Saponification of ethyl acetate (CH_(3)COOC_(2)H_(5)) by NaOH (Saponification of ethyl acetate by NaOH is second order reaction) is studied by titration of the reaction mixture initially having 1:! molar ratio of the reactants. If 10 mL of 1 N HCl is required by 2 mL of the solution at the start and 8 mL of 1 N HCl is required by another 5 mL after 10 minutes, then rate constant is :

Saponification of ethyl acetate (CH_(3)COOC_(2)H_(5)) by NaOH (Saponif

1.	Saponification of ethyl acetate (CH_(3)COOC_(2)H_(5)) by NaOH (Saponification of ethyl acetate by NaOH is second order reaction) is studied by titration of the reaction mixture initially having 1:! molar ratio of the reactants. If 10 mL of 1 N HCl is required by 2 mL of the solution at the start and 8 mL of 1 N HCl is required by another 5 mL after 10 minutes, then rate constant is :
Answer» `K=(2.303)/(10)"LOG"(10)/(8)` `k=(2.303)/(10)"log"(10)/(2)` `k=(1)/(10)[(1)/(8)-(1)/(10)]` `k=(1)/(10)[(1)/(2)-(1)/(10)]` Answer :C