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An involatile solute A dimerizes in a solvent as $2A \rightleftharpoons A_2$ . If m is the molarity of solute in the solution , then the equilibrium constant of the reaction is given as

$\begin{array}{1 1} \large\frac{K_b(K_bm +\Delta T_b)}{(2 \Delta T_b- K_bm)^2} \\\large\frac{K_b(K_bm -\Delta T_b)}{(2 \Delta T_b- K_bm)^2} \\ \large\frac{K_b(K_bm - \Delta T_b)}{(2 \Delta T_b- K_bm)^2} \\\large\frac{K_b(K_bm +\Delta T_b)}{(2 \Delta T_b- K_bm)^2} \end{array} $

1 Answer

Answer : $\large\frac{K_b(K_bm -\Delta T_b)}{(2 \Delta T_b- K_bm)^2}$
$2A \rightleftharpoons A_2$
$(1- \alpha)( \alpha/2) $;
Total amount $=(1-\alpha/2)$
The Van't Hoff factor is $i =\large\frac{Total\;amount\;in\;solution}{Amount\;of\;A\;to\;start\;with}$
$\qquad= \large\frac{(1-\alpha/2)}{1}$
$\qquad= 1- \alpha/2$
$\alpha =2 (1-i)$
Hence $K_{eq} =\large\frac{m(\alpha)}{m^2(1- \alpha)^2}$
$\qquad= \large\frac{(1-i)}{m(2i -1)^2}$
Where $i = \large\frac{\Delta T_b }{(\Delta T_b)_0}$
$\qquad= \large\frac{\Delta T_b}{K_b m}$
Thus $K_{eq} =\large\frac{K_b(K_bm -\Delta T_b)}{(2 \Delta T_b- K_bm)^2}$
answered Aug 6, 2014 by meena.p
 

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