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Question

The equilibrium constant for the reaction, CH3COOH(l)+C2H5OH(l)CH3COOC2H5(l)+H2O(l) is 4. What will be the composition of the equilibrium mixture when 1 mole of acetic acid is taken along with 4 moles of ethyl alcohol?

Solution
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CH3COOH(l)+C2H5OH(l)CH3COOC2H5(l)+H2O(l)
Initial moles 1 4 0 0
Equilibrium moles 1x 4x x x
Kc=[ester][water][acid][alcohol]=x2(1x)(4x)=4
or x2=4(1x)(4x)=1620x+4x2
3x220x+16=0
This is quadratic equation of type ax2+bx+c=0 with solution
x=b±b24ac2a
x=(20)±(20)24(3)(16)2(3)
x=20±4001922(3)
x=0.93 or 5.7366
The value 5.7366 is not possible as it will give negative number of moles, hence x=0.93
Thus, the composition of mixture at equilibrium is
[CH3COOH]=(1x)=(10.93)=0.07mole
[C2H5OH]=(4x)=(40.93)=3.07mole
[CH3COOHC2H5]=x=0.93mole
[H2O]=x=0.93mole

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Q1
The equilibrium constant for the reaction, CH3COOH(l)+C2H5OH(l)CH3COOC2H5(l)+H2O(l) is 4. What will be the composition of the equilibrium mixture when 1 mole of acetic acid is taken along with 4 moles of ethyl alcohol?
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Q2
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Q3
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