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Q.

In a reaction at equilibrium 'X' mole 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 of A, then the value of 'X' is:

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a

2

b

1

c

3

d

4

answer is C.

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Detailed Solution

$\underset{a (1 - α)}{\underset{a}{XA}} ⇌ \underset{\frac{aα}{x}}{\underset{0}{C}} + \underset{\frac{aα}{x}}{\underset{0}{D}}$
$\begin{array}{l} ∵ K = \frac{[C] [D]}{[A]^{X}} = [\frac{a^{2} α^{2}}{x^{2} V^{2}}] / {[\frac{a (1 - α)}{V}]}^{X} \\ = \frac{α^{2} \cdot a^{2 - x}}{x^{2} (1 - α)^{X} \cdot V^{2 - X}} \end{array}$
If α is independent of a, then 2- X = 0
or X = 2

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