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Question: Acid catalysed hydrolysis of ester is first-order reaction and rate constant is given by\(k = \frac{...

Acid catalysed hydrolysis of ester is first-order reaction and rate constant is given byk=2.303tlog⁡V∞−V0V∞−Vtk = \frac{2.303}{t}\log\frac{V_{\infty} - V_{0}}{V_{\infty} - V_{t}}where V0, Vt and V∞V_{0}\text{, }\text{V}_{t}\text{ and }\text{V}_{\infty} are the volume of standard NaOH required to neutralise acid present at a given time ; if ester is 50% hydrolysed then :

A

V∞ = VtV_{\infty}\text{ = }\text{V}_{t}

B

V = (Vt-V0)\text{V = }(V_{t}\text{-}\text{V}_{0})

C

V∞ = 2Vt - V0V\infty\text{ = 2}\text{V}_{t}\text{ - }\text{V}_{0}

D

V∞ = 2Vt + V0 V_{\infty}\text{ = 2}\text{V}_{t}\text{ + }\text{V}_{0}\

Answer

V∞ = 2Vt - V0V\infty\text{ = 2}\text{V}_{t}\text{ - }\text{V}_{0}

Explanation

Solution

If easter 50% hydrolysed then

x=a02x = \frac{a_{0}}{2} (a0−x)=a02\left( a_{0} - x \right) = \frac{a_{0}}{2}

We can write a0∝ (V∞- V0)a_{0} \propto \ (V_{\infty}\text{- V0})

(a0 - x ) → (V∞- Vt)(a_{0}\text{ - x })\ \rightarrow \ (V_{\infty}\text{- }\text{V}_{t})

V∞−V02=(V∞−Vt)\frac{V_{\infty} - V_{0}}{2} = \left( V_{\infty} - V_{t} \right)

V∞ - V0 = 2 V∞- 2 VtV_{\infty}\text{ - }\text{V}_{0}\text{ = 2 }\text{V}_{\infty}\text{- 2 }\text{V}_{t}

Vt=(V∞−V0)2V_{t} = \frac{\left( V_{\infty} - V_{0} \right)}{2}

or V∞= 2 Vt - V0V_{\infty}\text{= 2 }\text{V}_{t}\text{ - }\text{V}_{0}