Question

The freezing point of $0.02$ mole fraction of acetic acid in benzene is $277.4$ K. Acetic acid exists partly as dimer. Calculate the equilibrium constant for dimerisation. Freezing point of benzene is $278.4$ K and heat of fusion of benzene is $10.042$ kJ mol${}^{-1}$.

Answer 1

For benzene, $K_f=\frac{R{T}^2}{1000L_f\left(cal/g\right)}$

$=\frac{8.314\times (278.4{)}^2}{1000\times \frac{10.042\times {10}^3}{78}}$

$K_f=5.0$ K molality${}^{-1}$

Also, $\Delta T=278.4-277.4=1.0$

For acetic acid in benzene;
$2C{H}_{3}COOH\rightleftharpoons (C{H}_{3}COOH{)}_2$

Before association	$C$	0
After association	$C(1-\alpha )$	$C\alpha /2$

$(\because \text{total particles at equilibrium}=1-\alpha +\frac{\alpha }{2}=1-\frac{\alpha }{2})$

$\therefore K_e=\frac{\frac{C\alpha }{2}}{C^2(1-\alpha {)}^2}$ .....(i)

Where, $\alpha$ is degree of association

Also, $\Delta T=K_f\times molality\times [1-(\alpha /2\left)\right]$ ....(ii)

Given, mole fraction of acetic acid $=0.02=\frac{n}{n+N}$

$\therefore$ Mole fraction of benzene $=0.98=\frac{N}{n+N}$

Now, molality of acetic acid in benzene $=\frac{w\times 1000}{m\times W}$

$=\frac{w\times 1000}{m\times \frac{W}{M}\times M}$

$=\frac{n\times 1000}{N\times M}$

$=\frac{0.02\times 1000}{0.98\times 78}=0.2862M$

$\therefore$ From eq. (ii),
$1=5\times 0.262\times (1-\alpha /2)$
$\therefore \alpha =0.48$
From eq. (i), Assuming molarity = molality

$K_c=\frac{0.262\times 0.48}{2\times (0.262{)}^2\times (1-0.48{)}^2}=3.39$

Hence, the equilibrium constant for dimerisation is 3.39.