Question

The equilibrium constant for the reaction, ${CH}_{3}COOH\left(l\right)+C_2{H}_{5}OH\left(l\right)\rightleftharpoons {CH}_{3}COO{C}_2{H}_5\left(l\right)+H_{2}O\left(l\right)$ 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?

Answer 1

${CH}_{3}COOH\left(l\right)+C_2{H}_{5}OH\left(l\right)\rightleftharpoons {CH}_{3}COO{C}_2{H}_5\left(l\right)+H_{2}O\left(l\right)$
Initial moles $1$ $4$ $0$ $0$
Equilibrium moles $1-x$ $4-x$ $x$ $x$
$K_c=\frac{\left[ester\right]\left[water\right]}{\left[acid\right]\left[alcohol\right]}=\frac{x^2}{(1-x)(4-x)}=4$
or $x^2=4(1-x)(4-x)=16-20x+4x^2$
$3x^2-20x+16=0$
This is quadratic equation of type $a{x}^2+bx+c=0$ with solution
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-(-20)\pm \sqrt{(-20{)}^2-4(3\left)\right(16)}}{2\left(3\right)}$
$x=\frac{20\pm \sqrt{400-192}}{2\left(3\right)}$
$x=0.93$ 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
$\left[{CH}_{3}COOH\right]=(1-x)=(1-0.93)=0.07mole$
$\left[C_2{H}_{5}OH\right]=(4-x)=(4-0.93)=3.07mole$
$\left[{CH}_{3}COOH{C}_2{H}_5\right]=x=0.93mole$
$\left[H_{2}O\right]=x=0.93mole$