Question

4.6 g of ethyl alcohol and 6.0 g of acetic acid were kept at constant temperature until equilibrium was established 2.0 g of acetic acid remained unused. calculate kc for the reaction . give proper answer on paper... plz i want some one help


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Answer 1

Answer:

We assess the equilibrium reaction…

H3C−C(=O)OH(l)+HOCH2CH3(l)⇌H3C−C(=O)OCH2CH3(l)+H2O

For which … Keq=[H3C−C(=O)OCH2CH3][H2O][H3C−C(=O)OH][HOCH2CH3]

Now these are terms of CONCENTRATION, not molar quantities…. Nevertheless, because the VOLUME of the reaction mixture was a CONSTANT … we can remove this from the expression, and simply work with the NUMBER of moles…i.e.

Keq=nH3C−C(=O)OCH2CH3nH2OnH3C−C(=O)OHnHOCH2CH3

And so initially, nacetic acid=6.0∙g60.05∙g∙mol−1=0.0999∙mol=0.10∙mol

nEtOH=4.6∙g46.07∙g∙mol−1=0.0999∙mol=0.10∙mol

And FINALLY, there were nacetic acid=2.0∙g60.05∙g∙mol−1=0.0333∙mol

And THUS 0.10∙mol−0.0333∙mol=0.0667∙mol acetic acid REACTED...and so DID 0.0667∙mol with respect to ethyl alcohol...and we got 0.0667∙mol as products with respect to water, and ethyl acetate…

And finally, we can assess Keq, by interpolating the molar quantities…

Keq=(0.0667∙mol)2(0.0333)2=4.0

Explanation: