Kinetics of Hydrolysis of ethyl acetate | Chemical Kinetics | Physical Chemistry

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Ethyl acetate is a colorless liquid with a sweet smell. Ethyl acetate is used as a solvent and diluent.
CH_3COOC_2H_5 \ + \ H_2O \ \overset { H_2SO_4 }{ \longrightarrow } \ CH_3COOH \ + \ C_2H_5OH
Ethyl acetate on hydrolysis produces acetic acid and ethyl alcohol. This reaction is catalyzed by dilute sulphuric acid. In this reaction as seen from the equation acetic acid is formed, so the rate of reaction can be measured by measuring the concentration of acid formed at different intervals. To find the concentration of acetic acid we take a small amount of reaction mixture and titrate it against NaOH a base. But in this case, there is a small problem, that the reaction mixture contains acetic acid and also sulphuric acid, so the equivalence point of the titration is for both acids.
C{ H }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 } + { H }_{ 2 }O + { H }_{ 2 }S{ O }_{ 4 } longrightarrow C{ H }_{ 3 }COOH + { C }_{ 2 }{ H }_{ 5 }OH At the start of reaction, (t = 0) amount of acetic acid will be zero. So the volume of NaOH for equivalence point is only because of { H }_{ 2 }S{ O }_{ 4 } present in reaction mixture. left( { Volume of NaOH at zero time ({ V }_{ 0 }) } right) propto left( { Amount of { H }_{ 2 }S{ O }_{ 4 } } right) ...(1) Lets assume after time t the volume of NaOH for equivalence point is { V }_{ t }. at this time as the reaction is started, reaction mixture will have both { H }_{ 2 }S{ O }_{ 4 } and C{ H }_{ 3 }COOH. left( { Volume of NaOH at time t ({ V }_{ t }) } right) propto left( { Amount of { H }_{ 2 }S{ O }_{ 4 } + Amount of C{ H }_{ 3 }COOH } right) ...(2) Substracting equation (1) from equation (2) left( { V }_{ t } - { V }_{ 0 } right) propto { left( { Amount of C{ H }_{ 3 }COOH } right) } But amount of acetic acid produced depends on amount of ethyl acetate reacted at that time. which is x. So above equation can be written as, boxed { left( { V }_{ t } - { V }_{ 0 } right) propto x } ...(3) At infinite time all the reactant (ethyl acetate) is reacted and is converted into acetic acid. Therefore at this time the amount of acetic acid formed is dependent on total quantity of ethyl acetate taken initially. i.e. a left( { Volume of NaOH at infinite time ({ V }_{ infty }) } right) propto left( { Amount of { H }_{ 2 }S{ O }_{ 4 } + Initial Amount of C{ H }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 } } right) ...(4) Substracting equation (1) from equation (4) left( { V }_{ infty } - { V }_{ 0 } right) propto { left( { Initial Amount of C{ H }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 } } right) } or boxed { left( { V }_{ infty } - { V }_{ 0 } right) propto { a } } ...(5) Now we got the value of a from equation (5) and value of x from equation (3). Lets substitute these values in first order rate equation, i.e. k = frac { 2.303 }{ t } log { frac { a }{ a-x } } we get, k = frac { 2.303 }{ t } log { frac { left( { V }_{ infty } - { V }_{ 0 } right) }{ left[ left( { V }_{ infty } - { V }_{ 0 } right) - left( { V }_{ t } - { V }_{ 0 } right) right] } } k = frac { 2.303 }{ t } log { frac { left( { V }_{ infty } - { V }_{ 0 } right) }{ left[ { V }_{ infty } - { V }_{ 0 } - { V }_{ t } + { V }_{ 0 } right] } } boxed { k = frac { 2.303 }{ t } log { frac { left( { V }_{ infty } - { V }_{ 0 } right) }{ left( { V }_{ infty } - { V }_{ t } right) } } } Thus hydrolysis of ethyl acetate can be tested with this equation and in was proved that the reaction is of first order.
Numerical: 1 ml of ethyl acetate was added to 25ml of 2N HCl. 2ml of the mixture were withdrawn from time to time during the progress of the hydrolysis of the ester and titrated against standard NaOH solution. The amount of NaOH required for titration at various intervals is given below: Time (min): 0 20 75 119 183 infty NaOH used (ml): 20.24 21.73 25.20 27.60 30.22 43.95 The value at infinite time was obtained by completing the hydrolysis on boiling. Show that it is a reaction of the first order and find the average value of the velocity constant. Solution: In this case { V }_{ 0 } = 20.24ml and { V }_{ infty } = 43.95ml So lets calculate rate constant at each time interval using first order equation, k = frac { 2.303 }{ t } log { frac { a }{ a-x } } or boxed { k = frac { 2.303 }{ t } log { frac { left( { V }_{ infty } - { V }_{ 0 } right) }{ left( { V }_{ infty } - { V }_{ t } right) } } } 1) at time t= 20 min volume of NaOH used { V }_{ t }= 21.73ml k = frac { 2.303 }{ 20 } log { frac { left( 43.95 - 20.24 right) }{ left( 43.95 - 21.73 right) } } = 0.115 times log { frac { left( 23.71 right) }{ left( 22.22 right) } } = 0.115 times log { left( 1.067 right) } = 0.00324{ min }^{ -1 } 2) at time t= 75 min volume of NaOH used { V }_{ t }= 25.20ml k = frac { 2.303 }{ 75 } log { frac { left( 23.71 right) }{ left( 43.95 - 25.20 right) } }

[@omjoglekar3677]

its kinda fast. But really nice explanation out there ! :D

[@yd2763]

Nice video
Keep on making new videos

[@مهديمحمدكاظممحمد]

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[@prakhartomar8660]

Amazing sir. Really good explanation. Keep going. All the best.

[@Edmerls]

Thanks a lot

[@saisrivardhanreddylingala6967]

super excellent explanation thank you sooooooooooooooooo much

[@Edmerls]

You are most welcome

[@iitbombay454]

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Its very helpful to me ❤❤❤❤

[@Edmerls]

All the best

[@kadalimutyaveni6709]

K is itself constant how it can be 1st order

[@rajvinder4279]

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[@nikhilpundir1433]

Can you just explain how did you put the values of a and a-x because in the expression they were proportional to not equal to that value.

[@Footaball45718]

Take ratio they will give you tbe same BTW how are you doing now??? In which field you are because it's been 3 years of your comment

[@VarenyaMaheshwari]

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[@rajvinder4279]

Very good

[@naiyyagupta615]

good explanation

[@Edmerls]

Thanks for liking

[@JulioMartinez-pf8qw]

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[@Edmerls]

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