Theinitial concentrationof ethyl acetaneis0.85 mol L^(-1) . Followingthe acid catalysed hydrolysis the , concentrations of ester after 30 min and 60 minof the reactionare respectively0.8 and 0.754 mol L^(-1) .Calculatethe rateconstantand pseudo rateconstant.

Theinitial concentrationof ethyl acetaneis0.85 mol L^(-1) . Followingt

1.	Theinitial concentrationof ethyl acetaneis0.85 mol L^(-1) . Followingthe acid catalysed hydrolysis the , concentrations of ester after 30 min and 60 minof the reactionare respectively0.8 and 0.754 mol L^(-1) .Calculatethe rateconstantand pseudo rateconstant.
Answer» Solution :Acid catalysed ester follows PSEUDO first ORDER kinetics. The rate constant k is given as `k=(2.303)/(t)log.(a)/(a-x)` `k=(2.303)/(30)log.(0.85)/(0.05)=2.020xx10^(-3)"min"^(-1)("or")` `k=(2.303)/(30)log.(0,80)/(0.046)=1.997xx10^(-3)"min"^(-1)` The rate constant (k) is the product of pseudo first order rate constant (k.). Concentration of water, TAKEN as constant as `55.5` "MOL"L^(-1)` `k=k.[H_(2)O]` Substituting the values, `1.997xx10^(-3)=k.[55.5]` Pseudo rate constant=k. `3.6xx10^(-5)L "mol"^(-1)"min"^(-1)`