Question

The equilibrium constant at a certain temp. for the reaction $A _{2}+ B _{2} \rightleftharpoons 2 AB$ 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)

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