Question

The freezing point of a solution of acetic acid (mole fraction = $0.02$) in benzene is $277.4K$. Acetic acid exists partly as a dimer $2A\rightleftharpoons A_2.$ Determine the equilibrium constant for dimerisation. Freezing point of benzene is $278.4K$ and $K_f$ for benzene is $5$.

• A. $3.19kgmo{l}^{-1}$
• B. $31.9kgmo{l}^{-1}$
• C. $1.6kgmo{l}^{-1}$
• D. $16.0kgmo{l}^{-1}$

Answer 1

The correct option is A $3.19kgmo{l}^{-1}$

Let acetic acid $=A$; Benzene $=B$
Assume, $a$ part of $A$ forms dimer.

$\begin{array}{cccc}2A& \rightleftharpoons & A_2& \\ 1& & 0& \mathrm{initial}\mathrm{moles}\\ 1-a& & a/2& \mathrm{moles}\mathrm{after}\mathrm{dimer}\mathrm{is}\mathrm{formed}\end{array}$

$\therefore i=\frac{(1-a)+a/2}{1}=1-a/2$
$\text{Mole fraction of}A=X_A=0.02;$
$\text{Mole fraction of}B=X_B=1-0.02=0.98$

$\text{Molality of}A\text{in}B=\frac{X_A}{M_B}\times \frac{1000}{X_B}=\frac{0.02}{78}\times \frac{1000}{0.98}=0.262molk{g}^{-1}\text{of Benzene}$

Since, $\Delta T_f=K_f\times i\times \text{molality}$

$278.4-277.4=5\times i\times 0.262$
$1=5\times i\times 0.262$
$i=\frac{1}{5\times 0.262}=0.763$

$1-a/2=0.763\therefore a=0.47$

$\text{Hence, the molality of}A\text{after dimer is formed}=(1-a)\times \text{initial molality}=(1-0.47)\times 0.262=0.53\times 0.262=0.13886$

$\text{Molality of}{A}_2\text{after dimer is formed}=\frac{a}{2}\times \text{molality}$

$\Rightarrow \frac{0.47}{2}\times 0.262=0.235\times 0.262=0.0616$

The equilibrium constant, $K_{eq}=\frac{\left[A_2\right]}{[A{]}^2}=\frac{0.0616}{(0.13886{)}^2}=3.19kgmo{l}^{-1}$