Question

In the reaction $AB(g) \rightleftharpoons A(g) + B(g)$ at $30^{\circ} C, K_p$ for the dissociation equilibrium is $1.6 \times 10^{-3}$ atm. If the total pressure at equilibrium is $1$ atm, then calculate the percentage dissociation of $AB$.

Answer 1

$AB(g) \rightleftharpoons A(g) + B(g)$
Applying law of mass action
$K_p = \frac{\alpha^2p}{1 - \alpha^2}$(given $p = 1$ atm)
$\therefore \frac{\alpha^2}{1 - \alpha} = 1.6 \times 10^{-3}$
$\Rightarrow \alpha^2 = 1.6 \times 10^{-3}$
$(\because \alpha < < < 1)$
$\Rightarrow \alpha = \sqrt{1.6 \times 10^{-3}}$
$\Rightarrow \alpha = 0.04$
$\%$ age dissociation $= 4\%$.