2 mole of an equimolar mixture of alchols ROH and R'OH are taken in! L flask. One mole of acetic acid is added to it. At equilibrium , 80% of acetic acid is found to be reacted and the ratio of `RCOOCH_(3) and R' COOCH_(3)` formed is 3 : 2 , Calculate the equilibrium constant for the esterification of ROH.

Explanation

As the mixture is equimolar , we have ROH = 1 mole , R' OH = 1 mole
Suppose x mole of acetic acid react with ROH and y mole acetic acid react will R'OH. Then
`{:(,ROH,+,CH_(3)COOH,hArr,RCOOCH_(3),+,H_(2)0),("Intial moles",1,,,,,,),("Moles at eqm.",1-x,,1-x-y,,x,,x+y),(,,,,,,,"Total" CH_(3)COOH " reacted" =x+y "moles"):}`
`{:(,ROH,+,CH_(3)COOH,hArr,RCOOCH_(3),+,H_(2)0" "...(ii)),("Intial moles",1,,,,,,),("Moles at eqm.",1-y,,1-x-y,,y,,x+y),("Total acetic acid reacted = 80% of 1 mole" = 0*80 "mole",,,,,,,):}`
`:. x+y = 0* 80 " Also " x/y = 3/2 or y= 2/3 x `
` :. x + 2/3 x = 0*8 or 5/3 x = 0*8 or x = (2*4)/ 5 = 0*48 `
` y = 0*8 - x = 0* 8 - 0* 48 = 0*32 `
For reaction (i), ` K = ( x(x+y))/((1-x)(1-x-y))= (0*48 xx 0*8 )/(0*52 xx 0*2 ) = 3* 69 approx 3*7.`