Question

The dissociation equilibrium of a gas$A{B}_2$ can
be represented as :
$2A{B}_2\left(g\right)\rightleftharpoons 2AB\left(g\right)+B_2\left(g\right)$
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. $(2K_p/P{)}^{1/2}$
• B. $\left(K_p/P\right)$
• C. $\left(2K_p/P\right)$
• D. $(2K_p/P{)}^{1/3}$

Answer 1

Answer: (D)

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

Given reaction is :-

${2AB}_2\left(g\right)\rightleftharpoons 2AB\left(g\right)+B_2\left(g\right)$

Initial $2$ $0$ $0$

moles

At eqm. degree of dissociation is $x$ so amount left at equilibrium of the given species are as above.

So, $Kp=\frac{{\left(P_{AB}\right)}^2.\left(P_{B2}\right)}{{\left(P_{AB}\right)}^2}$

$=\frac{{\left(2x\right)}^2.x}{{(2-2x)}^2}\times \frac{P}{(2+x)}$

[$\because P_{AB}=\frac{2x}{2+x}\times P$ ; P is the total pressure at eqm.

$P_{B2}=\frac{x}{2+x}\times P;P_{AB2}=\frac{2(1-x)}{2+x}\times P$]

$\Rightarrow Kp=\frac{{4x}^{3}P}{(2+x).4.(1-x)}$

$=\frac{{4x}^{3}P}{2\times 4\times 1}[As,x<<1;1-x\approx 1\&2+x\approx 2]$

$\Rightarrow Kp=\frac{x^{3}P}{2}\Rightarrow x={\left(\frac{2Kp}{P}\right)}^{1/3}$