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VITEEE 2013 Paper

Section: Chemistry

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Question : 61 of 120

Marks: +1, -0

Acid hydrolysis of ester is first order reaction and rate constant is given by k =

\frac{2.303}{t}

\frac{V_\infty - V_0}{V_\infty - V_t}

V_0, V_t

V_\infty

are the volumle of standard NaOH required to neutralise acid present at a given time, if ester is 50% neutralised then

$V_\infty$ = $V_t$
$V_\infty$ = $(V_t - V_0)$
$V_\infty$ = $2V_t - V_0$
$V_\infty$ = $2V_t + V_0$

Solution:

RCOOR' +

\mathrm{H}_2\mathrm{O}

\xrightarrow{\mathrm{H}^+}

RCOOH + R'OH

\begin{array}{lccc} \text{At } t = 0 & a & 0 & 0 \\ \text{At time } t & a - x & x & x \\ \text{At time } \infty & a-a & a & a \end{array}

At t = 0 ,

V_0

= volume of NaOH due to

\mathrm{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

∴

V_\infty

=

2V_t - 2V_0+V_0

=

2V_t - V_0

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