Acid hydrolysis of ester is first order reaction and rate constant is given by k=(2.303/t) log (V∞-V0/V∞-Vt) where, V0, Vt and V∞ are the volume of standard NaOH required to neutralise acid present at a given time, if ester is 50% neutralised then

Question

Acid hydrolysis of ester is first order reaction and rate constant is given by k=(2.303/t) log (V∞-V0/V∞-Vt) where, V0, Vt and V∞ are the volume of standard NaOH required to neutralise acid present at a given time, if ester is 50% neutralised then (A) V∞ = Vt (B) V∞ = (Vt-V0) (C) V∞ = 2Vt-V0 (D) V∞ = 2Vt+V0

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/ff4e3cdb53e6eefdc02cdbfdb1c10057-.png" /> At t=0, V0= volume of NaOH due to H +(catalyst) Vt=x+V0 V∞=a+V0 If ester is 50 % hydrolysed then, x=(a/2) or Vt=(a/2)+V0 or a=2 Vt-2 V0 ∴ V∞=2 Vt-2 V0+V0 =2 Vt-V0

$V_{\infty} = V_{t}$

$V_{\infty} = (V_{t} - V_{0})$

$V_{\infty} = 2 V_{t} - V_{0}$

$V_{\infty} = 2 V_{t} + V_{0}$

At $t = 0, V_{0} =$ volume of $N a O H$ 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 $a = 2 V_{t} - 2 V_{0}$
$∴ V_{\infty} = 2 V_{t} - 2 V_{0} + V_{0}$
$= 2 V_{t} - V_{0}$

Q. Acid hydrolysis of ester is first order reaction and rate constant is given by k=t2.303​logV∞​−Vt​V∞​−V0​​ where, V0​,Vt​ and V∞​ are the volume of standard NaOH required to neutralise acid present at a given time, if ester is 50% neutralised then

V∞​=Vt​

V∞​=(Vt​−V0​)

V∞​=2Vt​−V0​

V∞​=2Vt​+V0​

At t=0,V0​= volume of NaOH due to H+(catalyst) Vt​=x+V0​ V∞​=a+V0​ If ester is 50% hydrolysed then, x=2a​ or Vt​=2a​+V0​ or a=2Vt​−2V0​ ∴V∞​=2Vt​−2V0​+V0​ =2Vt​−V0​

$V_{\infty} = V_{t}$

$V_{\infty} = (V_{t} - V_{0})$

$V_{\infty} = 2 V_{t} - V_{0}$

$V_{\infty} = 2 V_{t} + V_{0}$

At $t = 0, V_{0} =$ volume of $N a O H$ 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 $a = 2 V_{t} - 2 V_{0}$
$∴ V_{\infty} = 2 V_{t} - 2 V_{0} + V_{0}$
$= 2 V_{t} - V_{0}$