Chapter 10: Problem 27
The ester, ethyl acetate is formed by the reaction between ethanol and acetic acid and equilibrium is represented as : \(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{l})}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}_{(\mathrm{l})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\) (a) Write the concentration ratio (reaction quotient), \(Q_{\mathrm{e}}\), for this reaction. Note that water is not in excess and is not a solvent in this reaction. (b) At \(293 \mathrm{~K}\), if one starts with \(1.00\) mole of acetic acid and \(0.180\) of ethanol, there is \(0.171\) mole of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (c) Starting with \(0.500\) mole of ethanol and \(1.000\) mole of acetic acid and maintaining it at \(293 \mathrm{~K}, 0.214\) mole of ethyl acetate is found after some time. Has equilibrium been reached?
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