Question

The freezing point of $0.02$mol fraction of acetic acid in benzene is $277.4K$. Acetic acid exists partly as a dimer. Calculate the equilibrium constant for dimerization. (approximately)
(Freezing point of $C_6{H}_6=278.4K$ and $K_f$ of $C_6{H}_6=5K{m}^{-1}$)

• A. $2.6$
• B. $3.2$
• C. $4.3$
• D. $5.8$

Answer 1

Answer: (B)

$3.2$

Freezing point of acetic acid (mole fraction $=0.02$) in benzene is $277.4k$. Freezing point of benzene is $278.4k$. molal freezing point constant $=5mg/mole$

equilibrium constant for dimerisation $=?$

Let acetic acid $=A$ and

Benzene $=B$

Assume that

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

$\therefore 2A\rightleftharpoons A_2$

$1-\alpha$ $\alpha /2$ moles after formation of dimer.

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

mole fraction of Acetic acid $=X_A=0.02$

mole fraction of Benzene $=X_e=1-0.02=0.98$

Molality of $A$ in $B$.

$m=\frac{X_A\times 1000}{X_B\times M_B}=$

where $X_A=$ Mole fraction of Acetic acid

$X_B=$ Mole fraction of Benzene

$M_B=$ Molecular mass of Benzene

$=\frac{0.02\times 1000}{0.98\times 78}=0.262mol/kg$

Since $△T_F=K_{F}Xi\times m\left(molality\right)$

where

$△T_F=$ depression in freezing point

$K_F=$ The freezing point constant of a solvent.

$i=$ Vant Hoff factor

$m=$ molality

$\therefore$ The $△T_F$ is change in freezing point

$\Rightarrow 278.4-277.4=△T_F$

$△T_F=1$

$△T_F=K_{F}Xi\times molality$

$1=5\times i\times 0.262$

$\therefore i=\frac{1}{5\times 0.262}$

$i=0.763$

Substitute ${}^{\prime }{i}^{\prime }$ in eqn (1)

$0.763=1-\frac{\alpha }{2}$

$A=0.763-1=\frac{-\alpha }{2}$

$=-0.237=\frac{-\alpha }{2}$

$=0.237=\frac{\alpha }{2}$

$\Rightarrow 2\times 0.237=\alpha$

$0.474=\alpha$

$\therefore A\approx 0.48$

Hence the molality of $A$ after formation of dimer $=(1-\alpha )\times$ initial molality.

$\Rightarrow (1-0.48)\times 0.262$

$=0.52\times 0.262$

$=0.13624$

Now, the molality of $A_2$ after formation of dimer $=\left(\alpha /2\right)\times molality$

$=\frac{0.48}{2}\times 0.262$

$=0.24\times 0.262$

$=0.06288$

$\therefore$ The equilibrium constant.

$=K_{eq}=\frac{\left[A_2\right]}{{\left[A_1\right]}^2}=\frac{0.6288}{{\left(0.13624\right)}^2}=3.39kg/mol$