Question

The dissociation equilibrium of a gas $A{B}_2$ can be represented as
$2A{B}_2(g)\rightleftharpoons 2AB(g)+B_2(g)$
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)

Moles at equilibrium $=2(1-x)+2x+x$
$=2-2x+2x+x=x+2$,
$K_p=\frac{{\left[P_{AB}\right]}^2\left[P_{B_1}\right]}{\left[P_{Al{l}_1}\right]}=\frac{{\left(\frac{2x}{x+2}\times p\right)}^2\left(\frac{x}{2+x}\times p\right)}{{\left[\frac{2(1-x)}{x+2}\times p\right]}^2}$
$=\frac{\frac{4x^3}{x+2}\times p}{4(1-x{)}^2}=\frac{4x^2\times p}{2}\times \frac{1}{4}$
$x={\left(\frac{2K_p}{p}\right)}^{1/3}($ as $1-x\approx 1,2+x\approx 2)$