Calculating K Value for Fischer Esterification Homework

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In the discussion about calculating the K value for Fischer esterification, a participant seeks clarification on how a yield of 59% is derived from an equilibrium constant (K) of 2 for a 1:1 ratio of reagents. The formula K = [ester][H2O] / [acid][alcohol] is referenced, but the participant is unsure of the calculations leading to the yield percentage. Another contributor suggests that the yield is likely based on experimental data rather than theoretical calculations, emphasizing the importance of incorporating K into yield assessments. The conversation highlights the complexities of relating equilibrium constants to practical yields in chemical reactions. Understanding these calculations is essential for accurate predictions in lab proposals.
epyfathom
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Homework Statement


Hi.

In I'm doing a proposal for a lab and in it, it says that with an equilibrium constant value of 2 (K=2), the best yield would be about 59% of theoretical for a 1:1 ratio of reagents. I'm guessing it has to do with this equation, but I'm still sure how they calculated this value.

K = [ester][H2O] / [acid][alcohol]


Please show me how they got this value..


Thank you.
 
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I think it is an experimental value, it doesn't fit results of obvious approaches. Besides, I would say theoretical yield should take K (if known) into account.

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