Question

3. Methyl acetate is hydrolyzed at 25 oC in acidic environment. Aliquots of equal volume are removed and titrated with NaOH solution. The hydrolysis reaction is irreversible. The results are Time/s 339 1242 2745 4546 infinite Volume of base/ mL 26.34 27.80 29.70 31.81 39.81 Write the reaction of hydrolysis of methyl acetate. (3 pts.) Neglect the reverse reaction. Find the order of the hydrolysis reaction and the value of the rate constant at this temperature. (12 pts.) Total 15 pts.

Answer 1

Solution :

Time (sec)       Volume of NaOH (mL)

339                           26.23

1242                         27.80

2745                        29.70

4546                         3.81

$\infty$                               39.81

Now the example of the first order kinetics w.r.t volumetric analysis is :

$k=\frac{2.303}{t} \log \left(\frac{v_{\infty}-v_0}{v_{\infty}- v_t}\right)$

Here, $v_{\infty}= \text{ volume at }\infty = 39.81$

$v_{t}= \text{ volume at time 't' } = 27.80$

$v_0$ = volume at time 0 = 0

Since the interval is not constant, we take the time interval as

$=\frac{903+1503+1801}{3}$

$=\frac{4207}{3}$

= 1402.3333

≈  1402 seconds

$k=\frac{2.303}{1402} \log \left(\frac{39.81-0}{39.81-27.80}\right)$

  $=(0.001643) \log \left(\frac{39.81}{10.01}\right)$

  = 0.001643 x 0.52045

  = 0.00082

  $= 8.55 \times 10^{-4} \ sec^{-1}$

Therefore, the first order rate constant is k $= 8.55 \times 10^{-4} \ sec^{-1}$.