Question

The freezing point of 0.02 mol fraction solution of acetic acid $\left(A\right)$ in benzene $\left(B\right)$ is 277.4 K. Acetic acid exists partly as a dimer $2A\rightleftharpoons A_{2.}$ Calculate the equilibrium constant for the dimerisation? The freezing point of benzene is 278.4 K and its heat of fusion is $10.042KJmo{l}^{-1}.$

• A. $K=3.36$
• B. $K=6.4$
• C. $K=10.08$
• D. $K=12.56$

Answer 1

The correct option is A $K=3.36$

Assuming $\alpha$ part of $A$ forms dimer

$\underset{1-\alpha }{2A}\rightleftharpoons \underset{\alpha /2}{A_2}$

$i=(1-\alpha )+\left(\alpha /2\right)=1-\alpha /2$

Mole fraction of acetic acid $X_A=0.02$

Mole fraction of Benzene $X_B=1-0.02=0.98$

Molality of Acetic acid in Benzene

$m=\frac{X_A\times 1000}{X_B\times M_{Benzene}}=\frac{0.02\times 1000}{0.98\times 78}=0.26molK{g}^{-1}$

$\Delta T_f=i{K}_{f}m$

$278.4-277.4=i\times 5\times 0.26$

$\Rightarrow i=0.763$

$\Rightarrow 0.763=1-\alpha /2$

$\alpha =0.48$

Hence the molality of acetic acid after dimerization

$(1-\alpha )\times$ initial molality

$=(1-0.48)\times 0.26=0.136$

The molality of dimer $A_2=\alpha /2\times$ molality

$=\left(0.48/2\right)\times 0.26=0.0628$

The equlibrium constant $K=\frac{\left[A_2\right]}{[A{]}^2}$

$\Rightarrow K=\frac{0.0628}{(0.136{)}^2}=3.36kgmo{l}^{-1}$