Question

The equilibrium constant at a certain temp. for the reaction $A_2+B_2\rightleftharpoons 2AB$ is $2$ . Calculate the degree of dissociation of either $A_2$ or $B_2$ :

• A. 0.2
• B. 0.5
• C. $\frac{1}{(1+\sqrt{2})}$
• D. $\frac{\sqrt{2}}{(1+\sqrt{2})}$

Answer 1

Answer: (C)

$a-a\alpha a-a\alpha 2a\alpha \alpha =\frac{x}{a}$
$x=a\alpha$
$K=\frac{(2a\alpha {)}^2}{(a-a\alpha )(a-a\alpha )}=$
$2=\frac{4a^2{\alpha }^2}{a^2(1-\alpha {)}^2}=2$
$\frac{{\alpha }^2}{(1-\alpha {)}^2}=\frac{2}{4}=\frac{1}{2}$
taking squire root $\frac{\alpha }{1-\alpha }-\frac{1}{\sqrt{2}}$
$\sqrt{2}\alpha =1-\alpha$
$\alpha =\frac{1}{(\sqrt{2}+1)}$