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The dissociation equilibrium of a gas $AB_2$ can be represented as :
$2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g)$
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 -
$2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g)$
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
$(2K_p/P)^{1/2}$
B
$(K_p/P)$
C
$(2K_p/P)^{1/3}$
D
$(2K_p/P)$
Detailed Solution
$2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g)$
Initial moles: 2, 0, 0
Moles at equilibrium: $2(1 - x)$, $2x$, $x$
Total amount of moles at equilibrium $= 2(1 - x) + 2x + x = 2 + x$
Partial pressures: $p_{AB_2} = \dfrac{2(1 - x)}{2 + x}P$, $p_{AB} = \dfrac{2x}{2 + x}P$, $p_{B_2} = \dfrac{x}{2 + x}P$
$K_p = \dfrac{(p_{AB})^2\,(p_{B_2})}{(p_{AB_2})^2}$
$K_p = \dfrac{\left(\dfrac{2x}{2 + x}P\right)^2 \left(\dfrac{x}{2 + x}P\right)}{\left(\dfrac{2(1 - x)}{2 + x}P\right)^2} = \dfrac{4x^3 P}{4(1 - x)^2 (2 + x)}$
As $x$ is small, $1 - x \approx 1$ and $2 + x \approx 2$:
$K_p = \dfrac{4x^3 P}{4 \times 2} = \dfrac{x^3 P}{2}$
$x^3 = \dfrac{2K_p}{P}$
$x = \left(\dfrac{2K_p}{P}\right)^{1/3}$
Initial moles: 2, 0, 0
Moles at equilibrium: $2(1 - x)$, $2x$, $x$
Total amount of moles at equilibrium $= 2(1 - x) + 2x + x = 2 + x$
Partial pressures: $p_{AB_2} = \dfrac{2(1 - x)}{2 + x}P$, $p_{AB} = \dfrac{2x}{2 + x}P$, $p_{B_2} = \dfrac{x}{2 + x}P$
$K_p = \dfrac{(p_{AB})^2\,(p_{B_2})}{(p_{AB_2})^2}$
$K_p = \dfrac{\left(\dfrac{2x}{2 + x}P\right)^2 \left(\dfrac{x}{2 + x}P\right)}{\left(\dfrac{2(1 - x)}{2 + x}P\right)^2} = \dfrac{4x^3 P}{4(1 - x)^2 (2 + x)}$
As $x$ is small, $1 - x \approx 1$ and $2 + x \approx 2$:
$K_p = \dfrac{4x^3 P}{4 \times 2} = \dfrac{x^3 P}{2}$
$x^3 = \dfrac{2K_p}{P}$
$x = \left(\dfrac{2K_p}{P}\right)^{1/3}$