Determining the equilibrium constant for the hydrolysis of Ethyl Ethanoate

September 23, 2017 September 1st, 2019 Free Essays Online for College Students

Chemical equilibrium is when, in a the rate of the forward reaction is equal to the rate of the reverse reaction. When a chemical reaction has reached equilibrium, collisions are still occurring: the reaction is now happening in each direction at the same rate. This means that reactants are being formed at the same rate as products are being formed, and this is indicated by double arrows,. At equilibrium, the reaction can lie far to the right, meaning that there are more products in existence at equilibrium, or far to the left, meaning that at equilibrium there are more reactants. The concentration of the reactants and products in a reaction at equilibrium can be expressed by an equilibrium constant, symbolised Kc

Kc = [ethanoic acid][ethanol] [ethyl Ethanoate][water]

CH3COOC2H5 + H2O CH3COOH + C2H5OH

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ETHYL ETHANOATE + WATER ETHNOIC ACID + ETHANOL

CH3COOC2H5

Ester group

The functional group -COOR, R here is a ethyl group C2H5. In hydrolysis of ethyl Ethan ate, the reaction produces ethanoic acid (CH3COOH) and ethanol (C2H5OH).

Hydrolysis of esters – Esters break down into their respective organic acid and alcohol from which they are formed. Hydrolysis means to ‘split with water’. A number of organic compounds undergo hydrolysis, such as amides, esters and halogeno-alkanes. In the overall process of hydrolysis, a bond in an organic molecule is broken, and an O-H bond in a water molecule also breaks. Then, from the water molecule, an O-H group adds to one part from the organic molecule, and an H atom to the other. Hydrolysis is therefore the reaction of an organic compound with water. The hydrolysis of an ester forms a carboxylic acid and an alcohol. Few organic compounds react readily with water.

When water is added to an ester, a carboxylic acid and alcohol are formed. Hydrolysis of ester with an alkaline solution like sodium hydroxide is known as saponification (soap making). This reaction is used in the preparation of soaps. Physical Properties of esters include a strong fruity smell, may be present as colourless volatile liquids or colourless waxy solids. They are soluble in water and in organic solvents and are neutral to litmus tests.

Esters are used as artificial perfumes or scents as they emit a sweet smell. Also they can be used in making artificial food flavours that are added in many edible items like ice creams, soft drinks, sweets, etc. Esters are also used as industrial solvents for making cellulose, fats, paints and varnishes and also as solvents in pharmaceutical industries. Some esters are used as softeners in plastic industries and moulding industries. Many esters have distinctive odours, which has led to their widespread use as artificial flavourings and fragrances. For example methyl butyrate smells of pineapple or apple, isopentyl acetate smells of pear or banana (it is used as the flavouring in the manufacturing of old fashioned Pear Drops) and methyl benzoate smells of fruity-ylang ylang. Ethyl formate smells of rum. Ester are also used in the biosynthesis of fats and fatty acids, and in the making of soaps.

General formula of an ester.

Ester + Water Acid + Alcohol (reversible)

EXPERIMENT DETAILS

For full experimental details and for the table of results please refer to handout and attached sheet.

EVALUATION

Ester + Water Acid + Alcohol

Initial a b – –

Eqm a-x b-x x x

Kc = x2 ‘

(a-x) (b-x)

Calculations of a and b

Density of ester = Mass .

Volume

0.9 = M

10

Therefore the mass of ester is 9g

To calculate the number of moles in the ester = Mass

RMM

= : 9 .

88.1

? a = 0.1022M

Mass of water is 25g

To calculate the number of moles in water = Mass

RMM

= 25

18

?b = 0.13889M

Ma Va = Mb Vb

CH3COOC2H5 + H2O CH3COOH + C2H5OH

a b x x

(a-x) (b-x)

?a=0.1022M b= 1.389M = x + x

To calculate x CH3COOH in

CH3COOH + NaOH = CH3COONa + H2O

Number of moles in NaOH = 1.06M

Number of moles in NaOH in titration = 1.06M x 15.15

1000

No. of moles CH3COOH in titration (x5) = 1.06 x 15.15 x 5

1000

? x = 0.0802

Therefore we were able to calculate a-x and b-x to get the equilibrium constant Kc

a-x b-x

0.1022-0.0795 = 0.023 1.3889-0.0795 = 1.309

To find Kc = [0.023][1.309] [0.102][1.389]

= 0.213

HEALTH AND SAFETY

General laboratory health and safety precautions were considered. These included wearing of an over coat to protect clothes from spillages, placing a mate on work surfaces to protect it from heat as well as wearing of goggles to protect ones eyes from irritation. Also when handling hot test tubes, forceps and gloves need to be used to prevent one from burning their hands. Hands also need to washed after every experiment to wash off any excess chemicals.

CONCLUSION

I was able to perform the hydrolysis of Ethyl Ethanoate as well as to determine the equilibrium constant. The standard value is 0.23 and due to my results my experiment was very successful. My average Kc value was 0.22 and this was even better. The experiment was very successful.

ANALYSIS AND EVALUATION

For the calculations of the titrations, when determining the end point I had to be very careful not to pass the end point. Also we had to leave the reagents for a week so that they could reach equilibrium. My experiment was highly successful with all the results proving the success and accuracy of the experiment.

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