Question

In equilibrium reaction, $x$ moles of the reactant $A$ decompose to give 1 mole each of $C$ and $D$. If the fraction of $A$ decomposed at equilibrium is independent of initial concentration, then the value of $x$ would be:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

The fraction of A decomposed at equilibrium is independent of initial concentration.

This implies that the concentration term is not present in the equilibrium constant expression.

The equilibrium reaction is $xA\rightleftharpoons C+D.$

Let the inital concentration of $A$ be $cM$.

The equilibrium concentrations of $A,C$ and $D$ are $c(1-a),\frac{c\alpha }{x}$ and $\frac{c\alpha }{x}$ respectively.

The expression for the equilibrium constant is $K_C=\frac{\left[C\right]\left[D\right]}{[A{]}^x}=\frac{\left(\frac{c\alpha }{x}\right)\times \left(\frac{c\alpha }{x}\right)}{\left[c\right(1-\alpha ){]}^x}=\frac{(c\alpha {)}^2}{x^2\left[c\right(1-\alpha ){]}^x}$.

When $x=2$, the equilibrium constant expression will be free from the concentration term.

Hence, $K_C=\frac{c^2{\alpha }^2}{4\left[c\right(1-\alpha ){]}^2}=\frac{{\alpha }^2}{4(1-\alpha {)}^2}$.

Thus, option B is correct.