Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below : {:("t/min",,,0,,,30,,,60,,,90),("C/mol L"^(-1),,,0.8500,,,0.8004,,,0.7538,,,0.7096):} Rate =k'["CH"_(3)"COOCH"_(3)]["H"_(2)O]

Hydrolysis of methyl acetate in aqueous solution has been studied by t

1.	Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below : {:("t/min",,,0,,,30,,,60,,,90),("C/mol L"^(-1),,,0.8500,,,0.8004,,,0.7538,,,0.7096):} Rate =k'["CH"_(3)"COOCH"_(3)]["H"_(2)O]
Answer» Solution :For the GIVEN reaction, rate `=k'["CH"_(3)"COOCH"_(3)]["H"_(2)O]` But as `["H"_(2)O]`remains constant, we put `k'["H"_(2)O]=k` so that rate `=k["CH"_(3)"COOCH"_(3)]` Now, it is a pseudo first ORDER reaction. Hence, for the given data, `k=(2.303)/(t)log""(C_(0))/(C_(t))` We are given `C_(0)=0.8500" mol L"^(-1)` `{:("t/min",,,C_(t)//"mol L"^(-1),,,k=(2.303)/(t)log""(C_(0))/(C_(t))),(30,,,0.8004,,,k=(2.303)/(30min)log""(0.8500)/(0.8004)=2.004xx10^(-3)min^(-1)),(60,,,0.7538,,,k=(2.303)/(60min)log""(0.8500)/(0.7538)=2.002xx10^(-3)min^(-1)),(90,,,0.7096,,,k=(2.303)/(90min)log""(0.8500)/(0.7096)=2.005xx10^(-3)min^(-1)):}` Average value of `k=2.004xx10^(-3)min^(-1)` Thus, `k'["H"_(2)"O"]=k" or "k'[55" mol L"^(-1)]=2.0074xx10^(-3)min^(-1)` `:.k'=(2.004xx10^(-3)min^(-1))/(55" mol L"^(-1))=3.64xx10^(-5)" L mol"^(-1)" min"^(-1).`