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

Options

A. 1
B. 3
C. 2
D. 4

Answer

C

Explanation

`{:(xA,,hArr,,C,+,D,),(a,,,,0,,0,"at intial"),(a(1-a),,,,(aalpha)/x,,(aalpha)/x,"at equilibrium"):}`
`K = ([C][D])/([A]^(x)) = (((alpha^(2)a^(2))/x^(2)v^(2)))/([(a(1-alpha))/(V)]^(x))`
`= (alpha^(2)a^(2-alpha))/(x^(2)(1-alpha)^(x)V^(2-x))`
`.: alpha` is independent of A `implies 2 - x =0` or x = 2