Question

Acid hydrolysis of ester is first order reaction and rate constant is given by $k=\frac{2.303}{t}\log \frac{V_{\infty }-V_0}{V_{\infty }-V_t}$ where, $V_0$, $V_t$ and $V_{\infty }$ are the volume of standard $NaOH$ required to neutralise acid present at a given time, if ester is $50$% neutralised then :

• A. $V_{\infty }=V_t$
• B. $V_{\infty }=(V_t-V_0)$
• C. $V_{\infty }=2V_t-V_0$
• D. $V_{\infty }=2V_t+V_0$

Answer 1

The correct option is C $V_{\infty }=2V_t-V_0$

$RCOO{R}^{\prime }+H_{2}O\stackrel{H^{+}}{\to }RCOOH+R^{\prime }OH$

At $t=0$,	$a$	$0$	$0$
At time $t$,	$a-x$	$x$	$x$
At time $\infty$,	$a-a$	$a$	$a$

At $t=0$, $V_0=$ volume of $NaOH$ due to $H^{+}$ (catalyst)
$V_t=x+V_0$
$V_{\infty }=a+V_0$
If ester is $50$% hydrolysed then, $x=\frac{a}{2}$

or $V_t=\frac{a}{2}+V_0$

or $a=2V_t-2V_0$

$\therefore V_{\infty }=2V_t-2V_0+V_0$

$=2V_t-V_0$