In a reaction at equilibrium ‘X’ moles of the reactant A decomposes to give 1 mole each of C and D. If the fraction of A decomposed at equilibrium is independent of initial concentration of A then the value ‘X’ is 

      xA   $\rightleftharpoons$   C  +  D

       a            0       0     at initial

a(1 - a)        $\frac{\mathrm{a\alpha }}{\mathrm{x}}$   $\frac{\mathrm{a\alpha }}{\mathrm{x}}$    at equilibrium

$\mathrm{K}=\frac{\left[\mathrm{C}\right] \left[\mathrm{D}\right]}{{\left[\mathrm{A}\right]}^{\mathrm{x}}} = \frac{{\displaystyle \frac{{\mathrm{\alpha }}^2 {\mathrm{a}}^2}{{\mathrm{x}}^2{\mathrm{v}}^2}}}{{\left[{\displaystyle \frac{\mathrm{a} \left(1-\mathrm{\alpha }\right)}{\mathrm{V}}}\right]}^{\mathrm{x}}}$

                $=\frac{{\mathrm{\alpha }}^2{\mathrm{a}}^{2-\mathrm{\alpha }}}{{\mathrm{x}}^2{\left(1-\mathrm{\alpha }\right)}^{\mathrm{x}}{\mathrm{V}}^{2-\mathrm{x}}}$

$\because \mathrm{\alpha } \mathrm{is} \mathrm{independent} \mathrm{of} \mathrm{A} \Rightarrow 2-\mathrm{x}=0 \mathrm{or} \mathrm{x}=2$