to determine the rate constant for the hydrolysis of ester using hcl as catalyst
Answers
The hydrolysis reaction of an ester in pure water is a slow reaction and when a mineral acid like hydrochloric acid is added, the rate of the reaction is enhanced since the H+ ions from the mineral acid acts as the catalyst. The acid catalysed hydrolysis of an ester follows pseudo first order kinetics. The reaction can be represented as :
CH3COOCH3 + H2O -- H+ -- > CH3COOH + CH3OH
Methyl acetate
The overall rate of the reaction depends on the concentrations of reactants and also on the catalyst concentration.
Rate = k3 [ester] [H2O] [H+]
k3 = rate constant of the third order reaction.
Therefore the true order of the reaction is 3.0. Since water is used as the solvent, its concentration is excess.
Weight of 1 lit (1000 cc) of water = 1000 gm = 1 kg (\ density of water = 1 gm/cc)
No. of moles of water in 1 lit = 1000 / 18
Concentration of pure water = 55.55 moles
If 1 molar aqueous solution of ester is used then, 1 mole of water will be consumed for its complete hydrolysis. After the completion of hydrolysis, 55.55 - 1.0 = 54.55 moles water will be present in the medium. Therefore change in the concentration of water considered as negligible and concentration of water is assumed to be constant. Since acid acts the catalyst, there will be, no change in the catalyst concentration before initial and after the completion periods of times of the reactions. Hence [H+] is considered as a constant value. Hence the expression can be rewritten as
Rate = k3' [ester] where k3' = pseudo first order rate constant = k3 [H+] [H2O]. In this rate expression rate of the reaction is directly proportional to ester concentration only.
Procedure
Initially to a definite volume of (100 ml) hydrochloric acid (0.5 N), 10 ml of ester is added and the start of the reaction corresponds to time of addition of ester. The rate of the reaction is followed by withdrawing a definite volume of the reaction mixture consisting of the ester and acid at various time intervals and arresting the further progress of reaction by adding ice. The whole cold mixture is titrated with standard NaOH (0.1 N) using phenolphthalein as the indicator.
Let the volume of alkali consumed at t = 0 be Vo cc which is equivalent to the amount of hydrochloric acid present in the definite volume of the reaction mixture drawn out at regular intervals of time. If Vt cc is the volume of alkali consumed for the same definite volume of the reaction mixture drawn out after reaction time ' t', then (V t - Vo) cc is equivalent to the acetic acid produced by the hydrolysis of ester in time ' t'. A final titration is done after about 8 hours or after refluxing the solution for 45 mins to complete the hydrolysis which is V� cc. (V�- V o) cc is equivalent to acetic acid produced from complete hydrolysis of ester.
Calculations
The initial concentration of ester = a a (V�- V o) cc Concentration of ester reacted at 't' = x a (Vt - V o) cc
Concentration of ester remaining at time 't' = (a - x) a (V�- V t)
A / (a-x) = ((V�- V o)) / ((V�- V t))
The first order rate expression for the hydrolysis of ester can be
written as
k = (2.303/t ) log ((V�- V o)) / ((V�- V t))
By substituting Vt values for various 't' values, k is determined. These values are found to be constant indicating k as the rate constant of the reaction.
Answer:
Explanation:
Procedure:-
Initially to a definite volume of (100 ml) hydrochloric acid (0.5 N), 10 ml of ester is added and the start of the reaction corresponds to time of addition of ester. The rate of the reaction is followed by withdrawing a definite volume of the reaction mixture consisting of the ester and acid at various time intervals and arresting the further progress of reaction by adding ice. The whole cold mixture is titrated with standard NaOH (0.1 N) using phenolphthalein as the indicator.
Let the volume of alkali consumed at t = 0 be Vo cc which is equivalent to the amount of hydrochloric acid present in the definite volume of the reaction mixture drawn out at regular intervals of time. If Vt cc is the volume of alkali consumed for the same definite volume of the reaction mixture drawn out after reaction time ' t', then (V t - Vo) cc is equivalent to the acetic acid produced by the hydrolysis of ester in time ' t'. A final titration is done after about 8 hours or after refluxing the solution for 45 mins to complete the hydrolysis which is V� cc. (V�- V o) cc is equivalent to acetic acid produced from complete hydrolysis of ester.
Calculations:-
The initial concentration of ester = a a (V�- V o) cc Concentration of ester reacted at 't' = x a (Vt - V o) cc
Concentration of ester remaining at time 't' = (a - x) a (V�- V t)
A / (a-x) = ((V�- V o)) / ((V�- V t))
The first order rate expression for the hydrolysis of ester can be
written as
k = (2.303/t ) log ((V�- V o)) / ((V�- V t))
By substituting Vt values for various 't' values, k is determined. These values are found to be constant indicating k as the rate constant of the reaction.
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