Question

$1mL$ of methyl acetate was added to a flask containing $20mL$ of $N/20HCl$ maintained at ${25}^{o}C$. $2mL$ of the reaction mixture were withdrawn at different intervals and titrated with a standard alkali solution. The following results were obtained:

Time (min)	0	75	119	183	$\infty$
Alkali used $\left(V_t\right)$ in mL	19.24	24.20	26.60	29.32	42.03

where V is the volume of alkali required to neutralise the acidity of the solution.

Find the half life period of the reaction if it follows first order kinetics.

• A. $330min$
• B. $230min$
• C. $72min$
• D. $170min$

Answer 1

The correct option is B $230min$

Reaction mixture,
$C{H}_{3}COOC{H}_3+HCl\to C{H}_{3}COOH+C{H}_{3}OH$

Titration reaction is the neutralisation of acid and alcohols by alkali.

The volume of alkali used, evidently corresponds to the reacted methyl acetate.

But at time $t=0$, some volume of alkali is used which is not corresponds to the reactant.
$\therefore$
$\text{Initial concentration of methyl acetate (a)}=V_{\infty }-V_0$

$\text{Concentration of methy acetate left unreacted (a-x) at time 't'}=V_{\infty }-V_t$

Since the hydrolysis follows fist order kinetics, then

$k=\frac{2.303}{t}{\log }_{10}\frac{a}{a-x}k=\frac{2.303}{t}{\log }_{10}\frac{V_{\infty }-V_0}{V_{\infty }-V_t}$

(i) When, $t=75min$, $V_{\infty }=42.03,V_0=19.24,V_t=24.20$

$k=\frac{2.303}{75}{\log }_{10}\frac{(42.03-19.24)}{(42.03-24.20)}=\frac{2.303}{75}{\log }_{10}\frac{22.79}{17.83}$

$\approx 0.003{min}^{-1}$

(ii) $t=119min$, $V_{\infty }=42.03,V_0=19.24,V_t=26.60$

$k=\frac{2.303}{119}{\log }_{10}\frac{(42.03-19.24)}{(42.03-26.60)}=\frac{2.303}{119}{\log }_{10}\frac{22.79}{15.43}$

$\approx 0.003{min}^{-1}$

Since, the values of k are constant, it follows first order kinetics.
$\therefore$
$\text{Half life,}{t}_{1/2}=\frac{0.693}{k}$

$t_{1/2}=\frac{0.693}{0.003}$

$t_{1/2}\approx 230min$