Question

At 444 c the equilibrium constant k for the reaction 2ab(g) gives a2(g+b2(g) is 1/64. The degree of dissociation of ab will be

Answer 1

Answer : The degree of dissociation of 'ab' will be, 0.1

Solution :  Given,

Equilibrium constant = $k=\frac{1}{64}$

The given equilibrium reaction is,

                            $2ab\rightleftharpoons a_2+b_2$

initially conc.      c          0     0

At eqm.       $c(1-2\alpha)$     $c\alpha$   $c\alpha$

where,

$\alpha$ is degree of dissociation

The expression used for equilibrium constant is,

$k=\frac{(c\alpha)\times (c\alpha)}{(c(1-2\alpha))^2}$

$\frac{1}{64}=\frac{(c\alpha)\times (c\alpha)}{(c(1-2\alpha))^2}$

By solving the terms, we get :

$\alpha=0.1$

Hence, the degree of dissociation of 'ab' will be, 0.1