Answer: The concentration of ethyl ethanoate at equilibrium is [tex]0.653mol/dm^3[/tex]
Explanation:
Given values:
Equilibrium concentration of ethanol = [tex]0.42mol/dm^3[/tex]
Equilibrium concentration of ethanoic acid = [tex]0.42mol/dm^3[/tex]
[tex]K_c=0.27[/tex]
The given chemical equation follows:
[tex]CH_3COOC_2H_5+H_2O\rightleftharpoons C_2H_5OH+CH_3COOH[/tex]
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[C_2H_5OH][CH_3COOH]}{[CH_3COOC_2H_5]}[/tex]
Putting values in above expression, we get:
[tex]0.27=\frac{0.42\times 0.42}{[CH_3COOC_2H_5]}[/tex]
[tex][CH_3COOC_2H_5]=\frac{0.42\times 0.42}{0.27}=0.653mol/dm^3[/tex]
Hence, the concentration of ethyl ethanoate at equilibrium is [tex]0.653mol/dm^3[/tex]
