Question

At temperature $T$, a compound $A{B}_2\left(g\right)$ dissociates according to the reaction,
$2A{B}_2\left(g\right)\rightleftharpoons 2AB\left(g\right)+B_2\left(g\right)$

With degree of dissociation $x$, which is small compared to unity. The expression for $x$ in terms of the equilibrium constant $K_p$ and total pressure $P$ is:

• A. $x={\left(\frac{{2K}_p}{P}\right)}^{\frac{1}{3}}$
• B. $x={\left(\frac{2P}{K_P}\right)}^{\frac{1}{3}}$
• C. $x={\left(\frac{2K_p}{P}\right)}^{\frac{1}{2}}$
• D. $x={\left(\frac{2K_p}{P}\right)}^3$

Answer 1

The correct option is A $x={\left(\frac{{2K}_p}{P}\right)}^{\frac{1}{3}}$

Given reaction:

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

Initial pressure $P_o$ $0$ $0$

Equilibrium pressure $P_O(1-x)$ $P_{O}x$ $\frac{P_{O}x}{2}$

Where x = degree of dissociation

Now total pressure for the reaction at equilibrium can be given as

$P=P_O(1-x)+P_{O}x+\frac{P_{o}x}{2}$

Simplifying the above we get,

$P=P_O+\frac{P_{o}x}{2}$

$P_o$ in terms of P will be $P_O=\frac{2P}{2+x}$

Now, since it is given that the degree of dissociation is small compared to unity, that is, $x<<<1$

Therefore, $P_O=\frac{2P}{2+x}$

$\Rightarrow P_O=P$ (Since,$2+x\approx 2$)

Now the value of $K_p$ for this reaction will be:

$K_p=\frac{P_{B_2}\times {\left(P_{AB}\right)}^2}{{\left(P_{A{B}_2}\right)}^2}=\frac{\frac{P_x}{2}\times {\left(P_x\right)}^2}{{\left(P_O\right(1-x\left)\right)}^2}$

Now $x<<<1$ and $P_O=P$ therefore simplifying the above we get,

$K_p=\frac{P{x}^3}{2}$

Rearranging the above expression to write x in terms of $K_p$ we get,

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

Hence, the correct answer will be option A.