastic film around the stoppers?

This is because the stoppers are made up of polymers, but the reaction consist of organic solvent (ethanol and ethyl ethanoate) which may dissolve part of the stopper and stick the stopper to the boiling tube. Wrapping plasic film can prevent this from happening.

Sources of error in this experiment

We assumed that the amount of HCl(aq) added to each tube were the same, but this may not be true as there may be bubbles in the pipette during the transfer of HCl. This may cause error as we used the average amount of HCl in tube 1A and 1B as the amount of HCl in tubes 2, 3 and 4.
The burette has a maximum error of 0.05 cm3 and the electronic balance has a maximum error of 0.0005g and there are also air movements during weighing which affect the weighing results.
Ethyl ethanoate and ethanol are volatile as they have low boiling points. During the titration, there may have some of them vaporized and escaped from the conical flask, which makes the experiment result inaccurate.

To improve the accuracy of the result, we should perform the titration as fast as possible to minimize the loss of volatile liquids during the titration.

What is equilibrium constant (Kc) and equilibrium equation?

The larger the value for the equilibrium constant the more the reaction goes to completion. Irreversible reactions can be thought to have an infinite equilibrium constant so there are no reactants left.

The number values for equilibrium constants are tied to the nature of reactants and products in a reaction. The number values for "K" are gotten from experiments measuring equilibrium concentrations. The number value tells the equilibrium ratio of products to reactants. In an equilibrium mixture both reactants and products coexist.

The value for K is large when products dominate the mixture.

The value for K is small when the reactants dominate the mixture.

The equilibrium constant may be expressed in the form

where [C] represents the molar concentration of C at equilibrium.

For a given reaction, the concentrations at equilibrium would have to be determined experimentally.

The value of Kc is ONLY affected by temperature.

For gaseous reactants it is more convenient to express the equilibrium condition in terms of the partial pressures of the reactants and products. For this case the equilibrium constant is defined by

where P denotes the partial pressure, usually in atmospheres.

The two forms of the equilibrium constant are related by

where ∆n is the sum of the coefficients of the gaseous products in the chemical equation minus the sum of the coefficients of the gaseous reactants.

What is chemical equilibrium?

Chemical equilibrium means a reaction has reached a state that the forward reaction rate and the backward reaction rate are the same and there is no net gain or loss of a reactant or product.

Chemical equilibrium can be changed or shifted by changing the temperature, pressure, amount of reactants/products and volume of the system.

Reference: