Ester Hydrolysis Reaction Lab Report
Determination of the Equilibrium Constant of an Unknown
Ester Hydrolysis Reaction
Abstract The experiments to follow determined that the equilibrium concentrations of the reaction: ester + water ↔ alcohol + acid, are equal to 0.0363 moles of ester, 0.2852 moles of water, and 0.0268 moles each of alcohol and acid. Using this information the equilibrium constant was determined to be 0.06938.
1. Introduction In this lab the equilibrium constant, Kc, for the acid catalyzed reaction between an unknown ester and water to form an unknown alcohol and an unknown carboxylic acid was determined. The equation for the reaction is the following: R1COOR2 + H2O ↔ R2OH + R1COOH …show more content…
The equilibrium concentrations were determined through a series of experiments where a known base, Sodium Hydroxide (NaOH), was titrated into different reaction mixtures until an endpoint was reached. The equilibrium concentrations were then placed into the following equation to determine the constant:
Kc = [R2OH] [ R1COOH] [R1COOR2 ] [H2O] (eq.2)
Equation one and two can be generalized to the following: ester + water ↔ alcohol + acid (eq.3)
Kc = [alcohol]eq [ …show more content…
5.00 mL HCl soln. x 1.217g HCl soln = 6.085g HCl soln. 1 mL HCl soln
3.0 mol HCl x 36.5g HCl x 1mL soln. = 0.0899g HCl x 5mL = 0.4495 g HCl
1000mL soln mol HCL 1.217g HCl soln mL
6.085g HCl soln. 0.4495 g HCl = 5.63 g H2O
5.63 g H2O x 1 mole H2O = 0.312 moles H2O from HCl solution 18.02 g H2O
Then the product amounts at equilibrium were calculated. Using the titration value of the NaOH used for bottle 2, and its molar concentration, the total amount of acid present can be determined.
58.85mL NaOH soln x 0.7103 mol NaOH x 1 mole mixed acids = 0.04180 moles of 1000mL NaOH soln 1 mol NaOH mixed acid
To determine the amount of R1COOH acid present, the amount of HCl must be subtracted.
5.00 mL HCl soln x 3.00 moles HCl = 0.015 moles HCl 1000 mL HCL soln
0.4180 moles of mixed acid 0.015 moles of HCl = 0.0268 moles of R1COOH
Ester Hydrolysis Reaction
Abstract The experiments to follow determined that the equilibrium concentrations of the reaction: ester + water ↔ alcohol + acid, are equal to 0.0363 moles of ester, 0.2852 moles of water, and 0.0268 moles each of alcohol and acid. Using this information the equilibrium constant was determined to be 0.06938.
1. Introduction In this lab the equilibrium constant, Kc, for the acid catalyzed reaction between an unknown ester and water to form an unknown alcohol and an unknown carboxylic acid was determined. The equation for the reaction is the following: R1COOR2 + H2O ↔ R2OH + R1COOH …show more content…
The equilibrium concentrations were determined through a series of experiments where a known base, Sodium Hydroxide (NaOH), was titrated into different reaction mixtures until an endpoint was reached. The equilibrium concentrations were then placed into the following equation to determine the constant:
Kc = [R2OH] [ R1COOH] [R1COOR2 ] [H2O] (eq.2)
Equation one and two can be generalized to the following: ester + water ↔ alcohol + acid (eq.3)
Kc = [alcohol]eq [ …show more content…
5.00 mL HCl soln. x 1.217g HCl soln = 6.085g HCl soln. 1 mL HCl soln
3.0 mol HCl x 36.5g HCl x 1mL soln. = 0.0899g HCl x 5mL = 0.4495 g HCl
1000mL soln mol HCL 1.217g HCl soln mL
6.085g HCl soln. 0.4495 g HCl = 5.63 g H2O
5.63 g H2O x 1 mole H2O = 0.312 moles H2O from HCl solution 18.02 g H2O
Then the product amounts at equilibrium were calculated. Using the titration value of the NaOH used for bottle 2, and its molar concentration, the total amount of acid present can be determined.
58.85mL NaOH soln x 0.7103 mol NaOH x 1 mole mixed acids = 0.04180 moles of 1000mL NaOH soln 1 mol NaOH mixed acid
To determine the amount of R1COOH acid present, the amount of HCl must be subtracted.
5.00 mL HCl soln x 3.00 moles HCl = 0.015 moles HCl 1000 mL HCL soln
0.4180 moles of mixed acid 0.015 moles of HCl = 0.0268 moles of R1COOH