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

Answer: 4

$AB(g)\rightleftharpoons A(g)+B(g)$
Applying law of mass action
$K_p=\frac{{\alpha }^{2}p}{1-{\alpha }^2}$(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 $=4\%$.