soma
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mole fractions and equilibrium constants
A book I'm reading says that with a mixture of 1 mole of acetic acid and 1 mole of ethanol, "the equilibrium is found experimentally to contain 0.66
mole of ethyl acetate (and an equimolar amount of water). Thus, the mole-fractions of ester and water are 0.66/2."
If there's .66 moles of ethyl acetate and .66 moles of water, and .34 moles of acetic acid and .34 moles of ethanol, wouldn't the mole fraction of
ethyl acetate be .66/(.66+.68)?
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Sulaiman
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As the stoichiometry is 1+1=1+1
wherever equilibrium lies, there will always be 2 moles total.
If there is 0.66 mole ethyl acetate
then its molar ratio is 0.66 / 2 = 0.33 mole/mole
(moles of ethyl acetate / total moles)
CAUTION : Hobby Chemist, not Professional or even Amateur
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soma
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Thanks.
I wasn't counting the ethyl acetate as part of the whole.
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soma
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The next part of this has to do with the equilibrium constant.
The Ke for this is 3.77. They are saying that by increasing the amount of either alcohol or acid by 2x that the amount of acetate produced is
increased by 80% based on calculations from the Ke.
I was wondering how this would be calculated?
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