Question

In a gaseous reaction at equilibrium. ‘n’ mole of reactant ‘A’ decomposes to give 1 mole each of C and D. It has been found that degree of dissociation of A at equilibrium is independent of total pressure. Value of ‘n’ is.

• A. 2
• B. 3
• C. 0.0
• D. None of these

Answer 1

The correct option is A 2

$nA\rightleftharpoons C+D$

Initially only A will be present in the reaction mixture and its concentration is given by "c".

At equilibrium, the concentration of A will be $(c-c\alpha {)}^n$
the concenrations of C and D will be $c\alpha$

$K_c$ = $\frac{{\alpha }^2{c}^2}{(c-c\alpha {)}^n}$

Since $\alpha$ << 1

$K_c$ = $c^{2-n}{\alpha }^2$

$K_c$ is a constant

$\alpha$ is a constant

Therefore ${\alpha }^2$ = $\frac{K_c}{c^{2-n}}$

For $\alpha$ to be constant, $c^{2-n}$ should be equal to 1

$c^{2-n}$ = 0

2-n = 0

n= 2