Question

$1.0mole$ of ethyl alcohol and $1.0mole$ of acetic acid are mixed. At equilibrium, $0.666moles$ of the ester is present. The value of equilibrium constant is:

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $2$
• D. $4$

Answer 1

Answer: (D)

$4$

The given reaction is :-

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

$1$ $1$ $0$ $0$

At eq. : $(1-x)$ $(1-x)$ $x$ $x$

Given that, at equilibrium-

$Molesof{CH}_{3}COO{C}_2{H}_5=0.666$

$\Rightarrow x=0.666$

Now, number of moles of $C_2{H}_{5}OH=0.334$

number of moles of ${CH}_{3}COOH=1-0.666=0.334$

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

$=\frac{0.666\times 0.666}{0.334\times 0.334}\approx 2\times 2\approx 4$