1. Tardigrade
  2. Question
  3. Chemistry
  4. In one litre container, ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as mathrmCH3 mathrmCOOH(l)+ mathrmC2 mathrmH5 mathrmOH(l) leftharpoons mathrmCH3 mathrmCOOC2 mathrmH5(l)+ mathrmH2 mathrmO(l) At 293 K, 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.

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Q. 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 $293\, K$, 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.

NTA AbhyasNTA Abhyas 2020Equilibrium

Solution:

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Given, $\left[ CH _{3} COOC _{2} H _{5}\right]_{\text {equilibrium }}=0.171 mol =x$
$K _{ c }=\frac{\left[ CH _{3} COOC _{2} H _{5}\right]\left[ H _{2} O \right]}{\left[ CH _{3} COOH \right]\left[ C _{2} H _{5} OH \right]}$
$( K )_{ c }=\frac{0.171 \times 0.171}{(1-0.171) \times(0.180-0.171)}$
$K _{ c }=\frac{0.171 \times 0.171}{0.829 \times 0.009}=3.919 \approx 3.92$