When methyl acetate is hydrolyzed in $0.05 \, M$ $HCl$,the reaction occurs as follows: $CH_3COOCH_3 + H_2O \rightarrow CH_3COOH + CH_3OH$. $A$ $25 \, mL$ sample of the reaction mixture is taken at different time intervals,added to ice to stop the reaction,and then titrated with $0.05 \, M$ $NaOH$ solution. Prove that the reaction is first order using the data provided below:

Time (minute)	$0$	$20$	$75$	$120$	$\infty$
Volume of $0.05 \, M$ $NaOH$ ($V_t$ in $mL$)	$24.40$	$25.82$	$29.35$	$31.75$	$47.50$

(N/A) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{V_{\infty} - V_0}{V_{\infty} - V_t}$.
Here,$V_0 = 24.40 \, mL$,$V_{\infty} = 47.50 \, mL$,and $V_t$ is the volume at time $t$.
$1$. At $t = 20 \, min$: $k = \frac{2.303}{20} \log \frac{47.50 - 24.40}{47.50 - 25.82} = \frac{2.303}{20} \log \frac{23.10}{21.68} \approx 0.00308 \, min^{-1}$.
$2$. At $t = 75 \, min$: $k = \frac{2.303}{75} \log \frac{47.50 - 24.40}{47.50 - 29.35} = \frac{2.303}{75} \log \frac{23.10}{18.15} \approx 0.00315 \, min^{-1}$.
$3$. At $t = 120 \, min$: $k = \frac{2.303}{120} \log \frac{47.50 - 24.40}{47.50 - 31.75} = \frac{2.303}{120} \log \frac{23.10}{15.75} \approx 0.00318 \, min^{-1}$.
Since the value of $k$ remains approximately constant at different time intervals,the reaction follows first-order kinetics.