$1$ mole of $C H_{3} C O O H$ and $1$ mole of $C_{2} H_{5} O H$ are taken in $1 L$ flask, $50 %$ of $C H_{3} C O O H$ is converted into ester as:
$C H_{3} C O O H (l) + C_{2} H_{5} O H (l) ⇌ C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l)$

There is $33 %$ conversion of $C H_{3} C O O H$ into ester, if $C H_{3} C O O H$ and $C_{2} H_{5} O H$ have been taken initially in molar ratio $x ∶ 1,$ find $x .$

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Solution

$\begin{matrix} C H_{3} C O O H & + & C_{2} H_{5} O H & ⇌ & C H_{3} C O O C_{2} H_{5} & + & H_{2} O 1 & 1 & 0 & 0 1 - 0.5 & 1 - 0.5 & 0.5 & 0.5 \end{matrix}$

So, $K_{c} = \frac{Conc. of products}{Conc. of reactants} = \frac{0.5 \times 0.5}{0.5 \times 0.5} = 1$

Now, let $a$ moles of $C H_{3} C O O H$ and $b$ moles of $C_{2} H_{5} O H$ are taken then:
$\begin{matrix} C H_{3} C O O H & + & C_{2} H_{5} O H & ⇌ & C H_{3} C O O C_{2} H_{5} & + & H_{2} O a & b & 0 & 0 a - \frac{a}{3} & b - \frac{a}{3} & \frac{a}{3} & \frac{a}{3} \end{matrix}$

So, $K_{c} = \frac{(\frac{a}{3}) \times (\frac{a}{3})}{\frac{2 a}{3} \times (b - \frac{a}{3})} \Rightarrow 2 (b - \frac{a}{3}) = \frac{a}{3}$

$or 2b = a$
or $\frac{a}{b} = \frac{2}{1}$
$Given: \frac{a}{b} = \frac{x}{1} = \frac{2}{1}$
$\Rightarrow a = x = 2$

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