Question

In a reaction at equilibrium, '$x$' moles 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 the initial concentration of $A$. Calculate the value of $x$.

Answer 1

$XA\rightleftharpoons D+C$
$a(1-\alpha )$ $\frac{a\alpha }{x}$ $\frac{a\alpha }{x}$
$K_C=\frac{a^2{\alpha }^2}{V^2.x^2{\left[\frac{a(1-\alpha )}{V}\right]}^x}=\frac{a^2{\alpha }^{2-x}}{x^2{(1-\alpha )}^x.V^{2-x}}$
Since $\alpha$ is independent of $a\therefore 2-x=0,x=2$