6. If 1 mole of acetic acid and 1 mole of ethyl alcohol are mixed and the reaction proceeds to equilibrium, the
concentrations of acetic acid and water are found to be 2/3 and 2/3 mole respectively. If 1 mole of ethyl
acetate and 3 moles of water are mixed, how much ester is present when equilibrium is reached ?
Answers
The given reaction is :-
{ C }_{ 2 }{ H }_{ 5 }OH+{ CH }_{ 3 }COOH\rightleftharpoons { CH }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 }+{ H }_{ 2 }OC
2
H
5
OH+CH
3
COOH⇌CH
3
COOC
2
H
5
+H
2
O
11 11 00 00
At eq. : (1-x)(1−x) (1-x)(1−x) xx xx
Given that, at equilibrium-
Moles\ of\ { CH }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 } = 0.666Moles of CH
3
COOC
2
H
5
=0.666
\Rightarrow \quad x=0.666⇒x=0.666
Now, number of moles of { C }_{ 2 }{ H }_{ 5 }OH=0.334C
2
H
5
OH=0.334
number of moles of { CH }_{ 3 }COOH=1-0.666=0.334CH
3
COOH=1−0.666=0.334
\therefore \quad { K }_{ C }=\dfrac { \left[ { CH }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 } \right] \left[ { H }_{ 2 }O \right] }{ \left[ { C }_{ 2 }{ H }_{ 5 }OH \right] \left[ { CH }_{ 3 }COOH \right] } ∴K
C
=
[C
2
H
5
OH][CH
3
COOH]
[CH
3
COOC
2
H
5
][H
2
O]
=\dfrac { 0.666\times 0.666 }{ 0.334\times 0.334 } \approx 2\times 2\approx 4=
0.334×0.334
0.666×0.666
≈2×2≈4