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In an experiment starting with 1 mol C(2...

In an experiment starting with `1` mol `C_(2)H_(5)OH, 1` mol `CH_(3)COOH`, and `1` mol of water, the equilibrium mixture mixture of analysis shows that `54.3%` of the acid is esterified. Calculate `K_(c)`.

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The correct Answer is:
4
`CH_(3)COOH(I)hArrCH_(3)COOC_(2)H_(5)(l)+H_(2)O(l)`
`{:("initial",1,1,0,1),("at equilibrium",1-x,1-x,x,1+x):}`
`1-0543 1-0.543 0.543 1+0.543`
`(54.3% "of" 1 mol e=(1xx54.3)/(100)=0.543 mol e)`
Hence given `x=0.543 mol e`
Applying law of mass action
`K_(C)=(["ester"]["water"])/(["acid][alcohol])=(0.543xx1.543)/(0.457xx0.457)=4.0`
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