Question

Ethyl ethanoate undergoes following reaction [on the picture]

Answer 1

Answer: The concentration of ethyl ethanoate at equilibrium is $0.653mol/dm^3$

Explanation:

Given values:

Equilibrium concentration of ethanol = $0.42mol/dm^3$

Equilibrium concentration of ethanoic acid = $0.42mol/dm^3$

$K_c=0.27$

The given chemical equation follows:

$CH_3COOC_2H_5+H_2O\rightleftharpoons C_2H_5OH+CH_3COOH$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[C_2H_5OH][CH_3COOH]}{[CH_3COOC_2H_5]}$

Putting values in above expression, we get:

$0.27=\frac{0.42\times 0.42}{[CH_3COOC_2H_5]}$

$[CH_3COOC_2H_5]=\frac{0.42\times 0.42}{0.27}=0.653mol/dm^3$

Hence, the concentration of ethyl ethanoate at equilibrium is $0.653mol/dm^3$