Question

In a reaction at equilibrium, $x$ mole of the reactant $A$ decompose to give 1 mole each of $C$ and $D$. It has been found that the fraction of $A$ decomposed at equilibrium is independent of initial concentration of $A$. $x$ is.

Answer 1

$XA\rightleftharpoons C+D$
Before dissociation $a00$
At equilibrium $a(1-\alpha )\left(a\alpha \right)/x\left(a\alpha \right)/x$
where, $\alpha$ is degree of dissociation of $A$.
$\therefore K_c=\frac{a^2{\alpha }^2\times v^x}{a^x{(1-\alpha )}^x\times v^2\times x^2}$
$=\frac{{\alpha }^2\times a^{2-x}\times v^{x-2}}{{(1-\alpha )}^x\times x^2}$ ($v$ is total volume)
Since, $\alpha \propto {\left(a\right)}^0$
$\therefore 2-x=0$
or $x=2$