Chemistry - Liquid Solution Question with Solution | TestHub

Let's analyze the dimerization of solute A and find the equilibrium constant.

The elevation in boiling point, $\Delta T_b=i{K}_{b}m$, where $i$ is the van't Hoff factor, $K_b$ is the ebullioscopic constant, and $m$ is the molality. Since A dimerizes, $2A\rightleftharpoons A_2$.

Initial concentration of A is $m$, and at equilibrium, let $\alpha$ be the degree of association. Then, $[A]=m(1-\alpha )$ and $[A_2]=\frac{m\alpha }{2}$.

The total molality at equilibrium is $m^{\prime }=m(1-\alpha )+\frac{m\alpha }{2}=m(1-\frac{\alpha }{2})$.

The van't Hoff factor, $i=\frac{m^{\prime }}{m}=1-\frac{\alpha }{2}$. Therefore, $\alpha =2(1-i)$.

Also, $\Delta T_b=i{K}_{b}m$, so $i=\frac{\Delta T_b}{K_{b}m}$. Substituting this into the expression for $\alpha$: $\alpha =2(1-\frac{\Delta T_b}{K_{b}m})$.

The equilibrium constant $K=\frac{[A_2]}{[A{]}^2}=\frac{m\alpha /2}{[m(1-\alpha ){]}^2}=\frac{\alpha }{2m(1-\alpha {)}^2}$.

Substituting $\alpha =2(1-\frac{\Delta T_b}{K_{b}m})$:

$K=\frac{2(1-\frac{\Delta T_b}{K_{b}m})}{2m(1-2(1-\frac{\Delta T_b}{K_{b}m}){)}^2}=\frac{1-\frac{\Delta T_b}{K_{b}m}}{m(2\frac{\Delta T_b}{K_{b}m}-1{)}^2}=\frac{\frac{K_{b}m-\Delta T_b}{K_{b}m}}{m(\frac{2\Delta T_b-K_{b}m}{K_{b}m}{)}^2}=\frac{K_{b}m-\Delta T_b}{K_{b}m}\cdot \frac{(K_{b}m{)}^2}{m(2\Delta T_b-K_{b}m{)}^2}=\frac{K_b(K_{b}m-\Delta T_b)}{(2\Delta T_b-K_{b}m{)}^2}$.

Therefore, $K=\frac{K_b(K_{b}m-\Delta T_b)}{(2\Delta T_b-K_{b}m{)}^2}$.