Methyl acetate is hydrolysed in approximately N-HCl at 25°C. 5.0 mL portions of the reaction mixture were removed at intervals and titrated with 0.185 N-NaOH. From the data given below prove that hydrolysis of methyl acetate is a first order reaction.

t/sec	1242 sec	2745 sec	4546 sec
At Conc.	-27.80ml	-29.70ml	-31.81ml

Here [ A]₀ is proportional to the alkali consumed between t = &&& and t = 339 sec, i.e., [A]₀ = (39.81 mL – 26.34 mL) = 13.47 mL [A]_t is proportional to the alkali consumed at &&& minus the alkali consmed at the specified time. Thus,
[A], at 1242 sec = 39.81 mL – 27.80 mL = 12.01 mL
[A], at 2745 sec = 39.81 mL – 29.70 mL = 10.11 mL
[A], at 4546 sec = 39.81 mL – 31.81 mL = 8.00 mL
Substituting the values of t, [A]₀ and [A]_t in the first-order rate equation

$\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{{\left[\mathrm{A}\right]}_0}{{\left[\mathrm{A}\right]}_{\mathrm{t}}}$

we get the following values of k

(i)                  $\mathrm{k}=\frac{2.303}{(1242-339)\sec }\log \frac{13.47 \mathrm{mL}}{12.01 \mathrm{mL}}=1.27\times {10}^{-4}{\mathrm{s}}^{-1}$

(ii)                 $\mathrm{k}=\frac{2.303}{(247.5-339)\sec }\log \frac{13.47 \mathrm{mL}}{10.11 \mathrm{mL}}= 1.19 \times {10}^{-4}{\mathrm{s}}^{-1}$

(iii)                $\mathrm{k}=\frac{2.303}{(4546-339)\sec }\log \frac{13.47 \mathrm{mL}}{8.0 \mathrm{mL}}=1.24\times {10}^{-4}{\mathrm{s}}^{-1}$

Since the three values of k derived from first-order reaction are close to being identical, hydrolysis of methyl acetate is a first order reaction.