Question

Given,
$X\left(g\right)\rightleftharpoons nY\left(g\right)$
If degree of dissociation is $\alpha$, then $K_c$ of the reaction in a vessel of $1$ litre is:

• A. $\frac{n{\alpha }^n}{1-\alpha }$
• B. $\frac{n\alpha }{1-\alpha }$
• C. $\frac{{\left(n\alpha \right)}^n}{1+\alpha }$
• D. $\frac{{\left(n\alpha \right)}^n}{1-\alpha }$

Answer 1

The correct option is D $\frac{{\left(n\alpha \right)}^n}{1-\alpha }$

Given,
$X\left(g\right)\rightleftharpoons nY\left(g\right)$
If degree of dissociation is $\alpha$, then $K_c$ of the reaction in a vessel of $1$ litre is $\frac{{\left(n\alpha \right)}^n}{1-\alpha }$.
Since total volume is 1 L, the number of moles is molar concentration.
Let 1 mole of X is present initially.
$\alpha$ moles of X will dissociate to form $n\alpha$ moles of Y.
$(1-\alpha )$ moles of X will remain.
The equilibrium constant $K_c=\frac{[Y{]}^n}{X}$
$K_c=\frac{[n\alpha {]}^n}{(1-\alpha )}$