pHYSICAL cHEMISTRY - I

Experiment 5

Aim of the Experiment

Study the second order velocity constant of the hydrolysis of ethyl acetate by sodium hydroxide using conductivity measurements

Principle

The hydrolysis of ethyl acetate ester by NaOH is represented as

CH3COOC2H5 + (Na+ + OH-) → (CH3COO- + Na+ ) + C2H5OH

This reaction is called as saponification of ester. The velocity of saponification is proportional to the concentration of OH- ions.

(dx/dt) = k [CH3COOC2H5] [NaOH]

For equal initial concentrations of the reactants, the velocity constant of this second order reaction is given by the expression.

k= (1/t) (x/a (a-x))

This reaction can be monitored by measuring the electrical conductivity of the reaction mixture. During the progress of the reaction, OH- ions with higher value of ions conductivity are replaced by CH3COO- ions which have lower value of ionic conductivity. The decrease of conductivity is proportional to the amount of the reaction that has taken place at that time interval. The above expression can be written in terms of conductivity in the following manner. k = (1/at) (Co-Ct/Ct-C∞)

At t = 0 there are only reactants mainly ethyl acetate and sodium hydroxide. The resultant conductance (Co) at t = 0 is due to the contribution of both. However contribution towards conductance of ethyl acetate is zero as it is non-ionised solvent. Therefore sodium hydroxide (0.01M) contributes towards Co. As time passes ester is hydrolyzed and NaOH is utilized to form products. Conductance (Ct) during the progress of the reaction decreases. At t = ∞ the contribution towards C∞ is due to sodium acetate alone as the other product ethyl alcohol is non-ionised. Therefore C∞ is the conductance due to 0.01 M sodium acetate solution.

Materials required

Chemicals Required:- 0.02M ethyl acetate (to be freshly prepared: 2ml ethyl acetate dilute it to 100ml), 0.05M NaOH, 0.01 M sodium acetate.

Apparatus Required:- 250 ml Erlenmeyer flask with stopper X (2), 500 ml beaker, magnetic stirrer with magnetic bid, 100 ml measuring cylinder, 25 ml measuring cylinder, dropper, stop watch.

Procedure

1) Standardize the conductivity with the standard conductivity water.

2) Clean the conductivity cell and fill it with 0.01 M NaOH. Measure the conductance of the solution and call it as C0.

3) Wash the conductivity cell and fill it with 0.01 M CH3COONa. Measure the conductance of the solution and call it as C∞.

4) In two stoppered bottles take the reactants as given below:

Bottle 1: 0.02 M NaOH (50 ml )

Bottle 2: 0.02 M CH3COOC2H5 (50 ml )

5) Mix the two solutions thoroughly and start the stopwatch. Call the time of mixing as t=0. Fill the conductivity cell with the reaction mixture and stir it continuously with the help of magnetic stirrer. Record the conductance values (Ct) at 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, minutes.

Observations:

Temperature =

Co = conductance of the reaction mixture at t= 0

= conductance of 0.01 M NaOH = ________

C∞ = conductance of the reaction mixture at t= ∞

= conductance of 0.01 M CH3COONa = _______

Calculations

Given system is under equal initial concentrations of the reactants. The velocity constant k can be evaluated using the expression k=1/ (at)*(x/ (a-x))

Where ‘a’ is the initial concentration of the reactant and is equal to 0.01.

Graph:-

Plot (Co – Ct)/ (Ct - C∞) against t.

Find the slope of the straight line.

Slope = ak = 0.01 * k.

Calculate k.

Result

1.k by calculations = lt mol -1 min-1

2.k by graph = lt mol -1 min-1

Precautions

In this experiment you will be working with base (0.05 N NaOH) and sodium hydroxide is extremely corrosive. The lab coats and eye protection is must throughout the experiments and in case of chemical spill in eye or any other part of the body, immediately wash with lots of water.

Questions

1.What is the principle of Conductometric titration

2.What is the order of reaction for Base catalysed Hydrolysis of ester.Explain