Question

For a reaction at equilibrium
$A ( g )\rightleftharpoons B ( g )+\frac{1}{2} C ( g )$
the relation between dissociation constant (K), degree of dissociation $(\alpha)$ and equilibrium pressure $(p)$ is given by :

• A. $K =\frac{\alpha^{\frac{1}{2}} p ^{\frac{3}{2}}}{\left(1+\frac{3}{2} \alpha\right)^{\frac{1}{2}}(1-\alpha)}$
• B. $K =\frac{\alpha^{\frac{3}{2}} p ^{\frac{1}{2}}}{(2+\alpha)^{\frac{1}{2}}(1-\alpha)}$
• C. $K =\frac{(\alpha p )^{\frac{3}{2}}}{\left(1+\frac{3}{2} \alpha\right)^{\frac{1}{2}}(1-\alpha)}$
• D. $K =\frac{(\alpha p )^{\frac{3}{2}}}{(1+\alpha)(1-\alpha)^{\frac{1}{2}}}$

Answer 1

Answer: (B)

Now, equilibrium pressure (p),
$P = P _{ i } \times\left(1+\frac{\alpha}{2}\right)$
$\therefore P _{ A }=\left(\frac{1-\alpha}{1+\frac{\alpha}{2}}\right) P$
$P _{ B }=\left(\frac{\alpha}{1+\frac{\alpha}{2}}\right) P$
$P _{ C }=\left(\frac{\frac{\alpha}{2}}{1+\frac{\alpha}{2}}\right) P$
$\therefore K =\frac{ P _{ c }^{\frac{1}{2}} \times P _{ B }}{ P _{ A }}$
$K =\frac{\alpha^{\frac{3}{2}} p ^{\frac{1}{2}}}{(2+\alpha)^{\frac{1}{2}}(1-\alpha)}$