Please Explain! Thank you so much. An aqueous solution of ethyl acetate at room temperature has...

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Please Explain! Thank you so much.
An aqueous solution of ethyl acetate at room temperature has anitial concentration of 0.15 M. Calculate the equilibriumconcentrations of ethyl acetate (CH3COOC2H5), ethanol (C2H5OH) andacetic acid if the equilibrium constant at this temperature is Kc=3.00.
Part B. Repeat this calculation for initial ehtyl acetateconcentrations of 0.25 M and 0.35 M. Are your answers in agreementwith Le Chatelier's Principle?

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4.4 Ratings (547 Votes)

a)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â CH3COOC2H5 (aq)Â Â Â <-------------> C2H5OH + CH3COOHÂ Â Â Â Â Â Â Â Kc = 3.00

Â Â Â InitialÂ Â Â Â Â Â Â Â Â Â Â Â Â 0.15 MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

At equilibriumÂ Â Â Â Â 0.15-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x

Â Â Â Â Â Â Â Â Kc = x.x / 0.15-x

Â Â Â Â Â Â 3 = x²/ 0.15-x

Â Â Â Â 0.45 - 3x = x²

^Â Â Â x² + 3x -0.45 = 0

On solving,

x = 0.1432

Therefore, equilibrium concentrations are

[ CH3COOC2H5 (aq)] = 0.15- x= 0.15 -0.143 = 0.0068 M

[C2H5OH] = x = 0.1432 M

[CH3COOH] = x = 0.1432 M

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