SolutionHard

Question

A dilute solution contains n mole of solute ‘A’ in 1 kg of a solvent with molal elevation constant Kb. The solute A undergoes dimerization as 2A$\rightleftharpoons$A2. If ΔTb is the elevation in boiling point of a given solution and molality = molarity, the equilibrium constant KC for dimer formation is

Options

A.$\frac{K_{b}\left( n.K_{b} - \Delta T_{b} \right)}{\left( 2.\Delta T_{b} - n.K_{b} \right)}$
B.$\frac{K_{b}\left( n.K_{b} - \Delta T_{b} \right)}{\left( 2.\Delta T_{b} - n.K_{b} \right)^{2}}$
C.$\frac{K_{b}.n.\Delta T_{b}}{\left( 2.\Delta T_{b} - n.K_{b} \right)^{2}}$
D.$\frac{\left( n.K_{b} - \Delta T_{b} \right)}{\left( n.\Delta T_{b} - 2.K_{b} \right)^{2}}$

Solution

$2A \rightleftharpoons A_{2}$

Initial concentration n M 0

Equilibrium concentration $(n - x)\text{ M}$ $\frac{x}{2}\text{ M}$

Now, $\Delta T_{b} = K_{b}.m = K_{b}\left\lbrack n - x + \frac{x}{2} \right\rbrack$

$\therefore x = 2n - \frac{2\Delta T_{b}}{K_{b}}$

Now, $K_{c} = \frac{\left\lbrack A_{2} \right\rbrack}{\lbrack A\rbrack^{2}} = \frac{(x/2)}{(n - x)^{2}} = \frac{n - \frac{\Delta T_{b}}{K_{b}}}{\left( \frac{2\Delta T_{b}}{K_{b}} - n \right)} = \frac{K_{b}\left( n.K_{b} - \Delta T_{b} \right)}{\left( 2\Delta T_{b} - n.K_{b} \right)^{2}}$

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