Question

$K_c$ for the esterification reaction :

$C{H}_{3}COOH\left(\ell \right)+C_2{H}_{5}OH\left(\ell \right)\rightleftharpoons H_{3}COO{C}_2{H}_5\left(\ell \right)+H_{2}O\left(\ell \right)$ is 4. If 4 moles each of acid and alcohol are taken initially, what is the moles of the acid at equilibrium?

• A. $\frac{2}{3}$
• B. $\frac{4}{3}$
• C. $\frac{3}{4}$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

$\frac{4}{3}$

The given reaction is :-

${CH}_{3}COOH\left(l\right)+C_2{H}_{5}OH\left(l\right)\rightleftharpoons H_{3}COO{C}_2{H}_5\left(l\right)+H_{2}O\left(l\right)$

Initial conc. : $4$ $4$ $0$ $0$

At eqm : $4-x$ $4-x$ $x$ $x$

Given, initial conc. of acid & alcohol is & moles / litre. Let, $x$ moles of each acid & alcohol has reacted.

$\Rightarrow K_C=\frac{\left[H_{3}COO{C}_2{H}_5\right]\left[H_{2}O\right]}{\left[{CH}_{3}COOH\right]\left[C_2{H}_{5}OH\right]}$

$\Rightarrow 4=\frac{x.x}{(4-x)(4-x)}$

$\Rightarrow \frac{x^2}{{(4-x)}^2}=4\Rightarrow \frac{x}{4-x}=2$ (Taking square root)

$\Rightarrow \frac{x}{4-x}=2\Rightarrow x=8-2x$

$\Rightarrow x=8/3$

So, no. of moles of acid at equilibrium $=4-\frac{8}{3}=\frac{4}{3}$