Question

The dissociation equilibrium of a gas $AB_2$ can be represented as $2 A B_2(g) \rightleftharpoons 2 A B (g) + B_2 (g)$ The degree of dissociation is $x$ and is small compared to $1$. The expression relating the degree of dissociation (x) with equilibrium constant $K_p$ and total pressure $p$ is

• A. $(2K_p/p)$
• B. $(2K_p/p)^{1/3}$
• C. $(2K_p/p)^{1/2}$
• D. $(K_p/p)$

Answer 1

Answer: (B)

$(2K_p/p)^{1/3}$

Answer 2

Moles at equilibrium $=2(1-x)+2 x+x$
$=2-2 x+2 x +x=x+2$,

$K_{p}= \frac{\left[P_{A B}\right]^{2}\left[P_{B_{1}}\right]}{\left[P_{A l l_{1}}\right]}=\frac{\left(\frac{2 x}{x+2} \times p\right)^{2}\left(\frac{x}{2+x} \times p\right)}{\left[\frac{2(1-x)}{x+2} \times p\right]^{2}}$
$= \frac{\frac{4 x^{3}}{x+2} \times p}{4(1-x)^{2}}=\frac{4 x^{2} \times p}{2} \times \frac{1}{4}$
$x=\left(\frac{2 K_{p}}{p}\right)^{1 / 3} ($ as $1-x \approx 1,2+x \approx 2)$