Question

In one litre container, ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as ${\mathrm{CH}}_3\mathrm{COOH}(l)+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}(l)\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5(l)+{\mathrm{H}}_2\mathrm{O}(l)$
At $293K$, if one starts with $1.00$ mole of acetic acid and $0.18$ mole of ethanol, there is $0.171$ mole of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

• A. $K_c=3.92$
• B. $K_c=2.56$
• C. $K_c=4.89$
• D. $Kc=6.23$

Answer 1

Answer: (A)

Given, ${\left[C{H}_{3}COO{C}_2{H}_5\right]}_{\text{equilibrium }}=0.171mol=x$
$K_c=\frac{\left[C{H}_{3}COO{C}_2{H}_5\right]\left[H_{2}O\right]}{\left[C{H}_{3}COOH\right]\left[C_2{H}_{5}OH\right]}$
$(K{)}_c=\frac{0.171\times 0.171}{(1-0.171)\times (0.180-0.171)}$
$K_c=\frac{0.171\times 0.171}{0.829\times 0.009}=3.919\approx 3.92$