Kinetic data for hydrolysis of a 0.3 M solution of ethyl acetate in presence of HCl as catalyst is given in the following table. A constant volume of reaction mixture is taken at different time and ti

Question

Kinetic data for hydrolysis of a 0.3 M solution of ethyl acetate in presence of HCl as catalyst is given in the following table. A constant volume of reaction mixture is taken at different time and titrated against standard alkali solution at different time.

Time / min 0 10 t=∞

Vol. of Alkali / ml 25.00 28.00 40.00

If the rate law is given by rate = $k [E s t er]^{1} [H^{+}]^{1}$ , where H⁺ is a catalyst. The rate of ester hydrolysis in a solution which is 0.1 M each in ester and acid HCl is found to be $p \times 1 0^{- 4}$ M/min.

Calculate the value of $p$ .

TestHub · Accepted Answer

Ester + H 2 ​ O H + ​ Acid + Alcohol t = 0 0.3 × v - - t = t ( 0.3 − x ) × v t = ∞ − ​ xv 0.3 v ​ xv 0.3 v ​ R v [ H + ] v [ H + ] + v − x ​ = K [ Ester ] ′ [ H + ] ′ = K ′ [ Ester ] K ′ = K [ H + ] ∝ 25 × C … .. ( 1 ) ∝ 28 × C … .. ( 2 ) ​ v [ H + ] + 0.3 v ∝ 40 × C (3)-(1) 0.3 × v ∝ 15 × C (3)-(2) ( 0.3 − x ) v ∝ 12 × C K ′ = 10 1 ​ ln ( 0.3 − x ) × v 0.3 × v ​ ( 1 ) ( 3 ) ​ ( H + ) H + + 0.3 ​ = 25 40 ​ ( H + ) = 0.5 M As K ′ = K [ H + ] 0.02 = K ( 0.5 ) K = 0.04 R = 0.04 × ( 0.1 ) ( 0.1 ) = 4 × 1 0 − 2

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