For a reaction at equilibrium A ( g ) B ( g )+1 2 C ( g ) the rel…

MCQ

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
$K =\frac{\alpha^{\frac{1}{2}} p ^{\frac{3}{2}}}{\left(1+\frac{3}{2} \alpha\right)^{\frac{1}{2}}(1-\alpha)}$
✓
$K =\frac{\alpha^{\frac{3}{2}} p ^{\frac{1}{2}}}{(2+\alpha)^{\frac{1}{2}}(1-\alpha)}$
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

Correct option: B.

$K =\frac{\alpha^{\frac{3}{2}} p ^{\frac{1}{2}}}{(2+\alpha)^{\frac{1}{2}}(1-\alpha)}$

b
$\quad\quad\quad\quad A ( g ) \rightleftharpoons B ( g )+\frac{1}{2} C ( g )$

Initial $:\quad \, P _{ i } \quad\quad\quad0\quad\quad\quad 0$

At eq.$: \,P _{ i }(1-\alpha)\quad P _{ i } \cdot \alpha \quad P _{ i } \frac{\alpha}{2}$

Now, equilibrium pressure (p),

$P = P _{ i } \times\left(1+\frac{\alpha}{2}\right)$

$\therefore P _{ A }=\left(\frac{1-\alpha}{1+\frac{\alpha}{2}}\right) P$

$P _{ B }=\left(\frac{\alpha}{1+\frac{\alpha}{2}}\right) P$

$P _{ C }=\left(\frac{\frac{\alpha}{2}}{1+\frac{\alpha}{2}}\right) P$

$\therefore K =\frac{ P _{ c }^{\frac{1}{2}} \times P _{ B }}{ P _{ A }}$

$K =\frac{\alpha^{\frac{3}{2}} p ^{\frac{1}{2}}}{(2+\alpha)^{\frac{1}{2}}(1-\alpha)}$

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