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The equilibrium constant for the reacti...

The equilibrium constant for the reaction :
`CH_(3) COOH + C_(2)H_(5)OH hArr CH_(3) COOC_(2)H_(5) +H_(2)O`
is 4.0 at `25^(@)C`. Calculate the weight of ethyl acetate that will be obtained when 120 g of acetic acid are reacted with 92 g of ethyl alcohol.

Text Solution

Verified by Experts

The correct Answer is:
`117*04 g`

Initially, `CH_(3) COOH = 120/60 "mol" = 2 "mol"`
`C_(2) H_(5) OH= 92/46 "mol" = 2 "mol"`
At equilibrium, `[CH_(3)COOH] = (2-x)//V " mol "L^(-1)`
`[C_(2)H_(5)OH]= (2-x)//V " mol "L^(-1),`
`[CH_(3)COOC_(2)H_(5)]=[H_(2)O]= x//V " mol " L^(-1)`
` K = (x xx x )/(2-x)^(2) = 4 " (Given)"`.
`" This gives "x== 1*33 " mol"`
` :. " Mass of ethyl acetate " = 1*33 xx 88 = 117*04 g " "("Molar mass of " CH_(3)COOC_(2)H_(5)= 88" g mol"^(-1))`
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The equilibrium constant for the reaction CH_(3)COOH+C_(2)H_(5)OH hArr CH_(3)COOC_(2)H_(5)+H_(2)O is 4.0 at 25^(@)C . Calculate the weight of ethyl acetate that will be obtained when 120g of acetic acid are reacted with 92 g of alcohol.

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The equilibrium constant for the reaction: CH_(3)COOH(l) +C_(2)H_(5)OH(l) hArr CH_(3)COOC_(2)H_(5)(l)+H_(2)O(l) has been found to be equal to 4 at 25^(@)C . Calculate the free energy change for the reaction.

The equilibrium constant for the reaction: CH_(3)COOH(l) + C_(2)H_(5)OH(l) iff CH_(3)COOC_(2)H_(5)(l)+ H_(2)O(l) has been found to be equal to 4 at 25^(@)C . Calcuclate the free energy chnage for the reaction.

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